Solution manual for introduction to electric circuits

Solution Manual

to accompany

Introduction to Electric Circuits, 6e

By R. C. Dorf and J. A. Svoboda
1

Table of Contents

Chapter 1 Electric Circuit Variables
Chapter 2 Circuit Elements
Chapter 3 Resistive Circuits
Chapter 4 Methods of Analysis of Resistive Circuits
Chapter 5 Circuit Theorems
Chapter 6 The Operational Amplifier
Chapter 7 Energy Storage Elements
Chapter 8 The Complete Response of RL and RC Circuits
Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements
Chapter 10 Sinusoidal Steady-State Analysis
Chapter 11 AC Steady-State Power
Chapter 12 Three-Phase Circuits
Chapter 13 Frequency Response
Chapter 14 The Laplace Transform
Chapter 15 Fourier Series and Fourier Transform
Chapter 16 Filter Circuits
Chapter 17 Two-Port and Three-Port Networks
2

Errata for Introduction to Electric Circuits, 6th Edition
Errata for Introduction to Electric Circuits, 6th Edition
Page 18, voltage reference direction should be + on the right in part B:
Page 28, caption for Figure 2.3-1: "current" instead of "cuurent"
Page 41, line 2: "voltage or current" instead of "voltage or circuit"
Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit.
Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..."
Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources,
then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read:
"Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab.
Then Rt = Vab/1."
Page 340, Problem P8.3-5: The answer should be
.
Page 340, Problem P8.3-6: The answer should be .
Page 341, Problem P.8.4-1: The answer should be
Page 546, line 4: The angle is instead of .
Page 554, Problem 12.4.1 Missing parenthesis:
Page 687, Equation 15.5-2: Partial t in exponent:
http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM

Errata for Introduction to Electric Circuits, 6th Edition
Page 757, Problem 16.5-7: H
b
(s) = V
2
(s) / V
1
(s) and H
c
(s) = V
2
(s) / V
s
(s) instead of H
b
(s) = V
1
(s) / V
2
(s) and H
c
(s) = V
1
(s) / V
s
(s).
http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM

Chapter 1 – Electric Circuit Variables
Exercises
Ex. 1.3-1
()
2
23 2 3
00 0
84 A
8 8
() (0) (84)0 2 2 C
33
ttt
it tt
qt idq d ttττ ττ ττ
=−
=+ = − +=− =−∫∫
2

Ex. 1.3-3
() () ()
0
00
44
0 4sin3 0 cos3 cos3 C
33
tt
t
qt idq d tττ ττ τ= + = +=− =− +∫∫
4
3

Ex. 1.3-4

()
()
()22
00
()2 02
2 2
t
t
dqt
it it t
dt
et
−−
 <

== 

−>
<<

Ex. 1.4-1
i1 = 45 µA = 45 × 10
-6
A < i2 = 0.03 mA = .03 × 10
-3
A = 3 × 10
-5
A < i3 = 25 × 10
-4
A

Ex. 1.4-2
( )( ) = 4000 A0.001s 4 Cqi t∆= ∆ =

Ex. 1.4-3
9
6
3
4510
910
510
q
i
t
−
−
−
∆×
== =×
∆×
= 9 µA

Ex. 1.4-4
19 9 19
10 19
9
electron C electron C
= 10billion 1.60210 = 1010 1.60210
se lectron s e
electronC
= 101.60210
electrons
C
1.60210 1.602nA
s
i
−−
−
−
   
×× ×
   
   
××
=× =
lectron

1-1

Ex. 1.6-1
(a) The element voltage and current do not adhere to the passive convention in
Figures 1.6-1B and 1.6-1C so the product of the element voltage and current
is the power supplied by these elements.
(b) The element voltage and current adhere to the passive convention in Figures
1.6-1A and 1.6-1D so the product of the element voltage and current is the
power delivered to, or absorbed by these elements.
(c) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
delivered by this element: (2 V)(6 A) = 12 W. The power received by the
element is the negative of the power delivered by the element, -12 W.
(d) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
supplied by this element: (2 V)(6 A) = 12 W.
(e) The element voltage and current adhere to the passive convention in Figure
1.6-1D, so the product of the element voltage and current is the power
delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the
element is the negative of the power delivered to the element, -12 W.

Problems
Section 1-3 Electric Circuits and Current Flow

P1.3-1
() ()
55
41 20
ttd
it e e
dt
− −
=− = A

P1.3-2
() () () ()
55
00 0 0
44
04 1 04 4 4
55
tt t t
t
qt idq ed d ed te
ττ
ττ τ τ τ
−−
=+ = − += − =+∫∫ ∫∫
5−
− C

0=
P1.3-3
() () 0
t t
qt id dττ τ
−∞ −∞
==∫ ∫
C for t ≤ 2 so q(2) = 0.
() () ()
2
22
22 2 2
tt
t
qt idq d tττ ττ=+ = = =∫∫
4− C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C.
() () ()
4
44
41 4 48
tt
t
qt idq d tττ τ τ=+ =−+=−+=∫∫
−
0=
C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C.
() () ()
88
80 0
t t
qt idq dττ τ=+ = +∫ ∫
C for 8 ≤ t .

1-2

P1.3-4

5
C
= 600A = 600
s
Cs mg
Silver deposited = 60020min60 1.118= 8.0510mg=805g
sm in C
i
×× × ×

Section 1-6 Power and Energy

P1.6-1

a.) ()()( )
4
= = 10A2hrs3600s/hr= 7.210Cqi dtit=∆ ×∫

b.) ( )()110V10A1100WPvi== =
c.)
0.06$
Cost = 1.1kW2hrs = 0.132$
kWhr
××

P1.6-2
()( )
3
= 6V10mA 0.06 W
200Ws
3.3310s
0.06W
P
w
t
P
=
∆⋅
∆= = = ×

P1.6-3

30
for 0t10s: = 30 V and = 2 A 30(2) 60 W
15
vi tt P t≤≤ = ∴= =t
() ()
()( )
2
25
for 10 15s: 1030 V 80 V
5
()580 and () 2 A 2580 10160 W
tv t tb v b
vt t it t P tt t t
≤≤ =−+⇒ = ⇒ =
=−+ = ⇒ = −+=−+

()( )
30
for 15 t 25s:5Vand() A
10
(25) 0 b = 75 () 375 A
5375 15375 W
vi t t
ii t t
Pt t
≤≤ = =−+
=⇒ ⇒ =−+
b
∴=− +=−+

1-3

( ) ()
10 15 25
2
01 0 15
15 25
10
22 3 2
0
10 15
Energy 60 16010 37515
10 15
30 80 375 5833.3 J
32
Pdt tdt ttdt tdt
tt t tt
== + − + −
=+ −+ − =
∫∫ ∫ ∫

P1.6-4

a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully
charged).
()
()53600
53600
2
0 0
0
3
0.5 0.5
211 22
36003600
= 44110J441 kJ
t
wPdt vid d t
τ
ττ

== = + =+


×=
∫∫∫
τ

b.)
1 hr10¢
Cost = 441kJ 1.23¢
3600skWhr
×× =

P1.6-5
()() ( )
11
cos3sin3 sin6
36
ptt t== t
()
1
0.5sin30.0235W
6
p ==
()
1
1 sin60.0466W
6
p== −

1-4

Here is a MATLAB program to plot p(t):

clear

t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time

v=4*cos(3*t); % device voltage
i=(1/12)*sin(3*t); % device current

for k=1:length(t)
p(k)=v(k)*i(k); % power
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

P1.6-6
() ( )( )( )16sin3sin38cos0cos688cos6ptt t t== − = t− W

Here is a MATLAB program to plot p(t):

clear

t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time

v=8*sin(3*t); % device voltage
i=2*sin(3*t); % device current

for k=1:length(t)
p(k)=v(k)*i(k); % power
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

1-5

P1.6-7
()( ) ( )
22 2
41 2 81
tt t 2t
pte e ee
− −−
=− × =−
−

Here is a MATLAB program to plot p(t):

clear

t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time

v=4*(1-exp(-2*t)); % device voltage
i=2*exp(-2*t); % device current

for k=1:length(t)
p(k)=v(k)*i(k); % power
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

P1.6-8

=3 0.2=0.6 W
0.6560=180 J
PVI
wP t
=×
=⋅= ××

1-6

Verification Problems

VP 1-1
Notice that the element voltage and current of each branch adhere to the passive convention. The
sum of the powers absorbed by each branch are:

(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W
= 0 W
The element voltages and currents satisfy conservation of energy and may be correct.

VP 1-2
Notice that the element voltage and current of some branches do not adhere to the passive
convention. The sum of the powers absorbed by each branch are:

-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A)
= -9 W + 6 W + 6 W + 12 W + 9 W -12 W
≠ 0 W

The element voltages and currents do not satisfy conservation of energy and cannot be correct.

Design Problems

DP 1-1
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust
the estimates of the maximum voltage and current and a Grade A device otherwise.

1-7

DP1-2
() () ( )
88 8
201 0.03 0.61
tt t 8t
pte e ee
− −−
=− × = −
−

Here is a MATLAB program to plot p(t):

clear

t0=0; % initial time
tf=1; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time

v=20*(1-exp(-8*t)); % device voltage
i=.030*exp(-8*t); % device current

for k=1:length(t)
p(k)=v(k)*i(k); % power
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

Here is the plot:

The circuit element must be able to absorb 0.15 W.
1-8

Chapter 2 - Circuit Elements
Exercises
Ex. 2.3-1
()
() ()
12 1 2
11
superposition is satisfied
homogeneity is satisfied
Therefore the element is linear.
miimimi
maiami
+= + ⇒
=⇒

Ex. 2.3-2
() ( )( )12 1 2 1 2
superposition is not satisfied
Therefore the element is not linear.
miibmimibmibmib++ =++≠ ++ +⇒

Ex. 2.5-1
()
2
2
10
1 W
100
v
P
R
== =
Ex. 2.5-2
22
2(10 cos )
10 cos W
10
vt
Pt
R
== =
Ex. 2.8-1

1.2A, 24V
4(1.2) 4.8A
cd
d
iv
i
=−=
=− =−

id and vd adhere to the passive convention so
(24)(4.8)115.2 W
dd
Pvi= =− =−
is the power received by the dependent source

2-1

Ex. 2.8-2

2 V, 4 8 A and 2.2 V
cd c d
vi v v=− = =− =
id and vd adhere to the passive convention so
(2.2)(8)17.6 W
dd
Pvi= =− =−
is the power received by the dependent source. The power supplied by the
dependent source is 17.6 W.

Ex. 2.8-3

1.25 A, 22.5 V and 1.75 A
cd c d
iv i i== = =
id and vd adhere to the passive convention so
(2.5)(1.75)4.375 W
dd
Pvi= ==
is the power received by the dependent source.

2-2

Ex. 2.9-1

45, 2 mA, 20k
p
IRθ
°
= == Ω
45
(20k) 2.5k
360 360
aa R
p
θ
= ⇒= Ω= Ω

33
(210)(2.510)5 V
m
v
−
=× ×=

Ex. 2.9-2
A
10 V, 280A, 1 for AD590
K
K
(280A)1 280 K
A
vi k
i
ikT T
k
µ
µ
µ
µ
== =
°
°
°
=⇒ == =



Ex. 2.10-1
At t = 4 s both switches are open, so i = 0 A.

Ex. 2.10.2
At t = 4 s the switch is in the up position, so = (2 mA)(3 k) = 6VviR= Ω .

i
At t = 6 s the switch is in the down position, so v = 0 V.

Problems
Section 2-3 Engineering and Linear Models

P2.3-1
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.

P2.3-2
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v0.12= . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
(c) When v = 4 V,
4
33
0.12
i= = A = 33 A.
2-3

P2.3-3
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v256.5i= . The element is indeed
linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
(c) When v = 12 V,
12
0.04678
256.5
i== A = 46.78 mA.

P2.3-4
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear.

Section 2-5 Resistors

P2.5-1

3 A and = 7 3 = 21 V
and adhere to the passive convention
21 3 = 63 W
is the power absorbed by the resistor.
sii vRi
vi
Pvi
== = ×
∴== ×

P2.5-2

33
3 mA and 24 V
24
8000 8 k
.003
= (310)24 = 7210 72 mW
sii v
v
R
i
P
−−
===
== = =Ω
×× × =

P2.5-3

=10 V and 5
10
2A
5
and adhere to the passive convention
210 20 W
is the power absorbed by the resistor
svv R
v
i
R
vi
pvi
= =Ω
== =
∴= =⋅=

2-4

P2.5-4

24V and 2 A
24
12
2
242 = 48 W
svv i
v
R
i
pvi
===
== =Ω
== ⋅

P2.5-5

12
12
11
1
1
1
150 V;
50 ; 25
and adhere to the passive convention so
150
3 A
50
svv v
RR
vi
v
i
R
===
=Ω =Ω
== =

2
22 2
2
150
and do not adhere to the passive convention so 6 A
25
v
vi i
R
=− =− =−

11 11
The power absorbed by is 1503450 WRP vi== ⋅=

2 2 2 2
The power absorbed by is 150(6) 900 WRP vi=− =−−=

P2.5-6

1 2
12
1 1
1 1 1
1

1 1 1
2 A ;
=4 and 8
and do not adhere to the passive convention so
42 8 V.
The power absorbed by is
(8)(2) 16 W.
sii i
RR
vi
vR i
R
Pvi
===
Ω= Ω
=− =−⋅=−
=− =−− =

2 2 2 2 2
2 2 2 2
and do adhere to the passive convention so 82 16 V .
The power absorbed by 16232 W.
vi vRi
RisPvi
= =⋅ =
== ⋅=

P2.5-7

22 2
22
Model the heater as a resistor, then
(250)
with a 250 V source: 62.5
1000
(210)
with a 210 V source: 705.6 W
62.5
vv
PR
RP
v
P
R
=⇒= = =
== =
Ω

2-5

P2.5-8

2
22
5000125
The current required by the mine lights is: A
1203
Power loss in the wire is :
Thus the maximum resistance of the copper wire allowed is
0.050.055000
0.144
(125/3)
now
P
i
v
iR
P
R
i
== =
×
== = Ω
6
6
2
since the length of the wire is 2100 200 m 20,000 cm
thus / with = 1.710cm from Table 2.51
1.71020,000
0.236cm
0.144
L
RL A
L
A
R
ρρ
ρ
−
−
=× = =
=× Ω⋅
××
== =
−

Section 2-6 Independent Sources

P2.6-1

(a) ()
2
215
3 A and 53 45 W
5
s
v
iP Ri
R
== = = = =
(b) and do not depend on .
siP i
The values of and are 3 A and 45 W, both when 3 A and when 5 A.
ssiP i i= =

P2.6-2

(a)
22

10
5210 V and 20 W
5
s
v
vRi P
R
== ⋅= ===
(b) and do not depend on .
svP v
The values of and are 10V and 20 W both when 10 V and when 5 V
ssvP v v= =

2-6

P2.6-3

Consider the current source:
and do not adhere to the passive convention,
so 312 36 W
is the power supplied by the current source.
ss
cs ss
iv
Piv== ⋅=

Consider the voltage source:
and do adhere to the passive convention,
so 312 36 W
is the power absorbed by the voltage source.
The voltage source supplies 36 W.
ss
vs ss
iv
Pi v== ⋅=
∴ −

P2.6-4

Consider the current source:
and adhere to the passive convention
so 312 36 W
is the power absorbed by the current source.
Current source supplies 36 W.
ss
cs ss
iv
Pi v== ⋅=
−

Consider the voltage source:
and do not adhere to the passive convention
so 31236 W
is the power supplied by the voltage source.
ss
vs ss
iv
Pi v== ⋅=

P2.6-5
(a)
2
(2 cos ) (10 cos )20 cos mWPvi t t t== =
(b)
1
11
2
00
0
11
20 cos =20 sin2 105 sin 2 mJ
24
wP dt tdt t t

== + =+


∫∫

2-7

Section 2-7 Voltmeters and Ammeters

P2.7-1

(a)
5
10
0.5
v
R
i
=== Ω

(b) The voltage, 12 V, and the
current, 0.5 A, of the voltage
source adhere to the passive
convention so the power

P = 12 (0.5) = 6 W

is the power received by the
source. The voltage source
delivers -6 W.

P2.7-2
The voltmeter current is zero
so the ammeter current is
equal to the current source
current except for the
reference direction:

i = -2 A

The voltage v is the voltage of
the current source. The power
supplied by the current source
is 40 W so

402 20 Vvv= ⇒=

2-8

Section 2-8 Dependent Sources

P2.8-1

8
4
2
b
a
v
r
i
=== Ω

P2.8-2

2 A
8 V ; 2 A ; 0.25
8V
a
bb a
b
i
vg vi g
v
== = ===

P2.8-3

32 A
8 A ; 32A ; 4
8A
a
bb a
b
i
id ii d
i
== = ===

P2.8-4

8V
2 V ; 8 V ; 4
2V
b
aa b
a
v
vb vv b
v
== = ===

Section 2-9 Transducers

P2.9-1

360
= , =
360
(360)(23V)
= 75.27
(100 k)(1.1 mA)
m
p
v
a
RI
=
θ
θ
θ °
Ω

P2.9-2

A
AD590 : =1 ,
K
=20 V (voltage condition satisfied)
k
v
µ
°

4 A<<13 A
4K< <13K

i
Ti
T
k
µµ
°°


⇒
=



2-9

Section 2-10 Switches

P2.10-1

At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.

3
10
= = 2mA
510
v
i
R
=
×

At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
3
15
= = 3mA
510
v
i
R
=
×

P2.10-2
At t = 1 s the current in the resistor
is 3 mA so v = 15 V.

At t = 4 s the current in the resistor
is 0 A so v = 0 V.

Verification Problems

VP2-1

=40 V and =(2)2 A. (Notice that the ammeter measures rather than .)
40 V
So 20
2A
Your lab partner is wrong.
os ss
o
s
vi i
v
i
−−= −
==
i

VP2-2

12
We expect the resistor current to be =0.48 A. The power absorbed by
25
this resistor will be = (0.48) (12) = 5.76 W.
A half watt resistor can't absorb this much power. You should n
s
s
v
i
R
Piv
==
=
ot try another resistor.

2-10

Design Problems

DP2-1

1.)
10
004
10
004
250
R
R>⇒ < =.
.
Ω

2.)
101
2
200
2
R
R<⇒ >Ω
Ω

Therefore 200 < R < 250 Ω. For example, R = 225 Ω.

DP2-2
1.) 240 20RR>⇒ >
2.) 21 5
15
4
375
2
RR<⇒ <=.Ω
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.

DP2-3

() ( )()()
22
130mA1000 .0310000.9W1WP=⋅ Ω= = <
() ( )()( )
22
230mA2000 .0320001.8W2WP=⋅ Ω= = <
() ( )()( )
22
330mA4000 .0340003.6W4WP=⋅ Ω= = <

2-11

Chapter 3 – Resistive Circuits
Exercises

Ex 3.3-1

Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 A

Apply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 A

Apply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A

Apply KVL to the loop consisting of elements A and B to get

- v2 – 3 = 0 ⇒ v2 = -3 V

Apply KVL to the loop consisting of elements C, E, D, and A to get

3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V

Apply KVL to the loop consisting of elements E and F to get

v6 – 6 = 0 ⇒ v6 = 6 V

Check: The sum of the power supplied by all branches is

-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0

3-1

Ex 3.3-2

Apply KCL at node a to
determine the current in the
horizontal resistor as shown.

Apply KVL to the loop
consisting of the voltages source
and the two resistors to get

-4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A

Ex 3.3-3
2
18012 0 30Vand 3 9 A
5
aa m a mvv i v i−+ −−=⇒ =− = +⇒ =

Ex 3.3-4
18
10480 6Vand 4 24 V
3
aa a m avv v v v−− +−=⇒ == = =

Ex 3.4-1

3
3
From voltage division
3
12 = 3V
39
= = 1A
3
then
v
v
i



=
+

The power absorbed by the resistors is: ()()()()()()
22 2
1613 13 12 W++ =
The power supplied by the source is (12)(1) = 12 W.

3-2

Ex 3.4-2

1
1
01
6 W and 6
62
= 1 or =1 A
6
=(1) (6)=6V
PR
P
ii
R
viR
= =Ω
==
=

from KVL: (2462) 0
14 14 V
vi
s
vi
s
−++ ++=
⇒= =

Ex 3.4-3 ()
25
From voltage division = 82V
m25+75
v⇒=

Ex 3.4-4 ()
25
From voltage division = 82V
m25+75
v⇒− =−

Ex. 3.5-1

3
3333 3
-3
1 1111 4 101
k
1010101010 44
11
By current division, the current in eachresistor (10) mA
44
R
eqR
eq
= +++ = ⇒ == Ω
==

Ex 3.5-2
()
10
From current division = 51A
10+40
i
m
⇒− =−

3-3

Problems

Section 3-3 Kirchoff’s Laws

P3.3-1

Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A

The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is
power received by element B. The power supplied by element B is 12 W.

Apply KVL to the loop consisting of elements D, F, E, and C to get

4 + v + (-5) – 12 = 0 ⇒ v = 13 V

The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W
is the power supplied by element F.

Check: The sum of the power supplied by all branches is

-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0

3-4

P3.3-2

Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = -4 A

Apply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 A

Apply KVL to the loop consisting of elements A and B to get

- v2 – 6 = 0 ⇒ v2 = -6 V

Apply KVL to the loop consisting of elements C, D, and A to get

- v3 – (-2) – 6 = 0 ⇒ v4 = -4 V

Apply KVL to the loop consisting of elements E, F and D to get

4 – v6 + (-2) = 0 ⇒ v6 = 2 V

Check: The sum of the power supplied by all branches is

-(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0

3-5

P3.3-3

2
22
1
1
1
KVL :12(3) 0 (outside loop)
12
123 or
3
12
KCL 3 0 (top node)
12 12
3 or
3
Rv
v
vR R
i
R
iR
Ri
−−+ =
−
=+ =
+− =
=− =
−

(a)
()123321 V
12
31 A
6
v
i
=+ =
=− =

(b)
21
21210 12
; 8
33 31.5
RR
−
== −Ω = =
−
Ω

(checked using LNAP 8/16/02)

(c)
1
2
24 12 , because 12 and adhere to the passive convention.
12
2A and 2.4
32
9 3, because 3 and do not adhere tothe passive convention
312
3 V and 3
3
ii
iR
vv
vR
=−
∴= − = = Ω
+
=
−
∴== =−Ω

The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.

3-6

P3.3-4

2
Power absorbed by the 4 resistor = 4 =100 W
2
2
Power absorbed by the 6 resistor = 6 = 24 W
1
2
Power absorbed by the 8 resistor = 8 =72 W
4
i
i
i
Ω⋅
Ω ⋅
Ω⋅

12
2A
16
20
5A
2 4
3 2
32
3A
42 3
i
i
ii
ii i
==
==
=− =−
=+ =
A

(checked using LNAP 8/16/02)

P3.3-5

(checked using LNAP 8/16/02)
1
2
3
8 V
8812 12 V
24 8 V
v
v
v
=
=−++=
=⋅=

2
3
2
2
2
1
4: = 16 W
4
6: 24 W
6
8: 8 W
8
v
P
v
P
v
P
Ω=
Ω= =
Ω= =

P3.3-6

( )
33
2mA 3210 610 6 mWP
− −
=−×× =−× =−


( )
33
1mA 7110 7107 mWP
− −
=−−×× =× =


(checked using LNAP 8/16/02)

3-7

P3.3-7

( )
33
2V 2110 2102 mWP
− −
=+×× =× =


( )
33
3V 32 10 610 6 mP
− −
=+×−× =−× =−

W

(checked using LNAP 8/16/02)

P3.3-8

KCL: 21 3 A
KVL: 0120 12 V
12
4
3
RR
RR
R
R
ii
vv
v
R
i
=+⇒ =
+− =⇒ =
∴ == =Ω

(checked using LNAP 8/16/02)

P3.3-9

KVL: 56240 80 V
KCL: 80 8 A
80
10
8
RR
RR
R
R
vv
ii
v
R
i
++= ⇒ =−
+= ⇒ =−
−
∴ == =Ω
−

(checked using LNAP 8/16/02)

3-8

P3.3-10

KCL at node b:

KCL at node a:

11
1
5.613.715.61125.61 1.9
0.801 1.278
75
1.9
3.9834
1.2780.801
RR
R
−− −
=+ ⇒ = +
⇒= = ≈
−
Ω

()
22
2
3.713.715.613.7112 8.29
0 1.8550.475 0
24
8.29
6.0076
1.8550.475
RR
R
−− −
++ =⇒ +− + =
⇒= = ≈Ω
−

(checked using LNAP 8/16/02)

3-9

Section 3-4 A Single-Loop Circuit – The Voltage Divider

P3.4-1
66
12 124 V
16354 18
35
122 V ; 12 V
2 318 18 3
48
12 V
418 3
v
vv
v
== =
+++
== = =
==
10

(checked using LNAP 8/16/02)

P3.4-2

()
() 632415
2828
() 1.867 A
15
28 =28(1.867)=52.27 W
(28 V and do not adhere
to the passive convention.)
aR
bi
R
cp i
i
=+++=Ω
===
=⋅

(checked using LNAP 8/16/02)

3-10

P3.4-3

2
11
1
8 V
12 8
4
iRv
iRviR
iR
==
=+= +
⇒=

()
()
()
1
2
2
1
12
88 44 100
; 50
100 8
44 88 100
; 200
100 4
48
1.212 0.1 A ; 40 ; 80
ai R
Ri
bi R
Ri
ci i R R
ii
⋅
== == =Ω
⋅
== == =Ω
=⇒ = ==Ω ==Ω

(checked using LNAP 8/16/02)

P3.4-4

1
3
31
Voltage division
16
12 8 V
168
4
12 4 V
48
KVL: 0
4 V
v
v
vvv
v
==
+
==
+
−−=
=−

(checked using LNAP 8/16/02)

P3.4-5

100
using voltage divider: 50 1
01002
with 20 V and 9 V, 61.1
0
60
with 28 V and 13 V, 57.7
0
v
s
vv R
sRv
o
vv R
s
R
vv R
s


= ⇒= −

+

=> < Ω

=Ω
=< > Ω



3-11

P3.4-6

()
()()
240
a.) 1812 V
120240
18
b.) 18 0.9 W
120240
c.) 182 1822120 15
120
d.) 0.2 0.21200.8 30
120
R
RR R
R
R
RR
R

=

+

=

+

=⇒ =+ ⇒ =

+
=⇒ = ⇒ =Ω
+
Ω

(checked using LNAP 8/16/02)
3-12

Section 3-5 Parallel Resistors and Current Division

P3.5-1

1
116
4 4
111 111236 3
632 1
1
23
4 A;
211 11 3
632 1
1
2
41 A
311 11
632 1
1
42 A
411 1
1
632
i
i
i
i
==
+++++ +
==
++ +
==
++ +
==
++ +
A=

P3.5-2

()
()
()
1111 1

61242
6212 V
61272 W
aR
R
bv
cp
2=++ =⇒=Ω
=⋅=
=⋅=

P3.5-3

1
1
22
2
88
or
88
8 (2) 2 or
2
iR
Ri
Ri i R
R i
==
=− ⇒=− =
−

()
()
1
2
84 8
2 A ;
4123
3
82 8
A ; 6
2123
2
3
ai R
bi R
=− = ==Ω
== = =Ω
−
6

3-13

()
12 12
12
12 1 1 1 2
12
1
will cause i=2 1 A. The current inboth and will be 1 A.
2
1
2 8 ; 2 8 8 8
2
cR R R R
RR
RR R R RR
RR
==
⋅= = ⇒ ⋅ =⇒ =∴==
+
Ω

P3.5-4

()
()
Current division:
8
62
1168
8
6 3 A
288
1 A
12
i
i
ii i
=− =−
+
=− =−
+
=− =+
A

P3.5-5

1
current division: and
2
12
Ohm's Law: yields
22
12

21
plugging in 4, > 9 V gives 3.15 A
1
and 6, 13 V gives
1
R
ii
sRR
vi R
o
vRR
o
i
s RR
Rv i
os
Rv i
os

=
+

=
 +
 =
 
 
=Ω >
=Ω < <3.47 A
So any 3.15 A 3.47 A keeps 9 V 13 V.iv
so
<< <<

3-14

P3.5-6

()
()()
24
) 1.81.2 A
1224
) 21.6 21.61.612 48
12
) 0.4 0.4120.6 8
12
a
R
bR R
R
R
cR
R
R
R

=

+

=⇒ = + ⇒ =

+
=⇒ = ⇒ =Ω
+
Ω

Section 3-7 Circuit Analysis

P3.7-1

()
()
()
2
4824
16 32
4824
3232
3232
24 16 V ;
3232
8
3232
161
A
322
4811
A
482423
aR
bv
i
ci
⋅
=+ =Ω
+
⋅
+
==
⋅
+
+
==
=⋅ =
+

3-15

P3.7-2

1
2
21 2 21
11
11 12
36
() 4 6
36
11 11
() 2.4then 8 10.4
1266
() KCL: 2 and246 0
246(2)10.4 0
36
= =2.2 A = =2.2(10.4)=22.88 V
16.4
pp
p
aR
bR R
R
ci i iRi
ii
iv iR
⋅
=+ =Ω
+
=+ +⇒ =Ω =+=
+= −++ =
⇒− +−+ =
⇒⇒
R Ω

()
()
2
2
2
32 3
1
6
() 2.20.878 A,
11 1
6612
0.878(6)5.3 V
6
() 0.585 A 3 1.03 W
36
di
v
ei i P i
==
++
==
== ⇒ = =
+

3-16

P3.7-3
Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with
a 2 Ω resistor by the equivalent 1 Ω resistor

This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is
equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:

()
()
1
11
1.50.75 A
211
i
+
==
++

3-17

P3.7-4
(a) 11
24
1
12
1
8
4
2
2
R
R=+ +⇒ =Ω and R
1
1089
1089
6=
+⋅
++
=
()
bg
Ω
(b)

First, apply KVL to the left mesh to get −++=⇒=27630 3ii i
aa aA
30 225ii i
ba b
. Next,
apply KVL to the left mesh to get 4−=⇒ =.A.
(c)

i
2
1
8
1
24
1
8
1
12
2251125=
++
=.. A and v
110
9
1089
310=−
++
L
N
M
O
Q
P=−bg
bg
V

3-18

P3.7-5

30
1030
68
11
+
=⇒ =vv V

R
R
R
2
2
2
10
128 20
+
=⇒ =Ω

20
1030
1030
40
1
1
1=
+
++
⇒=
R
R
R
bg
bg
Ω

Alternate values that can be used to change the numbers in this problem:

meter reading, V Right-most resistor, Ω R1, Ω
6 30 40
4 30 10
4 20 15
4.8 20 30

3-19

P3.7-6

P3.7-7

33
3
24
110 121012 k
1210
p
p
R
R
−
× =⇒ =×=
×+
Ω

( )
()
3
3
3
2110
1210 28 k
2110
p
R
RR
R
×
×== ⇒ =
×+
Ω

P3.7-8

()
130500
Voltage division 50 15.963
13050020020
100 10
15.963 12.279 V
10030 13
.12279 A
100
vV
vv
h
v
h
i
h

⇒= =

++

 
∴= = =
 
+ 
∴= =

3-20

P3.7-9

3-21

P3.7-10

()
()
()
152010
10
152010
60 30 60 20
6 A, 4 A, 6040 V
3015 2010
eq
ab c
eq eq
R
ii v
RR
+
== Ω
++

  
=− =− = = = −=−  

++  


a)
P3.7-11

(24)(12)
2412 8
24 12
eq
R= ==
+
Ω

b)
from voltage division:
100
20 100 53
40 V A
204 3 203
vi
xx

== ∴=

+
=

85
from current division: A
88 6
ii
x

==

+

3-22

P3.7-12

()
9101736
3618
a.) 12
36+18
++ =Ω
=Ω

()()
36R
b.) 18 18 1836 36
36+R
RR=⇒ = ⇒ =Ω

P3.7-13

()
2
.

2 2
23
240
1920 W
2
3
Thus =45
eq
deliv
tockt
eq
RR
RR
RR
v
P
R R
R
==
+
== =
Ω

P3.7-14

()()21 61222 3418 R
eq
=++ + =++=Ω
()()
()()
1
2
from current division
4040
5 A
8
6
51 5 A
33
612
2
515 A
22
22
eq
i
R
ii
ii
∴== =

== =

+

== =

+

3-23

Verification Problems

VP3-1

()
KCL at node a:
31 2
1.167 =0.833 +0.333
1.167= 1.166 OK
KVL loop consisting of the vertical
6 resistor, the 3 Ω and4 resistors,
and the voltage source:
63 12 0
32
yields 4.0 V not
ii i
ii v
v
=+
−− −
−−
ΩΩ
++ +=
=− 2.0 Vv=−

VP3-2

reduce circuit: 5+5=10 in parallel with 20 gives 6.67Ω Ω

6.67
by current division: 51.25 A
206.67
i

==

+
∴Reported value was correct.

VP3-3
()
320
246.4 V
o320650230
v

=

++
=

∴Reported value was incorrect.

3-24

VP3-4

KVL bottom loop: 140.11.20
KVL right loop: 120.051.2 0
KCL at left node:
This alone shows the reported results were incorrect.
Solving the three above equations yields:
16.8
AH
BH
AB H
A
ii
ii
ii i
i
−++ =
−+ + =
+=
= A 10.3 A
6.49 A
Reported values were incorrect.
H
B
i
i
=
=−
∴

VP3-5

Top mesh: () (
1
442 100.5122
2
aa a bii i i

=+ + +−=−+−−


)0

Lower left mesh: () ()102 0.5 102214 V
sa bvi i=+ +−=+ =

Lower right mesh: 412 124(0.5)14
sa svi v+= ⇒ =−−= V

The KVL equations are satisfied so the analysis is correct.

3-25

VP3-6
Apply KCL at nodes b and c to get:

KCL equations:

Node e: 160.54.5−+= +

Node a: 0.5 1 1.5 mA
c cii+=−⇒ =−

Node d: 44.5 0.5 mA
ccii+=⇒ =

That's a contradiction. The given values of ia
and ib are not correct.

Design Problems

DP3-1

Using voltage division:

()
22
12 1 2
24 24
1
pp
m
pp
RaR RaR
v
Ra RRaR RRR
p
+ +
==
+− ++ ++

vm = 8 V when a = 0 ⇒
2
12
1
3
p
R
RR R
=
++

vm = 12 V when a = 1 ⇒
2
12
1
2
p
p
RR
RR R
+
=
++

The specification on the power of the voltage source indicates
2
12
12
24 1
1152
2
p
p
RR R
RR R
≤⇒+ +≥
++
Ω

Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives
and . Solving these equations gives
and .
21 23 2000RR R=+ +
16000R=Ω
2R
()21 22 2000 2000RR R+= ++
=Ω4000

With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate
24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.
3-26

DP3-2
Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage
division,
2
2
200
124 400
200
R
R
=⇒ =
+
Ω. The power required to be dissipated by R2
is
2
8
0.16W W
400 8
= <
1
. To reduce the voltage across any one resistor, let’s implement R2 as the
series combination of two 200 Ω resistors. The power required to be dissipated by each of these
resistors is
2
41
0.08W W
200 8
=< .
Now let’s check the voltage:

190 210
11.88 12.12
0190420 210380
v<<
++

03.700 4.314v<<

047.5% 47.85%v− << +

Hence, vo = 4 V ± 8% and the design is complete.

DP3-3

2
200 mV
10 10
120 (120) (0.2)
10 10
240
let 16 5
10
16
P = = 25.6W
10
ab
ab
V
vV
RR
vR
R
≅
==
++
== ⇒=Ω
+
∴

DP3-4

()()
11
where
1
912
18 bulbs
6
N
N
iGv v G N
TTR RR
n n
iR
N
v

== = =∑ 

=
∴== =
3-27

28

Chapter 4 – Methods of Analysis of Resistive Circuits
Exercises

Ex. 4.3-1

KCL at a: 30 5 3 18
32
vv v
aa b
vv
ab
−
+ += ⇒ − =−

KCL at b: 310 8
2
vv
ba
vv
ba
−
−−= ⇒ −=

Solving these equations gives:

va = 3 V and vb = 11 V

Ex. 4.3-2
KCL at a:

aa b
30 3 2 12
ab42
vv v
vv
−
+ += ⇒ − =−

KCL at a:
40
32
35
vv v
ba b
vv
ab
24
−
−− =
⇒− + =

Solving:
va = −4/3 V and vb = 4 V

Ex. 4.4-1
Apply KCL to the supernode to get

10
25
2030
vv
bb
+
++ =

Solving:

30 V and 1040 Vvv v
ba b
== +=
4-1

Ex. 4.4-2
() ()81 2
3 8 V and 16 V
10 40
v v
b b
vv
ba
+−−
+= ⇒ = =

Ex. 4.5-1
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
and solve for v4
bv=
ai b .
9
6
44 4.5
812 12
vv
bb
iv i v
ab a b
+
+= ⇒ == ⇒ =


V

Ex. 4.5-2
The controlling voltage of the dependent source is a node voltage so it is already expressed as a
function of the node voltages. Apply KCL at node a.

64
02
20 15
vv v
aa a
v
a
−−
+= ⇒ =− V

Ex. 4.6-1

Mesh equations:
1263 80 93 20
11 2 12
ii i ii

−+ + −−=⇒ − =


83 6 0 39 8
12 2 1 2
ii i ii

−− + =⇒−+ =

−

Solving these equations gives:
13 1
A and A
1266
ii== −

The voltage measured by the meter is 6 i2 = −1 V.
4-2

Ex. 4.7-1

Mesh equation: ()
31
24 0 324 93 A
49
ii i i i
−
++ ++=⇒ ++=−−⇒=


2
V
93
The voltmeter measures 34 i=−

Ex. 4.7-2

Mesh equation: () () ()
33 2
3630 36 1563 =3 A
93
ii i i
−
++ +=⇒ +=−− ⇒= −15

Ex. 4.7-3

Express the current source current in terms of the mesh currents:
12 1
33
44
ii i i
2=−⇒ =+.
Apply KVL to the supermesh:
12 2 2 2 2
3
94 320 4 59 96
4
ii i i i i

−+ ++=⇒ ++=⇒ =



so
2
2
A
3
i= and the voltmeter reading is
2
4
2 V
3
i=
4-3

Ex. 4.7-4

Express the current source current in terms of the mesh currents:
12 133ii i i
2=−⇒ =+.
Apply KVL to the supermesh: ( )12 2 2 215630 63 315 9 3ii i i i−+ +=⇒ ++= ⇒ =−
Finally,
2
1
A
3
=−i is the current measured by the ammeter.

Problems

Section 4-3 Node Voltage Analysis of Circuits with Current Sources

P4.3-1

KCL at node 1:

44 211 2
01 .5
86 8 6
vv v
ii i i1.5 A
−
−− −
=+ +=+ +=−+⇒=

(checked using LNAP 8/13/02)

4-4

P4.3-2
KCL at node 1:

12 1
10 5 20
1220 5
vv v
vv
−
++= ⇒ −=−
KCL at node 2:
12 23
23
12 320 10
vv vv
vv v2 40
− −
+=⇒ −+ − =
KCL at node 3:
23 3
13 5
2310 15
vv v
vv 30
−
+=⇒ − + =
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.

(checked using LNAP 8/13/02)

P4.3-3

KCL at node 1:

415412 1
2 A
1152 0 520
vv v
ii
−
−
+= ⇒ = +=−

KCL at node 2:

12 23
251 5
4151518
2 A
2 51 5
vv vv
i
i
− −
+=
−−
⇒= − + =



(checked using LNAP 8/13/02)
4-5

P4.3-4

Node equations:
11 2
1
12 2
2
.003 0
500
.005 0
500
vv v
R
vv v
R
−
−+ + =
−
−+ − =

When v1 = 1 V, v2 = 2 V
1
1
2
2
11 1
.003 0 200
1500
.003
500
12 2
.0050 667
1500
.005
500
R
R
R
R
−
−+ + =⇒ = =
+
−
−+ − =⇒= =Ω
−
Ω

(checked using LNAP 8/13/02)

P4.3-5

Node equations:
1311 2
2312
23 1 3 3
0
500125 250
.001 0
125 250
0
250250500
vvvv v
vvvv
vv vvv
−−
+ +=
−−
− −+ =
−−
− −+ =

Solving gives:

12 30.261 V, 0.337 V, 0.239 Vvv v= ==

13Finally, 0.022 Vvv v=−=

(checked using LNAP 8/13/02)

4-6

Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources

P4.4-1

Express the branch voltage of the voltage source in terms of its node voltages:

06 6
aavv V−=⇒ =−
KCL at node b:

6
22 1 2 30
61 0 6 10 6 10
ab bc b b c b b c
bc
vv vv v vv v vv
vv
−− −− − −
+= ⇒ += ⇒−−+= ⇒ =−83

KCL at node c:
9
44 5
10 8 4
bc c
bc c b
vv v
vv v v
−
=⇒ −= ⇒ =
cv

Finally:
9
308 3 2 V
4
cc cvv v

=− ⇒ =



(checked using LNAP 8/13/02)

P4.4-2

Express the branch voltage of each voltage source in terms of its node voltages to get:

12 V, 8
ab cvv v
dv=−= =+
4-7

KCL at node b:
()12
0.002 0.002 1284000
4000 4000
bba
b
vvv
ii v
−−−
=+ ⇒ = +⇒ +=+ i

KCL at the supernode corresponding to the 8 V source:
0.001 4 4000
4000
d
d
v
i v=+ ⇒ =+ i
so ( )44 844 4 V
bd d d d
vv v v v+=− ⇒ ++=− ⇒ =−
Consequently
4
84 V and 2 mA
4000
d
bc d
v
vv v i
−
== += = =

(checked using LNAP 8/13/02)

P4.4-3

Apply KCL to the supernode:
10 8
.030 7 V
100100100
aa a
a
vv v
v
−−
++ −=⇒ =

(checked using LNAP 8/13/02)

P4.4-4

Apply KCL to the supernode:

( )81281 2
0
500 125 250500
aaa
vvv +−+−
av
+ ++ =

Solving yields
4 V
av=

(checked using LNAP 8/13/02)

4-8

P4.4-5

The power supplied by the voltage source is

()12
129.882125.294
12
46 4 6
12(0.52951.118)12(1.648)19.76 W
ab a c
aa
vv vv
viiv
−− −− 
+= + = +


=+ = =

(checked using LNAP 8/13/02)

P4.4-6
Label the voltage measured by the meter. Notice that this is a node voltage.

Write a node equation at the node at which
the node voltage is measured.

mm m12 8
0.002 0
6000 3000
vv v
=
R
−−
−+ + +



That is

m
m
6000 6000
31 6
16
3
v R
R

v

+= ⇒ =


−

(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.
4-9

Section 4-5 Node Voltage Analysis with Dependent Sources

P4.5-1

Express the resistor currents in terms of the
node voltages:
1
2
8.667101.333 A and
1
210
4 A
22
ac
bc
vv
i
vv
i
−
== −=−
−−
== =−

Apply KCL at node c:

()
12 1 1.3334 (1.333)
5.333
4
1.333
ii Ai A
A
+= ⇒− +−=−
−
⇒= =
−

(checked using LNAP 8/13/02)

P4.5-2
Write and solve a node equation:

64
0 12
100020003000
aa a a
a
vv vv
v V
− −
++ =⇒ =

4
12 mA
3000
aa
b
vv
i
−
== −

(checked using LNAP 8/13/02)

P4.5-3
First express the controlling current in terms of
the node voltages:
2

4000
b
a
v
i
−
=
Write and solve a node equation:

22
50 1.
40002000 4000
bb b
b
vv v
v
−− 
−+ − =⇒ =


5 V

(checked using LNAP 8/14/02)
4-10

P4.5-4

Apply KCL to the supernode of the CCVS to get

121014101
02
42 2
bbii A
− −
+− +=⇒ =−

Next
10121
2V
4 42
1 A
1214
2
a
a
i
r
ri
− 
== − −
⇒= =
 −=−


(checked using LNAP 8/14/02)

P4.5-5

First, express the controlling current of the CCVS in
terms of the node voltages:
2
2
x
v
i=

Next, express the controlled voltage in terms of the
node voltages:
2
22
24
12 33 V
25
x
v
vi v−= = ⇒=

so ix = 12/5 A = 2.4 A.

(checked using ELab 9/5/02)
4-11

Section 4-6 Mesh Current Analysis with Independent Voltage Sources

P 4.6-1
11 3 1 229( )3( )ii i ii0+ −+ −=
12 2 3153( )6( )0ii ii−−+ −=
23 1 36( )9( )210ii ii−−− −− =
or
12 314390ii i−−=
12 339 6 1ii i 5−+− =−
12 3961521ii i−−+ =
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.

(checked using LNAP 8/14/02)

P 4.6-2

Top mesh:
4(23)(2)10(24)0R−++ −=
so R = 12 Ω.

Bottom, right mesh:
28(43)10(42) 0v−+− +=
so v2 = −28 V.

Bottom left mesh
14(32)8(34)0v−+− +−=
so v1 = −4 V.
(checked using LNAP 8/14/02)

4-12

P 4.6-3

Ohm’s Law:
2
6
0.75 A
8
i
−
==−
KVL for loop 1:
( )11 243 18Ri ii 0+ −+ +=

KVL for loop 2
()
( )()
12
1
1
(6)34 0
94 0.750
3 A
ii
i
i
+−−−−=
⇒− −−− =
⇒= −

() ( )()34 30.75210 4 RR−+ −−− +=⇒ =Ω
(checked using LNAP 8/14/02)

P4.6-4

KVL loop 1:

25 2 250 75 4100 () 0
450 100 2
aa a a b
ab
ii i ii
ii
−++ ++ −=
−= −

KVL loop 2:

100( ) 4100 100 8200 0
100 500 4
6.5 mA , 9.3 mA
ab b b b
ab
ab
ii i i i
ii
ii
− −− + + ++ =
−+ =−
⇒= − =−

(checked using LNAP 8/14/02)

P4.6-5

Mesh Equations:

11 2
21 2 3
32 3
mesh 1 : 2 2 () 10 0
mesh 2 : 2() 4 () 0
mesh 3 : 10 4 () 6 0
ii i
ii ii
ii i
+ −+=
−+ −=
−+ −+ =

Solving:
2
5
0.294 A
17
ii i=⇒=−=−

(checked using LNAP 8/14/02)

4-13

Section 4-7 Mesh Current Analysis with Voltage and Current Sources

P4.7-1

1
1
mesh 1: A
2
i=
22
2
12
mesh 2:75 1025 0
0.1 A
= 0.6 A
b
ii
i
ii i
++ =
⇒= −
=−

(checked using LNAP 8/14/02)
P4.7-2

mesh a: = 0.25 A
ai−
mesh b: = 0.4 A
100() = 100(0.15) =15 V
b
ca b
i
vi i
−
=−

(checked using LNAP 8/14/02)

P4.7-3

Express the current source current as a function of the mesh currents:
12 1 2
0.5 0.5ii ii−= −⇒ =−
Apply KVL to the supermesh:

2
12 2 2
22
12
30 20 10 0 30(0.5)20 10
5
5015 10 .1 A
50
.4 Aand 20 2 V
ii i i
ii
iv i
++ =⇒ −+ =−
−= − ⇒ = =
=− = =

(checked using LNAP 8/14/02)
4-14

P4.7-4

Express the current source current in terms
of the mesh currents:

0.02
baii=−

Apply KVL to the supermesh:

250 100 (0.02)9 0
.02 A 20 mA
100(0.02) 4 V
aa
a
ca
ii
i
vi
+− +=
∴= − =−
=− =−

(checked using LNAP 8/14/02)

P4.7-5

Express the current source current in terms of the mesh currents:
31 1 322ii ii−=⇒ =−
Supermesh: ()13 2 3 1 2 363 5 80 658ii ii ii i+− −−=⇒ −+=8
3Lower, left mesh: ()23 21285 0 545ii i i−+ + −=⇒ =+
Eliminating i1 and i2 from the supermesh equation:
() ( )33 362 45 88 9ii i i−−+ +=⇒ =
324
The voltage measured by the meter is:
3
24
8
9
i

==


33 V
(checked using LNAP 8/14/02)
4-15

P4.7-6

Mesh equation for right mesh:
() ()
10 5
422630 128180 A A
12 6
ii i i i−+++=⇒ −+=⇒=− =−

(checked using LNAP 8/14/02)

P 4.7-7

()
() ( )
() () ()()
2
12 1
1
31 3 3 2
3 A
53 5
2 A
24 0
21241 13
5
i
ii i
i
ii iRii
R
R
0
=−
−= ⇒ −−=
⇒=
−+ + −=
⇒− −+−+−−−=
⇒= Ω

(checked using LNAP 8/14/02)

4-16

P 4.7-8

Express the controlling voltage of the
dependent source as a function of the
mesh current

21
50vi=

Apply KVL to the right mesh:

11 1 1
21
100 (0.04(50))5010 0 0.2 A
50 10 V
ii i i
vi
−− ++=⇒ =
==

(checked using LNAP 8/14/02)

P 4.7-9

1
4
3
1
100 2008 0
3
0.048 A
bb a b
aa
a
ii i i
ii
i
=− ⇒ =

−+ +


⇒= −

ai
=

(checked using LNAP 8/14/02)

P4.7-10

Express the controlling current of
the dependent source as a function
of the mesh current:

.06
baii=−

Apply KVL to the right mesh:

aa a100 (0.06)50 (0.06)250 0 10 mA
aii i i−− + −+ =⇒ =
Finally:
ob5050(0.060.01) 2.5 Vvi== − =

(checked using LNAP 8/14/02)

4-17

P4.7-11

Express the controlling voltage of
the dependent source as a function
of the mesh current:

100 (.006)
bavi= −
Apply KVL to the right mesh:

[ ]100 (.006)3100(.006) 250 0 24 mA
aa a aii i i−− + − + =⇒ =−

(checked using LNAP 8/14/02)

P4.7-12

( ) ()
() ()
()()
2
2
33 3 3
11 2 1
33 3
12 21 1
12
apply KVL to left mesh: 31010 2010 030102010 3 1
apply KVL to right mesh: 51010010 2010 0 8 2
63
Solving 1 & 2 simultaneously mA, mA
55 220
ii i i i
ii ii i i
ii
−+ × +× −=⇒× −× =
×+ × +× −=⇒ =
⇒= =

()() ()
()() ()
12 2
2
Power delevered to cathode 5 100
2
63 3
5 100 0.026 mW
55220 220
ii i=+
=+ =

() () ( )
5 3600 s
Energy in 24 hr. 2.610 W24 hr
hr
2.25 J
Pt
−
∴ == ×
=

4-18

P4.7-13

(a)
2L o
Li
12 i 1
and
o
2
2
R RRv
vg Rv v v g
RRv R
=− = ⇒=−
++ R

(b)
() ()
o
i

33
51010
170 0.0374 S
3
1.110
v
gg
v
∴
×
=− =− ⇒=
×

PSpice Problems

SP 4-1

4-19

SP 4-2

From the PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V1 -3.000E+00
V_V2 -2.250E+00
V_V3 -7.500E-01

The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh
currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current
in the voltage source labeled V3).

SP 4-3
The PSpice schematic after running the simulation:

The PSpice output file:

**** INCLUDING sp4_2-SCHEMATIC1.net ****
* source SP4_2
V_V4 0 N01588 12Vdc
4-20

R_R4 N01588 N01565 4k
V_V5 N01542 N01565 0Vdc
R_R5 0 N01516 4k
V_V6 N01542 N01516 8Vdc
I_I1 0 N01565 DC 2mAdc
I_I2 0 N01542 DC 1mAdc

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V4 -4.000E-03
V_V5 2.000E-03
V_V6 -1.000E-03

From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA.

SP 4-4
The PSpice schematic after running the simulation:

The PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V7 -5.613E-01
V_V8 -6.008E-01

The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of
the circuit. However this current is directed from right to left in the 2 Ω resistor while the current
i is directed from left to right. Consequently, i = +5.613 A.

4-21

Verification Problems

VP 4-1

Apply KCL at node b:

1
0
42 5
4.85.21 4.83.0
0
42 5
ba b cvv vv−−
−+ =
−− −−
−+≠

The given voltages do not satisfy the KCL
equation at node b. They are not correct.

VP 4-2

Apply KCL at node a:

2 0
42
204 4
2 4
42
ba avv v−
−− +=


−
0− −+ =−≠



The given voltages do not satisfy the KCL
equation at node a. They are not correct.

4-22

VP 4-3

Writing a node equation:

13 2
127.57.57.56
0
RR R
−−
− ++ =


so
13 2
4.57.51.5
0
RR R
− ++ =
There are only three cases to consider. Suppose
12 35 k and 10 k.RR R=Ω= =Ω Then

13 2
4.57.51.50.90.750.15
0
1000RR R
−++
−+ += =

This choice of resistance values corresponds to branch
currents that satisfy KCL. Therefore, it is indeed possible
that two of the resistances are 10 kW and the other
resistance is 5 kW. The 5 kW is R1.

VP 4-4
KCL at node 1:

()12 1 8208
01
205 20 5
vv v− −−− −
=+ +⇒ ++1=0

KCL at node 2:

() ()12 2 3 820 206
22
20 10 20 10
126
2010
vv vv−− −−− −−−
=+ ⇒ =+
⇒=

KCL at node 3:
()23 3 206 64
11
10 15 10 15 1015
vv v− −−− 6− −−
+= ⇒ += ⇒ =

KCL is satisfied at all of the nodes so the computer analysis is correct.

4-23

VP 4-5

Top mesh: 10 (24)12(2)4(23)0−+ +−=

Bottom right mesh 8(34)4(32)40−+ −+ =

Bottom, left mesh: (Perhaps the polarity of the 28 V source was
entered incorrectly.)
2810(42)8(43)0+− +−≠

KVL is not satified for the bottom, left mesh so the computer analysis is not correct.
4-24

Design Problems

DP 4-1
Model the circuit as:

a)
22 2
We need to keep across as 4.8 5.4
0.3 Adisplay is active
For
0.1 Adisplay is not active
vR v
I
≤≤

=


22
12
2
2
15
KCL at a: 0
Assumed that maximum results in minimum and visa-versa.
Then
4.8 Vwhen 0.3 A

5.4 Vwhen 0.1 A
vv
I
RR
Iv
I
v
I
−
++ =
=
=
=

2
12
12
12
Substitute these corresponding values of and into the KCL equation and solve for the resistances
4.8154.8
0.30
5.4155.4
0.10
7.89 , 4.83
vI
RR
RR
RR
−
++ =
−
++ =
⇒= Ω = Ω

b)
()
1max 1max
2max 2max
1max
154.8 2
1.292 A (1.292)(7.89) 13.17 W
7.89
2
5.45.4
1.118 A 6.03 W
4.83 4.83
maximum supply current 1.292A
RR
RR
R
IP
IP
I
−
== ⇒ = =
== ⇒ = =
==

c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop
below 4.8V or rise above 5.4V.

The power dissipated in the resistors is excessive. Most of the power from the supply is
dissipated in the resistors, not the display.
4-25

DP 4-2

Express the voltage of the 8 V source in terms of its node voltages to get vv . Apply KCL
to the supernode corresponding to the 8 V source:
8
ba−=

()
()
21
12
12
12
12
02 2 0
22 8
41 60
4
4
ba ab
ab
aa
a
a
vvvv vv
vv vv
RR R R
vv v v
vv v
vv
v
−−−
++ + =⇒ −++=
⇒− + ++
⇒− ++=
−
⇒= −
0=

Next set va = 0 to get
12
120 4 16 V
4
vv
vv
−
=− ⇒ −=

For example, v1 = 18 V and v2 = 2 V.

4-26

DP 4-3
a)

pply KCL to left mesh:
11550300 ()0ii I−+ + −=

Apply KCL to right mesh:
1
(2) 300 () 0RI Ii++− =

Solving for I:
150

1570 35
I
R
=
+

We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA ﬁ l amp so the lamp will not
light.

b) From the equation for I, we see that decreasing R increases I:

try = 50 = 45 mA (won't light)RIΩ⇒

try = 25 = 61 mA will lightRIΩ⇒⇒
Now check R±10% to see if the lamp will light and not burn out:

10% 22.5 = 63.63 mAlamp will
10% 27.5 = 59.23 mAstay on
I
I
−→ Ω→ 

+→ Ω→ 

DP 4-4

Equivalent resistance: ( )12 3 4||||RRR RR=+
Voltage division in the equivalent circuit: ()
1 25
10
R
v
R
=
+

We require vab = 10 V. Apply the voltage division principle in the left circuit to get:
4-27

( )( )
()()
12 3 4
44
1
34 34 12 3 4

10 25
10
RR RR
RR
v
RR RR RR RR
+
== ×
++ ++
×
This equation does not have a unique solution. Here’s one solution:

12 3 4choose 25 and 20RR RR==Ω +=Ω
( )
()
4
4
12.520
then 10 25 18.4
2010 12.520
R
R=× ×⇒ =
+
Ω
34 3
and 20 1.6RR R+=⇒= Ω

DP 4-5

Apply KCL to the left mesh:

( )13 1 32 1 0RR iRiv+ −− =

Apply KCL to the left mesh:

( )31 2 32 2 0Ri RRiv− ++ +=

Solving for the mesh currents using Cramer’s rule:

()
() ( )
13
13 1
22 3 32
12
2
13 2 3 3

( )

and
where
vR
RRv
vR R R v
ii
RR RR R
− + 

 
−+ −− 
==
∆∆
∆= + + −


Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then ∆ = 3 MΩ. The mesh currents will be given by

[ ] [ ]
12 21 12
2166
1
2 1000 2 1000
and
310 310 3000
i
vv vv vv
ii i
−− + +
== ⇒=−
××
2i=
Now check the extreme values of the source voltages:

12
12
2
if 1 V mA okay
3
4
if 2 V mA okay
3
vv i
vv i
== ⇒=
== ⇒=

4-28

Chapter 5 Circuit Theorems

Exercises

Ex 5.3-1
R = 10 Ω and is = 1.2 A.
Ex 5.3-2
R = 10 Ω and is = −1.2 A.
Ex 5.3-3
R = 8 Ω and vs = 24 V.
Ex 5.3-4
R = 8 Ω and vs = −24 V.
Ex 5.4-1
() ()
20 10 2
1520 2620()2 V
102020 10(2020) 5
m
v

=+ − =+−

++ ++
=−
Ex 5.4-2
()
25 3
55 32 A
3223
mi=− =−=
++

Ex 5.4-3
() ()
33
35 1856
3(33) 3(33)
mv

=− =−

++ ++
1 A=−
5-1

Ex 5.5-1

5-2

Ex 5.5-2

212
3 A
6
26 V
a
aa
oc a
i
ii
vi
−
=⇒ =
== −
−

()
1262 3 A
2
32 32
3
aa a
sc a sc
ii i
ii i
+= ⇒ =−
=⇒ =−=− A

6
3
2
tR
−
==Ω
−

Ex 5.6-1

5-3

Ex 5.6-2

212
3 A
6
26 V
a
aa
oc a
i
ii
vi
−
=⇒ =
== −
−

()
1262 3 A
2
32 32
3
aa a
sc a sc
ii i
ii i
+= ⇒ =−
=⇒ =−=− A

6
3
2
tR
−
==Ω
−

5-4

Ex 5.6-3

()
12241224
8
122436
24
30 20 V
1224
t
oc
R
v
× ×
= ==
+
==
+
Ω

20
8
i
R
=
+

Ex 5.7-1

()
6
1812 V
63
ocv==
+

()()36
24
36
tR=+=
+
Ω

For maximum power, we require

4
LtRR==Ω
Then
()
2 2
max
12
9 W
44 4
oc
t
v
p
R
== =

5-5

Ex 5.7-2
() ()
1
503
5.6 5.65 A
11 1 5015
315030
sci==
++
++
=
()15030
3 32528
15030
tR=+ =+ =Ω
+

()
2 2
max
285
175 W
44
tsc
Ri
p== =

Ex 5.7-3
()
()
2
10010
10
LL
tL tL
tL
RR
piv
RR RR
RR
  
== = 
++ +  

The power increases as Rt decreases so choose Rt = 1 Ω. Then

()
()
max 2
1005
13.9 W
15
pi v== =
+

Ex 5.7-4
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:

Rt = 20 Ω
and
()
2
max max45 42020
4
oc
oc t
t
v
pv pR
R
=⇒ = = = V
5-6

Problems
Section 5-3: Source Transformations

P5.3-1
(a)

5-7

= 2
= 0.5 V
t
t
R
v
∴ Ω
−

(b) 94 2(0.5)0
9(0.5)
1.58A
42
ii
i
−−−+−=
−+−
== −
+

9494(1.58)2.67Vvi=+ =+− =
(c) 1.58 A
aii==−
(checked using LNAP 8/15/02)

P5.3-2

Finally, apply KVL:
16
1034 0 2.19 A
3
aa aii i−+ +−= ∴=

(checked using LNAP 8/15/02)

5-8

P5.3-3

Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:

Equivalents for series resistors, series voltage source at left; series resistors, then source
transformation at top:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:
5-9

Finally, apply KVL to loop
o6 (919)36 0iv−+ +−−=
o5/2 42 28 (5/2)28Viv=⇒ =−+ =

(checked using LNAP 8/15/02)

P5.3-4

4 2000 400010 200030
375 A
aa a
a
ii i
i µ
−− − +− −=
∴=

(checked using LNAP 8/15/02)
5-10

P5.3-5

126 24 3 3 0 1 A
aa aii i−−+− −=⇒=

(checked using LNAP 8/15/02)

P5.3-6
A source transformation on the right side of the circuit, followed by replacing series resistors
with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:
5-11

Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources
with an equivalent current source:

Finally,
()
() ()
50100 100
0.21 0.217 V
50100 3
a
+
v== =

(checked using LNAP 8/15/02)

5-12

Section 5-4 Superposition

P5.4–1
Consider 6 A source only (open 9 A source)

Use current division:

1
1
15
6 40 V
20 15 30
v
v

=⇒

+
=

Consider 9 A source only (open 6 A source)

Use current division:

2
2
10
9 40 V
20 10 35
v
v

=⇒

+
=

12
40 40 80 Vvv v∴=+ = + =
(checked using LNAP 8/15/02)

P5.4-2
Consider 12 V source only (open both current sources)

KVL:

11 1
1
20 124120
1/ 3 mA
ii i
i
+++ =
⇒=−

Consider 3 mA source only (short 12 V and open 9
mA sources)

Current Division:

2
16 4
3 mA
1620 3
i

==

+

5-13

Consider 9 mA source only (short 12 V and open 3
mA sources)

Current Division:

3
12
9 3
2412
i

=− =−

+
mA

12 3
1/3 4/3 3 2 mAii ii∴=+ +=− + −=−

(checked using LNAP 8/15/02)

P5.4–3
Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to
the 30 mA current source.

1
26
30 6 mA 2 mA
28 612
aaii i
  
== ⇒ = =
  
++  

Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
due to the 15 mA current source.

2
46
15 6 mA 2 mA
46 612
bbii i
  
== ⇒ = =
  
++  

5-14

Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V
voltage source.

()
3
6||6 3
2.5 10 0.5 mA
6||612 312
i
 
=− =− =− 
++ 

Finally,
12 3
220.5 3.5 mAii ii=+ +=+− =
(checked using LNAP 8/15/02)

P5.4–4
Consider 10 V source only (open 30 mA source and
short the 8 V source)

Let v1 be the part of va due to the
10 V voltage source.

()
()
()
1
100||100
10
100||100100
50 10
10 V
150 3
v=
+
==

Consider 8 V source only (open 30 mA source and
short the 10 V source)

Let v2 be the part of va due to the
8 V voltage source.

()
()
()
1
100||100
8
100||100100
50 8
8 V
150 3
v=
+
==

5-15

Consider 30 mA source only (short both the 10 V
source and the 8 V source)

Let v2 be the part of va due to the
30 mA current source.

3
(100||100||100)(0.03)
100
(0.03)1 V
3
v=
==

Finally,
12 3
108
17 V
33
avv vv=+ +=++=

(checked using LNAP 8/15/02)

P5.4-5
Consider 8 V source only (open the 2 A source)

Let i1 be the part of ix due to the 8 V
voltage source.

Apply KVL to the supermesh:

()()()11 163 3 8iii 0+ + −=

1
82
A
123
i==
Consider 2 A source only (short the 8 V source)

Let i2 be the part of ix due to the 2 A
current source.

Apply KVL to the supermesh:

()( )22 263 23ii i0+ ++ =

2
61
A
122
i
−
==−

Finally,
12
21 1
A
32 6
xii i=+ =−=

5-16

Section 5-5: Thèvenin’s Theorem

P5.5-1

(checked using LNAP 8/15/02)
5-17

P5.5-2
The circuit from Figure P5.5-2a can be reduced to its Thevenin equivalent circuit in four steps:

(a)

(b)

(c)

(d)

Comparing (d) to Figure P5.5-2b shows that the Thevenin resistance is Rt = 16 Ω and the open
circuit voltage, voc = −12 V.

5-18

P5.5-3
The circuit from Figure P5.5-3a can be reduced to its Thevenin equivalent circuit in five steps:

(a)

(b)

(c)

(d)

(e)

Comparing (e) to Figure P5.5-3b shows that the Thevenin resistance is Rt = 4 Ω and the open
circuit voltage, voc = 2 V.

(checked using LNAP 8/15/02)

5-19

P5.5-4
Find Rt:

( )
()
12102
6
12102
tR
+
= =Ω
++

Write mesh equations to find voc:

Mesh equations:

( )
()
11 2 1
21 2
12106 0
63 18
ii ii
ii i 0
+ −− =
−+ −=

12
21
286
96 1
ii
ii 8
=
−=

11
2
1
3618 A
2
1417
= A
32 3
ii
i
=⇒ =

=



Finally,
21
71
3103 10 12 V
32
ocvi i
 
=+ = + =
 
 

(checked using LNAP 8/15/02)
5-20

P5.5-5
Find voc:

Notice that voc is the node voltage at node a. Express
the controlling voltage of the dependent source as a
function of the node voltage:

va = −voc

Apply KCL at node a:

63
0
84 4
oc oc
oc
vv
v
−  
−+ +−


=
V

62 6 0 2
oc oc oc ocvv v v−+ + − =⇒ =−

Find Rt:

We’ll find isc and use it to calculate Rt. Notice that
the short circuit forces

va = 0

Apply KCL at node a:

60 0 3
00
84 4
sci
−  
−+ +− +
  
  
=

63
A
84
sci==

28
34 3
oc
t
sc
v
R
i
−
== =−Ω

(checked using LNAP 8/15/02)

5-21

P5.5-6
Find voc:

Apply KCL at the top, middle node:
2
30 18 V
36
aa a
a
vv v
v
−
=+ +⇒ =
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V

Find isc:

Apply KCL at the top, middle node:
2
31
36 3
aa a a
a
vv v v
v 8 V
−
=+ + ⇒ =−
Apply Ohm’s law to the right-hand 3 Ω resistor :
18
6 V
33
a
sc
v
i
−
== =−
Finally:
18
3
6
oc
t
sc
v
R
i
== =−
−
Ω

(checked using LNAP 8/15/02)

5-22

P5.5-7

(a)
()
12
10
sa a
vRidRi−++ + =
()
12
1
s
a
v
i
RdR
=
++

()
()
2
12
1
1
s
oc
dR v
v
RdR
+
=
++

1
s
a
v
i
R
=
()
()
1
1
1
s
sc a
dv
id i
R
+
=+ =

2
0
T
aa T
v
idi i
R
−−+ −=
1aTRiv=−
()
()
21
12 12
1
1
TT
TT
Rd Rvv
id
RR RR
++
v=++ = ×
()
12
12 1
T
t
T
RRv
R
iR dR
==
++

(b) Let R1 = R2 = 1 kΩ. Then
1000 1000
625 20.4A/A
2 625
tRd
d
Ω= = ⇒ = −=−
+

and
()1 0.42
5 513.33V
20 .41
s
oc s
dv
vv
d
+ −+
== ⇒ = =
+− +

(checked using LNAP 8/15/02)

5-23

P5.5-8

oc
t
R
vv
RR
=
+

From the given data:
2000
6
2000 1.2 V
16004000
2
4000
oc
t oc
t
oc
t
v
R v
R
v
R

=

+ =
⇒
=− Ω

=
+


When R = 8000 Ω,

()
8000
1.21.5 V
16008000
v==
−+

P5.5-9

oc
t
v
i
RR
=
+

From the given data:
0.004
2000 24 V
4000
0.003
4000
oc
t oc
toc
t
v
R v
Rv
R

=

+ =
⇒
=Ω

=
+


(a) When i = 0.002 A:
24
0.002 8000
4000
R
R
= ⇒=
+
Ω

(b) Maximum i occurs when R = 0:
24
0.0066 mA 6 mA
4000
i== ⇒≤

P5.5-10
The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA.
The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V.

The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so
100.002
150
30
t
t
R
R
−
−= ⇒ =−
−−
Ω
5-24

P5.5-11

126000200010000
43000 A
4
1000 V
3
aa a
a
oc a
ii i
i
vi
−++ +
=
==
=

ia = 0 due to the short circuit

1260000 2mA
4
3
667
.002
sc sc
oc
t
sc
ii
v
R
i
−+= ⇒ =
== =Ω

4
3
667
bi
R
=
+

ib = 0.002 A requires

4
3
667 0
0.002
R = −=

(checked using LNAP 8/15/02)

5-25

P5.5-12

10 0 10 A
42 0
22 0
oc
oc
ii
vi i
vi V
=+⇒ =
+− =
⇒= −=−

10 10
scsii i i
c+=⇒ =−

40 20 0 10 A
sci iii+−= ⇒=⇒ =

20
2
10
oc
t
sc
v
R
i
−
=== −Ω

20
2
2
LL
L
iR
R
12
−
−== ⇒ =Ω
−

(checked using LNAP 8/15/02)
5-26

Section 5-6: Norton’s Theorem

P5.6-1

When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in
Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of
each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will
be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will
cool off.

P5.6-2

(checked using LNAP 8/16/02)

5-27

P5.6-3

P5.6-4

To determine the value of the short circuit current, isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit
current. Also, the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently,
2 scii=. The controlling current of the CCVS is expressed in terms of the mesh currents as
12 1asii iii
c=−= −
Apply KVL to mesh 1 to get

() ( )11 2 1 2 1 232 6 100 741ii i ii ii−− +− −=⇒ −=0 (1)

Apply KVL to mesh 2 to get

()21 2 1 2 1
11
56 0 6110
6
ii i i i i− −=⇒ −+ =⇒ =
2i

Substituting into equation 1 gives

22 2
11
7 410 1.13 A 1.13 A
6
scii i i

−= ⇒ = ⇒ =



5-28

Figure (a) Calculating the short circuit current, isc, using mesh equations.

To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the
terminals of the circuit and then label the voltage across that current source as shown in Figure
(b). The Thevenin resistance will be calculated from the current and voltage of the current source
as
T
t
T
v
R
i
=

In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current.
Consequently, i . The controlling current of the CCVS is expressed in terms of the mesh
currents as
2 Ti=
12 1aTii iii=−= +
Apply KVL to mesh 1 to get

() ()11 2 1 2 1 2 1
4
32 6 0 740
7
ii i ii ii i−− +− =⇒ −=⇒ =
2i (2)

Apply KVL to mesh 2 to get

( )21 2 1 256 0 611
TTiv ii i iv+− −=⇒−+ =−

Substituting for i1 using equation 2 gives

22 2
4
61 1 7.57
7
TTii v i

−+ =−⇒ =


v−
Finally,
2
7.57
TTT
t
TT
vvv
R
ii i
− −
====
−
Ω

5-29

Figure (b) Calculating the Thevenin resistance,
T
t
T
v
R
i
=, using mesh equations.

To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the
open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing
mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently,
. The controlling current of the CCVS is expressed in terms of the mesh currents as
2
0 Ai=

12 10
aii ii
1i=−= −=
Apply KVL to mesh 1 to get

() () ()()11 2 1 2 1 1 1
1
32 6 100 32 06 0100
10
1.43 A
7
ii i ii i i i
i
−− +− −=⇒ − −+ −−=
⇒= =

Apply KVL to mesh 2 to get

() ()()
21 2 15 6 0 6 61.438.58 V
oc ociv ii v i+− −=⇒ = = =

Figure (c) Calculating the open circuit voltage, voc, using mesh equations.

As a check, notice that()()7.571.138.55
tsc oc
Riv== ≈

(checked using LNAP 8/16/02)
5-30

P5.6-5
To determine the value of the short circuit current, Isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit
current. Also, the nodes have been identified and labeled in anticipation of writing node
equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v . The voltage at node 3 is equal to the voltage across a short,
1
24 V=−
30v=.
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The
voltage at node 3 is equal to the voltage across a short, i.e.
2v=
av
30v=.
Apply KCL at node 2 to get

12 2 3
13 223 483 1
36
aa
vv vv
vv v v v
−−
=⇒ += ⇒−= ⇒ =−6 V

Apply KCL at node 3 to get

()
23
2
49 9
1624 A
63 6 6
sc a sc sc
vv
vi vi i
−
+= ⇒ = ⇒ =−=−

Figure (a) Calculating the short circuit current, Isc, using mesh equations.

To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across
the terminals of the circuit and then label the voltage across that current source as shown in
Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current
source as
T
th
T
v
R
i
=

Also, the nodes have been identified and labeled in anticipation of writing node equations. Let
v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
5-31

In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. . The
controlling voltage of the VCCS, v
1
0v=
a=a, is equal to the node voltage at node 2, i.e. v . The
voltage at node 3 is equal to the voltage across the current source, i.e. vv
2v
3 T=.
Apply KCL at node 2 to get

12 2 3
13 223
36
Ta
vv vv
vv v v v
−−
=⇒ += ⇒ =3

Apply KCL at node 3 to get

23
22 3
4
09 60
63
96 0
36 0 2
TT
aT T
TT T T
vv
vi vv i
vv i
vv i v i
−
++ =⇒ −+=
⇒− +=
⇒− +=⇒ =−6
T

Finally,
3
T
t
T
v
R
i
==−Ω

Figure (b) Calculating the Thevenin resistance,
T
th
T
v
R
i
=, using mesh equations.

To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the
open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing
node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v . The controlling voltage of the VCCS, v
1
24 V=−
2avv=
a, is equal to the node voltage
at node 2, i.e. . The voltage at node 3 is equal to the open circuit voltage, i.e. .
3 ocvv=
Apply KCL at node 2 to get

5-32

12 2 3
13 223 48
36
oc a
vv vv
vv v v v
−−
=⇒ += ⇒−+=3

Apply KCL at node 3 to get

23
22 3
4
09 09
63
ao
vv
vv v v
−
+= ⇒ −=⇒ =
cv

Combining these equations gives

()348 9 72 V
oc a oc oc
vv v v−+ = = ⇒ =

Figure (c) Calculating the open circuit voltage, voc, using node equations.

As a check, notice that
()()32472
thsc oc
RIV=−− ==

(checked using LNAP 8/16/02)

Section 5-7: Maximum Power Transfer

P5.7-1

a) For maximum power transfer, set RL equal
to the Thevenin resistance:

1001101
LtRR== +=Ω

b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin
equivalent circuit:
5-33

The voltage across RL is ()
101
10050 V
101101
Lv==
+

Then
2 2
max
50
24.75 W
101
L
L
v
p
R
== =

P5.7-2
Reduce the circuit using source transformations:

Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value
of that maximum power is
2 2
() (0.03)(60) 54mW
max
RPi R== =

5-34

P5.7-3

L
LS
SL
22
LS L
L 2
LS L

()
R
vv
RR
vv R
p
RR R

=
+
∴= =
+

By inspection, pL is max when you reduce RS to get the
smallest denominator.
∴ set RS = 0

P5.7-4
Find Rt by finding isc and voc:

The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix.
Apply KCL at the top-left node to get

0.9
0.910 0.1 A
9
xx xii i+= ⇒ ==
so
isc = 10 ix = 1A
Next

Apply KCL at the top-left node to get
5-35

0.9
0.910 0.1 A
9
xx xii i+= ⇒ ==

Apply Ohm’s law to the 3 Ω resistor to get

() ()310 300.13 V
oc x
vi== =

For maximum power transfer to RL:
3
3
1
oc
Lt
sc
v
RR
i
=== =Ω

The maximum power delivered to RL is given by

()
2 2
max
3 3
W
44 34
oc
t
v
p
R
== =

P5.7-5

The required value of R is

() ( )
() ( )
201201050
85
201201050
tRR
++
== + =
++ +
0Ω

() ()
170 30
2010 2050
17030 17030
170(20)(10)30(20)(50)4000
20 V
200 200
ocv
 
=−
 
++ 
−
== =

The maximum power is given by
()
2 2
max
20
2 W
44 50
oc
t
v
p
R
== =

5-36

PSpice Problems

SP5-1

a = 0.3333

b = 0.3333

c =33.33 V/A

(a)
1 20.33330.333333.33
ovv v=+ +
3i

(b) () ()
33
18
7
33
70.3333100.3333833.33 30 mA
100100
3
ii
−
=+ + ⇒ = = =
5-37

SP5-2
Before the source transformation:

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V1 -3.000E-02
V_V2 -4.000E-02

After the source transformation:

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V2 -4.000E-02

5-38

SP5-3

voc = −2 V

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V3 -7.500E-01
V_V4 7.500E-01

isc = 0.75 A

Rt = −2.66 Ω

5-39

SP5-4

voc = 8.571 V

VOLTAGE SOURCE CURRENTS

NAME CURRENT

V_V5 -2.075E+00
V_V6 1.132E+00
X_H1.VH_H1 9.434E-01

isc = 1.132 A

Rt = 7.571 Ω

5-40

Verification Problems

VP5-1

Use the data in the first two lines of the table to determine voc
and Rt:

0.0972
0 39.9 V
410
0.0438
500
oc
t oc
toc
t
v
R v
Rv
R

=

+ =
⇒
=Ω

=
+


Now check the third line of the table. When R= 5000 Ω:
39.9
7.37 mA
4105000
oc
t
v
i
RR
== =
++

which disagree with the data in the table.

The data is not consistent.

VP5-2

Use the data in the table to determine voc and isc:
12 V(line 1 of the table)
3 mA (line 3 of the table)
so 4 k
oc
sc
oc
t
sc
v
i
v
R
i
=
=
== Ω

Next, check line 2 of the table. When R = 10 kΩ:
() ()
33
12
0.857 mA
1010510
oc
t
v
i
RR
== =
+ +

which agrees with the data in the table.
To cause i = 1 mA requires
()
3
12
0.001 8000
1010
oc
t
v
iR
RR R
=== ⇒ =
+ +
Ω
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.

5-41

VP5-3

()
60 60
11 11
54.55 mA
6 6040
11040
11
oc
t
v
i
RR
== = =
++
+

The measurement supports the prelab calculation.

Design Problems

DP5-1
The equation of representing the straight line in Figure DP 5-1b is
tvR iv
oc=−+. That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
open circuit voltage. Therefore:
05
625
0.0080
tR
−
=− =Ω
−
and voc = 5 V.
Try RR . (R
12 1k== Ω 1 || R2 must be smaller than Rt = 625 Ω.) Then
2
12
1
51
2
ss s
R
vv v
RR
== ⇒=
+
0V
and
12
33 3
12
625 500 125
RR
RR R
RR
=+ =+ ⇒ =
+
Ω
Now vs, R1, R2 and R3 have all been specified so the design is complete.

oc
DP5-2
The equation of representing the straight line in Figure DP 5-2b is
tvR iv=−+. That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
open circuit voltage. Therefore:
()03
500
0.0060
tR
−−
=− =Ω
−−
and voc = −3 V.
From the circuit we calculate
()31 2
12
t
3
RRR
R
RRR
+
=
++
and
13
s
12 3
oc
RR
vi
RRR
=−
++

so
()31 2
12
500
3
RRR
RRR
+
Ω=
++
and
13
s
12 3
3V
RR
i
RRR
−=−
++

5-42

Try R= and . Then
31kΩ
12 1kRR+= Ω 500
tR= Ωand
11
s1
1000
36
2000 2
s
RR
ii−=− =− ⇒=
sRi

This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω =
400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.

DP5-3
The slope of the graph is positive so the Thevenin resistance is negative. This would require
12
3
12
0
RR
R
RR
+
+
<, which is not possible since R1, R2 and R3 will all be non-negative.
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.

oc
DP5-4
The equation of representing the straight line in Figure DP 5-4b is
tvR iv=−+. That is, the
slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open
circuit voltage. Therefore:
50
625
00.008
tR
−−
=− =− Ω
−
and voc = −5 V.

The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt,
of this circuit are given by
()
()
2
12
1
1
oc s
Rd
vv
Rd R
+
=
++
,
()
1
1
scs
d
iv
R
+
=

and
()
12
12 1
t
RR
R
RdR
=
++

Let R1 = R2 = 1 kΩ. Then
1000 1000
625 23.6A/A
2 625
tRd
d
−Ω == ⇒ = −=−
+−
and
()
()
1 3.62
5 53.077V
2 3.61
s
s
dv
v
d
+ −+
−= ⇒ = −=−
+− +

Now vs, R1, R2 and d have all been specified so the design is complete.

5-43

Chapter 6: The Operational Amplifier

Exercises

Ex. 6.4-1

12
2
1
0 0
1
ss o
o
s
vv v
RR
vR
vR
−
+ +=
=+

Ex. 6.4-2

a)
2
12
0
34
00
as
aa
R
vv
RR
vv v
RR
=
+
−
+ +=

42
31 2
11
oo
as
vv
4
3
R RR
vR vRR R

=+ ⇒ = +
+ 

b)
22
21
12 2
When then 1 and 1
o
s
4
3
R Rv
RR
R
RRR v
>> − = −+
+

R

6-1

Ex. 6.5-1

21
2
12
00
o
os
o
s
vvv
RR
vR
vR R
−
+ +=
=
+

Ex. 6.6–1

11
1
3
3
0 0 1
when 100 k and 25 k then
10010
1 5
2510
fin out in
out in
f
f
out
in
Rvv v
vv
RR R
RR
v
v
−
++ =⇒ =+


=Ω = Ω
 ⋅
=+ =
⋅

6-2

Ex. 6.7-1

33
21
32 33 3
2
21
2133
2
1010 1010
11
1010 1010 1010 1010
21
1010 1010
RR
vv
R
RR
vv
R
   ××
=− ++ +   
×+ × × ×  
 
=− ++ 
+× ×
13
v

We require ()
31
1
4
5
vv

=−


2
v, so

1 3
13
421 101010 k
1010
R
R

=+ ⇒ =×=
×
Ω
and
2 3
22 23
2
1
10105 2.5 k
51 010
R
RR R
R
=⇒ +×= ⇒ =
+×
Ω

6-3

Ex. 6.7-2
As in Ex 6.7-1
21
32 33
2
21
1010 1010
RR
vv
R
 
=− ++ 
+× ×
1
v
We require ()
31
4
6
5
vv

=−


2
v, so

1 3
13
621 201020 k
1010
R
R

=+ ⇒ =×=
×
Ω
and
2 3
22 23
2
4
44 0105 40 k
51 010
R
RR R
R
=⇒ +×= ⇒ =
+×
Ω

Ex. 6.8-1
3
1
1
3 3
1
Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 (5010)
For a A741 op amp, 1 mV and 80nA so
3
output offset voltage 6 (5010)6 (10)(50.10
os b
os b
os b
vi
vi
vi
µ
−
+×
≤≤
=+ × ≤ +
9
)(8010) 10mV
−
×=

Ex. 6.8-2

22
21
11
1
oi n os
RR
vv v
RR

=− ++ +

b
Ri

() ( )( )
21 1
33 9 3
When 10 k, 2 k, 5 mV and 500 nA then
output offset voltage 6510 1010 500.103510 35 mV
os bRR v i
− − −
=Ω =Ω ≤ ≤
≤× +× ≤×=

6-4

Ex. 6.8-3
()
3
1
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 5010
For a typical OPA1O1AM, 0.1 mV and 0.012 nA so
os b
os b
vi
vi
+×
==

( )
33 9
36 3
output offset voltage 60.110 50100.01210
0.610 0.610 0.6100.6 mV
−−
−− −
  ≤× +× ×
  
≤× +× −×=

Ex. 6.8-4

Writing node equations
0
0
0
so
ab i s
o
o
b
vv vv v
RR RR
R
i
vA v
RR
vvis
RR
−− −
−
−
−−
++ =
+

−−
 +
−
+ =

After some algebra
( )
() () ()
0
00
is ifo
v
s f is a f is i
RR RARRv
A
v
a
RRR RRRRRRARR
++
==
++ + +++−

For the given values, 2.00006 V/V.
vA=−

6-5

Problems

Section 6-4: The Ideal Operational Amplifier

P6.4-1

(checked using LNAP 8/16/02)

P6.4-2

Apply KVL to loop 1:

11
1
12300002000 0
12
2.4mA
5000
ii
i
−++ + =
⇒= =

The currents into the inputs of an ideal op
amp are zero so
()
o1
21
2
2.4mA
2.4mA
1000 02.4V
a
ii
ii
vi
==
=− =−
=+ =−

Apply Ohm’s law to the 4 kΩ resistor
()
() ()
o
3
4000
2.42.410400012V
oavv i
−
=−
=−−× =−

(checked using LNAP 8/16/02)
6-6

P6.4-3
The voltages at the input nodes of an ideal op amp
are equal so 2 V
av=−.

Apply KCL at node a:

() ()2122
03
8000 4000
o
o
v
v
−− −−
+= ⇒=−0 V

Apply Ohm’s law to the 8 kΩ resistor

2
3.5 mA
8000
o
o
v
i
−−
==

(checked using LNAP 8/16/02)
P6.4-4
The voltages at the input nodes of an ideal
op amp are equal so . 5 Vv=

Apply KCL at the inverting input node of
the op amp:
35
0.1100 = 0 = 4 V
a10000
a
v
v
−−
−− ×− ⇒



Apply Ohm’s law to the 20 kΩ resistor
1
mA
20000 5
av
i==

(checked using LNAP 8/16/02)

P6.4-5
The voltages at the input nodes of
an ideal op amp are equal so
. Apply KCL at node a: 0 V
av=

30120
2100
3000 4000
15V
o
o
v
v
−−− 
−− −⋅


⇒= −
=

Apply KCL at the output node of
the op amp:
0 7.5 mA
60003000
oo
oo
vv
ii++ =⇒=
(checked using LNAP 8/16/02)

6-7

P6.4-6
The currents into the inputs of an ideal op amp
are zero and the voltages at the input nodes of
an ideal op amp are equal so .
Apply Ohm’s law to the 4 kΩ resistor:
2.5 V
av=
2.5
0.625 mA
40004000
a
a
v
i===

Apply KCL at node a:
0.625 mA
baii==
Apply KVL:
() ( )
33
80004000
12100.62510 7.5 V
ob avi i
−
=+
=× × =

(checked using LNAP 8/16/02)

P6.4-7

2
12 1
23 23
23 23 13
42 4
43 3 1
00
00
00
00
00
o
sa
as
aa
oa
a
oa
Rvv
vv
RR R
RR RRvv
iv
RR RR RR
RR Rvv
vv
RR R RR
−−
−− +=⇒ =−

 
++−−
=+ =− =


  −−
−− +=⇒ =− =  
  
   3
s
s
v
v

6-8

P6.4-8

The node voltages have been
labeled using:

1. The currents into the
inputs of an ideal op amp are
zero and the voltages at the
input nodes of an ideal op
amp are equal.

2. KCL

3. Ohm’s law

Then

011.81.810 Vv=− =
and
10
2.5 mA
4000
o
i==

(checked using LNAP 8/16/02)
P6.4-9
KCL at node a:

()18
00 12 V
40008000
a a
a
v v
v
−−
++ =⇒ =−

The node voltages at the input nodes of
ideal op amps are equal so .
bavv=

Voltage division:

8000
8 V
40008000
ob
vv==
+
−

(check using LNAP 8/16/02)

6-9

Section 6-5: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers

P6.5-1

KCL at node b:
25
0 V
200004000040000 4
bb b
b
vv v
v
−+
++ =⇒ =−
1

The node voltages at the input nodes of an ideal op amp are equal so
1
V
4
eb
vv==− .
KCL at node e:
10
01 0
10009000 4
ee d
de
vv v
vv V
−
+= ⇒ = =−

(checked using LNAP 8/16/02)
P6.5-2
Apply KCL at node a:

12 0
0 4
600060006000
aa a
a
vv v
v
−−
=+ + ⇒ =V

Apply KCL at the inverting input of the
op amp:

o
o
00
0
6000 6000
4 V
a
a
vv
vv
−−  
−+ +
  
  
⇒= −=−
0=

Apply KCL at the output of the op amp:

o
o
oo
o
0

60006000
1.33 mA
3000
vv
i
v
i
−
−+ =


⇒= − =
0

(checked using LNAP 8/16/02)
6-10

P6.5-3

Apply KCL at the inverting input of
the op amp:

2 1
2
1
00

as
as
vv
RR
R
vv
R
  −−
0− −=  
  
⇒= −

Apply KCL at node a:
0 23 24 34

04
43 2 4 3 2 23
23 24 34
13
01 11
0

aa a
aa
s
vvvv RRRRRR
vR v v
RR R RRR RR
RRRRRR
v
RR
−− + +
++ =⇒ = ++ =

++
=−

o3090030
Plug in values yields 200 V/V
4.8
s
v
v
++
⇒= − =−

P6.5-4

Ohm’s law:
12
2
vv
i
R
−
=

KVL:

() ()
12 3
01 2 3 1
2
RR R
vR RRi vv
R
++
=+ + = −
2

6-11

P6.5-5

11 2 1
12
17 7
21 2 2 2
21
27 7 7
0 0 1
0 0 1
a
a
b
b
vv vv R R
vv
RR R R
vv vv R R
vv v
RR R R
−−
++ = ⇒ =+ −

−−
−+ =⇒=+ −

1
7
v

6
46 4 6
05
0
35 3
0
0 0
0 0 (1)
bc c
cb
ac c
ac
vv v R
vv
RR RR
vv vv R R
vv
RR R R
−−
−+ +=⇒ =
+
  −−
−+ += ⇒=− ++  
  
5
3
v

()
()
( )
()
63 5 63 5 251 521
02
37 34 6 7 3 7 3 4 6 7
0
0

(1) (1)

5
c
RR R RRRRRR RRR
vv
RRRRR R RRRRRR
vv
i
R
 ++
=+ + − ++ 
++
 
−
== "
1v




6-12

P6.5-6

KCL at node b:
33
5
0
20102510 4
ac
ca
vv
vv+= ⇒ =−
××

KCL at node a:
()
33 3 3
5
12 0 124
0 V
401010102010 1010 13
aa
a aa
a
vv
v vv
v

−−

−− + 
++ + =⇒ =−
×× × ×

So
51
41
ca
5
3
vv=−= −.
6-13

P6.5-7
Apply KCL at the inverting input
node of the op amp

()600
0
10000 30000
1.5V
aa
a
vv
v
+−−
−+ −
 
⇒= −
0=

Apply KCL to the super node
corresponding the voltage source:

()
()
06 0
1000030000
6
0
3000010000
3 6
36 0
26 3 V
aa
abab
aa a b
ab
ba
vv
vvvv
vv vv
vv
vv
−+ −
+
+−−
++ =
⇒+ ++−
++ − =

⇒= +=

Apply KCL at node b:

()
() ( )()
0
0
0
6
0
100003000030000 10000
33 6
8418 12 V
abbb a b
bb a b a b
ba
vvvv v vv
vv vvv v v
vv v
+−−−
+− − =
 
⇒ +−− −− +−=

⇒= −−=
0

Apply KCL at the output node of the op amp:

00
00
0 0.7 mA
3000030000
b
vv v
ii
−
++ =⇒=−
6-14

P6.5-8
Apply KVL to the bottom mesh:

00
0
(10000)(20000)5 0
1
mA
6
ii
i
−− +
⇒=
=

The node voltages at the input nodes of an
ideal op amp are equal. Consequently

0
10
10000 V
6
a
vi==
Apply KCL at node a:

0
0
0 35
1000020000
aa
a
vv v
vv
−
+= ⇒ = =V

P6.5-9

KCL at node b:
12
04
4000020000
bb
b
vv
v
+
+= ⇒ =− V
The node voltages at the input nodes of an ideal op amp are equal, so 4 V
cbvv==− .

The node voltages at the input nodes of an ideal op amp are equal, so .
4
010 4 V
dc
vv=+ ×=−

KCL at node g:
33
2
0
20104010 3
fg g
g f
vv v
vv
−
−+ =⇒ =
××

6-15

The node voltages at the input nodes of an ideal op amp are equal, so
2
3
eg
vv v==
f
.
KCL at node d:
33 3 3
2
62 43
0 V
2010201020102010 5 5
df
df dfde
fd
vv
vv vvvv
vv
−
−− −
=+ =+ ⇒ = =−
×× × ×

Finally,
21 6
V
35
eg f
v== =−vv .

P6.5-10
By voltage division (or by applying KCL at
node a)

0
10

as
R
vv
RR
=
+

Applying KCL at node b:

()
0
10
0
0
1
0

b sb
bs b
vv vv
RR R
RR
vv vv
R
−−
+=
+∆
+∆
⇒− +=

The node voltages at the input nodes of an
ideal op amp are equal so vv.
ba=

00 0 0

0
1 1 0 1 10 10 0

1
ss s
RR R RR RR R
vv v v
R RR R RR RRR
 +∆ +∆ ∆ ∆
=+ −= − =− 
++ 
+

6-16

Section 6-6: Design Using Operational Amplifier

P6.6-1
Use the current-to-voltage converter, entry (g) in Figure 6.6-1.

P6.6-2
Use the voltage –controlled current source, entry (i) in Figure 6.6-1.

P6.6-3
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.

6-17

P6.6-4
Use the difference amplifier, entry (f) in Figure 6.6-1.

P6.6-5
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.

P6.6-6
Use the negative resistance converter, entry (h) in Figure 6.6-1.

6-18

P6.6-7
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the
current iin to be zero.

P6.6-8

( )
12
12
Summing Amplifier: 62
62
Inverting Amplifier:
a
o
oa
vv v
vv
vv
=− + 
⇒= +
=−

v

P6.6-9

6-19

Using superposition,
12 3916327 V
ovv vv=++ =−−+=

P6.6-10

R1 6 12 24 6||12 6||24
R2 12||12||24 6||12||24 6||12||12 12||24 12||12
-vo/vs 0.8 0.286 0.125 2 1.25

R1 12||12 12||24 6||12||12 6||12||24 12||12||24
R2 6||24 6||12 24 12 6
-vo/vs 0.8 0.5 8 3.5 1.25

6-20

Section 6-7: Operational Amplifier Circuits and Linear Algebraic Equations

P6.7-1

6-21

P6.7-2

6-22

Section 6-8: Characteristics of the Practical Operational Amplifier

P6.8-1

The node equation at node a is:
133

10010 1010
outos os
b
vv v
i
−
= +
××

Solving for vout:
() ()
() ( )()
3
33
113
33 9
10010
1 10010 11 10010
1010
110.0310 100101.210 0.45 mV
out os b os b
vv i v
−−
×
=+ + × = + ×

×
=× +× × =
i

P6.8-2

The node equation at node a is:
0
1

10000 90000
os os
b
vv
i
v−
+=
Solving for vo:

() ( )
()
3
33
o 1 os b13
33 9 3
9010
1 9010 10 9010 i
1010
10(510)+9010(.0510) 50.00451050 mV
os b
vv i v
−− −
×
=+ +× = +×

×
=× × × = ×−
6-23

P6.8-3

11 01
12 00 2

01 01 1 0 2 1
02
0
()

() ( ) (1
0
in
in in
in in in
vv vvv
RR R vR RAR
vAvvv v RRRRRR A
RR
−− 
++ =

−
⇒=
+− + ++ +
+=


)

P6.8-4
a)
3
02
3
1
4910
= 9.6078
5.110
in
v R
vR
×
=− =− −
×

b)
() () ( )()
63
0
36 3 3 6
21075200,0005010
9.9957
(510210)(755010)(510)(210)(1200,000)
in
v
v
×− ×
==
×+× +×+× × +
−

c)
63
0
36 3 3 6
210(75(200,000)(4910))
= 9.6037
(5.110+210)(75+4910)+(5.110)(210)(1+200,000)
in
v
v
×− ×
=−
×× × × ×

6-24

P6.8-5

Apply KCL at node b:
3
23
()
bc m
R
vv
RR
=−
+
pv
Apply KCL at node a:
0
41
( )
0
aa cm nvv vvv
RR
− −+
+ =
The voltages at the input nodes of
an ideal op amp are equal so

abvv=.
44
0
11
()
cm n a
1R RR
vv v
RR
v
+
=− ++
4
0
1
41 3
12 3
()
()
()
()
cm n
cm p
R
vv v
R
RRR
vv
RRR
=−+ +
+
−
+

4
34 13 34 41
312 12 3 2 1
2
44 4
0
11 1
1
()
when then
()
1
so
() ( ) (
cm n cm p n p
R
RR RR RRR R
RRR RRR R R
R
RR R
vv v vv v
RR R
+
+
== ×
+
+
=− ++ − =− +)v
=

6-25

PSpice Problems

SP6-1:
(a)
zw xvavbvcv
y= ++
The following three PSpice simulations show

1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V

4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V

-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V

Therefore 45
zw xvv vv
y=+−

(b) When vw= 2 V: 45 2
zx yvv v=− +

1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:

6-26

4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V:

-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V:

6-27

SP6-2

a) Using superposition and recognizing the inverting
and noninverting amplifiers:

()
os s
80 80
12 3.2
25 25
vv v

=− ++ −=− −


8.4

b) Using the DC Sweep feature of PSpice produces the
plot show below. Two points have been labeled in
anticipation of c).

c) Notice that the equation predicts

()()3.258.47.6−− −=
and
()()3.208.48.4− −=−
Both agree with labeled points on the plot.

6-28

SP6-3

VOLTAGE SOURCE CURRENTS
NAME CURRENT

V_V1 -3.000E-04
V_V2 -7.000E-04

()
6
34
23
50
4
o
1.51210 1.5 V
4.51.56 V
12012 V
710 0.7 mA
v
v
v
i
−
−
=−−−× ≅−
=− −=
=− =
=−× =−

6-29

SP6-4
V4 is a short circuit used to measure io.

The input of the VCCS is the voltage of the
left-hand voltage source. (The nominal value
of the input is 20 mV.) The output of the
VCCS is io.

A plot of the output of the VCCS versus the
input is shown below.

The gain of the VCVS is

()
()
66
3
33
5010 5010 1A
10
2V10010 10010
gain
−−
−
−−
×− −×
==
×− −×
×

(The op amp saturates for the inputs larger than
0.1 V, limiting the operating range of this
VCCS.)

6-30

Verification Problems

VP6-1
Apply KCL at the output node of the op
amp

()
oo
o
5
0
100004000
vv
i
−−
= +=

Try the given values: io =−1 mA and
vo = 7 V

()
33
757
1103.710
100004000
−−
−−
−× ≠× = +

KCL is not satisfied. These cannot be the
correct values of io and vo.

VP6-2

() ( )
()
33
3

o 3
o
410210 8 V
1210
1.28 9.6 V
1010
So 9.6 V instead of 9.6 V.
a
a
v
vv
v
−
=× × =
×
=− =− =−
×
=−

6-31

VP6-3
First, redraw the circuit as:

Then using superposition, and recognizing of the inverting and noninverting amplifiers:

() ()
o
64 4
31 218612
22 2
v
 
=− −−++ =−+=−
 
 
V
c

The given answer is correct.

VP6-4
First notice that vv is required by the ideal op amp. (There is zero current into the input
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f,
hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op
amp are equal.)
ef v==

The given voltages satisfy all the node equations at nodes b, c and d:

node b:
0(5)0 02
0
10000400004000
−−−
+ +=

node c:
02 25
0
40006000
− −
= +

node d:
25 5511
600050004000
− −
=+

Therefore, the analysis is correct.

6-32

VP6-5
The given voltages satisfy the node equations at nodes b and e:

node b:
().255.252.25
0
200004000040000
−−−−− −
+ +=

node e:
( )2.50.250.25
0
9000 1000
−−− −
≠ +

Therefore, the analysis is not correct.

Notice that
( )2.50.250.25
0
9000 1000
−−+ +
= +

So it appears that v instead of
e0.25 V=+
e0.25 V.v=−

Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation

() ()()
12
4
11
2.5 102 5
24
da cvK KvKv

−= = + = +−



None-the-less, the analysis is not correct.

6-33

Design Problems

DP6-1
From Figure 6.6-1g, this circuit

is described by . Since
o fvR i=
in
o
out
5000
v
i= , we require
out f
in
1
4 5000
iR
i
== , or Rf = 1250 Ω
Notice that
()
in
oa in in
1250 5
5000 4
i
i=+ =ii . To avoid current saturation requires
in sat
5
4
ii<or
in sat
4
5
i<i. For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation.

DP6-2
33 312 3 3 12
31 5 1
44 435 4 4122
oi i ivv v v
  
=− +=−++ −++
  
+  
5
3

6-34

DP6-3
(a) ()
41 2
11 1 1
624 5 24, ,and .
220 2 20
iivv K K K

+= + ⇒ = = =


12 Take Ra = 18 kΩ and
Rb = 1 kΩ to get

(b)

(c)

(d)

6-35

DP6-4

6-36

DP6-5
We require a gain of
3
4
200
2010
−
=
×
. Using an inverting amplifier:

Here we have
2
3
1
200
1010
R
R
=−
×+
. For example, let R1 = 0 and R2 = 1 MΩ. Next, using the
noninverting amplifier:

Here we have
2
1
2001
R
R
=+ . For example, let R1 = 1 kΩ and R2 = 199 kΩ.

The gain of the inverting amplifier circuit does not depend on the resistance of the microphone.
Consequently, the gain does not change when the microphone resistance changes.

6-37

Chapter 7 Energy Storage Elements

Exercises

Ex. 7.3-1
() ()
22 4
1 14
0otherwise
Cs
t
d
it vt t
dt
<<

== − <


8< 8
e
and () ()
24 2 4
18 4
0otherwis
Rs
tt
it vt t t
− <<

= =− <<



so () ()
22 2 4
() 7 4 8
0otherwis
CR
tt
ititit t t
−< <

=+= − <<


e

Ex. 7.3-2
() ()
0
0
0
11
() () 12
1
3
tt
ss
t
vt idvt id
C
ττ ττ=+ =∫∫
−
<
<

0
()34 121212for0 4
t
vt d t tτ=− =− <∫
In particular, v(4) = 36 V.

()
4
()32 36606for4 10
t
vt d t tτ=− +=− <<∫
In particular, v(10) = 0 V.

10
()30 00for10
t
vt d tτ=+ =∫

Ex. 7.4-1
() ()
() ()
2
2
41
210100 1 J
22
0 0 100V
cc
Cv
vv
−
+−
== × =
==
W

7-1

Ex. 7.4-2
a)

b)
() ()
() ()
()()
() () ()
0
44
00
44
0
0
First, 0 0 since 0 0
1
Next, 0 102 210
210 2 210

t
tt
t
tv idt
v
vtv idt dt t
C
tt dt t
=+
==
=+ = =×
2
∴ =× =×
∫
∫∫
∫
WW
W
W

()
4
1s 210J = 20 kJ=×W
() ()
2
48
100s 210100 210J = 200 MJ=× =×W

Ex. 7.4-3

a)

b)

() ()
55 5
00
We have (0) (0) 3 V
1
() (0) 53 3 313 3V, 01
tt
tt t
cc
vv
vt itdtv edt e e t
C
+−
==
=+ = += −+=∫∫
<<

()()()()
55 5
() 5 15 3 18 V, 0 1
tt t
Rc c
vtvtvt itvt e e e t=+ = + = + = <<

() () ()
()
2
0.225 10
0.8
() 6.65 J
11
0.23 0.9 J
22
2.68 kJ
tstt
c
ts
t
tC vt e e
t
=
=
 =

== × = ⇒
=
W
W
W

Ex. 7.5-1

7-2

Ex. 7.5-2
12 1 2 1
12 1
12 2

dvdv ii C
vv i i
dtdt CC C
=⇒ = ⇒ =⇒ =
2
KCL:
12
12 2 2
21
1
CC
iii i i
CC

=+ = + ⇒ =
+ 2
i
C

Ex. 7.5-3

(a) to (b) :
1 1
mF
111 9
+ +
111
333
= , (b) to (c) :
110
= mF
99
1+ ,

(c) to (d) :
11 11 10
= + + = mF
1022 19
9
eq
eq
C
C
⇒

7-3

Ex. 7.6-1
() ()
22 4
1 14
0otherwis
Ls
t
d
vt it t
dt
<<

== − <


8
e
< 8
e
and () ()
24 2 4
18 4
0otherwis
Rs
tt
vt it t t
− <<

= =− <<



so () ()
22 2 4
() 7 4 8
0otherwis
LR
tt
vtvtvt t t
−< <

=+ =− <<


e

Ex. 7.6-2
() ()
0
0
0
11
() () 12
1
3
tt
ss
t
it vdit vd
L
ττ ττ=+ =∫∫
−
<
<

0
()34121212for0 4
t
it d t tτ=− =− <∫
In particular, i(4) = 36 A.
()
4
()32 36606for4 10
t
it d t tτ=− +=− <<∫
In particular, i(10) = 0 A.

10
()30 00for10
t
it d tτ=+ =∫

Ex. 7.7-1
() ()
() () ()
()
2
2
22 2
1
4 1 V
4
1 4 41 W
11 1
42 J
22 4
tt
tt
tt
di d
vL te te
dt dt
Pvi te te tte
Li te te
−−
−− −
−−

== =−


== − = −


== =


W
t
7-4

Ex. 7.7-2
() ()
()
()
0 0 00
2 01 101
an d
22 1221
02 0 2
t t
tt tdidi
vtL it vt
ttdtdt t
tt
< <
 
<< 1
12
<<
== = ⇒ =
−− << − <<

 >> 

() ()()
()
0
2 0 1

22 1
0
t
tt
pt vtit
tt
t
<

<<
== 
−< <


>
0
2
2

() () ()
() ()
0
0
0

() 0 for 0 0 for 0 0
t
t
tt ptdt
it t pt t t
=+ ∫
=< ⇒ = <⇒
WW
W =

()
() () ()
() ()
2
0
2
1
01 : 2
12 : 122 4
2 : 2 0
t
t
tt tdtt
tt tdtt t
tt
<< = =
<< = + − =−+
>= =
∫
∫
W
WW
WW
4

Ex. 7.8-1

7-5

Ex. 7.8-2

Ex. 7.8-3

() () () ()
00
1 10 2 20 10 20
12
11
, but 0 and 0
tt
tt
i vdtit i vdtit it it
LL
=+ = + =∫∫ =
00 0 0
0
0
12
11 2
12 11
12
12
11 1
+
11

11 1
+
tt t t
tt t t
P
t
t
t
t
P
iii vdtvdt vdt vdt
LL L
vdt
iLLL
iL L
vdt
LL L

=+ = + = =∫∫ ∫ ∫

∫
1
L
∴== =
+
∫

7-6

Problems

Section 7-3: Capacitors

P7.3-1

() () ()
t
0
1
0 vtv idandqCv
C
ττ=+ =∫

In our case, the current is constant so ()
t
0
idττ∫
.
() ()
() () ()
66
3
0
15010151050
3 ms
2510
CvtCv it
qCv
t
i
−−
−
∴ =+
×− ×−
∴== =
×

P7.3-2
() () () ()()() (
11
12cos230 122sin2 303cos2 120 A
88
dd
itCvt t t t
dt dt
°°
== + = − + = + )
°

P7.3-3
() ( ) () ()( )()
() ()
3
310cos50045 12cos50045 12500sin50045
6000cos50045
d
tC t C t
dt
Ct
−° °
°
×+ = − = −
=+
°
−

so
3
6
3
3101 1
10 F
6102 2
C µ
−
−×
== × =
×

7-7

P7.3-4

() () () ()
3
12
00
11
01
210
tt
vt idv id
C
ττ ττ
−
−
=+ =
×
∫∫
0−
9
02 10t
−
<<×
() ()
33
12
0
1
00 1
210
t
sit vt dτ0 10
− −
−
=⇒ = − =−
×
∫

99
210 310t
−−
×< <× ()
() () (
6
63 3
12
2ns
410 A
1
410 10 510210
210
s
t
it
vt d tτ
−
−− −
−
=×
⇒= × − =−×+×
×
∫ )
6

In particular, ( ) ( )( )
93 6 9
310 510210310 10
3
v
− −−
×= −×+× × =
−

99
310 510t
−−
×< <× ()
() () (
6
63 3
12
3ns
210 A
1
210 1041010
210
s
t
it
vt d tτ
−
−− −
−
=−×
⇒= −× + =×−
×
∫ )
6

In particular, ( ) ()( )
93 6 9
510 41010510 10 V
−− −
×= ×− ×= −
3−
v
9
510t
−
×<
() ()
33
12
5ns
1
0 0 10 10 V
210
t
sit vt dτ
−−
−
=⇒ = − =−
×
∫

P7.3-5
(b) 00
41 2
() ()
42 3
03
t
td
itCvt
tdt
t
1<<

<<
== 
−< <

 <

(a)
() () () ()
00
1
0
tt
vt idv id
C
τττ=+ =∫∫
τ
V

For 0 < t < 1, i(t) = 0 A so ()
0
00 0
t
vt dτ=+ =∫
For 1 < t < 2, i(t) = (4t − 4) A so
() ( ) ()
22
1
44 02 4=242 V
1
t t
vt d ttττ ττ=− += − −+∫

() ()
2
(2)224222 Vv=− +=
() ( )
. For 2 < t < 3, i(t) = (−4t + 12) A so
() ()
22
212+2=21214 V
1
t
t tττ−+ −+−
2
412 2
t
vt dττ=− + +=∫

() ()
2
(3)23123144 Vv=− + −=
For 3 < t, i(t) = 0 A so ()
0
04 4
t
vt dτ=+ =∫
V

7-8

P7.3-6
(a) 00
() ()0.12 6
06
t
d
itCvt t
dt
t
2<<

== <

<
<


(b)
() () () ()
00
1
02
tt
vt idv id
C
τττ=+ =∫∫
τ
For 0 < t < 2, i(t) = 0 A so ()
0
20 00 V
t
vt dτ= +=∫

For 2 < t < 6, i(t) = 0.2 t − 0.4 V so
() ( ) ()
2 2
1
20.20.4 00.20.8=0.20.80.8 V
2
t t
vt d t tττ ττ=− += − − +∫

() ()
2
(6)0.260.860.83.2 Vv=− +=
()
6
20.83.21.66.4 V
t
vt d tτ=+ = −∫
. For 6 < t, i(t) = 0.8 A so

P7.3-7

()() () ()
43 6
00
6
0
66
0
1
02 52.510610
25 150
1
25150 50 25V
6
tt
t
t
t
vtv id ed
C
ed
ee
τ
τ
τ
ττ τ
τ
−−
−
−−
=+ =+× ×
=+

=+ − = −


∫∫
∫

P7.3-8

() ()
() ()()
23 2
3
22
22
2
1
12 102512 A
2001040
6
1010 210 200 A
200 25 50
25 150 A
tt
R
tt
C
tt
RC
t
v
ie e
dv
iC e e
dt
ii i e e
e
µ
µ
µ
−− −
−−
−−
−
== − × = −
×
−
== × −− =
=+ = +−
=+

7-9

Section 7-4: Energy Storage in a Capacitor

P7.4-1
Given
() ()
02
0.222 6
0.8 6
t
it t t
t
<

= −<

>

<
The capacitor voltage is given by

() () () () ()
00
1
02 0
0.5
tt
vt idv idvττ ττ=+ =∫∫
+
For 2t<
()
0
20 00
t
vt dτ= +=∫

In particular, For ()20v=.26t<<

() () () () (
22 2
2 2
222 00.20.8 0.20.80.8 V0.244 V
tt
vt d t t ttττ ττ=− += − = − + = −+∫ )

In particular, For ()63.2 V.v= 6t<

() () (
2
6
20.83.21.63.21.66.4 V1.64 V
t
t
vt d t tττ= += +=− =−∫ )
Now the power and energy are calculated as

()()() ()
()
2
0
2
0.0422 6
1.284 6
t
ptvtit t t
tt
 <

== − <

−<

<
<

and
() () ()
()
4
0
2
0
2
0.012 2 6
6
0.840.64
t
t
tp d t t
t
t
ττ

<

== − <

<
−−
∫
W

7-10

7-11

These plots were produced using three MATLAB scripts:

capvol.m function v = CapVol(t)
if t<2
v = 0;
elseif t<6
v = 0.2*t*t - .8*t +.8;
else
v = 1.6*t - 6.4;
end

capcur.m

function i = CapCur(t)
if t<2
i=0;
elseif t<6
i=.2*t - .4;
else
i =.8;
end

c7s4p1.m

t=0:1:8;
for k=1:1:length(t)
i(k)=CapCur(k-1);
v(k)=CapVol(k-1);
p(k)=i(k)*v(k);
w(k)=0.5*v(k)*v(k);
end

plot(t,i,t,v,t,p)
text(5,3.6,'v(t), V')
text(6,1.2,'i(t), A')
text(6.9,3.4,'p(t), W')
title('Capacitor Current, Voltage and Power')
xlabel('time, s')

% plot(t,w)
% title('Energy Stored in the Capacitor, J')
% xlabel('time, s')

7-12

P7.4-2

() ()( )
()
()
6 4000 4000
19
0 0.2
1010 54000 0.2 A
10 8.510
c
tt
c
c
iA
dv
iC e e
dt ims
−− −
−
=

== × −− = ⇒
=×

A
() () () ()
() ()
20
34 0 18
1
and 055 0 0 0
2
1010 55 521.2105 10 1.2510 J
tC vt v e
ve
−− −
== −=⇒ =
×= − =− × ≅⇒ = ×
WW
W
4−

P7.4–3

() so read off slope of () to get ()
() ()() so multiply () & () curves to get ()
c
c
cc
dv
itC vt it
dt
ptv tit vtit p
=
= t

P7.4-4
() () () ()
() () ()
() ()
max
00
2
6
2
max
511 5
00 50 cos10 0 sin sin
26 262
55
Now since 0 0 sin0 sin10 V
26 2 6
2102.51
6.25 J
22
Fi
tt
cc c c
cc cave
c
vtv idv t d v t
C
vt v vt t
Cv
10
6
π ππ
ττ
ππ
µ
−
   
=+ =+ + = − + +
 
   

=⇒ − =⇒ = +


×



∴ == =
∫∫
W
rst non-negative for max energy occurs when:10 0.1047
62 30
tt t
ππ π
+= ⇒== s

7-13

P7.4-5
( )()
() ()
6
6
6
2
26
Max. charge on capacitor = 1010 6 60 C
6010
6 sec to charge
1010
11
stored energy 1010 6 180 J
22
Cv
q
t
i
Cv
µ
µ
−
−
−
−
=× =
∆×
∆= = =
×
== = × =W

Section 7-5: Series and Parallel capacitors

P7.5-1

6
2F 4F = 6F
6F3F
6F in series with 3F = 2F
6F+3F
() 2F (6cos100) (210) (6) (100) (sin100)A 1.2 sin100 mA
d
it t t t
dt
µµ µ
µµ
µµ µ
µµ
µ
−
⋅
=
== × − =−

P7.5-2

6 250 6 250 250
4F4F
4F in series with 4F 2F
4F+4F
2 F 2F = 4F
4F in series with 4F = 2F
()(210)(53 ) (210)(03(250) ) A 1.5 mA
ttd
it e e e
dt
t
µµ
µµ µ
µµ
µµ µ
µµ µ
−− − − −
×
==
=× + =× +− =−

P7.5-3

3
36
in series with
2
5

22
5
55
2
in series with
527
2
55
(2510)cos250 (14sin250) (14)(250)cos250
77
so 2510 2500 1010 10 µF
CCC
CC
CC
C
CC C
CC
CCC
CC
d
tC t C
dt
CC
−
−−
t
⋅
==
+
=
⋅
==
+
 
×= = 
 
×= ⇒=× =

7-14

Section 7-6: Inductors

P7.6-1
[]
6
6 6
max
6
Find max. voltage across coil:() = 200 100(400)cos 400 V
810
VV
v 810 Vthus have a field of =410
mm
2
which exceeds dielectric strength in air of 310 V/m
We get a discharge a
di
vt L t
dt
=
×
∴ =× ×
×
∴ s the air is ionized.

P7.6-2

(.1) (4 4)10(4) 0.4 39.6 V
tt t t tdi
vL Ri ete te e te
dt
−−− − −
=+ = − + = +

P7.6-3
(a) 00
41 2
() ()
42 3
03
t
td
vtLit
tdt
t
1<<

<<
== 
−< <

 <

(b)
() () () ()
00
1
0
tt
it vdi vd
L
τττ=+ =∫∫
τ
A

For 0 < t < 1, v(t) = 0 V so ()
0
00 0
t
it dτ=+ =∫
For 1 < t < 2, v(t) = (4t − 4) V so
() ( ) ()
2 2
0
t
44 02 4=242 A
1
t
it d ttττ ττ=− += − −+∫

() ()
2
(2)424222 Ai=− +=

For 2 < t < 3, v(t) = −4t + 12 V so
() ( ) () ()
22
2
t
412 2212+2=21214 A
2
t
it d t tττ ττ=− + +=−+ −+−∫

() ()
2
(3)23123144 Ai=− + −=
For 3 < t, v(t) = 0 V so ()
3
04 4
t
it dτ=+ =∫
A

7-15

P7.6-4
33
()(25010)(12010)sin(50030) (0.25)(0.12)(500)cos(50030)
15 cos(50030)
d
vt t t
dt
t
−− °
°
=× × − = −
=−
°

P7.6-5
()
6
3
0
3
36 6 6
33
0
3
66
3
1
() () 210
510
for 0 1 s () 4 mV
14 10
() 410 210 2100.8210 A
510 510
410
(1µs) 110 210
510

t
Ls
s
t
L
L
it vd
tv t
it d t
i
ττ
µ
τ
−
−
−
−− − −
−−
−
−−
−
=− ×
×
<< =
×
=× −×= −×=−×

×× 
×
=× −×=−

×
∫
∫
() ()
6
3
36 6 6
33
1
3
66
3
6
10 A1.2 A
5
for 1 3s () 1 mV
16 110 6
() 110 10 (110)10 0.210 A
510 5 510 5
110
( 3 ) 310 110 1.6 µA
510
for 3s
s
t
L
s
L
st vt
it d t t
is
t
µ
µµ
τ
µ
µ
−
−
−− − − −
−−
−
−−
−
×= −
<< =−
×
= −× −×=− −× −×= −−
××
×
=− +× −× =−

×
<
∫
()0 so () remains 1.6 µA
sL
vt it=−
6

7-16

P7.6-6
() ( )
()
() () ( )
33
3
33
6
33 3 3 6
() 210 () 410 () (in general)
110
for 0<<1 s () 1 10 () = 110
110
() (210)110 410110 210 4 V
for 1s << 3s () = 1 m
ss
ss
s
d
vt it it
dt
d
ti t t t it
dt
vt t t
ti t
µ
µµ
−
−
−
−
=× +×
×
== ⇒ ×

×
=× × +× × =× +
() ( )
() ( ) () ( )
33 3
33
33
66
33 3 3 3 6
A () 0
() = (210)110 + 4100 = 2 V
110 110
for 3s< t < 5s () = 410 () = 10
110 110
() = 21041010410104 210
when 5s <t< 7s
s
ss
d
it
dt
vt
d
it t it
dt
vt t t
µµ
µµ
−−
−−
−
−−
−−
⇒=
×× × ×
××
×− ⇒ − =−

××
×× − +× − =−×
() ()
() ( )( )() ( )
3
33
3
33
6
33 3 3 3 6
() = 110 and () = 0
() = 21010 2 V
110
when 7s< t < 8s () = 810 () = 110
110
() 21010810 41010 12210
when 8s < , then ()
ss
ss
s
d
it it
dt
vt
d
it t it
dt
vt t t
ti t
µµ
µ
−
−
−
−
−
−−
−×
×= −
×
−× ⇒ ×

×
=× −× +× =−+×
= 0 () 0
()= 0
s
d
it
dt
vt
⇒=

P7.6-8
(a) 00
() ()0.12 6
06
t
d
vtLit t
dt
t
2<<

== <

<

<
(b)
() () () ()
00
1
02
tt
vdi vdit
L
τττ=+ =∫∫
τ
For 0 < t < 2, v(t) = 0 V so it ()
0
20 00 A
t
dτ= +=∫

For 2 < t < 6, v(t) = 0.2 t − 0.4 V so
() ( ) ()
2 2
2
t
20.20.4 00.20.8=0.20.80.8 A
2
t
d t tττ ττ=− += − − +∫
it
() ()
2
(6)0.260.860.83.2 A=− +=i .
For 6 < t, v(t) = 0.8 V so
7-17

it () ( )
6
20.83.21.66.4 A
t
d tτ=+ = −∫

Section 7-7: Energy Storage in an Inductor

P7.7-1
() ()
()()()
() ()
3
2
0
0 t<0
10010 0.40t1
0 t>1
0 0
1.601
0 1
0
0.801
0.8 1
t
d
vt it
dt
t
ptvtit tt
t
t
tp d t t
t
ττ
−


=× = ≤≤


<

== ≤≤

>

0<

= =<

>

∫
W <

7-18

P7.7-2
()() ()5(4sin2)(4sin2)
5 (8cos2) (4sin2)
80 [2cos2 sin2]
80 [sin(22) sin(22)] 80 sin4 W
d
ptvtit t t
dt
tt
tt
tt tt t

==


=
=
=+ + −=

0
00
80
() () = 80sin4 d [cos4|]20 (1cos4)
4
tt
t
tp dττ ττ τ== − =∫∫
W t−

P7.7-3
3
0
t
03
1
()= 6 cos 100 + 0
2510
6
[sin 100 |]= 2.4sin100
(2510)(100)
t
it d
t
ττ
τ
−
−
×
=
×
∫

()() ()(6cos100)(2.4sin100)
7.2 [ 2(cos100)(sin100) ]
7.2 [sin200sin0]7.2sin200
ptvtit t t
tt
tt
==
=
=+ =

00
0
() () 7.2 sin 200
7.2
cos 200|
200
0.036[1cos200]J 36 [1cos200 ] mJ
tt
t
tp d d
tt
ττ ττ
τ
==
=−

=− = −
∫∫
W

Section 7-8: Series and Parallel Inductors

P7.8-1
63
6H 3H 2Hand2H + 2H = 4H
6+3
×
==
0
0
16
() 6cos100 sin100|0.015sin100 A15sin100 mA
4 4100
t
t
it d t tττ τ== = =

×
∫

P7.8-2
( )( )
33
33
3 250 3 250 250
810810
4mH + 4mH = 8mH,8mH 8mH = 4mH
810810
and4mH + 4mH = 8mH
() (810) (53) (810)(03(250) ) 6 V
ttd
vt e e e
dt
−−
−−
−− − − −
×× ×
=
×+ ×
=× + =× +− =−
t

7-19

P7.8-3
()()
33
5
and
22 2
55
25cos250 1410sin250 (1410)(250)cos250
22
LL L L
LL LL L
LL
d
tL t L
dt
−−
⋅
== ++ =
+
 
=× = × 
 
t

3
25
so = 2.86 H
5
(1410)(250)
2
L
−
=
×

Section 7-9: Initial Conditions of Switched Circuits

P7.9-1

Then
() () ()()0 00and 0 012 V
LL C C
ii v v
+− + −
== = =
Next

7-20

P7.9-2

Then
() () ()()00 1 mAand 0 0
LL C C
ii v v
+− + −
== = =6 V
Next

P7.9-3

Then
() () ()()00 0and 0 00
LL C C
ii v v
+− + −
== = = V
Next

7-21

P7.9-4
at t = 0
−

KVL: (0)32150 (0)(0)17 V
ccvv v
−−
−+ −=⇒ = =
c
+

at t = 0
+

Apply KCL to supernode shown above:
() ()
15915
0 0.0030 06 mA
40005000
cc
ii
++−−
+− + =⇒ =
Then
()
3
6
0
0 610 V
3000
210 s
c
c
t
idv
dt C+
+
−
−
=
×
== =
×

7-22

Section 7-10: Operational amplifier Circuits and Linear Differential Equations

P7.10-1

P7.10-2

7-23

P7.10-3

P7.10-4

7-24

Verification Problems

VP7-1
We need to check the values of the inductor current at the ends of the intervals.
()
()
()
1
?
at 1 0.025 0.0650.025Yes!
=
25
33
?
at 3 0.065 0.115
=
25 50
0.055 0.055 Yes!
9
?
at 9 0.115 0.065
=
50
0.065 0.065Yes!
t
t
t
=− + =
=− + −
−= −
=−
=

The given equations for the inductor current describe a current that is continuous, as
must be the case since the given inductor voltage is bounded.

VP7-2
We need to check the values of the inductor current at the ends of the intervals.
()
()
11
?
at 1 0.025 0.03Yes!
=
200 100
44
?
at 4 0.03 0.03No!
=
100 100
t
t
=− + − +
=− + −

The equation for the inductor current indicates that this current changes
instantaneously at t = 4s. This equation cannot be correct.

Design Problems

DP7-1
a) ()
3
6
td
vt e
dt
−
=− is proportional to i(t) so the element is a capacitor.
()
()
0.5 F
it
d
vt
dt
==C .
b) ()
3
6
td
it e
dt
−
=− is proportional to v(t) so the element is an inductor.
()
()
0.5 H
vt
L
d
it
dt
= = .
c) v(t) is proportional to i(t) so the element is a resistor.
()
()
2
vt
R
it
= =Ω.
7-25

DP7-2
() ()() ()()1.131cos2451.131cos45cos2sin45sin2
0.8cos20.8sin2
tt
tt
+°= ° − °

=−
t

The first term is proportional to the voltage. Associate it with the
resistor. The noticing that
()
()
4cos2 2sin2
4cos28sin2
tt
vd td
dd
vt t t
dt dt
ττ τ
−∞ −∞
==
== −
∫∫
t

associate the second term with a capacitor to get the minus sign.
Then
1
4cos24cos2
5
()0.8cos2
t t
R
it t
== =Ω and
2
() 0.8sin2
0.1F
8sin2
4cos2
it t
C
d t
t
dt
−
== =
−

() ()() ()()1.131cos2451.131cos45cos2sin45sin2
0.8cos20.8sin2
tt
tt
−°= −° −−°

=+
t

The first term is proportional to the voltage. Associate it with the
resistor. Then noticing that
()
()
4cos2 2sin2
4cos28sin2
tt
vd td
dd
vt t t
dt dt
ττ τ
−∞ −∞
==
== −
∫∫
t

associate the second term with an inductor to get the plus sign. Then

1
4cos24cos2
5
()0.8cos2
t t
R
it t
== =Ω and
2
4cos2
2sin2
2.5H
() 0.8sin2
t
td
t
L
it t
τ
−∞
==
∫
=

7-26

DP7-3
a)

() ()() ()( )11.31cos24511.31cos45cos2sin45sin2
8cos28sin2
tt
tt
+°= ° − ° t 
 
=−

The first term is proportional to the voltage. Associate it with the resistor. The noticing that
()
()
4cos2 2sin2
4cos28sin2
tt
id td
dd
it t t
dt dt
ττ τ
−∞ −∞
==
== −
∫∫
t

associate the second term with an inductor to get the minus sign. Then
1
()8cos2
2
4cos24cos2
vt t
R
tt
== =Ω and
2
() 8sin2
1H
8sin2
4cos2
vt t
L
d t
t
dt
−
==
−
=
b)

() ()() ()( )11.31cos24511.31cos45cos2sin45sin2
8cos28sin2
tt
tt
+°= −° −−° t 
 
=+

The first term is proportional to the voltage. Associate it with the resistor. The noticing that
()
()
4cos2 2sin2
4cos28sin2
tt
id td
dd
it t t
dt dt
ττ τ
−∞ −∞
==
== −
∫∫
t

associate the second term with a capacitor to get the minus sign. Then
1
()8cos2
2
4cos24cos2
vt t
R
tt
== =Ω and
2
4cos2
2sin2
0.25F
() 8sin2
t
td
t
vt t
τ
−∞
== =
∫
C
7-27

DP7-4
at t=0
−

()00
L
i
−
=

By voltage division: ()
B
0
4
C
V
v
−
=
We require ()03
C
−
=Vv so

VB = 12 V
at t=0
+

Now we will check
0
C
t
dv
dt +
=

First:
()()00
LL
ii
+−
== 0
and
()()00
CC
vv
+ −
== 3 V

Apply KCL at node a:
() ()
()B 0
00
3
C
LC
Vv
ii
+
++
−
+=

() ()
123
00 03
3
CC
ii
++ −
+= ⇒ = A
Finally
()
0
0 3V
24
0.125 s
C
C
t
idv
dt C+
+
=
== =
as required.

DP7-5
We require
221 1
22
L C
Li C= v v where i are the steady state inductor current and capacitor
voltage. At steady state,
and
CL

C
L
v
i
R
=. Then
2
2
2 42
26
10
10 10
10
C
C
vL L
LC v C R
RR C
−
−

=⇒ = ⇒ = = = =


Ω

so . = 100 R Ω
7-28

Chapter 8 – The Complete Response of RL and RC Circuit
Exercises

Ex. 8.3-1
Before the switch closes:

After the switch closes:

Therefore ()
2
8so 80.050.4 s
0.25
tR τ== Ω = = .
Finally,
2.5
() ((0)) 2 Vfor 0
oc oc
tt
vv ve e t
τ−−
=+ − =+ >vt

Ex. 8.3-2
Before the switch closes:

After the switch closes:
8-1

Therefore
26
8so 0.75
0.25 8
t
R τ== Ω == s.
Finally,
1.3311
() ((0)) Afor 0
412
sc sc
tt
iiie e t
τ−−
=+ − =+ >it

Ex. 8.3-3
At steady-state, immediately before t = 0:

()
10 12
0 0.1 A
10401640||10
i

==
 
++ 

After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be

22
20 1
so 0.3 A, 40, s
40 2
Finally: () (0.10.3) 0.3 0.30.2 A
sc t
t
tt
L
IR
R
it e e
τ
−−
== Ω = = =
=− +=−

8-2

Ex. 8.3-4
At steady-state, immediately before t = 0

After t = 0, we have:

6
so 12 V, 200 , (200)(2010) 4 ms
oc t t
VR RCτ
−
== Ω = = ⋅ =
4
Finally: () (1212) 12 12 V
t
vt e
−
=− +=

Ex. 8.3-5
Immediately before ti 0, (0)0.==

After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the
inductor current:

255
0.2 A, 45 ,
459
sc t
th
L
IR
R
τ= =Ω ===

So it
1.8
() 0.2 (1 ) A
t
e
−
=−
Now that we have the inductor current, we can calculate v(t):

1.8 1.8

1.8
() 40 ()25()
8(1 )5(1.8)
8 V for 0
tt
t
d
vt it it
dt
ee
et
−−
−
=+
=− +
=+ >

8-3

Ex. 8.3-6
At steady-state, immediately before t = 0

so i(0) = 0.5 A.

After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get

= 93.75 mA, 640 ,
.1 1
s
6406400
sc t
t
IR
L
R
τ
=Ω
== =

6400
()406.25 93.75 mA
t
it e
−
=+
Finally:

6400 6400
6400
()400 () 0.1 ()400 (.40625 .09375)0.1(6400)(0.40625 )
37.597.5 V
tt
t
d
vt it it e e
dt
e
−−
−
=+ = + + −
= −

Ex. 8.4-1

( )( )
()
36 3
500
210110 210 s
()51.55 V
t
cvt e
τ
−−
−
=× × =×
=+ −

0.5
(0.001)53.5 2.88 V
cve
−
=− =
So () will be equal to at 1 ms if 2.88V.
cTvt vt v
T
==

8-4

Ex. 8.4-2

()
()

500
500

(0) 1 mA, 10 mA
10 9 mA
500 ,
500
() = 300 = 32.7e V
Ls c
t
L
L
t
t
L
RL
iI
it eL
R
vt it
τ
−
−
== 

⇒= −
=Ω =


−

()We require that 1.5 V at = 10 ms = 0.01s
Rvt t= . That is
500
(0.01) 5
1.532.7 8.5 H
0.588
L
eL
−
=− ⇒ = =
Ex. 8.6-1
10 tt<<

/
/(1)(.1) 10
() ()+ where () = (1 A)(1 ) = 1 V
() 1 1
tRC
tt
vtvAe v
vt Ae Ae
−
−−
=∞ ∞ Ω
=+ =+

10
Now (0)(0)0 1 1
() 1 V
t
vv A A
vt e
−+
−
= == +⇒=−
∴= −

1 tt>

.5
10(.5)(1)(.1)
11
10(0.5)
0.5 s,()() (0.5)
Now (0.5) 1 0.993 V
t
t
tv tvte v e
ve
−
−
−−
−
== =
−=

10(0.5)
() 0.993 V
t
vt e
−−
∴ =

8-5

Ex. 8 6-2
0 no sources (0)(0)0tv
−+
<∴ =v=

1
0 tt<<

57
210(10)
/
() () ()
t
tRC
vt v Ae v Ae
−
−
×
−
∞=+ =∞+

where for t = ∞ (steady-state)
∴capacitor becomes an open ⇒ v(∞) = 10 V

50
50
() 10
Now (0) 0 10 10
() 10(1 ) V
t
t
vt Ae
v AA
vt e
−
−
=+
== +⇒=−
∴ =−

11
, .1 stt t>=

50(.1)
50(.1)
50(.1)
()(.1)
where (.1) 10(1 ) 9.93 V
() 9.93 V
t
t
vtve
ve
vt e
−−
−
−−
=
=− =
∴ =

Ex. 8.6-3
For t < 0 i = 0.

For 0 < t < 0.2 s

10
10
KCL: 5 /2 0
10 50
also: 0.2
() 5
(0) 0 5 5
so we have () 5 (1 ) A
t
t
vi
di
idi
dtv
dt
it Ae
iA A
it e
−
−
−+ +=

+=
=


∴= +
== +⇒=−
=−

For t > 0.2 s
10(.2)
(0.2) = 4.32 A () 4.32 A
t
ii t e
−−
∴=

8-6

Ex. 8.7-1

()
()
()
()
()
t10sin20 V
KVL: 10sin2010.01 0
10 100sin20
s
vt
dvt
tv
dt
dvt
vt t
dt
=

−+ +

⇒+ =
t=
10
12
1
Natural response: () where ()
Forced response: try () cos20 sin 20
Plugging () into the differential equation and
equating like terms yields: 4 &
tt
nt
f
f
vtAe RC vtAe
vt B tB t
vt
B
τ
τ
−−
== ∴
=+
=−
n=
2
10
10
2
Complete response: () () ()
() 4cos202 sin 20
Now (0) (0) 0 4 4
() 4 4cos 20 2sin 20 V
nf
t
t
B
vt vtvt
vtAe t t
vv A A
vt e t t
−
−+
−
=
=+
=− +
== =−∴=
∴= − +

Ex. 8.7-2

()
()()
()
() ()
()
5
5
5
10 for 0
KCL at top node: 10 /10 0
Now 0 .1 100 1000
t
s
t
t
it e t
ei tvt
dit dit
vt it e
dt dt
−
−
−
=>
−+ + =
=⇒ + =

100
5
55 5
1
Natural response: () where ()
Forced response: try () & plug into the differential equation
5 100 1000 10.53
Complete response: ()
tt
nt n
t
f
tt t
itAe LR itAe
it Be
Be Be e B
itAe
τ
τ
−−
−
−− −
−
== ∴ =
=
−+ = ⇒ =
=
00 5
51 00
10.53
Now (0) (0) 0 10.53 10.53
() 10.53 ( ) A
tt
tt
e
ii A A
it ee
−
−+
−−
+
== =+ ⇒=−
∴= −

Ex. 8.7-3
When the switch is closed, the inductor current is /
Lsiv Rv
s= =. When the switch opens, the
inductor current is forced to change instantaneously. The energy stored in the inductor
instantaneously dissipates in the spark. To prevent the spark, add a resistor (say 1 kΩ) across the
switch terminals.
8-7

Problems

Section 8.3: The Response of a First Order Circuit to a Constant Input

P8.3-1

Here is the circuit before t = 0, when the switch is
open and the circuit is at steady state. The open
switch is modeled as an open circuit.

A capacitor in a steady-state dc circuit acts like an
open circuit, so an open circuit replaces the
capacitor. The voltage across that open circuit is the
initial capacitor voltage, v (0).

By voltage division
() ()
6
01 2
666
v==
++
4 V

Next, consider the circuit after the switch closes. The
closed switch is modeled as a short circuit.

We need to find the Thevenin equivalent of the part
of the circuit connected to the capacitor. Here’s the
circuit used to calculate the open circuit voltage, Voc.

()
oc
6
126 V
66
V==
+

Here is the circuit that is used to determine Rt.
A short circuit has replaced the closed switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
()()
t
66
3
66
R= =Ω
+

Then
()
t
30.250.75 sRCτ== =

Finally,
() ()()
oc oc
1.33/
0 62 V for > 0
tt
vtVv Ve e t
τ −−
=+ − =−

8-8

P8.3-2

Here is the circuit before t = 0, when the switch is
closed and the circuit is at steady state. The closed
switch is modeled as a short circuit.

An inductor in a steady-state dc circuit acts like an
short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the
initial inductor current, i(0).

()
12
02
6
i== A

Next, consider the circuit after the switch opens.
The open switch is modeled as an open circuit.

We need to find the Norton equivalent of the part
of the circuit connected to the inductor. Here’s the
circuit used to calculate the short circuit current,
Isc.

sc
12
1 A
66
I==
+

Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by an short circuit.
()()
()
t
666
4
66 6
R
+
= =Ω
++

Then
t
8
2 s
4
L
R
τ== =

Finally,
() ()()
sc sc
0.5/
0 1 A for > 0
tt
itIi Ie e t
τ −−
=+ − =+
8-9

P8.3-3
Before the switch closes:

After the switch closes:

Therefore ()
6
3so 30.050.15 s
2
tR τ
−
== Ω = =
−
.
Finally,
6.67
() ((0)) 618 Vfor 0
oc oc
tt
vv ve e t
τ−−
=+ − =−+ >vt

P8.3-4
Before the switch closes:

After the switch closes:
8-10

Therefore
66
3so 2
23
tR τ
−
== Ω ==
−
s.
Finally,
0.510
() ((0)) 2 Afor 0
3
t
t
sc sciiie e t
τ
−
−
=+ − =−+ >it

P8.3-5
Before the switch opens, () ()5 V 05 V
o o
vt v=⇒ =. After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get:

Therefore . () ( )
36
2010410 0.08 sτ
−
=× × =
Next,
12.5
() ((0)) 105 Vfor 0
t
t
Co c ocvv ve e t
τ
−
−
=+ − =− >vt

Finally, vt
12.5
0
()()105 Vfor 0
t
C
vt e t
−
== − >

8-11

P8.3-6
Before the switch opens, () ()5 V 05 V
o o
vt v=⇒ =. After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Norton equivalent circuit to get:

Therefore
3
5
0.25 ms
2010
τ==
×
.
Next,
4000
LL() ((0)) 0.50.25 mAfor 0
t
t
sc scii ie e t
τ
−
−
=+ − =− >it

Finally, () ()
4000
5 5 V for 0
t
oL
d
ite t
dt
−
==vt >

P8.3-7
At = 0 (steady-state)t
−

Since the input to this circuit is constant, the
inductor will act like a short circuit when the
circuit is at steady-state:

for t > 0

()()
() 2 0
0 6
RLt t
LLit ie e
−−
== A

8-12

P8.3-8
Before the switch opens, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the capacitor voltage, will have constant values. Opening the switch disturbs the circuit.
Eventually the disturbance dies out and the circuit is again at steady state. All the element
currents and voltages will again have constant values, but probably different constant values than
they had before the switch opened.

Here is the circuit before t = 0, when
the switch is closed and the circuit is at
steady state. The closed switch is modeled as
a short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the capacitor voltage, vo(t).

Because the circuit is at steady state, the value of the capacitor voltage will be constant.
This constant is the value of the capacitor voltage just before the switch opens. In the absence of
unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor
voltage immediately after the switch opens is equal to the value immediately before the switch
opens. This value is called the initial condition of the capacitor and has been labeled as vo(0).
There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal
to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore

()
os
0vV=

The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives

()
0
o
02 810 Vve=+ =
Consequently,
s10 VV=

8-13

Next, consider the circuit after the switch
opens. Eventually (certainly as t →∞) the
circuit will again be at steady state. Here is
the circuit at t = ∞, when the switch is open
and the circuit is at steady state. The open
switch is modeled as an open circuit. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the steady-state capacitor voltage,
vo(∞). There is no current in the horizontal
resistor and vo(∞) is equal to the voltage
across the vertical resistor. Using voltage
division,
() ()
o
10
10
10
v
R
∞=
+

The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives

()
o
28 2 Vve
−∞
∞=+ =
Consequently,
()
10
2 10 220100 40
10
RR
R
=⇒ += ⇒ =
+
Ω
Finally, the exponential part of vo(t) is known to be of the form
t
e
τ−
where
tRCτ= and
Rt is the Thevenin resistance of the part of the circuit connected to the capacitor.

Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.

()()
t
4010
10 18
4010
R=+=
+
Ω
so
t18RCCτ= =

From the equation for vo(t)
0.5 2 s
t
t τ
τ
−= −⇒ =
Consequently,

218 0.111111 mFCC= ⇒= =
8-14

P8.3-9:
Before the switch closes, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting
out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All
the element currents and voltages will again have constant values, but probably different constant
values than they had before the switch closed.

The inductor current is equal to the current in the 3 Ω resistor. Consequently,

o
0.35
()63 0.35
() 2 A when 0
33
t
vt e t
it e t
−
− −
== =− >

In the absence of unbounded voltages, the current in any inductor is continuous. Consequently,
the value of the inductor current immediately before t = 0 is equal to the value immediately after
t = 0.
Here is the circuit before t = 0, when the
switch is open and the circuit is at steady
state. The open switch is modeled as an open
circuit. An inductor in a steady-state dc
circuit acts like a short circuit, so a short
circuit replaces the inductor. The current in
that short circuit is the steady state inductor
current, i(0). Apply KVL to the loop to get

() ()( )
()
12
12
00 3024
24
0
3
Ri Ri i
i
RR
0+ +−
⇒=
++
=

The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives

()
0
02 1 Aie=− =
Consequently,

12
12
24
12
3
RR
RR
1= ⇒+ =
++

Next, consider the circuit after the switch
closes. Here is the circuit at t = ∞, when the
switch is closed and the circuit is at steady
state. The closed switch is modeled as a
short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R1.
8-15

An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to
the loop to get
() () ()
2
2
24
32 40
3
Ri i i
R
∞+ ∞− =⇒ ∞=
+

The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives

()22i e
−∞
∞=−= A
Consequently
2
2
24
29
3
R
R
= ⇒=
+
Ω
Then
112 R=Ω
Finally, the exponential part of i(t) is known to be of the form
t
e
τ−
where
t
L
R
τ=and
Rt is the Thevenin resistance of the part of the circuit that is connected to the inductor.

Here is shows the circuit that is used to determine Rt.
A short circuit has replaced combination of resistor
R1 and the closed switch. Independent sources are set
to zero when calculating Rt, so the voltage source
has been replaced by an short circuit.

t2 39312RR=+=+= Ω
so
12
t
LL
R
τ==
From the equation for i(t)
0.35 2.857 s
t
t τ
τ
−= −⇒ =
Consequently,
2.857 34.28 H
12
L
L=⇒ =
8-16

P8.3-10
First, use source transformations to obtain the equivalent circuit

for t < 0:

for t > 0:

()
()
() ()
24
24
1
1
2
So 0 2 A, 0, 3912 , = s
1224
and 2 0
Finally 9 18 0
Ls c t
t
t
L
t
L
L
iI R
R
it e t
vt it e t
τ
−
−
== =+=Ω==
=>
== >

8-17

Section 8-4: Sequential Switching

P8.4-1
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to
get:

Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc =
5 V and Rt = 4 Ω so 40.050.2 sτ=× = . Therefore

5
() ((0)) 55 Vfor 0 1.5 s
oc oc
tt
vtvv ve e t
τ−−
=+ − =+ <<

At t =1.5 s,
()0.051.5
(1.5)55 = 5 Vve
−
=+
0.4 s
. For 1.5 s < t, voc = 10 V and Rt = 8 Ω so
80.05τ=× = . Therefore
( ) ( )1.5 2.51.5
() ((1.5)) 105 Vfor 1.5 s
oc oc
tt
vtvv ve e t
τ−− − −
=+ − =− <

Finally
()
5
2.51.5
55 V for 0 1.5 s
()
for 1.5 s 105 V
t
t
e t
vt
te
−
−−
+ <<
=
<−

P8.4-2
Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:

Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2
A and Rt = 6 Ω so
12
2 s
6
τ== . Therefore
0.5
() ((0)) 2 Afor 0 1.5 s
sc sc
tt
itiiie e t
τ−−
=+ − =+ <<

8-18

At t =1.5 s,
()0.51.5
(1.5)2 = 2.47 Aie
−
=+ . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so
12
1.5 s
8
τ== .
Therefore
( ) ( )1.5 0.6671.5
() ((1.5)) 30.53 Vfor 1.5 s
sc sc
tt
itii ie e t
τ−− − −
=+ − =− <
Finally
()
0.5
0.6671.5
2 A for 0 1.5 s
()
for 1.5 s
30.53 A
t
t
e t
it
t
e
−
−−
 + <<
=
<
−

P8.4-3
At t = 0
−
:

KVL : 5218(128)0
(0) =10439 A
6
2 A (0)
62
LL
ii
i
ii i
−

+
−++ =
⇒

∴== =

+

For 0 < t < 0.051 s

t
() (0) where
LL
t
it ie LR
τ
τ
−
==
t
6
61226
()2A
t
L
R
it e
−
= += Ω
=

6612
() () 6 A
6
t
Litit e
−+
∴ ==



For t > 0.051 s

()
()(0.051)
6(.051)
14(0.051)
14(0.051)
() (0.051)
(0.051) 2 1.473 A
() 1.473 A
() 1.473 A
RLt
LL
L
t
L
t
L
it i e
ie
it e
itit e
−−
−
−−
−−
=
==
=
==

8-19

P8.4-4
At t = 0
-
: Assume that the circuit has reached steady state so that the voltage across the 100 µF
capacitor is 3 V. The charge stored by the capacitor is

()( )()
66
0 10010330010 Cq
−−
=× =×
−

0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at
t = 0
+
). The voltage across the parallel capacitors is determined by considering charge
conservation:

() ()( )
()
() ()
()
6
66 6
0(100 F) 0 (400 F) 0
00 30010
0
10010400105001050010
0 0.6 V
qv v
qq
v
v
µµ
++ +
+−
6
−
+
− −−
=+
×
== =
×+×××
+
=
−

10 ms < t < l s: Combine 100 µF & 400 µF in parallel to obtain

()
(.01)
34
(.01)(10) (510)
() 0
0.6
tR C
tx
vt v e
e
+− −
−
−−
=
=

2(.01)
() 0.6 V
t
vt e
−−
=

P8.4-5

For t < 0:
Find the Thevenin equivalent of the
circuit connected to the inductor (after
t >0). First, the open circuit voltage:

1
11
40
1 A
2020
20515 V
oc
i
vi i
==
+
=− =

8-20

Next, the short circuit current:

11 1205 0ii i=⇒ =
40
02
20
sc scii+= ⇒ = A
Then
15
7.5
2
oc
t
sc
v
R
i
== =Ω
Replace the circuit connected to the
inductor by its Norton equivalent
circuit. First
3
1510 1
7.5500
t
L
R
τ
−
×
== =
Next
()
500
22 A 0
t
it e t
−
=− >
After t = 0, the steady state inductor
current is 2 A. 99% of 2 is 1.98.

500
1.9822 9.2 ms
t
e t
−
=− ⇒=

P8.4-6

()05 Vv= , ()0v∞= and
56
10100.1 sτ
−
=× =

()
10
5 V for 0
t
vt e t
−
∴ =>

110
12.55 0.0693 s
t
et
−
==

()
()1
1 33
2.5
25A
1001010010
vt
it µ===
××

8-21

Section 8-5: Stability of First Order Circuits

P8.5-1
This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the
inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor
is calculated as

100
() (400)100
=100400
100400
400()()
TT
t
T
T
vit i R
R
i
vi tRit

= −
⇒=+ 
+
=− 

The circuit is stable when R < 400 Ω.

P8.5-2
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as

Ohm’s law: ()(1000
T
vt it= )

KVL: ()()()(4000 0
TT
Avtvtvt it)+ −− =

()()( )11000 4000
TT
vt A it it()
T
∴ =− +
()
()
()5 1000
T
t
T
vt
RA
it
== −×
The circuit is stable when A < 5 V/V.

P8.5-3
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as

Ohm’s law: ()
()
6000
Tvt
it=−
KCL: () ()()
()
3000
T
T
vt
itBitit++ =
()()
() ()
() ()
1
60003000
3
6000
TT
T
T
vt vt
it B
Bv t

∴ =−+− +

+
=

()
()
6000
3
T
t
T
vt
R
it B
==
+

The circuit is stable when B > −3 A/A.
8-22

P8.5-4
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as

()()
()
10004000
() ()800()
10004000
() () ()1 ()
()
800(1)
()
TT
T
T
t
T
vt it it
vt vtAvt Avt
vt
RA
it
==
+
=− =−
== −

The circuit is stable when A < 1 V/V.

8-23

Section 8-6: The Unit Step Response

P8.6-1
The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be

() for 0
at
o
vt ABe t
−
=+>

where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.

The steady-state circuit for t < 0.
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.

The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage before
the input changes. At time t = 0 the output voltage is
()
()0
0
a
ovA Be A
−
B=+= +

Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives

8 VAB+=

The steady-state circuit for t > 0.
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.

The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the input
changes. At time t = ∞ the output voltage is

()
()a
ovA Be
−∞
A∞=+ =

Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives

A = -7 V
Therefore
B = 15 V

8-24

The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.

t
1
RC
a
τ==

Here is the circuit used to calculate Rt.

Rt = 6 Ω
Therefore
()()
3
11
2.5
s666.710
a
−
==
×

(The time constant is ()( )
3
666.710 0.4 sτ
−
=× = .)

Putting it all together:
()
2.5
8 V for 0
715 Vfor 0
o t
t
vt
et
−
≤
=
−+≥

P8.6-2
The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be

() for 0
at
o
vt ABe t
−
=+>

where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.

The steady-state circuit for t < 0.
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.

The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage
before the input changes. At time t = 0 the output
voltage is
()
()0
0
a
ovA Be A
−
B=+= +

Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
8-25

()
6
32 V
36
AB+= =
+

The steady-state circuit for t > 0.
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.

The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the
input changes. At time t = ∞ the output voltage is

()
()a
ovA Be
−∞
A∞=+ =

Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives

()
6
64 V
36
A==
+

Therefore
B = −2 V

The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.

t
1
RC
a
τ==

Here is the circuit used to calculate Rt.

()()
t
36
2
36
R= =Ω
+

Therefore
()()
11
1
2.5 s
a==

(The time constant is()()20.51 sτ= =.)

Putting it all together:
()
2 V for 0
42 Vfor 0
o t
t
vt
et
−
≤
=
− ≥

8-26

P8.6-3
The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be

() for 0
at
o
vt ABe t
−
=+>

where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.

The steady-state circuit for t < 0.
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.

The value of the inductor current at time t = 0,
will be equal to the steady state inductor current before
the input changes. At time t = 0 the output current is

()
()0
o0
a
iA Be A
−
B=+= +

Consequently, the inductor current is labeled as A + B.
Analysis of the circuit gives

7
1.4 A
5
AB
−
+= =−

The steady-state circuit for t > 0.
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.

The value of the inductor current at time t = ∞, will be
equal to the steady state inductor current after the
input changes. At time t = ∞ the output current is

()
()
o
a
iA Be
−∞
A∞=+ =

Consequently, the inductor current is labeled as A.
Analysis of the circuit gives

6
1.2 A
5
A==
Therefore
B = −2.6 V
8-27

The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit
connected to the inductor.

t
1 L
aR
τ==

Here is the circuit used to calculate Rt.

()()
t
54
2.22
54
R= =Ω
+

Therefore
2.22 1
1.85
1.2 s
a==

(The time constant is
1.2
0.54 s
2.22
τ== .)

Putting it all together:
()
1.85
1.4 A for 0
1.22.6 Afor 0
o t
t
it
et
−
− ≤
=
− ≥

P8.6-4
() 4()(1)(2)(4)(6)vt utut ut ut ut= −−−−+ −− −

P8.6-5

()
0 1
4 1 2
0 2
s
t
vt t
t
<

=<

>

<

( )( )
56
510210 1 sRCτ
−
== × × =
Assume that the circuit is at steady
state at t = 1
−
. Then

()
(1)
44 V for 1 2
t
vt e t
−−
=−≤ ≤

so ()
(21)
244 2.53 V
−−
=− =ve
and

()
(2)
2.53 V for 2
t
vt e t
−−
= ≥

(1)
(2)
0
()44 1 2
2.53 2
t
t
t
vt e t
et
−−
−−
≤

1
∴ =− ≤≤

≥


8-28

P8.6-6
The capacitor voltage is v(0
−
) = 10 V immediately before the switch opens at t = 0.

For 0 < t < 0.5 s the switch is open:

() ()
11
010 V, 0 V, 3 s
62
vv τ=∞ = =×=

so ()
2
10 V
t
vt e
−
=

In particular, ()
()20.5
0.510 3.679 Vve
−
= =

For t > 0.5 s the switch is closed:

() ()03.679 V, 10 V, 6||32,
t
vv R= ∞= = =Ω
11
2 s
63
τ=×=
so
() ( )
( )
()
30.5
30.5
103.67910 V
106.321 V
t
t
vt e
e
−−
−−
=+ −
=−

P8.6-7
Assume that the circuit is at steady state before t = 0. Then the initial inductor current is
i(0
−
) = 0 A.

For 0 < t < 1 ms:

The steady state inductor current will be
() () ()
30
4024 Alim
3020t
ii t
→∞
∞= = =
+

The time constant will be
3
35010 1
10 s
3020 1000
τ
−
−×
== =
+

The inductor current is () ()
1000
241 A
t
e
−
=−it
In particular, ( ) ()
1
0.001241 15.2 Aie
−
=− =
For t > 1 ms

Now the initial current is i(0.001) = 15.2 A and
the steady state current is 0 A. As before, the time
constant is 1 ms. The inductor current is

()
( )10000.001
15.2 A
t
it e
−−
=

8-29

The output voltage is

() ()
()
()
1000
10000.001
4801 V 1 ms
20
303 V 1 ms
t
t
et
vt it
et
−
−−
 −<

== 
>

P8.6-8
For t < 0, the circuit is:

After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent
circuit to get:

() ()( )
()/40000.00005
5
15615
159 V
t
c
t
vt e
e
−×
−
=−+−−−
=−+

8-30

Section 8-7 The Response of an RL or RC Circuit to a Nonconstant Source

P8.7-1
Assume that the circuit is at steady state before t = 0:

KVL : 123(3)38.5 = 0 1.83 A
Then (0) 12 22 V =(0)
xx x
cx c
ii i
vi v
−+
++ ⇒ =−
=− =

After t = 0:

()
()()
()
()
()
()
()
5
5
KVL : 12()8 0
1
KCL : 2 (136) 0
108
1
12 8 0
108
c
t
t
it e vt
xc
dvt dvt
cc
itit itxx x
dt dt
dvt
ev t
cdt
−
−
−+ =
−− + =⇒ =

∴ −+ =


59
5
9
()
9()72 ()
Try () & substitute into the differential equation 18
5
() 18
(0)22 18 4
95
() 4 18 V
tt
t
t
dvt
c
vt e vtAe
cc ndt
vt Be B
cf
t
vt Ae e
c
vA A
c
tt
vt e e
c
−−
−
−
+= ⇒ =
=⇒
−
=
∴ =+
== +⇒ =
−−
∴ =+

8-31

P8.7-2
Assume that the circuit is at steady state before t = 0:

12
(0)(0) 3 A
4
LL
ii
+−
== =

After t = 0:

()
()
()
()
()
()
()
()
()
2
2
2
12
KCL : 6
42
also: (2/5)
3
(2/5) 3 6
4
10
1020
3
t
L
L
L t
L
L t
L
vt vt
it e
dit
vt
dt
dit
it e
dt
dit
it e
dt
−
−
−
−
++ =
=

+=

+= +
+

()
()
10/3 2
10/3 2
() , try () , substitute into the differential equation,
2
and then equating like terms 3, 15 ()315
()()() 315,(0) 3 3 15
t t
nf
tt
Ln f L
itAe itBCe
t
BC it ef
itititAe e i A A
− −
−−
∴= =+
−
⇒= =⇒ =+
∴= + = ++ ==++⇒
()
(10/3) 2
(10/3) 2
15
() 15 3 15
Finally, () 2/5 20 12 V
tt
L
ttL
it e e
di
vt e e
dt
−−
−−
=−
∴= − ++
== −

P8.7-3
Assume that the circuit is at steady state before t = 0:

6
Current division: (0)5 1 mA
624
L
i
− 
=− =−

+

8-32

After t = 0:

()
()
()
()
L
L
KVL: 25sin4000 24 .008 0
di 25
3000i sin4000t
dt .008
L
L
dit
ti t
dt
t
t
−+ +
+=
=

3000
3000
() , try ()cos4000 sin4000, substitute into the differential equation
and equate like terms 1/2, 3/8 ()0.5cos40000.375sin4000
()()() 0.5cos400
t
nf
f
t
Ln f
itAe itB tC t
B Ci t t
itititAe
−
−
== +
⇒= − = ⇒ =− +
=+ = −
3
3000
3000
00.375sin4000
(0) (0) 10 0.5 0.5

() 0.5 0.5cos40000.375sin4000 mA
but () 24()12 12cos4000 9sin4000 V
LL
t
L
t
L
tt
ii A A
it e t t
vt it e t t
+− −
−
−
+
== − =−⇒≅
t
∴ =− +
== − +

P8.7-4
Assume that the circuit is at steady state before t = 0:

Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:

()
()
()
()
1KVL: 10cos215 0 2 20cos2
30
c c
cc
dvt dvt
tv t vt
dt dt

−+ + =⇒ + =

t

2
2
() , Try ()cos2 sin2 & substitute into thedifferential equation to get
5 ()5cos25sin2. () ()() 5cos2 5sin2
Now (0) 0 5 5 () 5
t
nf
t
fc n f
cc
vtAe vtB tC t
BCv t t tvtvtvtAe t
vA A vt
−
−
== +
== ⇒ = + ∴ = + = + +
== +⇒=−⇒ =−
t
2
5cos2 5sin2 V
t
et t
−
++

8-33

P8.7-5
Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) =
0 A. After t = 0, we have:

()
()
()
()
()
()
()
KVL :10sin100 5 0
Ohm's law :
8
18 160sin100
dit
tit vt
dt
vt
it
dvt
vt t
dt
− ++ + =
=
∴ +=

18
18
() , try ()cos100 sin100, substitute intothe differential equation
and equate like terms 1.55 & 0.279 ()1.55cos1000.279sin100
() ()() 1.55cos100
t
nf
f
t
nf
vtAe vtB tC t
B Cv t t
vtvtvtAe t
−
−
t
∴ == +
⇒= − = ⇒ =− +
∴ =+ = −
18
0.279sin100
(0) 8 (0) 0 (0) 0 1.55 1.55
so () 1.551.55cos100 0.279 sin100 V
t
t
vi v A A
vt e t t
−
+
== ⇒ ==− ⇒=
=− +

P8.7-6
Assume that the circuit is at steady state
before t = 0.
() ()
(0)(0)10 V
oc
CC
vt vt
vv
+−
=−
== −

After t = 0, we have
()
()
5
58
= 0.533 mA
1500015000
t
s t
vt e
it e
−
−
==

The circuit is represented by the differential
equation: ()
()()
C Cdvtvt
itC
dt R
= + . Then

() ( )
()
() ()
()
()
35 6 3
0.53310 0.2510 10 4000 4000
c ct t
cc
dvt dvt
ev t v
dtdt
−− − − −
×= × + ⇒ + =
5
t e
t−
=

() ()
4000 5
Then . Try .
t
n fvt Ae vtBe
−
= Substitute into the differential equation to get

()
()
55
5
4000
4000 4000 1.001251
3995
tt
t
dBe
Be e B
dt
−−
−
+ =⇒ = =−
−
≅−

8-34

()()
5 4000
()
tt
Cf n
vtvtvteAe
−−
=+ =+

2 4000
(0)101 11 () 1 11 V
tt
CCvA A vt e e
− −
=−=+⇒=−⇒ = −

Finally
4000 5
() () 11 1 V , t 0
tt
oCvt vt e e
−−
=− = − ≥

P8.7-7
1
From the graph () mA
4
L
it t= . Use KVL to get

()
11
() ()
1()0.4 () 2.5()2.5()
LL
LL
dit dit
it vt it vt
dt dt
+= ⇒ + =
Then
11
11
2.5 2.5() 0.10.25 V
44
di
tt vt v
dt
 
+= ⇒ =+
 
t

P8.7-8
Assume that the circuit is at steady state before
t = 0.

2
(0)(0) 3010 V
42
vv
+−
== =
+

After t = 0 we have

()
()
()
() ()
()
3
5 1
KVL: 4 30
22
1
24 30
2
t
dvt dvt
vt i
dt dt
dvt
it it e
dt
−

+ +−


+− +=

=

The circuit is represented by the differential equation

()
()
366 2
(10 )
19 19 3
tdvt
vt e
dt
−
+= +

Take ()
()6/19
n
t
Ae
−
=vt . Try ,vt()
3

t
f
BCe
−
=+ , substitute into the differential equation to get
8-35

3366 0
3( )
19 1919
tt
Ce BCe e
−−
−+ + = +
34
t−

Equate coefficients to get
()
(6/19)344
10 ,
51 51
f
tt
BC vt e Ae
−−
== − ⇒ = +
Then
() ()()
(6/19)34
10
51
nf
tt
vtvtvt e Ae
−−
=+ =− +
Finally
44
(0)10 V, 1010
51 51
cvA
+
=⇒ =−+⇒ =A
(6/19) 34
() 10( ) V
51
c
tt
vt e e
−−
∴ =+ −

P8.7-9
We are given v(0) = 0. From part b of the
figure:
()
5 0 2 s
10 2 s
s
tt
vt
t
≤≤
=
>

Find the Thevenin equivalent of the part of the
circuit that is connected to the capacitor:

The open circuit voltage:

The short circuit current:

(ix=0 because of the short across the right 2 Ω
resistor)

Replace the part of the circuit connected to the
capacitor by its Thevenin equivalent:

KVL:
()
()()
()() ()
20
22
s
s
dvt
vtvt
dt
vtvtdvt
dt
+ −=
+=

()
0.5t
nvt Ae
−
=
8-36

For 0 < t < 2 s, vt . Try . Substituting into the differential equation and
equating coefficients gives B = −10 and C =5. Therefore
()5
s t= ()
fvt BCt=+
()
/2
510
t
vt t Ae
−
=− + . Using v(0) = 0,
we determine that A =10. Consequently, ()
/2
1)
t−
510(tevt=+− .
At t = 2 s, . ()
1
2 10 3.68ve
−
= =

Next, for t > 2 s, vt . Try ()10 V
s= ()
fvt B=. Substituting into the differential equation and
equating coefficients gives B = 10. Therefore ()
()2/2
10
t
Ae
−−
=+vt . Using , we
determine that A = −6.32. Consequently,

()2 3.68v=
()
()2/2
106.32
t
e
−−
=−vt .

P8.7-10

()
()
()
()
KVL: 0
1

C
sC
C
C
ss
dvt
ktRC vt
dt
dvt k
vt t
RCRdt

−+ + =

⇒+ =
C

() () () () ()
[]
/
01
10 1 0 1
/
, where = . Try =
1
& plug into D.E. thus , .
Now we have () ( ). Use (0)0 to get 0
s
s
tRC
cn f c f
s
ss
tRC
cs c s
vt vtvt vt Ae vtBBt
k
BB Bt t B kRCBk
RC RC
vtAe ktRC v AkRC AkRC
−
−
=+ +
⇒+ + = =− =
=+ − = =− ⇒=
s
/
38 00
.
() [ (1 )]. Plugging in 1000 , = 625kΩ& 2000 pF get
() 1000[1.2510(1 )]
tRC
s
cs s
t
c
vt ktRCe k R C
vt t e
−
−−
∴ =− − = =
=− × −

v(t) and vC(t) track well on a millisecond time scale.

8-37

Spice Problems

SP 8-1

8-38

SP 8-2

8-39

SP 8-3

/
() for > 0
t
vtABe t
τ−
=+
0
7.2(0) 7.2
0.8 V
8.0() 8.0 V
vA Be AB
B
vA Be A
−∞
== + ⇒ =+
⇒= −
=∞=+ ⇒ = 

0.05/ 0.05 87.7728
7.7728(0.05)80.8 ln 1.25878
0.8
0.05
39.72 ms
1.25878
ve
τ
τ
τ
− −
== − ⇒− = =−


⇒= =

Therefore
/0.03972
()80.8 V for > 0
t
vt e t
−
=−
8-40

SP 8-4

/
() for > 0
t
itABe t
τ−
=+
3
33
0
0(0) 0
410 A
410 () 410 A
iA Be AB
B
iA Be A
−
−− −∞
== + ⇒ =+ 
⇒= −×
×= ∞=+ ⇒ =× 

() ()
( )
()
36 3 3
36
3
6
6
510/
2.451410 (510)410 410
42.451410510
ln 0.94894
410
510
5.269 s
0.94894
ve
τ
τ
τµ
−− − −
−−
−
−
−
−×
×= × =× −×
−××
⇒− = =−
×

×
⇒= =

Therefore
6
/5.26910
()44 mA for > 0
t
it e t
−
−×
=−
8-41

Verification Problems

VP 8-1
First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor
voltage is vc = 4 V.

We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot.

Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
the circuit connected to the capacitor is
( )( )
t
200040004
k
200040003
R= =
+
Ω so the time constant is
( )
3 6
t
4
100.510 ms
33
RCτ
−
== × × =


2
. Thus the capacitor voltage is
0.67
()4 4 V
c
t
vt e
−
=+

where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then

1
1.33
.67
() = 4 4 4.541 ~ 4.5398 V
cvt e

−


+=
So the plot is correct.

VP 8-2
The initial and steady-state inductor currents shown on the plot agree with the values obtained
from the circuit.

Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
the circuit connected to the inductor is
( )( )
t
200040004
k
200040003
R= =
+
Ω so the time constant is
3
t
5 15
ms
4 4
10
3
L
R
τ== =
×
. Thus inductor current is
3.75
()2 5 mA
L
t
it e
−
−+

where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then

1
3.75
3.75
()2 5 4.264 mA4.7294 mA
Lit e

−


=− += ≠

so the plot does not correspond to this circuit.
8-42

VP 8-3
Notice that the steady-state inductor current does not depend on the inductance, L. The initial
and steady-state inductor currents shown on the plot agree with the values obtained from the
circuit.

After t = 0

So 5 mA and
1333
sc
L
I τ==

The inductor current is given by
1333
()2 5 mA
tL
L e
−
=− +it , where t has units of seconds and L
has units of Henries. Let t 1 = 3.75 ms, then

(1333)(0.00375) 5
14.836()2 52 5
LL
Lit e e
−⋅ −
== − +=− +
so
54.8365

2
L
e
−−
=
−

and
5
2 H
4.8365
ln
2
L
−
==
−

−

is the required inductance.

VP 8-4
First consider the circuit. When t < 0 and the circuit is at steady-state:

For t > 0

So
21 2
12 12 12
( ) , and
oc t
12R RR RRC
VA BR
RRR R
τ=+ = =
+ + RR+

8-43

Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor
voltage is (Voc =) 4 V, so
() 6 4
t
Cvt e
τ−
=−+
At
1 1.333 mst=
0.001333
13.1874 () 6 4
Cvt e
τ−
= =− +
so
0.001333
0.67 ms
43.1874
ln
6
τ
−
==
−+

−

Combining the information obtained from the circuit with the information obtained from the plot
gives
22 12
12 12 12
2, ( )4, 0.67 ms
RR RRC
AA B
RR RR RR
=− += =
++ +

There are many ways that A, B, R1, R2, and C can be chosen to satisfy these equations. Here is
one convenient way. Pick R1 = 3000 and R2 = 6000. Then

2
2 3
3
2()
4 3 6 9
3
21
2000 ms F =
33
A
A
AB
BB
CC µ
=−⇒ =−
+
=⇒− =⇒ =
⋅= ⇒

Design Problems

DP 8-1
Steady-state response when the switch is open:
3
12
12 3
2
R
361 RRR
RR R
= ⇒+ =
++
.
Steady-state response when the switch is open:

3 3
1
13
12
5
R
10
R
R
RR
=⇒
+
=.
10 ms = 5 τ = ()
3
13||
6
R
RRC C=

Let C = 1 µF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ.

8-44

DP 8-2
steady state response when the switch is open :
12
12
12
12 kRR
RR
0.001= ⇒+ =
+
Ω.
steady state response when the switch is open:

1
1
12
3 kR
R
0.004= ⇒= Ω.
Therefore, R 2 = 9 kΩ.

10 ms = 5 τ =
12
5 240 H
2400
LL
L
RR

=⇒ =
+


DP 8-3
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
(a)
()
6
3
10
54
55010
ttRC C
−
∆= ⇒ = =
×
pF
(b) ∆= () ( )
36
55010210 0.5 st
−
× × =

DP 8-4
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
When the switch is open: () ()1)5 ln1 ln
tt
ek k t
τ
τ
τ
5( 1k
∆−∆
=−⇒ −=− ⇒∆=− −
When the switch is open: ()5 ln1
t
ek t55 k
τ
τ
−∆
−= ⇒∆=− −
(a)
() ()
6
3
10
6.67 pF
ln1.955010
−
==
−− ×
C
(b) ∆= () ( )
36
ln(1.95)5010210 0.3 st
−
−− × × =

DP 8-5

1
1 1
1
20
(0)
40 40
40
40
R
i
RR
R
=×
+
+
+

8-45

For
t > 0:

2
2
10
() (0) where =
40
t
t
L
itie
R R
τ
τ
−
−
==
+

() ()
2
2
2
At t200s we need > 60 mA and <180 mA
First let's find a value of to cause (0) 180 mA.
1
Try 40 . Then (0) A = 166.7 mA so ()0.1667 .
6
Next, we find a value of t
t
it it
Ri
R ii t
R
e
τ
µ
−
<
<
=Ω = =
2
2
350000.0002 31
o cause (0.0002) 60 mA.
10 1
Try R 10, then 0.2 ms = s.
50 5000
(0.0002) 166.710 166.710 61.3 mA
i
ie e
τ
−
−− × −−
>
=Ω = =
=× =× =

DP 8-6

The current waveform will look like this:

We only need to consider the rise time:

() (1 ) (1 )
22
t
t
s
L
VA
it e e
RR
τ
τ
−
−
=− =−
++

where
0.21
s
315
t
L
R
τ== =
15
()(1 )
3
t
L
A
it e
−
∴ =−

2
2
2 15(.25)2
Now find so that 10 W during 0.25 0.75 s
we want [(0.25)] 10 W (1 )(1)10 9.715 V
9
Lfuse
Lf use
Ai R t
A
iR e A
−
≥≤ ≤
∴ =⇒ − =⇒=

8-46

Chapter 9 - Complete Response of Circuits with Two Energy
Storage Elements
Exercises

Ex. 9.3-1

Apply KVL to right mesh:

()
() ()()()
()
()
()()()
21
2
dit
vt itit
s
dt
dit
vt itit
s
dt
0+ +− =
⇒= − − −
The capacitor current and voltage are related by

()
() ()
()()
() () ()
2
2
11 1 1
2
22 2 2
s
s
dvt dit ditditditd
it itit
dtdt dt dt dt dt

== − −+ = − −


() ()
()
()
2
2
11

22
s
dit dit dit
it
dt dt dt
∴ ++ =

Ex. 9.3-2

The inductor voltage is related to the inductor
current by
()
()
1
dit
vt
dt
=

Apply KCL at the top node:

()
()
()
()1
12
s
vt dvt
it it
dt
=+ +

Using the operator
d
s
dt
= we have
() ()
()()() ()
()()
()
()
1
1
2
2
s
s
vtsit
vt
itvt svt
sitvtit svt
=

⇒= + +
=+ +



Therefore
() () () ()
() ()
()
()
2
2
2
22 2 2 2 2
s
s
dvtdvt dit
sitsvt vtsvt vt
dt dt dt
=+ + ⇒ + + =

9-1

Ex. 9.3-3

Using the operator
d
s
dt
= , apply KVL to the
left mesh:
() ()()( ) ()
11 2
sit sitit vt+− =
Apply KVL to the right mesh:
() () ()()()
() () ()()
22 2 1
12 2 22
1
22
12
2
it itsitit
s
it it itit
ss
0+ +−
=+ +
=

Combining these equations gives:
() () () ()
() ()
()
()
22
2222
22 2 222
3 4 2 or 3 4 2
s
s
dit dit dvt
sitsit itsvt it
dt dt dt
++ = + + =

Ex. 9.4-1

Using the operator
d
s
dt
= , apply KCL at the
top node:
()
()
() ()
1
44
s
vt
it it svt=+ +

Apply KVL to the right-most mesh:

() ()( )( )60vtsitit− +=

Combining these equations gives:

() ()( ) ()
2
71 0 4
s
sitsit it it++=
The characteristic equation is: .
2
710ss++=0
The natural frequencies are: . 2 and 5ss=− =−

Ex. 9.4-2

Assume zero initial conditions . Write mesh
equations using the operator
d
dt
=s :
()() ()
12 1
1
710 100
2
sitit it− ++ −=


and
() ()()
12
1
70
2
vt sitit−−−

=

9-2

Now ()() ()
()
2
2 200
it
svtit vt
s
=⇒ =0.005 so the second mesh equation becomes:

()
()()
2
12
1
200 7 0
2
it
sitit
s
−−−

=

Writing the mesh equation in matrix form:

()
()
1
2
1
10
322
11200 7
22
s
s
it
it
ss
s

+−
  
=  
 
−+


Obtain the characteristic equation by calculating a determinant:

2
1,2
1
10
22
204000 1017.3
11200
22
s
s
ss s j
ss
s
+−
=+ +=⇒ =−±
−+

Ex. 9.5-1

2
o
2
22
,
12 o
After 0 , we have a parallel circuit with
11 7 1 1
and 6
2 2(6)(1/42)2 (7)(1/42)
77
6 1, 6
22
tR LC
RC LC
ss
αω
αα ω
=
== = = = =

∴ =−± −=−± −=−−



()
12 12
6
0
() .We need (0) and to evaluate &
n
nn
tt
t
dvt
vt AeAe v AA
dt
−−
=
∴ =+ .

At t = 0
+
we have:

()
()
0
10
0 10 A 420 Vs
1
42
C
t
dvt
i
dt
+
+
=
=− ⇒ ==
Then
12
12
12
0
(0) 0
84 , 84
420 6
n
n
t
vA A
AAdv
AA
dt +
+
=
== +

=− =
=−=−−


9-3

Finally
6
() 84 84 V
n
tt
vt e e
−−
∴= − +

Ex. 9.5-2

2211
0 401000ss s s
RCLC
+ +=⇒ ++=
Therefore
,
12 2.68 ,37.3ss=−−

()
12 1
2.68 37.3
, (0) 0
n
tt
vt Ae Ae v AA
−−
=+ ==
2+

KCL at t = 0
+
yields
(0) 1(0)
(0) 0
14 0
vd v
i
dt
++
+
++ = so
12
(0)
40(0) 40(0) 40(0)40(1) 2.7 37.3
dv
vi A
dt
+
++
=− − =− − =− − A
Therefore:
12
2.68 37.3
1.16 , 1.16 ()()1.16 1.16
n
tt
AA vtvt e e
−−
=− = ⇒ = =− +

Ex. 9.6-1

For parallel RLC circuits:
2
33
11 1 1
50, 2500
2 2(10)(10) (0.4)(10)
o
RC LC
αω
−−
== = == =

The roots of the characteristic equations are:
2
1,2
50 (50)250050,50s∴ =−± − =−−

The natural response is .
12
50 50
()
n
tt
vtAe Ate
−−
=+

At t = 0
+
we have:

()
(0)
(0) .8
10
(0)
800 /
0
c
c
v
iV
dvt i
Vs
dt c
t
+
+
+
+
−
== −
Ω
∴ == −
=

9-4

12
22
50 50
50 50
So (0)8 ()8
(0)
800 400 400
() 8 400 V
nn
n
tt
tt
vA vte At
dv
AA
dt
vt e te
+
+
−−
−−
== ⇒ = +
=− =− +⇒ =−
∴= −
e

Ex. 9.7-1

6
28
o 6
11
8000
2 2(62.5)(10)
11
=1
(.01)(10)
RC
LC
α
ω
−
−
== =
== 0

22 2 8
1
2
8000
8000 (8000)10 8000 6000
() [cos6000 sin6000 ]
o
n
t
sj
vt e A tA t
ααω
−
∴=−± −=− ± −=− ±
∴ =+

at t = 0
+

5
10
0.08 (0) 0
62.5
(0).24 A
(0)(0)
2.410 V/s
c
c
c
i
i
idv
dt C
+
+
++
+
++ =
⇒= −
∴ == −×

5
12
8000
(0)
(0)10 and 2.41060008000 26.7
() [10cos6000 26.7sin6000] V
n
n
n
t
dv
vA A A A
dt
vt e t t
+
+
−
== =−×= − ⇒ =−
1 2
∴ =−

Ex. 9.8-1
The differential equation is
() ()
() ()
2
2
56
s
dvtdvt
vtvt
dt dt
++ =
12,2 ,3ss=−−
so the characteristic equation is
. The roots are .
2
56ss++ =0
(a)
() ()
()
2
2
56
dvtdvt
vt
dt dt
++ 8=. Try ()
fvt B=. Substituting into the differential equation gives
()68 8/6 V
fBv t=∴ = .

9-5

(b)
() ()
()
2
4
2
56 3
t
dvtdvt
vt e
dt dt
−
++ = . Try ()
4t
fvt Be
−
= . Substituting into the differential
equation gives ()
2 433
(4)5(4)63 .
22
t
fB BB B vt e
−
−+ −+=⇒ =∴ = .
(c)
() ()
()
2
2
2
56 2
t
dvtdvt
vt e
dt dt
−
++ = . Try ()
2t
fvt Bte
−
= because –2 is a natural frequency.
Substituting into the differential equation gives

()
2
(44)5(12)6 2 2. 2
t
ftB B tBt B vtte
−
−+ −+ =⇒ =∴ = .

Ex. 9.8-2
() ()
()
2
2
92 0 361
ditdit
it t
dt dt
++ =+2. Try ()
f ABt=+
12t B+ ⇒
it . Substituting into the differential
equation gives 0920( )BA Bt++ + 36 0.6 and 1.53.A= = =

()1.530.6 A
fit t∴ =+

Ex. 9.9-1

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

()
2
12
1
C
R
v
RR
∞=
+

Next, represent the circuit by a 2nd order differential equation:

KCL at the top node of R2 gives:
()
() ()
2
C
CL
vt d
Cv tit
Rd t
+=
KVL around the outside loop gives: () () ()()
1sL L
d
vt LitRitvt
dt
=+ +
C
Use the substitution method to get

9-6

()
()
()
()
() ()
() () ()
1
22
2
1
12
22
1
CC
sC C
CC
vt vtdd d
vtL Cvt R Cvt vt
dtR dt R dt
RdL d
LCvt RCvt vt
dt R dt R
 
=+ + + + 
 
 
  
=+ + ++  
  
  
C
C

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
()
2
12
1
1 V
2
fC
R
vv
RR
=∞ = =
+

The characteristic equation is
()()
1
1222
2
1
1
68 2 4
R
RR
ss ss s s
RC L LC

+


++ + =++=+ +
 




0=

so the natural response is
24
1 2
V
tt
n
vA eAe
−−
=+

The complete response is
()
24
1 2
1
V
2
tt
cvt AeAe
−−
=+ +

()
()
()
24
1 21.236 3.236 0.3819
1.309
C tt
LC
vt d
it vt Ae Ae
dt
−−
=+ =− − +
At t = 0
+

() 120 0 0.5
cvA A
+
== ++
() 1 20 0 1.2363.2360.3819
LiA A
+
== − − +

Solving these equations gives A1 = −1 and A2 = 0.5, so

()
2411
V
22
tt
cvt e e
−−
=− +
(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
()
2
12
1
1 V
4
fC
R
vv
RR
=∞ = =
+

The characteristic equation is
9-7

()
1
21222
2
1
1
44 2
R
RR
ss ss s
RC L LC

+


++ + =++=+
 




0=
so the natural response is
( )
2
12 V
t
nvA Ate
−
=+
The complete response is
() ()
2
12
1
V
4
t
cvt AAte
−
=+ +

() () () ()( )
2
21 2
1
4
t
LC C
d
itvt vt AAAte
dt
−
=+ =+ −−
At t = 0
+

() 1
1
00
4
cvA
+
= =+
() 21
1
00
4
LiA
+
A= =+ −

Solving these equations gives A1 = −0.25 and A2 = −0.5, so

()
211 1
V
44 2
t
cvt te
− 
=− +
 


(c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Use the steady state response as the forced response:
()
2
12
4
1 V
5
fC
R
vv
RR
=∞ = =
+

The characteristic equation is
() ( )
1
1222
2
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=
so the natural response is
( )
2
12cos4 sin4 V
t
nve A tA t
−
= +
The complete response is
() ( )
2
120.8 cos4 sin4 V
t
cvt eA tA t
−
=+ +

()
()
()
21221
0.2 cos4 sin4
48 2 2
C tt
LC
AAvt d
it vt e te t
dt
−−
=+ =+ −
9-8

At t = 0
+

() 10 00.8
cvA
+
= =+
()
2
0 00.2
2
L
A
i
+
== +

Solving these equations gives A1 = -0.8 and A2 = -0.4, so

() ( )
2
0.8 0.8cos40.4sin4 V
t
c
vt e t t
−
=− +

Ex 9.9-2

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

() () ()
22
12 1 2 12
1
1, 1
CL o
RR
v i andv
RRR R R
∞= ∞= ∞=
++ R+

Next, represent the circuit by a 2nd order differential equation:

KVL around the right-hand mesh gives: () () ()
2CL
d
vt LitRit
dt
=+
L
KCL at the top node of the capacitor gives:
()()
() ()
1
sC
CL
vtvt d
Cv tit
Rd t
−
−=
Use the substitution method to get

() () () () () ()
()() ()() ()
12 2
2
11 2 1 22
sL L L L
LL L
dd d
vtRC Li tRit Li tRitRit
dtdt dt
dd
RLCitL RRCitR Rit
dt dt
 
=+ + +
 
 
=+ + ++
1L
+

Using ()
()
2
o
L
vt
it
R
= gives
() () () ()
2
11
12
22 2
so o
RRdL d
vt LCvt RCvt vt
Rd t R dt R
  +
=+ + +  
  
  
2
o
R

9-9

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
()
2
12
1
1 V
2
fo
R
vv
RR
=∞ = =
+

The characteristic equation is
()()
2
2122
1
1
1
68 2 4
R
RR
ss ss s s
RC L LC

+


++ + =++=+ +
 




0=

so the natural response is
24
1 2
V
tt
n
vA eAe
−−
=+
The complete response is

()
24
1 2
1
V
2
tt
ovt AeAe
−−
=+ +

()
() 12 241
V
1.3092.6181.309 1.309
o tt
L
AAvt
it e e
−−
== + +

() () ()
24
12
11
1.309 0.6180 0.2361
42
tt
CL L
d
vt it it Ae Ae
dt
− −
=+ =+ +
At t = 0
+

()
121
00
2.6181.3091.309
L
AA
i
+
== + +
() 12
1
0 0 0.61800.2361
2
CvA
+
== + + A

Solving these equations gives A1 = −1 and A2 = 0.5, so

()
2411
V
22
tt
ovt e e
−−
=− +

(b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω
Use the steady-state response as the forced response:
()
2
12
3
1 V
4
fo
R
vv
RR
=∞ = =
+

The characteristic equation is
()
2
22122
1
1
1
44 2
R
RR
ss ss s
RC L LC

+


++ + =++=+
 




0=
9-10

so the natural response is
( )
2
12 V
t
nvA Ate
−
=+
The complete response is
() ()
2
12
3
V
4
t
ovt AAte
−
=+ +

()
() 12 21
V
34 33
o t
L
AAvt
it te
−
== + +


() () ()
12 2 23
3
43 3 3
t
CL L
AA Ad
vt it it te
dt
−

=+ =+ + +



At t = 0
+

()
11
00
34
L
A
i
+
= =+
()
123
00
43 3
C
AA
v
+
== ++

Solving these equations gives A1 = -0.75 and A2 = -1.5, so

()
233 3
V
44 2
t
ovt te
− 
=− +
 


(c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
()
2
12
1
1 V
5
fo
R
vv
RR
=∞ = =
+

The characteristic equation is
() ( )
2
2122
1
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=

so the natural response is
( )
2
12cos4 sin4 V
t
nve A tA t
−
= +
The complete response is

() ( )
2
120.2 cos4 sin4 V
t
ovt eA tA t
−
=+ +

()
()
()
2
120.2 cos4 sin4 V
1
o t
L
vt
it eA tA t
−
== + +
9-11

() () ()
22
21
1
0.22 cos42 sin4
2
tt
CL L
d
vt it it Ae tAe t
dt
−−
=+ =+ −
At t = 0
+

() 10 00.2
LiA
+
= =+
() 20 00.22
CvA
+
== +

Solving these equations gives A1 = -0.8 and A2 = -0.4, so

() ( )
2
0.2 0.2cos40.1sin4 V
t
c
vt e t t
−
=− +

Ex. 9.10-1
At t = 0+
12no initial stored energy (0)(0)(0) = 0vv i
++ +
⇒= =

3(0) (0)
KVL : 0 00 0
10
di di
dt dt
++
−+ +=⇒ =
1
1
22
22
0( 0)
KCL at A : (0)0 0 0
1
5(0) (0)
KCL at B : 0(0)100 (0) 10 12 Vs
6
dv
i
dt
dv dv
ii
dt dt
+
+
++
++
++ = ⇒ =
−+ −=⇒ = = ⇒ =

For t > 0:

11
2
12
1
KCL at A : 0
112
5
KCL at B : 10
6
3
KVL: = 0
10
vd v
i
dt
dv
i
dt
di
vv
dt
+ +=
−+ =
−+ +

Eliminating i yields
12
1
2
2
12 2
15
10 0
12 6
35
0
106
dv dv
v
dt dt
dv
vv
dt
+ +− =

−+ +=


Next
9-12

22
21 2
12 22
11

44
dv dvdv dv
vv
dt dtdt dt
=+ ⇒ = +
3
2
3

Now, eliminating v
1
23
22 2 2
2 23
11 1 5
10
41 2 4 6
dv dv dv dv
v
dt dt dt dt
 
++ + + = 


Finally, the circuit is represented by the differential equation:

32
22 2
232
12 44 48 = 480
dv dv dv
v
dt dt dt
++ +

The characteristic equation is . It’s roots are
32
1244480ss s++ +=
1,2,32,4,6s=−−−. The
natural response is
24
12 3
tt
nvAeAeAe
6t− −−
=+ +

Try v as the forced response. Substitute into the differential equation and equate
coefficients to get B = 10. Then
f=B
3
24 6
21 2
()()() 1 0
tt t
nf
vtvtvtAeAeAe
−− −
=+ = + + +

We have seen that and
2(0)0v
+
=
2(0)
12 V/s
dv
dt
+
= . Also
2
2
122
(0)
4[(0)(0)]0
dv
vv
dt
+
+ +
= −= .
Then
21 2 3
2
12 3
2
2
122
(0) 0 10
(0)
12 246
(0)
0 41636
vA AA
dv
AA A
dt
dv
AA
dt
+
+
+
== +++
== −−−
== + +
3
A

Solving these equations yields so
12 315, 6, 1AA A=− = =−

( )
24 6
2()15 6 10 V
tt t
vt e ee
−− −
=− + −+

Ex. 9.11-1
211
0ss
RC LC
++ =
210
In our case 0.1, 0.1 so we have 100 0LC s s
R
= =+ + =

9-13

a)
2
1,2
0.4 251000
5, 20
Rs s
s
=Ω ⇒+ +=
=−−

b)
2
1,2
1 10100 0
55 3
Rs s
sj
=Ω⇒++ =
=−±

9-14

Problems

Section 9-3: Differential Equations for Circuits with Two Energy Storage Elements

P9.3-1
()
() ()
() ()
()
()
2
1
KCL:
KVL:
L
L
sL
vt dvt
it C
Rd t
dit
vt RitL vt
dt
=+
=+ +

()
() () () ()
()
() ()
() ()
2
1 2
22
2
1
1 2
22
1[
s
s
vt dvt dvt dvtL
vt R C LC vt
R dt Rdt dt
dvt dvtRL
vt vtRC LC
R Rdt dt

=+ + + +

  
=+ + + +
  
  
]

In this circuit
122, 100, = 1 mH, = 10 FRR L Cµ=Ω =Ω so
() ()
() ()
() ()
() ()
2
8
2
2
88
2
1.02 .00003 10
10 1.0210 3000
s
s
dvt dvt
vt vt
dt dt
dvtdvt
vt vt
dt dt
−
=+ +
=× + +

P9.3-2

Using the operator
d
s
dt
= we have

()
()
() ()
() () ()
1
2
KCL:
KVL:
s L
LL
vt
it itCsvt
R
vtRitLsit
=+ +
=+

() ()
()
22
2
11
Solving by usingCramer's rule for :
1
s
LL
it
it it
RLs
RCsLCs
RR
=
+++ +

() ()[] () ()
22
2
11
1
LL L
RL
it RCsitLCsit it
RR
  
++ + + =
  
  
s
9-15

In this circuit
12100, 10, = 1 mH, = 10 FRR L Cµ=Ω =Ω so

() () () ()
82
1.1 .00011 10
LL L
it sit sit it
−
++ =
s

()
() ()
()
2
88
2
1.110 11000 10
LL
Ls
ditdit
it it
dt dt
×+ + =

P9.3-3

After the switch closes, a source
transformation gives:

KCL:
()
()()()
2
0
cs c
L
dvtvtvt
itC
dt R
+
+ +=
KVL:
() ()
()
()()
11 0
L
sL c s
dit
RitRitL vtvt
dt
+ +− − =
() () ()
()
()
11
L
cs L s
dit
vt RitRitL vt
dt
=+ + −
Differentiating

() () () () ()
2
11 2
cs L Ldvt dit dit ditdvt
RR L
dt dt dt dt dt
=+ + −
s

Then
()
() () () () ()
() ()
()
()
2
11 2
2
11
2
1
0
sL L s s
L
L
sL s
dit dit ditdvtvt
itCR R L
dt dt dt dt R
dit
RitRitL vt
Rd t

++ + − +


+ ++ −

():
L
=
Solving for it

() ()
() ()
() ()
2
11 1 1
2
22 2
d 11
dt
LL ss
Ls
it R dit R R Rdit dvt
it it
LRCdt LRCLC LCR LdtLdt
   −
++ + + = − +  
    
1

9-16

Section 9-4: Solution of the Second Order Differential Equation - The Natural Response

P9.4-1

From Problem P 9.3-2 the characteristic equation is:
28
82
12
11000(11000)4(1.110)
1.11011000 0 , 55008930
2
ss ss j
−± −×
×+ +=⇒ = =− ±

P9.4-2

()()() ()
()
()
3
KVL: 40 10010
L
sL c
dit
itit vt
dt
−
−= × +

The current in the inductor is equal to the current in
the capacitor so
()
()
31
10
3
c
L
dvt
it
dt
−
=×

 

() ()
() ()
2
36
2
40 100
40 10 10
33
CC
Cs
dvt dvt
vt it
dt dt
−−  
=− × − ×  
  

() ()
() ()
2
2
400 30000 40
CC
Cs
dvt dvt
vt it
dt dt
++ =
2
1240030000 0 (100)(300)0 100, 300ss s s s s++ =⇒ + + =⇒ =− =−

P9.4-3

()()
()()
()
() ()()
()
6
3
KCL: 10 10 0
1
KVL: 2 110
s
L
L
L
vtvt dvt
it
dt
dit
vt it
dt
−
−
−
+ +× =
=+ ×

()()
()
()()() ()
()
() ()
()
2
36 6
02 110 10102 101010
LL
Ls L
dit dit dit
it vtit
dt dt dt
−− −
=+ × − + +× +×
3 L−

() ()
() () () ()
() ()
22
8 88
2
3 0.00102 10 102000 310 10
LL L L
sL Ls
dit dit dit dit
vtit it vt
dt dt dt dt
−
=+ + ⇒ + +× =

28
12102000310 0 3031, 98969ss s s++ ×=⇒ = =−
9-17

Section 9.5: Natural Response of the Unforced Parallel RLC Circuit

P9.5-1
The initial conditions are()
()0
06 V, 3000 V/s
dv
dt
= =−v . Using the operator ,
d
s
dt
= the node
equation is ()
()()()
() ()
2
0or 1
s
s
vtvtvt L
t LCssvtvt
Rs L R
− 
++ = ++ =


Csv
The characteristic equation is:
2211
0 50040,0000ss s s
RCLC
++ =⇒ + + =
The natural frequencies are:
2
1,2
25025040,000 100,400s=−± − =− −

The natural response is of the form . We will use the initial conditions
to evaluate the constants A and B.
()
100 400

t
vtAe Be
− −
= +
t

()
()
0 6
2 and 80
3000 100 400
vA B
ABdv
AB
dt
== + 

⇒= − =
=− =− − 


Therefore, the natural response is

()
100 400
2 8
tt
vt e e t
−−
=− + >0

P9.5-2
The initial conditions are () ()02 V, 0== 0vi .
The characteristic equation is:
2211
04ss ss
RCLC
++ =⇒ ++=30
3

The natural frequencies are:
12
, 1,ss=−−
The natural response is of the form ()
3

t
vtAeBe
t− −
= + . We will use the initial conditions to
evaluate the constants A and B.
Differentiating the natural response gives
()
3
3
t
dvt
Ae Be
dt
t− −
=− − . At t = 0 this becomes
()0
= 3
dv
A
dt
−−B. Applying KCL gives
()()
()0
dvtvt
Ci
dt R
t+ += or
() ()()
=
dvt vtit
dt RCC
− − .
At t = 0 this becomes
() ()()00
=
dv v i
dt RC C
− −
0
. Consequently

()()00 2
13 0
14
vi
AB
RC C
−−=− − =−−=−8
9-18

Also, . Therefore A = −1 and B = 3. The natural response is ()02 v== +AB

()
3
3 V
tt
vt ee
−−
=−+

P9.5-3

() ()
()
() ()
() ()
12
1
12
2
KVL : 5 3 0 1
KVL : 3 3 2 0 2
ditdit
i
dt dt
ditdit
it
dt dt
+− =
−+ + =

Using the operator
d
s
dt
= , the KVL equations are
() ()
() ( )
() () () ( )()
212
11
12
15 3 0 3
15 3 0 15 323 0
33 2 0 32
si si s
sis i ss si
si si s
++ − =
⇒+ − =⇒ + +− =
−+ += +
1
The characteristic equation is ()( )
2 2
1,2
1
15329 ,2 61320
6
ss s sss++ −= + ⇒ =−+= −
The currents are () ()
26
1 2 and
t
t
itAe Be itCe De
−
−
=+ = +
26
t
t
−
−
, where the constants A, B, C
and D must be evaluated using the initial conditions. Using the given initial values of the
currents gives
() ()
12
011 and 011 iA B i C== + ==+D

Let t = 0 in the KCL equations (1) and (2) to get

() ()1 2
0 033 143
2 and
26 6 6
di diAC
B D
dt dt
=−=−− =−=−−

So A = 3, B = 8, C = −1 and D = 12. Finally,

/6 2 /6 2
12()3 8 A and () 12 A
tt t t
it e e ite e
−− − −
=+ =−+
9-19

Section 9.6: Natural Response of the Critically Damped Unforced Parallel RLC Circuit

P9.6-1
After t = 0

Using KVL: ()
()
()0.025 0
c
cc
dit
it vt
dt
100 + +=

The capacitor current and voltage are related by:

()
()
5
10
c
c
dvt
it
dt
−
=

() ()
()
2
6
2
4000 410 0
cc
c
dvt dvt
vt
dt dt
∴ ++ × =
=

The characteristic equation is:
26
40004100ss++ ×

The natural frequencies are:
1,2 2000, 2000s=− −

The natural response is of the form: ()
2000 2000
12
tt
c Ae Ate
−−
=+vt

Before t = 0 the circuit is at steady state

(The capacitor current is continuous at t = 0
in this circuit because it is equal to the
inductor current.)

()
()
()( )
1
12 2
2000
03
0
0 2000 6000
36000 Vfor 0
c
c
t
c
vA
dv
AA A
dt
vt te t
+
+
−
==
==− +⇒ =
∴ =+ ≥

P9.6-2
After t = 0

Using KCL:
() ()
()
() ()
()
2
1
0
4
44
t
c
cc
cc
c
dvt
vd vt
dt
dvtdvt
vt
dt dt
ττ
−∞
++ =
⇒+ +
∫
0=
0

The characteristic equation is:
2
44ss++ =

9-20

The natural frequencies are:
1,22,2s=−−

The natural response is of the form: ()
22
12
tt
c AeAtevt
− −
=+

Before t = 0 the circuit is at steady state

()( ) ()()00 and 0
LL C C
ii v v0
+ −+
==
−

()( )
() ()
00
00
8 V
14
CL
C
ii
dv i
dt
++
++
=− =−
== −
2 A

()
()
()
2
12
0
00 and 8 8 V
c
t
c c
dv
vA A vt t
dt
+
+ −
== =−= ⇒ =−e

P9.6-3
Assume that the circuit is at steady state before t = 0. The initial conditions are

() ()
4
0 10 V & 00 A
cLvi
−−
==

After t = 0

KVL: ()
()
() ()
6
.01 10 0 1
L
c L
dit
vt it
dt
−+ + =

KCL:
()
() () ()
()
2
6
2
.01 10 2
CL L
L
dvt dit dit
it C C
dt dt dt

=− =− +


() ()
()
2
6
2
0.01 10 0
LL
L
dit dit
CC i
dt dt
t∴ ++ =

The characteristic equation is: () ( )
26
0.01 10 10Cs Cs+ +=

The natural frequencies are:
() ()
()
2
66
1,2
10 10 40.01
20.01
CC
s
C
−± −
=
C

For critically-damped response: 10 so .
122
.040 0.04 pFCC C−= ⇒ =
77
1,2
510,510s=−×−×
9-21

The natural response is of the form: it ()
77
510 510
12
tt
L Ae Ate
−× −×
=+
() () ()
()
()
()
() ()
7
7
66
66
12
61 2 510
ANow from (1) 0 100 0100 10
s
0
So 00 and 10 10 A
Now 10 10 V
L
cL
L t
LL
t
L
di
vi
dt
di
iA A it te
dt
vt it te
++ +
−×
−×
⇒= − =

== = = ∴ =
==
510

P9.6-4
The characteristic equation can be shown to be:
2211
50062.5100ss s s
RCLC
3
+ += + + ×=
The natural frequencies are:
1,2 250, 250s=− −
The natural response is of the form: ()
250 250tt
Ae Bte
−−
=+vt
()
()0
06 and 3000250 1500
dv
vA AB B
dt
== =− =− +⇒=−
()
250 250
6 1500 V
tt
vt e te
−−
∴ =−

P9.6-5
After t=0, using KVL yields:

()
() ()
()
()
0
24 6 1
tdit
Rit id
dt
vt
ττ++ + =∫
((

Take the derivative with respect to t:

() ()
()
2
2
4
ditdit
R it
dt dt
++ =0

()
()
2
2
22
The characteristic equation is 4 0
Let 4 for critical damping 2 0
So the natural response is
tt
sRs
Rs
itAteBe
− −
++ =
=⇒
=+
+=

()
()
()() ()
()
2
0
0 0 0 and 4 04 0 4
4A
t
di
iB Ri R
dt
it te
−
=⇒ = =− =− ==A
∴ =
9-22

Section 9-7: Natural Response of an Underdamped Unforced Parallel RLC Circuit

P9.7-1
After t = 0

KCL:
()
()
()
()
6
510 01
250
cc
L
vt dvt
it
dt
−
++ × =

KVL:
()
()
()0.8 2
L
c
dit
vt
dt
=

() ()
()
()
1,2
12
400
2
52
800 2.510 0 800250,0000, 400 300
2
The natural response is of the form v cos300 sin300
t
dvt dvt
cc
vt s s s j
cdt
dt
te A tA t
c
−
++ × = ⇒ + + = =−±
=+


Before t = 0 the circuit is at steady state:

() ()
() () ()
6
00 A
500
6
0 0250 63 V
500
LL
cc
ii
vv
+−
+−
−
==
−
== +=

From equation (1) :

()
() ()
5
0
21008000 0
c
Lc
dv
i v
dt
+
++
=−× − =

()
()
() []
1
12 2
400
03
0
0 400300 4
3cos3004sin300 V
c
c
c
t
vA
dv
AA A
dt
vte t t
+
+
−
==
==− + ⇒ =
∴ =+

9-23

P9.7-2
Before t = 0

()( )
() ()
00 0
00 2
vv
ii
+−
+−
==
==
V
A

After t = 0
KCL :

() ()
() ()
0
1
20
14
tvt dvt
vd i
dt
ττ 0+ ++∫
=

Using the operator
d
s
dt
= we have

() () ()() () ()
212
00 48
4
vt svt vti ssvt
s
 
++ + =⇒ ++
 
 
0=
2

The characteristic equation and natural frequencies are:
2
48 0 2ss s j++= ⇒=−±
The natural response is of the form: ()
2
12cos2 sin2
t
eB tB t
−
vt  =+
 

()
()
()() []
1 22
0
00 and 40 0 4282 or
dv
vB i v B B
dt
== =−− =− =−= =−

4
so
()
2
4sin 2 V
t
vt e t
−
=−

P9.7-3
After t = 0

()()
() ()
()
()
() ()
1
KCL : 01
42
4
KVL : 8 2
cc
L
L
cL
dvtvt
it
dt
dit
vt it
dt
++ =
=+

Characteristic Equation:
() ()
()
2
2
1,22
45 0 4 5 0
LL
L
ditdit
it ss s
dt dt
++ =⇒++=⇒ =−2i±

Natural Response: it ()
2
12 co s sin
t
L eA tA t
−
=+


9-24

Before t = 0

()
() ()
() ()
0 48
7 0 0
24 82
8
0 0 4 A
2
c
cc
LL
v
vv
ii
−
+−
+−

=⇒ = =

+

== −=−
8V

()()
() ()
00 8A
20 2410
44
Lc
L
di v
i
dt
++
+
=− =−−=
s

()
()
1
12 2
04
0
10 2 2
L
L
iA
di
AA A
dt
+
+
=−=
==− +⇒ =

() [ ]
2
4cos 2sinA
t
Lit e t t
−
∴ =− +

P9.7-4
The plot shows an underdamped response, i.e. () [ ]

12
cos sin
t
vtek tk tk
α
ωω
−
3
= ++ .
Examining the plot shows () ()
31
0 0, 0 0 0vk v∞= ⇒ = =⇒=k.
Therefore, vt()

2
sin
t
ke t
α
ω
−
= .

Again examining the plot we see that the maximum voltage is approximately 260 mV the time is
approximately 5 ms and that the minimum voltage is approximately −200 mV the time is
approximately 7.5 ms. The time between adjacent maximums is approximately 5 ms so
3
2
1257 rad/s
510
π
ω
−
≈=
×
. Then
( )
()() ()
()
()() ()
2
2
0.005
0.0075
0.26 sin1257 .005 1
0.2 sin1257 .0075 2
ke
ke
α
α
−
−
=
−=

To find α we divide (1) by (2) to get

() ( )
()
0.0025 0.0025sin6.29 rad
1.3 1.95 267
sin9.43 rad
ee
α α
α

−= ⇒ = ⇒=


From (1) we get . Then
2
544k=

()
267
544 sin1257 V
t
vt e t
−
=

9-25

P9.7-5
After t = 0

The characteristic equation is:

1122
0 or 25 = 0ss ss
RCLC
++ = ++

The natural frequencies are:
1,212sj=−±

The natural response is of the form:

12() cos2 sin2
t
vteB tB t
−
 =+
 

1(0)2v
+
==B. From KCL, ()
()
()
0 21 1V
00
55 10
c
v
+
++
=− − =−−=−
2s
ii so
()
12 2
0 13
10 2
22
dv
BB B
dt
+

=− =−+⇒ =−



Finally, ()
3
2cos2 sin2 V 0
2
tt
e te t t
−−
=−vt ≥

9-26

Section 9-8: Forced Response of an RLC Circuit

P9.8-1

After t = 0

()
()
()
()
()
()
KCL :
KVL :
sL
L
vt dvt
it itC
R dt
dit
vt L
dt
=+ +
=

()
()
()
()
2
2

LL
sL
dit ditL
it itLC
Rdt dt
=+ +

() ()
() ()
2
2
11 1
LL
Ls
dit dit
it it
dt RCdtLC LC
++ =

()
()
()
() ()()()
2
55
2
650 10 10
LL
Ls
dit dit
it it
dt dt
++ =

(a) Try a forced response of the form ()
fit A=. Substituting into the differential equations gives
()() ()()
3 3
11
00 1
.01110 .01110
AA
−−
++ = ⇒ =
××
. Therefore ()1 A
f=it .

(b) Try a forced response of the form ()
f AtBit=+. Substituting into the differential equations
gives
()( )
()
)()(
65
5
100 0.001
1
0.010.001
00 .+ . Therefore 0.5A= and
. Finally
3
3.2510B
−
=− × ()
3
53.2510 A
fit t
−
=− × .
A AtB t++ =

(c) Try a forced response of the form ()
250t
fit Ae
−
= . It doesn’t work so try a forced response of
the form . Substituting into the differential equation gives ()
250t
fitBte
−
=

() ( )
2
250 250 250 250 5 250 250
250 500 650250 10 2
tt t t t
Be Be Bte Be Bte e
−− − − −
  −− + − + + =
 
t−
B
.

Equating coefficients gives
() ( ) () ( ) []
2 2
55
250 65025010 0 25065025010 0 0 0BB B B+− + =⇒ + − + =⇒ =


and
5006502 0.0133BB B−+ =⇒ =

Finally ()
250
0.0133 A
t
fit te
−
= .
9-27

P9.8-2

After t = 0

() ()
()
()
2
1

s
dvt dvt vtR
vt
dtLdtLC LC
++ =

() ()
() ()
2
70 12000 12000
s
dvt dvt
vt vt
dt dt
++ =

(a) Try a forced response of the form ()
fvt A=. Substituting into the differential equations gives
001200024000 2A++ = ⇒=A. Therefore ()2 V
fvt= .

(b) Try a forced response of the form ()
f ABt=+vt . Substituting into the differential equations
gives7012000 120002400A At B++ = t. Therefore 0.2A= and
370
1.16710
12000
A
B
−−
== − ×.
Finally ()()
3
1.16710 0.2
f t
−
=− × + Vvt .

(c) Try a forced response of the form ()
30t
fvt Ae
−
=
30
12000
t
. Substituting into the differential equations
gives 900
30 30 30
2100 12000
tt t
AeA e
−−
−+ Ae
−
= e
−
. Therefore
12000
1.11
10800
A= =. Finally
()
250
1.11 V
t
e
−
=
fvt .

9-28

Section 9-9: Complete Response of an RLC Circuit

P9.9-1

First, find the steady state response for t < 0, when the switch is open. Both inputs are constant
so the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit. After a source transformation at the left of the circuit:

()
L
114
0 2.33 mA
3000
i
−
==
and
()
C
04 Vv =

After the switch closes

Apply KCL at node a:

C
CL 0
v d
Cv i
Rd t
++ =

Apply KVL to the right mesh:

Ls C C Ls0
dd
Li Vv vLiV
dt dt
+− =⇒ = +

After some algebra:

() ()
22
5s
LL L L L L22
11
500 1.610 320
Vdd d d
ii i i i i
dt RCdtLC RLC dt dt
+ + =− ⇒ + +× =−

The characteristic equation is

25
1,2
5001.6100 250312 rad/sss s j++ ×=⇒ =−±

9-29

After the switch closes the steady-state inductor current is iL(∞) = -2 mA so

() ( )12
250
0.002 cos312 sin312
L
t
it e A tA t
−
=− + +

() ()
() ()
() ( )
12 1 2
21 2 1
250
250
6.25 4
6.25 250cos312 sin312312sin312 cos312 4
6.25 312250cos312250312sin3124
CL
t
t
d
vt it
dt
e A tA t A tA t
eA A t A A t
−
−
=+
 =− − − − +
 
=− + + +


Let t = 0 and use the initial conditions:

() 1100.002330.002 0.00433
LiA
+
== − + ⇒ A=

() () ()
21 2 2
250 250
046.25312250 4 0.004330.00347
312 312
CvA A A A
+
== − +⇒ = = =

Then
() ( )
()
250
250
0.002 0.00433cos3120.00345sin312
0.0020.00555 cos31236.68 A
L
t
t
it e t t
et
−
−
=− + +
=− + − °

() ()
250
413.9 sin312 V
C
t
vt e t
−
=+

()
()
()
250
26.95 sin312 mA
2000
C t
vt
it e t
−
== +
(checked using LNAP on 7/22/03)

P9.9-2
First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.

()
1
00
14
i
−
== −
+
.2 A
and
() ()
4
01 0.
14
v=− =−
+
8 V

9-30

For t > 0

Apply KCL at node a:

s
1
0
vV d
Cv i
Rd t
−
++ =

Apply KVL to the right mesh:

22 0
dd
LRiL iv vRiL i
dt dt
+− =⇒ =+

After some algebra:
2 2
12 1 2 s
2 2
11 1
55
LRRC RR Vdd d d
ii i i i
dt RLCdtRLC RLC dt dt
++
++ = ⇒ + 1i+=

The forced response will be a constant, if = B so
2
2
5 5 0.2 A
dd
BB B B
dt dt
=+ + ⇒ =1 .

To find the natural response, consider the characteristic equation:

( )( )
2
0 55 3.621.38ss s s=+ +=+ +
The natural response is
n1 2
3.62 1.38tt
iAe Ae
−−
=+
so
()
12
3.62 1.38
0.2
tt
itAe Ae
− −
=+ +
Then
() () ()
12
3.62 1.38
4 4 10.48 1.52 0.8
ttd
vt it it Ae Ae
dt
−−
=+ =− − +



At t=0+
()
12
0.20 0.2iA A−= +=++
()
12
0.80 10.481.520.8vA A−= +=− − +

so A1 = 0.246 and A2 = −0.646. Finally

()
3.62 1.38
0.20.246 0.646 A
tt
it e e
−−
=+ −

9-31

P9.9-3
First, find the steady state response for t < 0. The input is constant so the capacitors will act like
an open circuits at steady state.

() ()
1
1000
01 0
10001000
v==
+
5 V

and
()
2
00 Vv=

For t > 0,

Node equations:

611
1
3
11
101
10 0
10006 1000
1
21 0 10
6
vd vv
v
dt
d
vv
dt
−
−
−−
+× + =



⇒+ × −=


2
2
v

612
2
3
12
1
10
100016
1
10
16
vv d
v
dt
d
vv v
dt
−
−
− 
=×



⇒− =×


2

After some algebra:
() ()
2
47
11 12
2.810 9.610 9.610
dd
vv v
dt dt
+× +× =×
8

The forced response will be a constant, vf = B so

() ()
2
47 8
2
2.810 9.610 9.610 10 V
dd
BB B B
dt dt
+× +× =× ⇒ = .

To find the natural response, consider the characteristic equation:

() ( )
24 7 3
1,22.810 9.6100 410,2.410ss s+× +× =⇒ =−×−×
4
+

The natural response is
n1 2
34
410 2.410tt
vA e Ae
−× −×
=+
so
()
11 2
34
410 2.410
10
tt
vt Ae Ae
−× −×
=+
At t = 0
9-32

(1) ()
() ()
11 2 1 2
34
4100 2.4100
5 0 10 10vA e Ae AA
−× −×
== + +=++
Next
3 4
11 2 1 1 2
1
2 10 10 120006000610
6
dd
vv v v v v
dt dt
−
+× −= ⇒ =− + +×



At t = 0
() () () () ()
44
11 2012000060000610120005600006100
d
vv v
dt
=− + +×=− + +×=
so
() () ( )
34
11 2
34
410 2.410
410 2.410
ttd
vt A e A e
dt
−× −×
=− × +−×
At t = 0+

() ()
()
()
()
() (
34 3
11 2 1 2
34
4100 2.4100
0 0 410 2.410 410 2.410
d
vA e A e A A
dt
−× −×
= =−× +−× =−× +−×)
4

so A1 = −6 and A2 = 1. Finally

()
1
43
2.410 410
10 6 V for 0
tt
vt e e t
−× −×
=+ − >

P9.9-4
For t > 0

KCL at top node:

()
()
()
()
1
0.5 5cos 0 1
12
L
L
dit dvt
tit
dt dt

−+ + =


KVL for right mesh:

() ()
() ()
1
0.5 2
12
Ldit dvt
vt
dt dt
=+

Taking the derivative of these equations gives:

()
() () ()
()
() () ()
()
22
22
22
22
1
of 1 0.5 5sin (3)
12
1
of 2 0.5 4
12
LL
L
ditdit dvt
d
t
dt
dt dt dt
dit dvtdvt
d
dt
dt dt dt
⇒+ + =−
⇒= +

9-33

()
()
()
() ()
() ()
()
2
2
2
2
Solving for in 4 and in 2 & plugging into 3 gives
71 2 30sin
LL
dit dit
dt dt
dvtdvt
vt t
dt dt
++ =−

The characteristic equation is: s .
2
7s+12 = 0+

The natural frequencies are .
1,23,4s=−−
The natural response is of the form
3
1 2
()
t
n
vt AeAe
4t− −
= + . Try a forced response of the form
()
1 2
cos sin
f
vt B tBt=+ . Substituting the forced response into the differential equation and
equating like terms gives
12
21 33
and
17 17
BB== −.

() ()
34
12
21 33
() cos sin
17 17
tt
nfvtvtvtAeAe t t
−−
=+ = + + −

We will use the initial conditions to evaluate A1 and A2. We are given ()0
L
=0i and .
Apply KVL to the outside loop to get
()01 Vv=

()() ()()()11 5c
CL Citit it vt t++ + −

os0=
At t = 0+
()
()()()5cos0 0 0501
02
22
L
C
iv
i
+− +−
== A=
() ()00 2
24 V/s
112112
Cdv i
dt
== =

12 1
2
12
21
(0)1 25
17
429
(0) 33
2434 17
17
vA A A
Adv
AA
dt
+
+

== ++ =


⇒
=−

==−−−


Finally,
4
3429 21cos33sin
()25 V
17
t
t et
vt e
−
− −+ t
∴ =−

9-34

P9.9-5
Use superposition. Find the response to inputs 2u(t) and –2u(t-2) and then add the two responses.
First, consider the input 2u(t):

For 0< t < 2 s

Using the operator
d
s
dt
= we have
KVL:
()() ()42 0
cL Lvtsit it++ −=

(1)

KCL:
() () () ()
13
(2)
3
L ccLit svt vt it
s
=⇒ =
Plugging (2) into (1) yields the characteristic equation:
2
(4 3)ss 0++=. The natural frequencies
are . The inductor current can be expressed as
1,21 ,3s=−−

( )
33
12 12()()() 0
tt tt
LnfitititAe Ae Ae Ae
− −−
=+ = + += +
−
.

Assume that the circuit is at steady state before t = 0. Then (0)0 and (0)0
cLvi
++
= =.
Using KVL we see that ()
(0)
14 2(0) (0)
L
LC
di
iv
dt
+
++
=− − =

8 A/s. Then
12
1 2
12
(0)0
4 , 4(0)
83
L
L
iA A
A Adi
AA
dt
== + 

= =−
==−−


.
Therefore it . The response to 2u(t) is
3
()4 4 A
t t
L e e
− −
=−
()
1 3
3
00
()84 ()
816 16 V 0
816 16 V
L tt
tt
t
vt it
ee t
ee ut
−−
−−
<
=− =
− +>
=− +

.

The response to –2u(t-2) can be obtained from the response to 2u(t) by first replacing t by t-2
everywhere is appears and the multiplying by –1. Therefore, the response to –2u(t-2) is

()
(2) 3(2)
2() 816 16 - 2V
tt
vt e e ut
−− −−
=−+ −

.

By superposition, . Therefore
1 2
() () ()vtvtvt= +

3 (2) 3(2)
()81616 () 816 16 (2) V
tt t t
vt e eut e e ut
−− −− −−
  =− + +−+ − −
  

9-35

P9.9-6

First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit.

()
5
0 1.25 A
4
i=−=−
and
()05 Vv=

After the switch opens

Apply KCL at node a:

0.125
2
vd
vi
dt
+ =

Apply KVL to the right mesh:

10cos 4 40
d
tv ii
dt
− ++ +=
After some algebra:
2
2
56 20co
dd
vv v
dt dt
++ = st
3
t

The characteristic equation is
2
1,2
56 0 2,ss s++ =⇒ =−−
Try
f
cos sinvA tB= +

() () ()
2
2
cos sin5 cos sin6cos sin20cos
dd
At Bt At Bt At Bt
dt dt
++ ++ + = t
9-36

() ( )( )cos sin5 sin cos6cos sin20cosAt Bt AtBt At Bt−− +− + + + = t
() ( )56 cos 56sin20cosAB A tBAB t−+ + +−− + = t
t

Equating the coefficients of the sine and cosine terms yields A =2 and B =2. Then

f
2cos2sinvt= +

()
23
12
2cos2sin
tt
vt t tAeAe
− −
=+ + +
Next
()
()() () () ()0.125 8 4
2
vt dd
vtit vt itvt
dt dt
+= ⇒ = −

() () () ()
5V
08 0408 4530
4s
d
vi v
dt

=− =−− =−



Let t = 0 and use the initial conditions:

()
00
12 1
50 2cos02sin0 2vA eAe
−−
== + + + =++
2
AA

()
23
122sin2cos2 3
ttd
vt t tAe Ae
dt
− −
=− + − −
()
00
12 130 02sin02cos02 3 22 3
d
vA e Ae
dt
−−
−= =− + − − =− −
2A A
>

So A1 = −23 and A2 = 26 and

()
23
2cos2sin23 26 V for 0
tt
vt t te e t
−−
=+ − +

9-37

P9.9-7

First, find the steady-state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.

()00 Ai=
and
()00 Vv=

After t = 0

Apply KCL at node a:
d
Cv
dt
=i

Apply KVL to the right mesh:
()82 42
12 2 8
d
iv i i
dt
d
iv i
dt
0+++ +=
++ =−

After some algebra: ()
2
2
14
6
2
dd
vv v
dt dt C C

+ +=

−
The forced response will be a constant, vf = B so

()
2
2
14
68
2
dd
BB B B
dt dt C C

++ =−⇒ =−

V

(a)
When C = 1/18 F the differential equation is () ()
2
2
69
dd
vv v
dt dt
72+ + =−.
The characteristic equation is
2
1,2
69 0 3,ss s 3++= ⇒ =−−
Then . ()( )
3
12 8
t
vt AAte
−
=+ −
Using the initial conditions:
9-38

() ()( )
() () ()()
0
12 1
00
12 2 2 1
00 0 8 8
00 0 3 0 3 2
vA A e A
d
iC v C AA eAe A A
dt
== + −⇒ =
== =− + + ⇒ = =

4

So
()( )
3
824 8 V for 0
t
vt te t
−
=+− >

(b)
When C = 1/10 F the differential equation is () ()
2
2
65
dd
vv v
dt dt
40+ + =−
The characteristic equation is
2
1,2
65 0 1,ss s 5++= ⇒ =−−
Then . ()
5
12
8
tt
vtAeAe
− −
=+ −
Using the initial conditions:
()
()
00
12 1 2
12
00
12 1 2
00 8 8
10 and 2
00 5 5 0
vA eAe AA
AAd
vA eAe AA
dt
== + −⇒ +=

⇒= =−
== − − ⇒−− =



So
()
5
10 2 8 V for 0
tt
vt e e t
−−
= −− >

(c)
When C = 1/20 F the differential equation is () ()
2
2
61 0
dd
vv v
dt dt
80+ += −
The characteristic equation is
2
1,2
6100 3ss s j++= ⇒ =−±
Then . () ()
3
12cos sin8
t
vteA tA t
−
= + −
Using the initial conditions:
() ( )
() () ( )
0
12 1
00
12 1 2 2 1
00 cos0 sin08 8
0 03 cos0 sin0 sin0 cos0 3 24
ve A A A
d
ve AA eA A A A
dt
== + −⇒ =
== − + +− + ⇒ = =

So
() ( )
3
8cos24sin8 V for 0
t
vte t t t
−
=+ − >

9-39

P9.9-8

The circuit will be at steady state for t<0:
so
iL(0
+
) = iL(0
−
) = 0.5 A
and
vC(0
+
) = vC(0
−
) = 2 V.

For t>0:

Apply KCL at node b to get:

() () () ()
11 11
44 44
dd
it vt it vt
LC Ldt dt
=+ ⇒ =−
C

Apply KVL at the right-most mesh to get:

() () () ()
1
42 8
4
dd
it it vtvtc
LL dt dt

+= +


c

Use the substitution method to get

() () () ()
11 11 1
42 8
44 44 4
dd d d
vt vt vtvtCC c
cdt dt dt dt
   
−+ − = +
   
   

or
() () ()
2
2
26 2
dd
vt vt vt
CCdt dt
=+ +
C

The forced response will be a constant, vC = B so
2
26 2
2
dd
BB B B
dt
dt
=+ + ⇒ =1 V.

To find the natural response, consider the characteristic equation:

( )( )
2
0 62 5.650.35ss s s=+ +=+ +
The natural response is
12
5.65 0.35
n
tt
vA e Ae
− −
=+
so
()
1 2
5.65 0.35
1
C
tt
vt Ae Ae
− −
= ++
Then
() ()
12
5.65 0.3511 1
1.41 0.0875
44 4
LC
ttd
it vt Ae Ae
dt
−−
=+ =+ +
9-40

At t=0+
()
1
2
20
C
vA A1= += ++
()
1
2
11
0 1.410.0875
24
LiA=+ =+ + A

so A1 = 0.123 and A2 = 0.877. Finally

()
5.65 0.35
0.123 0.877 1V
C
tt
vt e e
− −
=++

P9.9-9

After t = 0

The inductor current and voltage are related be

()
()
(1)
dit
vtL
dt
=

Apply KCL at the top node to get

()
()
()
5(
2
dvt vt
Ci t
dt
++= 2)

Using the operator
d
s
dt
= , and substituting (1) into (2) yields (4 ()
2
29 ) ss it5++= .
The characteristic equation is ss . The characteristic roots are .
2
429++ =0
1,2 2 5sj=−±
The natural response is of the form . () []
2
cos5 sin5
t
niteA tB t
−
=+
Try a forced response of the form ()
f Ait=. Substituting into the differential equation gives
5A=. Therefore it . ()5 A
f=
The complete response is [ ]
2
() 5 cos5 sin5
t
it eA tBt
−
=+ + where the constants A and B are
yet to be evaluated using the initial condition:

(0)0 5 5iA A==+ ⇒=−
()
(0) (0) 2
00 025
5
di di A
vL AB B
dt dt
== ⇒ ==−+ ⇒==2−

Finally, [ ]
2
() 5 5cos52sin5 A
t
e t t
−
=+ − −it .

P9.9-10

9-41

Assume that the circuit is at steady before t = 0.

2
(0)(0) 9 6 A
21
1
(0)(0) 9 1.5 4.5 V
21
ii
vv
+−
+−
== ×=
+
== ××=
+

After t = 0:

Apply KCL at the top node of the current source
to get

()
()()
() 0.5 (1)
1.5
s
dvtvt
it it
dt
++ =

Apply KVL and KCL to get

()
()() ()
()
5
0.5 0.5 (2)
1.5
dvtvt dit
vt it
dt dt

++ = +



Solving for i(t) in (1) and plugging into (2) yields

() ()
() ()
()
()
2
2
2
49 4 2
2 where 93 A
30 5 5
st
ss
dvt dvt dit
vtit it e
dt dt dt
−
++ = + =+

Using the operator
d
s
dt
=, the characteristic equation is
2494
0
305
ss+ += and the characteristic
roots are . The natural response has the form
1,2.817 .365s=− ±j

0.817
12() cos(0.365) sin(0.365)
t
nvte A tA t
−
 =+
 

Try a forced response of the form vt
2
01()
t
f BBe
−
=+ . Substituting into the differential equations
gives . The complete response has the form
0 1 4.5 and 7.04B B= =−
−

.817 2
12() cos(0.365) sin(0.365)4.57.04
t t
vte A tA t e
− −
=+ +


Next, consider the initial conditions:

11(0)4.5 4.57.04 7.04vA A== +− ⇒ =

9-42

(0) 4 4
2(0)-2(0)(0) 2(93)2(6)(4.5) 6
33
s
dv
ii v
dt
=− = +− − =

()
12 2
0
6 0.8170.365 14.08 6.38
dv
AA A
dt
== − + + ⇒ =−

So the voltage is given by

0.817 2
2() 7.04cos(0.365) sin(0.365) 4.57.04
t t
vt e tA t e
− −
=+ +

−

Next the current given by
() ()
() () 0.5
1.5
s
vt dvt
itit
dt
=− −

Finally
[ ]
0.817 2
() 2.37cos(0.365) 7.14sin(0.365) 6 0.65 A
t t
ite t t e
− −
=+ ++

P9.9-11

First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.

()(04 0
avi )
− −
=−

() ()( )( )()02 02 40 00 A
a
iv i i
−− − −
== − ⇒ =

and
()010 Vv
−
=

9-43

For t > 0

Apply KCL at node 2:

0
a
a
vd
KvCv
R dt
++ =

KCL at node 1 and Ohm’s Law:

avR=−i
so
1dK
vi
dt C
+
=
R

Apply KVL to the outside loop:
s
0
d
Li RivV
dt
++− =
After some algebra:

22
s22
11
40 1442304
dR d KR KR d d
vv v V v v v
dt Ldt LC LC dt dt
++
++ = ⇒ + + =

The forced response will be a constant, vf = B so

() ()
2
2
40 144 2304 16 V
dd
BB B B
dt dt
++ = ⇒ =

The characteristic equation isss .
2
1,2401440 4,36s++ =⇒ =−−

Then ()
4 36
12 16
tt
vtAeAe
− −
= ++ .

Using the initial conditions:

()
()
00
12 1 2
12
00
12 1 2
100 16 6
6.75 and 0.75
00 4 36 436 0
vA eAe AA
AA
d
vA e Ae A A
dt
+
+
== + + ⇒ +=−

⇒= − =
== − − ⇒− − =


So
()
36 4
0.75 6.75 16 V for 0
tt
vt e e t
−−
=− + >

(checked using LNAP on 7/22/03)
9-44

Section 9-10: State Variable Approach to Circuit Analysis

P9.10-1

At 0 the circuit is source free (0)0 and (0)0.
Lti
−
=∴ = v=

After t = 0

Apply KCL at the top node to get

()
()1
4(
5
L
dvt
it
dt
+= 1)

Apply KVL to the right mesh to get

()()
()
()16 0
L
L
dit
vt it
dt
−− = (2)

()
() ()
()
2
1 2
Solving for in (1) and plugging into (2) 6 5 120
dvt dvt
it vt
dt dt
⇒+ + =.

The characteristic equation is ss . The natural frequencies are . The natural
response has the form . Try
2
65++=
1 AeA
−
=+
0
−
1,21, 5s=−−
5
2()
tt
nvt e ()
f Bvt= as the forced response.
Substituting into the differential equation gives B = 24 so ()24= V.
fvt The complete response
has the form vt .
5
12 24
tt
Ae
− −
+() Ae=+

Now consider the initial conditions.
(0)
V
From (1) 205 (0) = 20
s
L
dv
i
dt
=− . Then

12
12
12
(0) 0 24
25,1
(0)
20 5
vA A
AA
dv
AA
dt
== ++ 

⇒= − =
== −−



Finally
5
() 25 24 V
tt
ee
−−
=− + +vt .

9-45

P9.10-2
Before t = 0 there are no sources in the circuit so (0)0 and (0)0
L
iv= =.

After t = 0 we have:

Apply KCL at the top node to get

()
()1
4(
10
L
dvt
it
dt
=− 1)

Apply KVL to the left mesh to get

()
()
()6 0(
L
L
dit
vt it
dt
−− = 2)

Substituting ()
Lti from (1) into (2) gives

() ()
()
2
2
6 10 2
dvt dvt
vt
dt dt
++ =40

The characteristic equation isss . The natural frequencies are. The
natural response has the formvt
2
6 100++ =
3
() c
t
n
eA
−
1,23 sj=−±
1 2
os sintAt +=
 
. Try () B
fvt= as the forced
response. Substituting into the differential equation gives B = 24 so ()
f 24 V.=vt The complete
response has the form .
3
12
cos
t
eAtA
−
=+()vt sin24t +


Now consider the initial conditions.
(0)
V
From (1) 4010 (0) = 40
s
L
dv
i
dt
=− . Then

11(0)0 24 24vA A==+ ⇒ =−
12 2 2
(0)
403 72 32
dv
AA A A
dt
==−+=+⇒ =−
Finally, [ ]
3
() 24cos32sin24 V
t
e t t
−
=− − +vt

9-46

P9.10-3
Assume that the circuit is at steady state before t = 0 so (0)3 A and (0)0 V
L
iv=−= .

After t = 0 we have

KCL: ()
()()
60
dvtvt
C
dtR
it+ ++ =

KVL: ()
()dit
vtL
dt
=

()
() ()1
60
dit ditd
itCL L
dtdtR dt
 
+ ++
 
 
=

() ()
()
() ()
()
22
22
11 6
100 250 1500
dit dit dit dit
it it
dtRCdtLC LC dt dt
−
++ = ⇒ + + =−

The characteristic equation isss . The natural frequencies are
. The natural response has the formit . Try
as the forced response. Substituting into the differential equation gives B = −6 so
. The complete response has the form
2
100 250 0++ =
1,22.57,97.4s=− −
()
fit B=
()6 A
fit=−
2.57 97.4
1 2()
tt
n Ae Ae
−−
=+
2.57 97.4
12() 6
tt
Ae Ae
−−
it= +− .

Now consider the initial conditions:

12
1
2
12
(0) 6 3
3 .081

(0)
0.081 0 2.5797.4
iA A
A
di
AAA
dt
=+ −=− 
=

=−==− −



Finally:
()
2.57 97.4
2.57 97.4
() 3. 081 .081 6 A
() .2 1.58 1.58 V
tt
tt
it e e
dit
vt e e
dt
−−
−−
=− −
== − +

9-47

P9.10-4

(Encircled numbers are node numbers.)
Apply KCL to the supernode corresponding to
the dependent voltage source to get

() ()
() ()
20 .01
2
x
xx
dvtvt
it it
dt
0− − + =

Apply KCL at node 1 to get

() ()
()
20
2
x
x
vt
itit− +=

Apply KVL to the top-right mesh to get

()()
()
0.1 0
x
dit
vt vt
dt
+ −=

Apply KVL to the outside loop to get () ()() 2
xxit vtvt=− − .
Eliminate ()
xit to get
()()
()
() () ()
() ()
()
5
0.01 0
2
9
20
2
0.01
x
x
x
dvt
vtvt
dt
it vtvt
dit
vt vt
dt
+ −=
++ =
=− +

Then eliminate v to get ()
xt
()
() ()
()()
()
1.5 0.01 0.25 0
2.5 0.45 0
dvt dit
vt
dt dt
dit
vtit
dt
−− +
−+ + =
=

Using the operator
d
s
dt
= we have
() ()
() ()
(1.5.01) (.25) 0
(2.5) (1.45) 0
svt sit
vt siy
−− + =
−+ + =

The characteristic equation isss . The natural frequencies
are . The natural response has the form
2
13.33 333.33 0++ =
12
, 6.67 17ss j=− ± [ ]
6.67
() cos17 sin17
t
nvt A tB te
−
=+ .
The forced response is vt . The complete response has the form ()
f 0=
[ ]
6.67

t
e
−
() cos17 sinA tB=+ 17tvt .
9-48

The given initial conditions are i(0)0 and (0)10 V.v= = Then

(0)
(0)10 and 111 6.67 17 2.6
dv
vA A B B
dt
== =−=− + ⇒=−

Finally
6.67
() [3.27 sin 17] A
t
te
−
=it .
(Checked using LNAP on 7/22/03)

P9.10-5
Assume that the circuit is at steady state before t = 0 so
(0)10
(0)10 V and (0) = A
33
L
v
vi== .

The switch is open when 0 < t < 0.5 s

For this series RLC circuit we have:

2
0
1
3and 12
2
R
LL
αω
C
=== =

22
01 ,2
33sjαα ω−± −==−±

The natural response has the form ( )
3
() cos1.73 sin1.72
t
n
vteA tB t
−
= + . There is no source so
. The complete response has the form ()0
fvt= ( )
3
() cos1.73
t
vteA t
−
=+ sin1.72B t.

Next
()
(0)10
10
0(0) 103
5.77 20 3 1.73
16
vA
A
idv
BAB
dt C
== 
=
⇒
==−=− =−=−+ 


so
( )
()
3
3
() 10cos1.73 5.77sin1.73 V
() 3.33cos1.73 5.77sin1.73 A
t
t
vte t t
ite t t
−
−
=+
=−

In particular,
()
1.5 1.73 1.73
(0.5) 10cos 5.77sin 0.22316.48644.39152.43 V
22
ve
−
=+ = × + =



and
()
1.5 1.73 1.73
(0.5) 3.33cos 5.77sin 0.22312.16004.39150.50 A
22
ie
−
=− = × − =


−
9-49

The switch is closed when t > 0.5 s

Apply KCL at the top node:

()
()
()
() ()
()
() () ()
2
2
30 1
0
66
1
5
6
1
6
L
L
L
vt dvt
it
dt
dvt
it vt
dt
dit dvtdvt
dt dtdt
−
++ =

⇒= − +


⇒= − +


Apply KVL to the right mesh:
() ()
()1
3
2
L
L
dit
vt it
dt
=+
The circuit is represented by the differential equation
() ()
()
2
2
71 8
dvtdvt
vt
dt dt
++ =180
18

The characteristic equation is . The natural frequencies are .
The natural response has the form
2
0 7ss=+ +
1,23.72.4sj=−±
( )cos2.4 sin2.4A tB t= +
3.5
()
t
n
e
−
vt . The forced response is
. The complete response has the form()10 V
fvt= ( ) sin2.410B t
3.5
() cos2
t
vte A
−
.4t= ++ .

Next
( )
3.50.5
(0.5) cos1.2 sin1.20.0630.162ve A B A
−×
=+ = + B

()
() ( )
() (
3.50.5
3.50.5 3.50.5
0.5
3.52.4cos1.23.52.4sin1.2
3.5cos1.22.4sin1.2 2.4cos1.23.5sin1.2
0.60910.4158
dv
eA B B A
dt
eA e
AB
−×
−× −×
=− + − +

=− − + −
=− −
)B

Using the initial conditions yields

()
2.43(0.5) 0.0630.162
20.65
0.5(0.5) 12
23.03 30.60910.4158
16
vA B
A
idv
BAB
dt C
== + 
=−
⇒− 
==− =− ==− −



Finally
()
3.5
() 20.65cos2.4 23.03sin2.410
t
vte t t
−
=− + +

In summary
()
()
3
3.5
10cos1.73 5.77sin1.73 V 00.5
()
20.65cos2.4 23.03sin2.410 V0.5
t
t
et t
vt
et t
−
−
 +<
=
−+ +
t
t
<
<

9-50

Section 9-11: Roots in the Complex Plane

P9.11-1
After t = 0

13
12
210 6
0
2000
di
dt
ii
−
×−
++=
1233
2210 3000210
di di
i
dt dt
−−
×= +×
Using the operator
d
dt
s= yields
3
1
33
2
62000210 2000
0210 3000210
is
iss
−
−−
+× 
= 
×+ ×  

26 12
56
1,2
3.5101.5100
510,310
ss
s
+× +×=
⇒= −×−×

P9.11-2
From P9.7-1

()() ()()
22
66
2
11 1 1
00
250510 0.8510
8002500000
ss s s
RCLC
ss
−−
++ =⇒ + + =
××
⇒+ + =

1,2400 300sj=±

P9.11-3

()
()()
()
()
()
6
L
L
1
KCL: 10
4 4000
KVL: 4
dvtvt
it
dt
dit
vt vt
s
dt
−
=× +
=+

()
()()
()
() ()
()
2
63
2
1 6
41 0 10 10
4 4000
dvt dvtdvtvtd
vt vt vt
s dt dt dt dt
−−

−
=× + + = + +


9-51

23 6
1,2
Characteristic equation: 10 100
Characteristic roots: 500 866
ss
sj
++ =
=− ±

P9.11-4
Before the voltage source voltage is 0 V so 0t= (0)(0) 0 V
bbvv+=− = and
. Apply KCL at node a to get (0) 0 A+= =(0)ii −

(0)36 (0) (0)
(0) 0 (0)2(0)36 (0) 12 V
12 6
aa b
aa a
vv v
iv v
+− +− +
−+ + =⇒ ++ += ⇒ =v

After t = 0 the node equations are:

()()
() ()()
()()
0
1
0
12 6
t
as b a
ba
vtvt vtvt
vv d
L
ττ τ
−−
−+ − +∫
=

() ()()
() ()()
0
1
0
6
t
bb a
ba
dvtvtvt
Cv
dt L
ττ τ
−
++ −∫
v d=

Using the operator
d
s
dt
= we have

() ()
()11 1 11

126 6 12
s
ab
vt
vt vt
ss
  
++ +−− =
  
  

() ()
11 111
0
61 86
abvt s vt
ss
  
−− + ++ =  
  

Using Cramer’s rule
9-52

() () ()
2
(5 6) ( 6) ( 6) 36
bs
ss vts vts++ =+ =+

The characteristic equation is . The natural
frequencies are . The natural response has the
form vt . Try
2
56ss++ =
2
tt
0
1,22,3s=−−
1
Ae A=+
23
()
n
e
−−
()
f
vt B= as the forced
response. Substituting into the differential equation gives
B = 36 so The complete response has the form
.
() 36 V.
f
vt=
23
b1 2

tt
Ae Ae
−−
=+() 36vt +

Next
12
12
(0)36
(0)2 3
b
b
vA
dv
AA
dt
+
+
=+ +
=−−
A

Apply KCL at node a to get
()()()
()
1
0
18 6
bb advtvtvt
it
dt
−
+ +=
At t = 0+
()()()
()12
00 011 12
(2 3) = 0= 02
18 18 6 6
ba bdv v v
AA i
dt
++ +
− −+
−− = − −=
0

So
12
12
12
0(0)36
72, 36
1
(2 3)2
18
b
vA A
AA
AA
+
== ++

⇒= − =
−− =


Finally
()
23
3672 36V for 0
tt
b
vt e e t
−−
=− + ≥
9-53

PSpice Problems

SP 9-1

Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)

9-54

V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.

9-55

SP 9-2

Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)

9-56

V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.

9-57

SP 9-3

9-58

SP 9-4

9-59

Verification Problems

VP 9-1

This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current
()25
() 250 mA
100100
svt
it== =

This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an
underdamped response. That requires

32 2 6
1210 4 4(100)(210)810LR C
−−
×= < = × =×
2−

This inequality is satisfied, which also agrees with the plot.

The damped resonant frequency is given by

() ( )
2
2
3
663
11 1 1
5.9510
2 2(100)(210)2101210
dLC RC
ω
−−−

=− = − = ×

××× 

The plot indicates a maxima at 550.6µs and a minima at 1078.7µs. The period of the damped
oscillation is

2(1078.7s550.6 s)1056.2 s
dT µ µµ=− =
Finally, check that
33
6
22
5.9510= = = = 5.94910
1056.210
d
d
T
π π
ω
−
××
×

The value of
dω determined from the plot agrees with the value obtained from the circuit.

The plot is correct.

VP 9-2

This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current.
()15
() 150 mA
100100
svt
it== =

9-60

This agrees with the value of 149.952 mA shown on the plot.

Next, the plot shows an underdamped response. This requires

32 2 6
810 4 = 4 (100)(0.210) = 810LR C
−−
×= < × ×
3−

This inequality is not satisfied. The values in the circuit would produce a critically damped, not
underdamped, response.

This plot is not correct.

Design Problems

DP 9-1

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

()
2
12
1
C
R
v
RR
∞=
+

The specifications require that ()
1
2
Cv∞=so
2
12
12
1
2
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:

KCL at the top node of R2 gives:
()
() ()
2
C
CL
vt d
Cv tit
Rd t
+=
KVL around the outside loop gives: () () ()()
1sL L
d
vt LitRitvt
dt
=+ +
C
Use the substitution method to get

9-61

()
()
()
()
() ()
() () ()
1
22
2
1
12
22
1
CC
sC C
CC
vt vtdd d
vtL Cvt R Cvt vt
dtR dt R dt
RdL d
LCvt RCvt vt
dt R dt R
 
=+ + + + 
 
 
  
=+ + ++  
  
  
C
C

The characteristic equation is
()()
1
1222
2
1
1
68 2 4
R
RR
ss ss s s
RC L LC

+


++ + =++=+ +
 




0=

Equating coefficients of like powers of s:
1
12
2
1
1
6and 8
R
RR
RC L LC
+
+==
Using
12RR==R gives
11
64
R
RCL LC
+=⇒ =
These equations do not have a unique solution. Try C = 1 F. Then
1
H
4
L= and
213 1
46 0 1.309 or 0.191
24
RR R R R
R
+= ⇒ −+=⇒ = Ω = Ω

Pick R = 1.309 Ω. Then
()
24
1 2
1
V
2
tt
cvt AeAe
−−
=+ +

()
()
()
24
1 21.236 3.236 0.3819
1.309
C tt
LC
vt d
it vt Ae Ae
dt
−−
=+ =− − +

At t = 0
+

() 1200 0
cvA A
+
== ++.5
() 1 20 0 1.2363.2360.3819
LiA A
+
== − − +

Solving these equations gives A1 = -1 and A2 = 0.5, so

()
2411
V
22
tt
cvt e e
−−
=− +

9-62

DP 9-2

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

()
2
12
1
C
R
v
RR
∞=
+

The specifications require that ()
1
4
Cv∞=so
2
21
12
1
3
4
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:

KCL at the top node of R2 gives:
()
() ()
2
C
CL
vt d
Cv tit
Rd t
+=
KVL around the outside loop gives: () () ()()
1sL L
d
vt LitRitvt
dt
=+ +
C
Use the substitution method to get

()
()
()
()
() ()
() () ()
1
22
2
1
12
22
1
CC
sC C
CC
vt vtdd d
vtL Cvt R Cvt vt
dtR dt R dt
RdL d
LCvt RCvt vt
dt R dt R
 
=+ + + + 
 
 
  
=+ + ++  
  
  
C
C

The characteristic equation is
()
1
21222
2
1
1
44 2
R
RR
ss ss s
RC L LC

+


++ + =++=+
 




0=
Equating coefficients of like powers of s:
1
12
2
1
1
4and 4
R
RR
RC L LC
+
+==

9-63

Using
2RR= and
13RR= gives
13 1
41
R
RCL LC
+=⇒ =

These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and

214 1
34 0 1 or
33 3
RR R R R
R
+=⇒ −+=⇒ =Ω =
1
Ω

Pick R = 1 Ω. Then and
13R=Ω
21R=Ω.

() ()
2
12
1
V
4
t
cvt AAte
−
=+ +

() () () ()( )
2
21 2
1
4
t
LC C
d
itvt vt AAAte
dt
−
=+ =+ −−
At t = 0
+

() 1
1
00
4
cvA
+
= =+
() 21
1
00
4
LiA
+
A= =+ −

Solving these equations gives A1 = -0.25 and A2 = -0.5, so

()
211 1
V
44 2
t
cvt te
− 
=− +
 


DP 9-3

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
()
2
12
1
C
R
v
RR
∞=
+

The specifications require that ()
4
5
Cv∞=so
9-64

2
12
12
4
4
5
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:

KCL at the top node of R2 gives:
()
() ()
2
C
CL
vt d
Cv tit
Rd t
+=
KVL around the outside loop gives: () () ()()
1sL L
d
vt LitRitvt
dt
=+ +
C
Use the substitution method to get

()
()
()
()
() ()
() () ()
1
22
2
1
12
22
1
CC
sC C
CC
vt vtdd d
vtL Cvt R Cvt vt
dtR dt R dt
RdL d
LCvt RCvt vt
dt R dt R
 
=+ + + + 
 
 
  
=+ + ++  
  
  
C
C

The characteristic equation is
() ( )
1
1222
2
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=

Equating coefficients of like powers of s:
1
12
2
1
1
4and 20
R
RR
RC L LC
+
+==

Using
1RR= and
24R R=gives
11
4and 16
4
R
RCL LC
+==

These equations do not have a unique solution. Try
1
F
8
C= . Then
1
H
2
L= and
22
24 210 1RR R R
R
+= ⇒ −+=⇒ =Ω

Then and . Next
11R=Ω
24R=Ω

() ( )
2
120.8 cos4 sin4 V
t
cvt eA tA t
−
=+ +

9-65

()
()
()
21221
0.2 cos4 sin4
48 2 2
C tt
LC
AAvt d
it vt e te t
dt
−−
=+ =+ −
At t = 0
+

() 100 0.8
cvA
+
= =+
()
2
00 0.2
2
L
A
i
+
== +

Solving these equations gives A1 = −0.8 and A2 = −0.4, so

() ( )
2
0.8 0.8cos40.4sin4 V
t
c
vt e t t
−
=− +

DP 9-4

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

()
2
12
1
C
R
v
RR
∞=
+

The specifications require that ()
1
2
Cv∞=so
2
12
12
1
2
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:

KCL at the top node of R2 gives:
()
() ()
2
C
CL
vt d
Cv tit
Rd t
+=
KVL around the outside loop gives: () () ()()
1sL L
d
vt LitRitvt
dt
=+ +
C
Use the substitution method to get

9-66

()
()
()
()
() ()
() () ()
1
22
2
1
12
22
1
CC
sC C
CC
vt vtdd d
vtL Cvt R Cvt vt
dtR dt R dt
RdL d
LCvt RCvt vt
dt R dt R
 
=+ + + + 
 
 
  
=+ + ++  
  
  
C
C

The characteristic equation is
() ( )
1
1222
2
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=

Equating coefficients of like powers of s:
1
12
2
1
1
4and 20
R
RR
RC L LC
+
+==

Using
12RR==R gives
11
4and 10
R
RCL LC
+==
Substituting
1
10
L
C
= into the first equation gives
() ()
()
2
2 0.40.440.141
0
10 10 2
RC RC RC
±−
−+ =⇒ =

Since RC cannot have a complex value, the specification cannot be satisfied.

DP 9-5

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

9-67

() () ()
22
12 12 12
1
1, and 1
CL o
RR
vi v
RRR R R
∞= ∞= ∞=
++ R+

The specifications require that ()
1
2
ov∞=so
2
12
12
1
2
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:

KVL around the right-hand mesh gives: () () ()
2CL
d
vt LitRit
dt
=+
L
KCL at the top node of the capacitor gives:
()()
() ()
1
sC
CL
vtvt d
Cv tit
Rd t
−
−=
Use the substitution method to get

() () () () () ()
()() ()() ()
12 2
2
11 2 1 22
sL L L L
LL L
dd d
vtRC Li tRit Li tRitRit
dtdt dt
dd
RLCitL RRCitR Rit
dt dt
 
=+ + +
 
 
=+ + ++
1L
+

Using ()
()
2
o
L
vt
it
R
= gives
() () () ()
2
11
12
22 2
so o
RRdL d
vt LCvt RCvt vt
Rd t R dt R
  +
=+ + +  
  
  
2
o
R

The characteristic equation is
()()
2
2122
1
1
1
68 2 4
R
RR
ss ss s s
RC L LC

+


++ + =++=+ +
 




0=

Equating coefficients of like powers of s:
2
21
1
1
1
6and 8
R
RR
RC L LC
+
+==

Using
12RR==R gives
11
64
R
RCL LC
+=⇒ =

These equations do not have a unique solution. Try C = 1 F. Then
1
H
4
L= and
9-68

213 1
46 0 1.309 or 0.191
24
RR R R R
R
+= ⇒ −+=⇒ = Ω = Ω

Pick R = 1.309 Ω. Then
()
24
1 2
1
V
2
tt
ovt AeAe
−−
=+ +

()
() 12 241
V
1.3092.6181.309 1.309
o tt
L
AAvt
it e e
−−
== + +

() () ()
24
12
11
1.309 0.6167 0.2361
42
tt
CL L
d
vt it it Ae Ae
dt
− −
=+ =+ +

At t = 0
+

()
121
00
2.6181.3091.309
L
AA
i
+
== + +
() 12
1
0 0 0.61670.2361
2
CvA
+
== + + A

Solving these equations gives A1 = -1 and A2 = 0.5, so

()
2411
V
22
tt
ovt e e
−−
=− +

DP 9-6

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

() () ()
22
12 1 2 12
1
1, 1
CL o
RR
v i andv
RRR R R
∞= ∞= ∞=
++ R+

The specifications require that ()
3
4
ov∞=so

2
12
12
3
3
4
R
RR
RR
= ⇒=
+

9-69

Next, represent the circuit by a 2nd order differential equation:

KVL around the right-hand mesh gives: () () ()
2CL
d
vt LitRit
dt
=+
L
KCL at the top node of the capacitor gives:
()()
() ()
1
sC
CL
vtvt d
Cv tit
Rd t
−
−=
Use the substitution method to get

() () () () () ()
()() ()() ()
12 2
2
11 2 1 22
sL L L L
LL L
dd d
vtRC Li tRit Li tRitRit
dtdt dt
dd
RLCitL RRCitR Rit
dt dt
 
=+ + +
 
 
=+ + ++
1L
+

Using ()
()
2
o
L
vt
it
R
= gives
() () () ()
2
11
12
22 2
so o
RRdL d
vt LCvt RCvt vt
Rd t R dt R
  +
=+ + +  
  
  
2
o
R

The characteristic equation is
()
2
22122
1
1
1
44 2
R
RR
ss ss s
RC L LC

+


++ + =++=+
 




0=

Equating coefficients of like powers of s:
2
21
1
1
1
4and 4
R
RR
RC L LC
+
+==

Using
1RR= and
23R R= gives
13 1
4and 1
R
RCL LC
+==

These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and

214 1
34 0 1 or
33 3
RR R R R
R
+=⇒ −+=⇒ =Ω =
1
Ω

Pick R = 1 Ω. Then and
11R=Ω
23R=Ω.
() ()
2
12
3
V
4
t
ovt AAte
−
=+ +

9-70

()
() 12 21
V
34 33
o t
L
AAvt
it te
−
== + +


() () ()
12 2 23
3
43 3 3
t
CL L
AA Ad
vt it it te
dt
−

=+ =+ + +



At t = 0+
()
11
00
34
L
A
i= += +
()
123
00
43 3
C
AA
v=+ =++

Solving these equations gives A1 = −0.75 and A2 = −1.5, so

()
233 3
V
44 2
t
ovt te
− 
=− +
 


DP 9-7

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

() () ()
22
12 1 2 12
1
1, 1
CL o
RR
v i andv
RRR R R
∞= ∞= ∞=
++ R+

The specifications require that ()
1
5
ov∞=so
2
12
12
1
4
5
R
R R
RR
=⇒ =
+

Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives: () () ()
2CL
d
vt LitRit
dt
=+
L
KCL at the top node of the capacitor gives:
()()
() ()
1
sC
CL
vtvt d
Cv tit
Rd t
−
−=
9-71

Use the substitution method to get

() () () () () ()
()() ()() ()
12 2
2
11 2 1 22
sL L L L
LL L
dd d
vtRC Li tRit Li tRitRit
dtdt dt
dd
RLCitL RRCitR Rit
dt dt
 
=+ + +
 
 
=+ + ++
1L
+

Using ()
()
2
o
L
vt
it
R
= gives
() () () ()
2
11
12
22 2
so o
RRdL d
vt LCvt RCvt vt
Rd t R dt R
  +
=+ + +  
  
  
2
o
R

The characteristic equation is
() ( )
2
2122
1
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=

Equating coefficients of like powers of s:
2
21
1
1
1
4and 20
R
RR
RC L LC
+
+==
Using
2RR=and
14R R= gives
11
4and 16
4
R
RCL LC
+==

These equations do not have a unique solution. Try
1
F
8
C= . Then
1
H
2
L= and
22
24 220 1RR R R
R
+= ⇒ −+=⇒ =Ω

Then and
14R=Ω
21R=Ω. Next

() ( )
2
120.2 cos4 sin4 V
t
ovt eA tA t
−
=+ +

()
()
()
2
120.2 cos4 sin4 V
1
o t
L
vt
it eA tA t
−
== + +

() () ()
22
21
1
0.22 cos42 sin4
2
tt
CL L
d
vt it it Ae tAe t
dt
−−
=+ =+ −
At t = 0
+

9-72

() 100 0.2
LiA
+
= =+
() 200 0.22
CvA
+
== +

Solving these equations gives A1 = −0.8 and A2 = −0.4, so

() ( )
2
0.2 0.2cos40.1sin4 V
t
c
vt e t t
−
=− +

DP 9-8

When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions

() () ()
22
12 1 2 12
1
1, 1
CL o
RR
v i andv
RRR R R
∞= ∞= ∞=
++ R+

The specifications require that ()
1
2
Cv∞=so
2
12
12
1
2
R
RR
RR
= ⇒=
+

Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives: () () ()
2CL
d
vt LitRit
dt
=+
L
KCL at the top node of the capacitor gives:
()()
() ()
1
sC
CL
vtvt d
Cv tit
Rd t
−
−=
Use the substitution method to get

() () () () () ()
()() ()() ()
12 2
2
11 2 1 22
sL L L L
LL L
dd d
vtRC Li tRit Li tRitRit
dtdt dt
dd
RLCitL RRCitR Rit
dt dt
 
=+ + +
 
 
=+ + ++
1L
+

Using ()
()
2
o
L
vt
it
R
= gives

9-73

() () () ()
2
11
12
22 2
so o
RRdL d
vt LCvt RCvt vt
Rd t R dt R
  +
=+ + +  
  
  
2
o
R

The characteristic equation is
() ( )
2
2122
1
1
1
420 24 24
R
RR
ss ss s js
RC L LC

+


++ + =++=+− ++
 




0j=

Equating coefficients of like powers of s:
2
21
1
1
1
4and 20
R
RR
RC L LC
+
+==
Using
12RR==R gives
11
41
R
RCL LC
0+=⇒ =

Substituting
1
10
L
C
= into the first equation gives
() ()
()
2
2 0.40.440.141
0
10 10 2
RC RC RC
±−
−+ =⇒ =

Since RC cannot have a complex value, the specification cannot be satisfied.
9-74

DP 9-9
Let’s simulate the three copies of the circuit simultaneously. Each copy uses a different value of
the inductance.

The PSpice transient response shows that when L = 1 H the inductor current has its maximum at
approximately t=0.5 s.

Consequently, we choose L = 1 H.

9-75

Chapter 10 – Sinusoidal Steady-State Analysis
Exercises
Ex. 10.3-1
(a)
2/ 2/4Tπωπ==
(b)
leads by 30(70)100vi −− =°

Ex. 10.3-2
() ()( )
122 4
= 3cos44sin4 (3)(4)cos4tan 5cos(453)
3
vt t t t t
− °
+= + − = −

Ex. 10.3-3
()
11222
5cos512sin5 (5)(12) cos5180tan 13 cos(5112.6)
5
it t t t t
− 
=− + =−+ − + = − ° 
−

Ex. 10.4-1

()
()
() ()
()
m
KCL: = cos
s
vt vtdd I
it Cvt vt t
Rd t dt RC C
ω=+ ⇒ +

() Try cos sin & plug into above differential equation to get
f
vt A tB tωω=+
()
m1
sin cos cos sin cos
I
At B t A tB t
RC C
tωωω ω ω ω ω−+ + + =
Equating sin & cos terms yieldsttω ω

2
mm
22 2 222
and
11
RIR
AB
CI
RCR
ω
ωω
==
++ C

Therefore

()
2
1mmm
22 2 222
22 2

cos sin cos tan()
11 1
f
RIRI RCI
vt t t t RC
RC RC RC
ω
ωω ω
ωω ω
−
=+ = −

++ +
ω

10-1

Ex. 10.4-2

10 100 10
KVL : 10 32 0 56.3 A
23 13 56.3 13
j
j
∠
−+ +=⇒ = = = ∠−
+ ∠
II I
D
D
D

Therefore
10
() cos(356.3) A
13
it t=−
D

Ex. 10.5-1
45
45
10
4.24 33
2.36
j
j
ej
e
−
= =−

Ex. 10.5-2
90
(90-111) 21
111
32 32 32
3.75
38 8.54
8.54
j
jj
j
je
ee
j
e
−
== =
−+

Ex. 10.6-1
(a) 80 80
4cos(80)Re{4 } 4 480 A
jt j j
it ee e
ω
ω
−° −°
=− = ⇒ = =∠−I
D
°
∠°
°
°

(b) 20 20
10cos(20) Re{10 } 10 1020
jtj j
it ee e
ω
ω
°° °
=+ = ⇒ = =I
(c) 110 110
8sin(20)8cos(110)8Re{ } 8 8110 A
jt j j
it t ee e
ω
ωω
−° −°
=− = − = ⇒ = =∠−I
DD

Ex. 10.6-2
(a) 140 140
1014010 V ()Re{10 }10cos(140) V
jj jt
ev t ee t
ω
ω
−° −°
=∠−°= ⇒ = = −°V
(b) 43.2
8075109.743.2109.7
j
je
°
=+ = ∠ °= ⇒V
43.2
()Re{109.7 }109.7cos(43.2) V
jj t
vt ee t
ω
ω
°
== +

10-2

Ex. 10.6-3

()( )
0.01 10cos100
0.01100 10
10
7.07145
1
7.071cos100V
d
vv t
dt
j
j
vt
+=
+=
== ∠−°
+
=
VV
V

Ex. 10.6-4

{ }
100
40cos100Re4
jt
svt== e
KVL:
3
3
() 1
()1010 ()
510
t
S
dit
itdtv
dt
−
−
−∞
it+×+
×
∫
=

100
Assume () where is complex number to be determined. Plugging into
the differential equation yields
jt
itAe A=

100 100 100 100 454
(2) 4 22
1
jt jt jt jt j
Ae jAe jAe e A e
j
°
++ − = ⇒ = =
−

In the time domain:

{ } { }
100 45 (100-45)
()Re22 Re22 22cos(10045) A
jt j jt
it ee e t
°°
== =
°
+
t

Ex. 10.7-1
(a) 10(5cos100) 50cos100vRi t== =
(b)
[]0.015(100)sin100 5sin1005cos(10090) V
di
vL t t t
dt
°
== − =− = +
(c)
31
105cos100 50sin10050cos(10090) Vvi dt tdt t t
C
== = = −°∫∫

Ex. 10.7-2
6
1010[100(500)sin(50030)]
0.5sin(50030) 0.5sin(500210) 0.5cos(500120) A
dv
iC t
dt
tt t
−
== × − +°
=− +°= +°= +°

10-3

Ex. 10.7-3
From Figure E10.7-3 we get

mm m
mm
() sin cos(90) A 90A
()cos 0 V
itI tI t I
vtV t V
ω ω
ω
== −° ⇔ =∠−
=⇔ =∠°
I
V
°

The voltage leads the current by 90° so the element is an inductor:

m m
eq
mm
0
90
90
VV
II
∠°
== =∠°
∠−°
V
Z
I
Ω
Also
mm
eq
mm
90
VV
jL L L L
I I
ωω ω
ω
== ∠°⇒ = ⇒ =Z

Ex. 10.8-1
ZR = 8 Ω, ZC =
12 .42.4
2.4
1
5
12
j
j
jj j
j
== =−
×
Ω, ZL1 = j 5 (2) = j 10 Ω,
ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V.

Ex. 10.8-2
ZR = 8 Ω, ZC =
14 4
4
1
3
12
j
j
jjj
j
== =−
×
Ω, ZL1 = j 3 (2) = j 6 Ω,
ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V.

10-4

Ex 10.9-1

()
90 51
1
10
5 3.9
810
jjj
ee
j
ω
−−
==
+
V

()
90 90
2
20
55 .68
202.4
jjj
ee
jj
ω
−−
==
−
V

() () ()
51 90
12
47
3.9 5.68
3.58
jj
j
ee
e
ωω ω
−−
=− = −
=
VV V

Ex 10.9-2

()
()
15 68
1
86
4 19.2
86
jj
j
ee
j
ω==
+
V

()
()
15 75
2
124
42 4
124
jj
jj
ee
jj
ω
−
−
==
−
V

() () ()
22
12
14.4
j
eωωω
−
=+ =VV V

10-5

Ex. 10.10-1

KCL at Va:
aa b
1
42 10jj
−
+ =
−−
VV V

a b(412)(42) 2040j jj− +−+ =−−VV
KCL at Vb:
ba b
a b0.5900(24)(26)1020
1024
j jj
jj
− °
++ ∠−=⇒−− +− =+
−+
VV V
VV

Cramer’s rule yields:
(2040)(42)
(1020)(26)200100
5296.5 V
a(412)(42) 8060
(24)(2-6)
jj
jj j
jj j
jj
−− −+
+− −+
== =
−− + −−
−−
V ∠ °
Therefore
()5 cos (100296.5)5 cos (10063.5) V
a
vt t t
°°
=+ =−

Ex. 10.10-2

The mesh equations are:

11 2 1 2
22 1 1 2
1510( ) 20 (1015)10 20
510( ) 3090 10(105) 30
jj
jj
+− = ⇒ + − =
− + −=−∠−°⇒−+− =
II I I I
II I I Ij

Cramer’s rule yields:
1
20 10
30105 200200
2.2638.1 A
101510 75100
10105
jj j
j j
j
−
− +
== = ∠
+− +
−−
I −°
Next
L1 (15) (1590)(2.2638.1) 24282 Vj== ∠° ∠−°= ∠VI °
Therefore
L() 242cos(82) Vvt tω= +°
10-6

Ex. 10.10-3

The mesh equations are:
12
12 3
23
(1050)10 30
10(1020) 20 50
20(3010) 0
jj
j jj
jj
+− =
−+ − + =
+− =
II
II I
II

Solving the mesh equations gives:

12 3 0.870.09 A, 1.321.27 A, 0.51.05 Ajj=− − =−+ =+II I j
Then
a1 2 b a 10( ) 14.372 Vand 50 36.6 83 Vj=− = ∠−° =+ = ∠°VI I VV

Ex 10.11-1

90 51
1
10
5 3.9
810
jjj
ee
j
−−
==
+
V

90 90
2
20
55 .68
202.4
jjj
ee
jj
−−
==
−
V

51 90
12
47
3.9 5.68
3.58
jj
t
j
ee
e
−−
=− = −
=
VV V

10-7

() ()810 2.420
810 2.420
t
jj j
jj j
−
= +=
+− +
Z 4.9 + j 1.2

Ex 10.11-2

()
90 51
1
10
5 3.9
810
jjj
ee
j
ω
−−
==
+
V

()
90 90
2
20
55 .68
202.4
jjj
ee
jj
ω
−−
==
−
V

() () ()
51 90
12
47
3.9 5.68
3.58
jj
j
ee
e
ωω ω
−−
=− = −
=
VV V

()
()
15 68
1
86
4 19.2
86
jj
j
ee
j
ω==
+
V

()
()
15 75
2
124
42 4
124
jj
jj
ee
jj
ω
−
−
==
−
V

() ()( )
22
12
14.4
j
eωωω
−
=+ =VV V

Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t - 22° )

10-8

Ex. 10.11-3
Use superposition. First, find the response to the voltage source acting alone:

eq
1010
5(1)
1010
j j
j
⋅
=−= −
−
Z Ω

Replacing the parallel elements by the equivalent impedance. The write a mesh equation :

11 1 1
10
105j155(1j)0 0.70745 A
10j10
−+ + +− =⇒ = = ∠−°
+
II I I
Therefore:
1
() 0.707cos(1045) Ait t
°
=−

Next, find the response to the dc current source acting alone:

Current division:
2
10
3 2A
15
I=−×=−

Using superposition:
() 0.707cos(1045)2 Ait t= −°−

Ex. 10.12-1
26
33
11
10 1000 radsec
(110)(110)LC
ωω
−−
== = ⇒ =
××

Ex. 10.12-2
Diagram drawn with relative magnitudes
arbitrarily chosen:

10-9

Ex. 10.12-3
Two possible phasor diagrams for
currents:

In both cases:
()()
22
CL LC
2515 20 A == − =II
In the first case:
LC L C C
620 14 A=− ⇒ =−=−II I I

That isn’t possible. Turning to the second case:

CL C L C
20626 A=− ⇒ =+=II I I

Ex. 10.14-1
()()
11 1 1
11 122 6
11 1
() 1 1
and 1 k , 1 k
100010
RXXjR
RX
RX Cω
−
−
== Ω = =
+
Z =Ω

12
(1)(1)(11)11
k and 1 k
11 22
j
jR
−
== − Ω = =
+
ZZ
2 Ω

o 2
s1
1
1
11
22
j
j
−
=−= =−−
−
V Z
VZ

10-10

Problems
Section 10-3: Sinusoidal Sources

P10.3-1
(a) () 2 cos(6120)4 sin(660)
2 (cos6 cos120sin6 sin120)4 (sin6 cos60cos6 sin60)
2.46 cos60.27 sin6 2.47 cos(66.26)
it t t
tt t t
tt t
°°
°° °
°
=+ + −
=− + −
=+ = −
°

(b)
() 52 cos810 sin(845)
52 cos810[sin8 cos45cos8 sin45]
102 cos852 sin8
() 250 cos(826.56) 510 sin(863.4) V
vt t t
tt t
tt
vt t t
°
°°
°°
=+ +
=+ +
=+
=− = +

P10.3-2
22
2 6283 radsec
3T
110
f
ππ
ωπ== = =
−
×

m
1
() sin( ) 100 sin(6283)
(0) 10 100 sin sin(0.1) 6
() 100 sin(62836) V
vt V t t
v
vt t
ωφφ
φφ
−
=+ =
== ⇒= =
=+ °
+
°
P10.3-3
1200
600 Hz
22
f
ω π
ππ
== =
33
(210)300cos(1200(210)55)3cos(2.455)i ππ
−−
×= × +°= +°
3180
2.4 432 (210)300 cos(432+55)300 cos(127)180.5 mAiπ
π
−°
×= °⇒ × = °°= °=−


P10.3-4

10-11

P10.3-5

18 VA=

18216 msT=−=

22
393 rad/s
0.016T
ππ
ω== =

1618cos() 27θ θ= ⇒= °

()( )18cos39327 Vvt t=+ °

P10.3-6

15 VA=

432132 msT=−=

22
196 rad/s
0.032T
ππ
ω== =

815cos() 58θ θ= ⇒= °

()( )15cos19658 Vvt t=+ °

10-12

Section 10-4: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function

P10.4-1
120 400cos300
is
di di
L Rv i
dt dt
+=− ⇒ + =− t
Try cos300 sin300 then 300sin300300cos300
f
f
di
iA tB t A t B
dt
=+ =− + t. Substituting and
equating coefficients gives
0.46 300120 0
1.15
300120 400
AAB
B
BA
=−−+ =
=−
+= −

Then
()0.46cos3001.15sin3001.24cos(30068) Ait t t t=− − = −°

P10.4-2
0 500500cos1000
2
s
vdv dv
iC v
dt dt
−++ =⇒ + = t
Try cos1000 sin1000 then 1000cos10001000cos1000
f
f
dv
vA tB t A t B
dt
=+ =− + t
°
.
Substituting and equating coefficients gives

10005000
0.2

0.41000500500
AB
A
BBA
−+ = =
⇒
=+= 

Then
()0.2cos10000.4sin10000.447cos(100063) Vvt t t t=+ = −

P10.4-3

(4) (.05)(0.2)jj =

45 45
12 12 453
~() (210) ()2cos(445) mA
6000(0.2)6000
jj
ee j
ei t t
j
ω
°°
°−
== ⋅ ⇒ =
+
I +°

10-13

Section 10.5: Complex Exponential Forcing Function

P10.5-1
(536.9) (1053.1)5016.21016.2
2510.36
(4j3)(6j8) 10j5 526.56
°° ° °
°
°
∠∠ − ∠− ∠−
== =∠
+− − ∠−

P10.5-2
32 45 3
581.87 43 581.87[43 36.87]
5528.13
581.87(4.483.36) 581.87(5.636.87)
2845 142142
jj
j
j
 ∠−°°
∠+ −+ =∠+ °−+∠− °
∠− °

=∠+ ° − =∠+ °∠− °
=∠+°= +

P10.5-3
2.3
15
(37)5
0.656.31
6
j
j
je
j
e
−
−
== −
**
AC
B

P10.5-4
15
(6120) (432)12.121.3 12.1 and 21.3
j
je j a b
°
∠ −++ =−− ⇒=− =−

P10.5-5
(a)
31
tan
120 2 2 4
1
22 2 2
4( 3)4(3)
3
120 tan 34tan (120) 3.93
4
4(3) 4(3(3.93))8.00
b
j
j
Ae jb be
b
b
Ab
−−

−
−°
=−+−=+−
−
=⇒ =++ =

−
=+ −= +−− =
−

(b)
120
1
48 cos(8 sin) 3 1.52.6
2.5
48 cos 1.5 cos 72
8
8 sin (72) 26 10.2
j
jb e j
bb
θθ
θθ
−
−°
°
−+ ++ = =−−
−+ =−⇒= =
+= −⇒=−

(c)
60
102 = cos60 sin60
10 20sin60
20 and 8.66
cos 60 2
j
jaAe A jA
Aa
−+ = °−
−−
== − = =−
°
°

10-14

P 10.5-6
50.1 cos2 22cos2
dd
vv t vv
dt dt

+= ⇒ +=


t

Replace the real excitation by a complex exponential excitation to get

2
22
jtd
vv e
dt
+=
Let
2jt
ee=vA so
2
2
jt
e
d
vjAe
dt
= and
22 2
22 2 2 2
2jtj t jt
ee
d
vv e jAe Ae e
dt
+= ⇒ + =
jt

()
22 21
22 2 45
22 2
jt jt
jA e e A
j
+= ⇒ = =∠−
+
°

so
()24545 211
22
jtjj t
eve e e
−°−°
==



Finally () () ()
1
Re cos245 V
2
ev t== −°vt

P 10.5-7
22
22
2080
0.45 0.15 4cos5 3 cos5
33
dd d d
vv v t v v v t
dt dt dt dt
++ = ⇒ + + =

Replace the real excitation by a complex exponential excitation to get

2
5
2
2080
3
33
jtdd
vv v e
dtdt
++ =
Let
5jt
ee=vA so
5
5
jt
e
d
vj Ae
dt
= , and
2
5
2
25
jt
e
d
vA e
dt
=−
() ()
2
55 5 5
2
2080 20 80
32 5 35
33 3 3
5jtj t jt jtdd
vv v e Ae jAe Ae e
dtdt
++ = ⇒− + + =
jt

55
80
20 80 803
2515 1.126141
203 3 5545
2515
3
jt jt
jA e e A
j
j

−+ + = ⇒ = = = ∠−

−+
−+ +

so ()
( )5141141 5
1.126 1.126
jtjj t
e
ve e e
−°−°
==
Finally vt () () ( )Re 1.126cos2141 V
e
v t== −°
10-15

Section 10-6: The Phasor Concept

P10.6-1

Apply KVL
62 15cos4
d
ii t
dt
+− =0
or
26 15cos
d
ii
dt
+= 4t
Now use
( )4
Re{ }
jt
m eiI
θ+
= and 15 to write
4
cos415Re{}
t
t= e

()
()
()
( )
44 4
2 Re{ }6Re{ }15Re{}
jt jt t
mm
d
I eI e
dt
θθ++
+= e

() ()
44
Re2 6 Re{15}
jtj jtj t
mm
d
4
Iee Iee e
dt
θθ
+=


() ( ){ }
44
Re24 6 Re{15}
jtj jtj t
mmjIee Iee e
θθ
+=
4

( )( )86
jj
m mjIe Ie
θθ
+= 15

15 15
1.553
681053
j
mIe
j
θ
°
= == ∠−
+∠
°

() ( )1.5cos453 Ait t=− °
Finally
() () ()() ()()
()()
()
2 21.5cos453 34sin453
12cos4143
12cos437 V
dd
vt it t t
dt dt
t
t
= = −° =− −°
=−−
=+ °
°

10-16

P10.6-2

Apply KCL at node a:
4cos2
0.25 0
1
vt d
vi
dt
−
+ +=

Apply KVL to the right mesh:

L44 0 44
dd
ii v vii
dt dt
+− =⇒ =+
After some algebra:
2
2
55 4cos
dd
ii i
dt dt
++ = 2t
Now use
( )2
Re{ }
jt
m eiI
θ+
= and to write
2
4cos24Re{}
t
t e=

() () ()
2
22 2 2
2
Re{} 5 Re{ }5Re{} 4Re{
jt jt jt t
mm m
dd
}I eI e I e
dt dt
θθ θ++ +
  ++ =
  
e

() () ()
2
22 2 2
2
Re 5 5 Re{4}
jt jt jt t
mm m
dd
IeI e Ie
dt dt
θθ θ++ +
  ++ =
  

e
( ){ }
22 2
Re4 52 5 Re{4}
jj t j jt jj t
mm meIe jeIe eIe e
θθ θ
−+ + =
2t

( )45 2 5
jj j
mmeI jeI eI
θθ θ
−+ + 4
m=

()
44 4
0.39884
452511010.0584
j
mIe
jj
θ
== = =
−+ + + ∠
∠−°

() ( )0.398cos285 Ait t=− °
10-17

P10.6-3

S290 V=∠−°V

RC 6
; =16000
(500)(0.12510)
jj
Rj
Cω
−
− −
= == −
×
ZZ Ω

()
( )( )160009029016000
() = 290 1.25141 V
2000016000 2561239
j
j
ω
∠−°∠−°−
∠−°= = ∠−°

−∠ −°
V
therefore
v(t) = 1.25 cos (500t141) V
°
−

Section 10-7: Phasor Relationships for R, L, and C Elements

P10.7-1

P10.7-2

10-18

P10.7.3

P10.7-4

10-19

P10.7-5
(a)
peak
L
peak
15cos(40030) V
i = 3 sin(400t+30) = 3 cos (400t60) V
leads i by 90 element is an inductor
15
5 400 0.0125 H = 12.5 mH
3
vt
v
v
LL L
i
ω
=+ °
°− °
°⇒
== == = ⇒=Z

(b)
peak
c
peak
leads by 90 the element is a capacitor
81 1
4 F
2 900
iv
v
C
iC C
277.77 µ
ω
°⇒
== == = ⇒=Z

(c)
peak
peak
v20cos(25060) V
5sin(250150)5cos(25060) A
Since & are in phase element is a resistor
20
4
5
t
it t
vi
v
R
i
=+ °
=+ °= +°
⇒
∴= ==Ω

P10.7-6
1
2
12
12
150cos(30)150sin(30) 13075 V
200cos60200sin60 100173 V
= 23098 25023.1 V
Thus () ()() 250cos (37723.1) V
jj
jj
j
vtvtvt t
=− °+ −°= −
=° + °=+
+= +=∠ °
=+ = +°
V
V
VV V

10-20

Section 10-8: Impedance and Admittance

P10.8-1
3
RR
R
6

L L
L
CC 6
C
eq R L C
2 2(1010) 62830 radsec
11
36 = 0.0278 S
36
1
(62830)(16010) 10 0.1 S
1
16 0.0625 S
(62830)(110)
0.0278j0.
f
R
jLj j j
jj
jj
C
ωπ π
ω
ω
−
−
== × =
== Ω⇔ = =
== × =Ω⇔ = =−
−−
== =− Ω⇔ ==
×
=+ += −
ZY
Z
ZY
Z
ZY
Z
YY YY
eq
eq
003750.027 9 S
1
36.59 36 j5.86
=∠ °
== ∠°=− ΩZ
Y

P10.8-2
6
10 40
5000155 45322113 = +
3
210 195
21132113
so =4532 and = 1.06 m H
210
jR j
RL
Lω
ω
−∠°
== =− ∠−°Ω= +
−−
×∠ °
Ω= =
×
V
Z
I

P10.8-3
2
2
2
2
2
()
()
1
()
1

1
11
1
jL R
RjL j
CC C
j
RjLRj L
C C
LR
jR jL
CC C
RL
C
RLR RL
Lj L
CC C CC C
RL
C
ω
ωω
ω
ω ω
ω ω
ω
ωω
ω
ω
ωω
ωω ω ω
ω
ω
−+ −
==

−+ + +−


  
−− −
  
   
=

+−


 
−− − + −
 
 
=

+−

Z

()ωZwill be purely resistive when

10-21

2
2
211
0
RLR
L
CC C CLL
ωω
ωω
 
+− =⇒ = −


when =6 , 22 F, and 27 mH, then 1278 rad/s.RC Lµ ωΩ= = =

P10.8-4
23 2
c
L 2
c
()
=
1+1 (
R
R
2
)
RjL RC RLCjC
jL
RR
R
jC
ωω ωω
ω
ω
ω
+− +
+= + =
+
+
Z
ZZ
Z C

Set real part equal to 100 to get CΩ

2
100 0.158 µF
1( )
R
C
RCω
=⇒ =
+

Set imaginary part of numerator equal to0 to get L(2 6283 radsecfωπ= = )

22 2 2
0 0.1587 HLRC RLC Lω−+ =⇒=

Ω
P10.8-5
66
L (6.2810)(4710) 300 jLj jω
−
== × × =Z
()
eq c R L
1
300300
||(+) 590.7
1
300300
j
jC
j
jC
ω
ω

+


== =
++
ZZ ZZ Ω
300300
590.7 590.7(590.7)(300)(590.7)(300) 300300
1300 300
j
Cj C j
jC C
ω ω
ωω
+
=⇒ − + =
+−
+

()
6
Equating imaginary terms =2f = 6.2810 radsecωπ ×
(590.7)(300)300 0.27 nFCCω= ⇒=

10-22

Section 10-9: Kirchhoff’s Laws Using Phasors

P10.9-1

(a)
12 34 553.1 and 88 82 45 jj=+ =∠°Ω =−= ∠−°ΩZZ

(b)
12
Total impedance 3488114 11.720.0 jj j=+ =++−=−= ∠−°ZZ Ω

(c)
12
1000 100 100
= 20.0 () 8.55 cos (125020.0)A
+ 11.7 20 11.7
it t
∠°
== ∠ °⇒ = +
∠−°
I
ZZ
°

P10.9-2
() () ()
() (
() ()
12 7.68471.59125
5.235.620.911.30
5.230.915.621.30
6.144.32
7.5135
j
j
j
ωω ω=− =∠°−∠°
=+ −−+
=+ + −
=+
=∠ °
sVV V
)

() ( )
1
7.51cos235 Vvt t=+ °

P10.9-3
( )( )
() (
12 0.7441180.54051000.3490.6570.0940.532
0.3490.094 0.6570.532
0.4430.125
0.460196
jj
j
j
=+ = ∠−°+ ∠=− − +− +
=− − +− +
=− −
=∠ °
II I
)

()460cos(2196) mAit t= +°

P10.9-4
s
2 30
2 30 Vand 0.185 26.3 A
6123/jj
∠°
=∠° = = ∠−°
++
VI

() 0.185 cos (426.3) Ait t= −°

10-23

P10.9-5

3
15(2796)(310)jj π
−
=⋅ ⋅

12
0.48 37 A
2015
() 0.48cos(279637) A
j
it tπ
== ∠−°
+
= ⋅− °
I

P10.9-6

12 s 8 , 3, 51.87and 2 15 ARj L B== Ω = =∠− ° =∠−°ZZ I I

1
22 1
s1 2
51.87 8 8 0
32 15 83
8(3) tan
8
B
LjL
L
−
∠− ° ∠°
== = =
∠−° + + 
+∠


IZ
IZ Z

Equate the magnitudes and the angle.
1
2
3
angles: 36.87tan 2 H
8
8
magnitudes: 1.6
2649
L
L
B
B
L
−
+= + ⇒=

=⇒ =
+

P10.9-7

The voltage V can be calculated using Ohm's Law.

(1.72-69)(4.2445)7.29-24 V=∠ ° ∠°= ∠°V

The current I can be calculated using KCL.
10-24

(3.05-77)-(1.72-69)1.34-87 A = ∠° ∠° =∠°I

Using KVL to calculate the voltage across the inductor and then Ohm's Law gives:

24-4(1.34-87)
24
3.05-77

jL = L H
∠°
⇒=
∠°

P10.9-8

10 s
10 10
200
1010 10245
0245
j
 
== ∠°
 
− ∠−°
=∠ °
VV

10
()102cos(10045) Vvt t= +°

P10.9-9
(a)

160 0 160 0
=
(1326) (30037.7)303 5.9
1326 300 37.7
0.53 5.9 A
j
jj
∠°∠
=
−+ ∠−°
−+ +
=∠ °
I
°

i() 0.53cos(1205.9) Att π= +°
(b) 1600 1600
(199)(300251)25659.9
199300251
0.62559.9 A
jj
jj
∠°∠
==
−+ ∠− °
−+ +
°
= ∠°
I

()0.625 cos (80059.9) Ait tπ= +°

10-25

Section 10-10: Node Voltage and Mesh Current Analysis Using Phasors

P10.10-1
Draw frequency domain circuit and write node equations:

ACA
AC
CA C
AC
KCL at A2 0 (2)2 20
10 5
KCL at C: (1) = 0 4 20 20
54
jj
j
jj
jj
−
−+ + =⇒ + − =
−
+− + ⇒ +=−
−
VVV
VV
VV V
VV

Solve using Cramers rule:

c
(2) 20
4202060100116.6 59
11.6 64.7 V
(2)2 10 101 5.7
41
jj
j j
j j
+
− −∠ −°
== = = ∠
+− + ∠°
V −°

P10.10-2

(100)
KCL: + 0 = 57.6 22.9 V
150 12580250jj
−
++ =⇒ ∠ °
−
VV VV
V

S
100
0.667 0.384 22.9 0.347 25.5 A
150
−
== − ∠ °= ∠−°
V
I

C 0.461 112.9 A
125 90
== ∠
∠−°
V
I °

10-26

L 0.720 67.1 A
8090
== ∠−
∠°
V
I °

R 0.23022.9 A
250
= =∠ °
V
I

P10.10-3
KCL at node A:

aa b
0(
200100j
1)
−
+=
VV V
KCL at node B:

ba b b
ab
1.2
0
100 50 80
13
(2)
42
jj j
−−
++ =
−
⇒= −
VV V V
VV
Substitute Eqn (2) into Eqn (1) to get

b2.21 144 V= ∠− °V
Then Eqn (2) gives
()
a0.551441.5 1.97171 V=∠ −°−= ∠−°V
Finally
()1.97cos(4000171) V and ()2.21cos(4000144) V
abvt t vt t=− ° = −°

10-27

P10.10-4

4

s
10rads
2053 A
ω=
=∠°I

The node equations are:
ab
ab
bc
11 1
KCL at a: 2053.13
204060 40
11 j
KCL at b: V 0
8040 404080
1
KCL at c: 0
80 4080
j
jj
jj
 
c
+++ − =∠
 
 
  
−+ −+ − =  
  
− 
++ =


VV
VV
VV
°

Solving thes equation yields

a224045 V ()339.4cos(45) V
avt tω=⋅ ∠° ⇒ = +°V

P10.10-5

R
S
sin (2400) V
100
40 mH
40 mH door opened
60 mH door closed
svt
R
L
L
π= ⋅
=Ω
=

=


R S
AB
LL
With the door open 0 since the bridge circuit is balanced.
With the door closed (800)(0.04)100.5 and (800)(0.06)150.8 .jj jjππ
−=
== Ω = =
VV
ZZ Ω

The node equations are:
R
BC B
BC
L
100.5
KCL at node B: 0
100.5100
j
Rj
−
+= ⇒ =
+
VV V
VV
Z

S
AC A
L
KCL at node A : 0
R
−
+ =
VV V
Z

10-28

Cs
Since 1 V==VV
BA
0.70944.86 V and 0.83333.55 V=∠ ° = ∠VV
Therefore

AB 0.83333.550.70944.86(0.694.460)(0.5030.500)0.1910.040
0.19511.83 V
jj j−= ∠ °− ∠ °= + − + = −
=∠ − °
VV

P10.10-6

The node equations are:
11 1(1)
0
22 2
j
jj
2−−+ −
++=
−
VV VV

21 2
C 0
j2 j2
−
+ −=
−−
VV V
I

Also, expressing the controlling signal of the dependent source in terms of the node voltages
yields
xC x
1 1
2 2 1
-2-2
j j
Aj
jj
−+ −+
=⇒ == =−−


II I
Solving these equations yields

22
3
2 135 V ()()2 cos (40135) V
12
j
vtvt t
j
−−
== ∠−° ⇒ = = −°
+
V

10-29

P10.10-7

2
3
0.757166.7 V
0.606469.8 V
= ∠°
= ∠− °
V
V

12 3
3
32
22
1
3
3

0.3032 20.2 A
yields 0.1267184 A
10
0.19536 A

2
j
j


=+
 = ∠°
− 
== 

=∠ °


= 
−
III
I
VV
II
I
V
I
∠−°
therefore
1
()0.195cos(236) Ait t= +°

P10.10-8
The mesh equations are

12
12
(46) 612123
-6 (82)0
jj j
jj
+− =+
++ =
II
II

Using Cramer’s rule yields
1
(12123)(82)
2.5292.21.2 A
(46)(82)(6)(6)
jj
j
jj j j
++
== ∠°=
++ −− −
I +
Then
2
66 90
(2.529) (2.529)1.82105 A
82 6814
j
j
∠°
=∠ °= ∠°=∠
+ ∠°
I °
and

L1 2
6( )(690)(2.5291.82105)(690)(2.7111.3) 16.378.7 Vj= −=∠ ° ∠°−∠°=∠ ° ∠−°= ∠ °VI I

Finally
2
4(490)(1.82105)7.2815 V
c
j=− =∠−° ∠°= ∠°VI

10-30

P10.10-9

The mesh equations are:

12 3
12 3

12 3
(10) () 010
0
0 ( 1) 1
jj
jj j
0j jj
− ++ =
−+ =
++ − =
II I
II I
II I

Solving these mesh equations using Cramer’s rule yields:

()
3
2
(10)10 0
0
0 1 0(1)90 20
8.3877.5 A ()8.38cos1077.5 A
(10) 0 11
0 (1)
j
jj
jj j
it t
jj j
jj j
jj
−
− −
== =∠ °⇒ =
− −
−
−
I +°
(checked using LNAPAC on 7/3/03)

P10.10-10

The mesh equations are:

1
2
3
(24) 1 4 1030
1(21/4) 1 0
41 (34) 0
jj
j
jj

+ −− ∠°  
  
−+ − =
  
  −− +
  
I
I
I

Using Cramer’s rule yields

()
3
28
= 10303.22544 A
12 22.5
j
j
+
∠°= ∠°
+
I

Then
()
5
3
2 23.22544= 6.4544 V () 6.45cos(10 44) Vvt t
°
== ∠° ∠°⇒ = +VI

10-31

P10.10-11

Mesh Equations:
12
12
75 100375
100(100100)0
jj
jj
− =
−+ +
II
II =

22 2Solving for yields 4.51.5 A () = 4.74 18.4 Aji t=+ ⇒ ∠°II

10-32

Section 10-11: Superposition, Thèvenin and Norton Equivalents and Source
Transformations

P10.11-1
Use superposition

1
1245
3.311.3 mA
30002000j
∠°
== ∠
+
I °

2
50
1.5153 mA
30001500j
−∠°
== ∠
+
I °

() 3.3cos(400011.3)1.5cos(3000153) mAit t t=+ °+ +°

P10.11-2
Use superposition

1
3
0.5 mA
6000
==I

3
2
145
() 0.1661045 A
60000.2j
ω
−−∠°
= =− ×∠°
+
I

21()()() 0.166cos(445)0.5 mA
0.166cos(4135)0.5 mA
ititit t
t
=+ =− +°+
=− °+

P10.11-3
Use superposition

1
12 45
() 2 45mA
6000 0.2j
ω
°
∠
== ∠
+
I °

2
590
() 0.83390 mA
6000 0.15 j
ω
°
∠−
== ∠−
+
I °

12()()()2cos(445)0.833cos(390) mAititit t t=− = +°− −°
10-33

P10.11-4

Find :
ocV

()
()
oc
80 80
5 30
80 8020
80221.9
5 30
10036.90
4221.9 V
j
jj
 +
=∠−°

+−
∠− °
=∠−°
∠°

=∠ −°
V

Find :
tZ

( )( )
t
208080
23 81.9
208080
jj
jj
−+
= =∠−°
−+ +
Z Ω

The Thevenin equivalent is

10-34

P10.11-5
First, determine V:
oc

The mesh equations are

11 2 1 2
600 300( )9 (600300) 30090jj j−− =⇒ − + =II I I I∠°
j

21 2 1 2 1 2
2300 300( ) 0and 300( ) 3(13)0jj j−+ − −= = − ⇒ +− =VI II V II I I

Using Cramer’s rule:
2
0.012416 A= ∠−°I
Then
oc 23003.7116 V= =∠ −°VI

Next, determine I:
sc

sc
90
20 0 0.0150
600
A
∠°
−−=⇒ =⇒ = = ∠°VV V I
The Thevenin impedance is
oc
T
sc
3.7116
24716
0.0150
∠−°
== =∠−°
∠°
V
Z
I
Ω

The Thevenin equivalent is

10-35

P10.11-6

First, determine V:
oc

The node equation is:

oc oc oc
(68) (68)3
0
42 2 2
jj
jj j
−+ −+
+−

− 
VV V
=

oc34553.1 Vj=+=∠°V

s105368 Vj=∠°=+V

Next, determine I:
sc

The node equation is:

(68)3 (68)
0
24 2 2 2
jj
jj j
−+ −+
++ − =

− 
VVV V

34
1
j
j
+
=
−
V

sc
34
22 2
j
j
+
==
−
V
I

s105368 Vj=∠°=+V

The Thevenin impedance is
oc
T
sc
22
34 22
34
j
jj
j
−
== + =−

+
V
Z
I
Ω

The Thevenin equivalent is

10-36

P10.11-7
LCG=+ +Y YY
LC
1
when 0 or 0Gj
jL
ω
ω
=+ = +YY Y C=
7
11 1
,
21
239.610
0.0799810Hz800 KHz
(80 on the dial of the radio)
OO f
LC LC
ω
π
π
== =
5−
×
=× =

P10.11-8
In general:

oc L
oc
tL tL
and
++
==
V Z
IV
ZZ ZZ
V

In the three given cases, we have
1
11
1
25
50 0.5 A
50
=Ω ⇒ = ==
V
ZI
Z

2
2 26
2
1 1 100
200 0.5 A
(2000)(2.510) 200
j
jC jω
−
== =− Ω⇒ = = =
×
V
ZI
Z

33
3 3
3
50
(2000)(5010)100 0.5 A
100
jLj jω
−
== × = Ω⇒ = = =
V
ZI
Z

Since |I| is the same in all three cases,
t1 t 2 t 3
++ +==ZZ ZZ ZZ. Let
tRjX=+Z . Then

22 2 2 2
(50) (200) (100)RX RX RX++ =+− =++
2
Ω
Ω

This requires
22
(200)(100) 50 XX X−= + ⇒=
Then
22 2 2
(50)(50) (150) 175 RR R++ =+− ⇒=
so
t17550 j=+ΩZ
and
oc
22
1t 1
(0.5)(17550)(50)115.25 VR=+ = ++ =VI Z

10-37

P10.11-9

1
(3)(4)
2.453.1
34
=1.441.92
j
j
j
−
= =∠−°
−+
−Ω
Z Ω

21
4
1.442.08
2.5355.3
j
j
=+
=+
=∠°
ZZ
Ω

33.5137.9
2.772.16 j
=∠− °Ω
= −Ω
Z

() ()
( )
()
3.5137.93.5137.9
2.8578.4 2.8578.4 1.992 A
2.772.162 5.2424.4j
°
∠− °∠−
=∠ −° =∠ −° =∠−°

−+ ∠−°
I

10-38

P10.11-10

2
(200)(4)
488.8
2004
j
j
−
= =∠−°Ω
−
Z

0.444
=4 4
41004jj
4 mA
∠−°
=∠−°
−+ +
I

()4cos(2500044) mAit t= −°

10-39

Section 10-12: Phasor Diagrams

P10-12-1

() ( )( )
**
12 3=334232 43j jj j=−+ +−+++ =−+VV VV

P10.12-2

100
0.7442 A
10110jj
∠°
= =∠ °
+−
I

R
LL
CC
S
= 7.442 V
= (190)(0.7442) = 0.74132 V
= (1090)(0.7442) = 7.448 V
100 V
R=∠ °
=∠ ° ∠° ∠°
=∠ −° ∠° ∠−
=∠ °
VI
VZ I
VZ I
V
°

P10.12-3
723363(14090)14421025 40.0824.2325
46.8331.1525
jφ φ
φ
=+ ∠ °−°+∠°+∠= − +∠
= ∠− °+∠
I

To maximize, require that the 2 terms on the right side have the same angle 31.15.φ⇒= −I °

10-40

Section 10-14: Phasor Circuits and the Operational Amplifier

P10.14-1

()
()
()
() ()
()
44
225
225 225
10||10 10
10
1000 1 2
10
22
2
10cos(1000225)V
o j
s
jj
so
o
jj
e
j
ee
vt t
ω
ω
ω
ωω
−
−−
−−
== − =− =
−

=⇒ = =


=− °
V
H
V
VV 10

P10.14-2

Node equations:

1S S
11 1
11
0
1
jC
1R jCR
ω
ω
−
+= ⇒ =
+
VV V
VV

3101
01
23 2
01
R
RR R
−
+= ⇒ =+


VVV
VV

Solving:
3
20
S1
1
1
R
R
jCRω
+
=
+
V
V
1

P10.14-3
Node equations:

()
1S1
11 S 1
11
0
1
jC
jC
1
R jCR
ω
ω
ω
+− =⇒ =
+
VV
VV V

3101
01
23 2
01
R
RR R
−
+= ⇒ =+


VVV
VV

Solving:
3
1
2
0
S1
1
1
1
R
jC
R
jCR
ω
ω

+


=
+
V
V

10-41

P10.14-4

Node equations:

1S S1
1
0
175 1.6 1109jj
−
+= ⇒ =
−+
VV VV
V

101
0101
100010000
1
−
+= ⇒ =
VVV
VV

Solving:

()
0S
11 11
0.0050
1109 11089.5
0.589.5 mV
j
= =∠
+∠ °
=∠ °
VV °

Therefore

0() 0.5cos(89.5) mVvt tω= −°

10-42

PSpice Problems
SP10-1

10-43

SP10-2

10-44

SP 10-3

10-45

SP 10-4 The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of
Vm and Im are Vm = 12.5 V and Im = -1.667 A.

10-46

Verification Problems

VP 10-1
Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this
case, it is more convenient to do the division in rectangular form.

Express V1 and V2 as:
1 20j=−V and
22040j=−V

KCL at node 1:
( )11 2 20204020
22 222
10 10 10 10
jjj
jj
jj
− −− −−
−− =− − =+−−=
VV V
20

KCL at node 2:

()
() ( )
12 2 1 2020402040 20
33 22
1010 10 10 10 10
jj jj
jj j
jj
− −− − −− 
−+ = − + =+−−−= 

VV V V
2460

The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1
and V2 are correct.

VP 10-2

10.39039= ∠I ° and
20.284180= ∠°I

Generally, it is more convenient to multiply complex
numbers in polar form. Sometimes, as in this case, it is more
convenient to do the multiplication in rectangular form.

Express I1 and I2 as:
10.3050.244j= +I and
20.284=−I

KVL for mesh 1:

( ) ( )80.3050.244100.3050.244(5)10jj j j++ +− −=j

Since KVL is not satisfied for mesh 1, the mesh currents are
not correct.

Here is a MATLAB file for this problem:

10-47

% Impedance and phasors for Figure VP 10-2
Vs = -j*5;
Z1 = 8;
Z2 = j*10;
Z3 = -j*2.4;
Z4 = j*20;

% Mesh equations in matrix form
Z = [ Z1+Z2 0;
0 Z3+Z4 ];
V = [ Vs;
-Vs ];
I = Z\V
abs(I)
angle(I)*180/3.14159

% Verify solution by obtaining the algebraic sum of voltages for
% each mesh. KVL requires that both M1 and M2 be zero.
M1 = -Vs + Z1*I(1) +Z2*I(1)
M2 = Vs + Z3*I(2) + Z4*I(2)

VP 10-3

119.268= ∠V ° and
224105 V=∠°V

KCL at node 1 :

19.26819.268
4150
2j 6
∠°∠ °
+ −∠ =
KCL at node 2:

2410524105
4150
j4 j12
∠°∠ °
+ +∠ =
−

The currents calculated from V1 and V2 satisfy
KCL at both nodes, so it is very likely that the V1
and V2 are correct.

Here is a MATLAB file for this problem:

% Impedance and phasors for Figure VP 10-3
Is = 4*exp(j*15*3.14159/180);
Z1 = 8;
Z2 = j*6;
Z3 = -j*4;
10-48

Z4 = j*12;

% Mesh equations in matrix form
Y = [ 1/Z1 + 1/Z2 0;
0 1/Z3 + 1/Z4 ];
I = [ Is;
-Is ];
V = Y\I
abs(V)
angle(V)*180/3.14159

% Verify solution by obtaining the algebraic sum of currents for
% each node. KCL requires that both M1 and M2 be zero.
M1 = -Is + V(1)/Z1 + V(1)/Z2
M2 = Is + V(2)/Z3 + V(2)/Z4

VP 10-4
First, replace the parallel resistor and capacitor by an equivalent impedance

P
(3000)(1000)
949 72 300900
30001000
j
j
j
−
== ∠−°=−
−
Z Ω

The current is given by
S
P
100 0
0.253 A
500 500300900jj j
°
∠
== =∠
++ −
V
I
Z
°
Current division yields
()
()
1
2
1000
0.2 53 63.3 18.5 mA
30001000
3000
0.2 53 19071.4 mA
30001000
j
j
j
−
=∠ °= ∠−

−

=∠ °=∠

−
I
I
°
°

The reported value of I1 is off by an order of magnitude.
10-49

Design Problems

DP 10-1

2
2
2
1
1
R
R
jCj CRωω
=
+

()
()
22
21
12
1
1
o
i
RR
jCR R
R jCR
ωω
ωω
+
=− =−
+
V
V

()
()
()
()
1
2
2
180tan
1
2
21
jC
o
i
R
R
e
CR
ωω
ω
ω
−
−
=
+
V
V
R

In this case the angle of
()
()
o
i
ω
ω
V
V
is specified to be 104° so
( )
2
tan18076
0.004
1000
CR
−
= = and the
magnitude of
()
()
o
i
ω
ω
V
V
is specified to be
8
2.5
so
2
1
1
8
132
2.5116
R
RR
R
=⇒ =
+
12, 1515 , 20kR
2
. One set of values
that satisfies these two equations is CR0.2Fµ= =Ω =Ω.

DP 10-2

2
2
2
1
1
R
R
jC jCRωω
=
+

()
()
2
2
2
1
2
1
1
1
o
ip
R
jCR K
R jCR
R
jCR
ωω
ωω
ω
+
==
+
+
+
V
V

11
12 12
where and
p
2R RR
KR
RRR
==
++ R

()
()
()
1
tan
2
1
pjCo
i
p
K
e
CR
ωω
ω
ω
−
−
=
+
V
V
R

10-50

In this case the angle of
()
()
o
i
ω
ω
V
V
is specified to be -76° so
()12
12
tan76
0.004
1000
p
RR
CR C
RR
−
== − =
+
and the magnitude of
()
()
o
i
ω
ω
V
V
is specified to be
2.5
12
so
2
12
2.5
0.859
12116
RK
K
RR
=⇒ ==
++
120.2F, 23.3k, 142kCR R
. One set of values that satisfies these two equations is
µ== Ω = Ω.

DP 10-3

()
()
2
21
2
1
2
1
o
i
p
jLR L
j
RjL R
jLR L
jR
RRj L
ω
ω
ωω
ωω
ω
ω
+
== −
++
+
V
V

12
12
where
p
RR
R
RR
=
+

()
()
1
90tan
1
2
1
p
L
j
Ro
i
p
L
R
e
L
R
ω
ω
ω
ω
ω
−

−


=

+


V
V

In this case the angle of
()
()
o
i
ω
ω
V
V
is specified to be 14° so
( ) ()12
12
tan9014
0.1
40
p
LR RL
RR R
+ −
== =
and the magnitude of
()
()
o
i
ω
ω
V
V
is specified to be
2.5
8
so
1
1
40
2.5
0.0322
8116
L
R L
R
=⇒ =
+
12H, 31 , 14.76LR R
. One
set of values that satisfies these two equations is 1= =Ω = Ω.

10-51

DP 10-4

()
()
2
21
2
1
2
1
o
i
p
jLR L
j
RjL R
jLR L
jR
RRj L
ω
ω
ωω
ωω
ω
ω
+
== −
++
+
V
V

12
12
where
p
RR
R
RR
=
+

()
()
1
90tan
1
2
1
p
L
j
Ro
i
p
L
R
e
L
R
ω
ω
ω
ω
ω
−

−


=

+


V
V

In this case the angle of
()
()
o
i
ω
ω
V
V
is specified to be -14°. This requires
( ) ()12
12
tan9014
0.1
40
p
LR RL
RR R
+ +
== =−
This condition cannot be satisfied with positive

DP 10-5

1 1
22
3
3
1
=10 S
10
1
1
jC
jC
RjL
RjL
ω
ω
ω
ω
Ω =
==
==+
+
Z Y
ZY
YZ

s
() 80 cos (1000) V 8 V
() 10 cos 100 A 100 A
S
vt t
it t
θ θ=− ⇒ =
=⇒ =∠°
V
I
∠−

so
()
211
80 100 1010 (10 )1.251.25
10
jC R LCjL RC Rj L
RjL
θ ωω ωω
ω

∠− + + =∠°⇒ +− + + = +

+
ω

2
72
Equate real part: 4040 where 1000 radsec
Equate imaginary part: 40
Solving yields 40(1410 )
LCR
RCL
RR C
ωω−= =
=
=− ×

10-52

72
5
Now try 20 12(1410(20))
which yields 2.510F25 F so 40 0.02 H=20 mH
RC
CL RCµ
−
=Ω ⇒−−×
=× = = =

Now check the angle of the voltage. First

1
2
3
= 1/10 0.1 S
= 0.25 S
= 1/(2020) .025.025 S
j
jj
=
+= −
Y
Y
Y

then
12 3 s 0.125 , so = (0.1250)(100)1.250 V=+ += = ∠°∠°=∠°YY YY VYI

So the angle of the voltage is =0θ°, which satisfies the specifications.

10-53

Chapter 11: AC Steady State Power

Exercises:
Ex. 11.3-1
4(2)
3 0.81.4
42
1.6 60.3
70
4.38 60.3A
1.660.3
j
jj
j
°
°
°
°
−
=+ =+
−
=∠ Ω
∠
∴ == ∠−
∠
Z
V
I =
Z

()4.38cos(1060.3) Ait t
°
=−
The instantaneous power delivered by the source is given by

[]
(7)(4.38)
()()()(7cos10)(4.38cos(1060.3)) cos(60.3)cos(2060.3)
2
7.615.3cos(2060.3) W
ptvtit t t t
t
=⋅ = −°= °+ −°
=+ −°

The inductor voltage is calculated as

LL
= = (4.38 60.3)(j3) = 13.12 29.69 V⋅∠ −° ∠VI Z °

L
() 13.12cos(1029.69) Vvt t= +°

The instantaneous power delivered to the inductor is given by

[]
LL ()()()(13.12 cos (1029.69)(4.38cos(l0t60.3)
57.47
cos(29.6960.3)cos(2029.6960.3)
2
28.7 cos (2030.6) W
ptvtit t
t
t
°
=⋅ = + ° −

=° +°+ + °
=− °
−°

11-1

Ex. 11.3-2
(a) When the element is a resistor, the current has the same phase angle as the voltage:

m()
() cos( ) A
vtV
it t
RR
ωθ== +

The instantaneous power delivered to the resistor is given by
2 22
m
2 mmm
Rm
()()() cos( )cos( ) c os( ) cos(2 )
22
VVVV
pt vtitV t t t t
RR RR
ωθω θ ωθ ω=⋅ = +⋅ += += + +θ

(b) When the element is an inductor, the current will lag the voltage by 90°.

()
mm
L 90 90
90
VV
jL L
LL
θ
ωω θ
ωω
∠
= =∠ °Ω⇒ = =∠−°
∠°
V
ZI =
Z

The instantaneous power delivered to the inductor is given by

()
2
mm
Lm() ()() cos( 90)cos( ) cos2290 W
2
VV
pt itvt t V t t
LL
ωθ ωθ ωθ
ωω
°
=⋅ = +−⋅ += +−°

11-2

Ex. 11.3-3

The equivalent impedance of the parallel resistor
and inductor is
()()
()
1 1
1
12
j
j
j
= =+
+
Z Ω. Then

()
100 20
18.4 A
1 10
1+1
2
j
°
∠
= =∠ −°
+
I
(a)
()
()
source
20
10
10
cos cos18.430.0 W
22
P θ



== −°=
IV

(b)
()
1
2
2
max 1
R
20
1
10
20 W
22
IR
P



== =

Ex. 11.4-1
2 3
22 2
eff 002
11
= () ( 10) (5) 8.66
3
t
Ii tdt dt dt
T

=+

∫∫∫
=

Ex. 11.4-2
(a)
max
eff
2
()2cos3 A 2 A
22
I
it t I=⇒ ==

(b) ()cos(390)cos(360) Ait t t
°°
=− + +
() ( )
13
190160 0.51815 A
22
jj=∠−°+∠°=−++ = ∠−°I
eff
0.518
()0.518 cos (315) A = 0.366 A
2
it t I=− °⇒ =

(c)
22
2
eff eff
23
2.55 A
22
II
 
=+ ⇒ =
 
 

11-3

Ex. 11.4-3
Use superposition:

1
50 V=∠°V

2
2.5 V (dc)=V

3 390 =∠−°V V

V1 and V2 are phasors having the same frequency, so we can add them:

() ( )
13 50390535.8331.0j V+=∠°+∠−°=−=∠−°VV
Then
()
R1 2 3() ()()()2.5 5.83cos(10031.0) Vvt vtvtvt t=+ + =+ −°
Finally
eff eff
2
22
RR
5.83
(2.5) 23.24 V 4.82 V
2
VV

=+ = ⇒ =



Ex. 11.5-1

Analysis using Mathcad (ex11_5_1.mcd):

Enter the parameters of the voltage source:A12:= ω2:=
Enter the values of R and LR10:= L4:=
The impedance seen by the voltage source is:ZR jω⋅L⋅+:=
The mesh current is:I
A
Z
:=

11-4

The complex power delivered by the source is:Sv
I

IZ⋅()⋅
2
:= Sv4.393.512i+=
The complex power delivered to the resistor is:Sr
I

IR⋅()⋅
2
:= Sr4.39=
The complex power delivered to the inductor is:Sl
I

Ij⋅ω⋅L⋅()⋅
2
:= Sl3.512i=
Verify Sv = Sr + Sl :SrSl+ 4.393.512i+= Sv4.393.512i+=

Ex. 11.5-2

Analysis using Mathcad (ex11_5_2.mcd):

Sv6.6061.982i+=SrSl+ Sc+ 6.6061.982i+=Verify Sv = Sr + Sl + Sc :
Sc3.303i−=Sc
I

I
1
jω⋅C⋅
⋅





⋅
2
:=The complex power delivered to the capacitor is:
Sl5.284i=Sl
I

Ij⋅ω⋅L⋅()⋅
2
:=The complex power delivered to the inductor is:
Sr6.606=Sr
I

IR⋅()⋅
2
:=The complex power delivered to the resistor is:
Sv6.6061.982i+=Sv
I

IZ⋅()⋅
2
:=The complex power delivered by the source is:
I
A
Z
:=The mesh current is:
ZR jω⋅L⋅+
1
jω⋅C⋅
+:=The impedance seen by the voltage source is:
C0.1:=L4:=R10:=Enter the values of R, L and C
ω2:=A12:=Enter the parameters of the voltage source:

11-5

Ex. 11.5-3

Analysis using Mathcad (ex11_5_3.mcd):

Sv86i+=SrSl+ 86i+=Verify Sv = Sr + Sl :
Sl6i=Sl
I

Ij⋅ω⋅L⋅()⋅
2
:=The complex power delivered to the inductor is:
Sr8=Sr
I

IR⋅()⋅
2
:=The complex power delivered to the resistor is:
Sv86i+=Sv
I

IZ⋅()⋅
2
:=The complex power delivered by the source is:
I
A
Z
:=The mesh current is:
L2.16=R5.76=L
ImZ()
ω
:=RReZ():=Calculate the required values of R and L
Z
A
2
2S

⋅
:=The impedance seen by the voltage source is:
SP jQ⋅+:=The complex power delivered to the RL circuit is:
Q6:=P8:=Enter the Average and Reactive Power delivered to the RL circuit:
ω2:=A12:=Enter the parameters of the voltage source:

Ex. 11.6-1
()
11 (377)(5)
cos( )costan costan 0.053
100
L
pf
R
ω−− 
=∠ = = =
 
Z

Ex. 11.6-2
() ()
1 180
cos( )costan costan 0.53
50
X
pf
R
− − 
=∠ = = =
    
Z lagging

22
CC 1
(50)(80)
111.25 111.25
50 tan (cos 1)80
Xj
−
+
== − Ω⇒ =−
−
Z Ω

11-6

Ex. 11.6-3
TT
T
plant
3086116 W and 51 VAR
= 11651 126.7 23.7 VA
cos23.70.915
TT
PQ
PjQ j
pf
=+= =
+= += ∠ °
=° =
S

Ex. 11.6-4
1
22 2 2

1 1
4000
cos 44.3 A
cos (110)(.82)
cos(0.82) 2.48 34.9 2.031.42
To correct power factor to 0.95 requires
(2.03)(1.42)

tan(cos ) (2.03) tan (18.19)
P
PVI I
V
V
jR jX
I
RX
X
Rp fcX
θ
θ
−°
−°
=⇒ = = =
=∠ = ∠ = + =+
++
==
−
Z
1
8.16
1.42
1
=325 FC
X
µ
ω
=−Ω
−
−
=

Ex. 11.7-1
(a) () ( )
()
12
2
0.47141351.414450.942845 A
0.9428
62.66 W
2
p
=+ = ∠°+ ∠−°= ∠−°
⇒= =
II I

(b)
()
()
2
11
2
2 1
12
1.2
1.253 A 64.32 W
2
0.4714
0.4714135 A 60.666 W
2
4.99 W
p
p
pp p
=∠ ° ⇒ = =
=∠ ° ⇒ = =
∴=+ =
I
I

Ex. 11.8-1
For maximum power, transfer

*
Lt
1014 j==−ZZ Ω
100
5 A
(1014)(1014)jj
==
+−
I

2
L
5
Re(1014)125 W
2
Pj

=− =



11-7

Ex. 11.8-2

If the station transmits a signal at 52 MHz
then
6
2 10410 rad/secfωπ π== ×

so the received signal is

6
s()4cos(10410) mVvt tπ=×

(a) If the receiver has an input impedance of
in
=300 ΩZ then

2
3
32in
in s in
in L
300 1 1 2.410
= 4102.4 mV 9.6 nW
200+300 2 2(300)
VV PV
RR
−
−
 ×
=× × = ⇒ = = 
+ 
Z
Z
=

(b) If two receivers are connected in parallel then
in
300||300150 = =Z Ω and

33in
in S
in
150
V ( 410)1.711
200150
V
R
−−
== × =×
++
Z
Z

0V

2 32
in
in
1 (1.7110)
total 9.7 nW or 4.85 nW to each set
2 2(150)
V
P
−
 ×
== =

Z

(c) In this case, we need , where R is the input
impedance of each television receiver. Then
in
||200 400RR R== Ω⇒ =Z Ω

23 2
m
total
in
(210)
10 nW 5 nW to each set
2 2(200)
V
P
−
×
== = ⇒
Z

11-8

Ex 11.9-1

Coil voltages:
11 224 16 40jj j= + =VI I I
21 216 40 56jj j= +=VII I
Mesh equation:
1224 40 56 96jj j=+= + =VV I I I
24 1
96 4
j
j
== −I
()
2
1
56 14
4
o jj

==− =


VV
vo = 14 cos 4t V

Ex 11.9-2

Coil voltages:
11 224 16 8j jj= −=VI I I
21 216 40 24jj j=−+ =VI I I
Mesh equation:
1224 8 24 32jj j=+= + =VV I I I
24 3
32 4
j
j
== −I
()
2
3
24 18
4
o jj

==− =


VV
vo = 18 cos 4t V

Ex 11.9-3

2101 6 4jj
20==+VI I
12
40
2.5
16
⇒= − =−II
2I
11 2
2
2
24 16
(24(2.5)16)
44
s jj
j
j
==+
=− +
=−
VV I I
I
I

2
24 6
4411
j
j
==
−
I
12 2
2
(2.51)
3.5
6
3.5 1.909
11
o
jj
=−=−−
=−

=− =−


II I I
I
io = 1.909 cos ( 4t - 90° ) A

11-9

Ex 11.9-4

2101 6jj
240==− +VI I
12
40
2.5
16
⇒= =II
2I
2

11
2
2
24 16
(24(2.5)16)
44
s jj
j
j
==−
=−
=
VV I I
I
I

2
24 6
44 11
j
j
== −I
12 2
2
(2.51)
1.5
6
1.5 0.818
11
o
jj
=−= −
=

=− =−


II I I
I
io = 0.818 cos ( 4t - 90° ) A

Ex. 11.10-1

()
() ( )
1
2
50 50
20 A
1007513 43
13
5
j jj
j
∠° ∠°
==
− ++ −
++
I =∠°

()
1
43201036.9 Vj=− ∠°=∠−°V

11-10

Ex. 11.10-2

()
1
2
50 50 50
0.6842 A
20.25.54.957.442
55
2
j j
j
∠° ∠° ∠°
== = =
+ +∠ −°
−+
I ∠°

21
1
0.3442 A
2
= =∠ °II

( )( )( )
22
20.2 2.015.70.34420.6847.7 Vj=+ = ∠° ∠°= ∠ °VI
so
22()0.68cos(1047.7) V and ()0.34cos(1042) Avt t it t=+ ° = +°

Ex. 11.10-3

12 32 222
3 123
1 1
, 9 and =(
4 4nn nn


== = = 



ZZZZ
Z Z Z+ Z+ Z Z+Z
2)
then
ab in 3
1
+ +9 4.0625
44
 
== = =
 

Z
ZZ Z+ZZ ZZ+ Z

11-11

Problems
Section 11-3: Instantaneous Power and Average Power

P11.3-1
10 14.643 V
2063 16jj
∠°=+ + ⇒=∠−°
−
VV V
V

0.23133 A
63j
= =∠ −°
V
I

33
33
3
3
()()()0.23(cos (210133))14.6 cos (21043)
3.36 cos (210133) cos (21043)
1.68 (cos (90)cos (410176))
1.68 cos (410176)
ptitvt t t
tt
t
t
ππ
ππ
π
π
°°
°°
°°
°
== ⋅− × ⋅−
=⋅ − ⋅−
=+ ⋅−
=⋅ −

P11.3-2
Current division:

18002400
45
18002400600
5
5 8.1 mA
2
j
j
 −
=
 
−+ 
=∠ −°
I

()
2
4
600
4
source
600 5
P 300(25) 1.87510 µW18.75 mW
22
cos15
= (600)5 45cos(8.1) 2.110 µW = 21 mW
22 2
P
θ
Ω

== = × =


°
=− =×

I
VI

P11.3-3

11-12

Node equations:
XX
XX
X
XX
20100 20
0 (2015)=50
10 5
20
3 0 (4030)(2)0
51 0
jj
j
jj
j
−−
++ =⇒ − −−
−
−
−+ =⇒ −+ +−−=
−
II V
II
VI V
II V
V

Solving the node equations using Cramer’s rule yields

X
50(2) 50563.4
2510.3 A
(4030)(2015)(2)2553.1
jj
jj j
−∠ °
== =
−− − − ∠°
I ∠°
Then

()
2
2
X
AVE (20)1025200 W
2
P== =
I

P11.3-4
A node equation:

(16)
(2245)0
48
2
=16 18.4 V
3
j
−
+−∠ °=

⇒∠ °


VV
V

Then
16
3.2 116.6 A
4j
−
= =∠ − °
V
I

2
2
AVE 8
2
16
511
6.4 W absorbed
28 2 8
P
Ω



=× =× =
V

() () ()
AVE current source
11 2
22cos = 16 22 cos 26.6 12.8 W absorbed
22 5
P θ

=− − °=−

V

AVE inductor0P =

AVE voltage source
11
(16) cos (16)(3.2)cos(116.6)6.4 W absorbed
22
P θ=− =− − °=I

11-13

P11.3-6

A node equation:
11 1
1
(3/2)
20 0 505 26.6 V
101520j
+
−+ + =⇒ = ∠−°
−
VV V
V
Then
() ()
11 1
3/2 5 /2
55 26.6 A
1520 2553.1j
+
== = ∠
−∠ −°
VV V
I °

Now the various powers can be calculated:

22
1
AVE 10
11 (505)
625 W absorbed
210210
P
Ω== =
V

AVE current source
11
(20)cos (505)(20)cos (26.6)1000 W absorbed
22
P θ=− =− −°=−V

()
()
()
2
2
AVE 15
55
15 15937.5 W absorbed
22
P
Ω
== − =
I

()() ()
AVE voltage source 1
13 1
cos 55755cos53.1 562.5 W absorbed
22 2
P θ=− =− −°=−IV

AVE capacitor0 WP =

P11.3-7
()
()
200 200 200 90200
45
200 1 2 45 2
j
j
∠°
== = ∠
+ ∠°
Z °Ω

R
1200 200
0.85 45 A, 0.60 A
200 200200
45
2
j
∠
== ∠−° = =∠

+
∠
I I
D
D
°I

()() ()()
22
= 0.6200 72 W and 72172 JPR w== =I =
11-14

Section 11-4: Effective Value of a Periodic Waveform

P11.4-1
(a) ( )
12
24cos2 Treat as two sources of different frequencies.it ii i=− =+

2
eff
1
2A source: lim (2) 2 A
T
oT
Id
T
→∞
== ∫
t
and
eff
4
4cos2 source: I A
2
t =
The total is calculated as
()
2
22
eff rms eff
4
21 2 A 122
2
II I

=+ = ⇒ == =


3 A

(b)
() ( ) () ( )3cos 90 2 cos 390 20
233.3264.8 A
it t t
j
ππ= −°+ ⇒ =∠−°+ ∠°
=−= ∠−°
I

rms
3.32
2.35 A
2
I==

(c) () ( ) ( )2cos2 42cos24512cos290it t t t=+ +°+ −°

() () () ()( )20 42451290 24 412=1053.1 Ajj=∠°+ ∠°+∠−°=++− ∠−°I

rms
10
52 A
2
I==

P11.4-2
(a)
( ) ( ) ()
25 2 5
22
rms
02 0 2
11 1 84
6 2 36 4 7212 4.10 V
55 5 5
Vd t dt dt dt=+ = + = += =∫∫ ∫∫

(b)
( ) ( ) ()
25 2 5
22
rms
02 0 2
1 1 1 116
2 6 4 36 8108 4.81 V
55 5 5
Vd t dt dt dt=+ = + = + = =∫∫ ∫∫

(c)
( ) ( ) ()
35 3 5
22
rms
03 0 3
11 1 84
2 6 4 36 1272 4.10 V
55 5 5
Vd t dt dt dt=+ = + = += =∫∫ ∫∫

11-15

P11.4-3
(a)
() ()
() ()(() )
()
2
44 4
2
2
rms
11 1
44
32
4
1
11
14 2 4 4
21 441
33 3 27 27
44 4
273 2
4
85.331.3321613
27
4
1174.16 V
27
Vt dt t dt ttdt
tt
t

=+ = + = ++



=+ +


= −+ −+
==
∫∫ ∫

(b)
() ()
() ()()
()
2
44 4
2
2
rms
11 1
44
32
4
1
11
14 22 4 4
211 444121
33 3 27 27
44 44
121
273 2
4
8422151213
27
4
1174.16 V
27
Vt dt t dt t t
tt
t

=− + = −+ = −+



=− +


=+ − +
==
∫∫ ∫
dt

(c)
() ()
()
()
2
33 3
2
2
rms
00 0
33
32
3
0
00
14 4 4
22 3 4129
33 27 27
44 12
9
273 2
4
365427
27
4
1174.16 V
27
Vt dt t dt t t
tt
t

=+ = + = ++



=+ +


=+ +
==
∫∫ ∫
dt

11-16

P11.4-4
(a)
()
dc ac
2
1cosvt tvv
T
π
=+=


+

2 2
dc eff ac eff
0
0
11
1 01 Vand V
2
T
T tT
vd t v
TT T
  
== = −= =
  
  
∫

2
22 2 2
eff dc eff ac eff
1
1 1.225 V
2
vv v

=+ = + =



(b)
()
2
rms
0
21
= ,
T
T
I itdt
T
π
ω =∫

() ()
22
/2 /2 /2 /222
rms
00 0 0
11
sin 1cos2 co s2 A
22
TT T TAA
I A tdt tdt dt tdt
TT T
ωω ω

== − = −
∫∫ ∫∫
2
4
A
=

2
rms rms = , where 10 mA = 5 mA
42
AA
IA == ⇒I
1

P11.4-5
() ( )
90 0 0.
900.2 0.1 0.2
0.2 0.30
t t
vt t t
t
≤≤

=− ≤≤

≤≤


() ( ) ()
2
0.1 0.20.1 0.2 22 2
2 2
rms 00 .100 .1
2
19 0
90 900.2 0.2
.3 .3
90.001.001
18 V
.333
V tdt tdt tdt tdt
 
=+ − = +

−
   

=+ =


∫∫∫∫

rms
18 4.24 VV==

11-17

Section 11-5: Complex Power

P11.5-1
()
*
23.67.22
0.61.21.34263.43 A
120 120
j
j
+
== =+ = ∠ °
∠° ∠°
S
I

120
4 8.9463.4348 4 and 2 H
1.34263.43
RjL j R L
∠°
+= =∠ =+ ⇒ =Ω =
∠−

P 11.5-2
( )
*
21892
31.53.3526.56 A
120 120
j
j
+
== =+ =∠ °
∠° ∠°
S
I

111 13.3526.56
0.279126.560.2500.125
44 120
jj
Rj LR L
∠−°
+= − = = ∠− = +
∠°

4 and 2 HRL⇒= Ω =

P11.5-3
Let
()
p
88 81 88
44 V
88 1 1 2
j jj j
j
jj j
−+
== × = =+
++ −
Z
Next
()
p
120 120
1.34226.6 A
44 88j
∠°∠ °
= == ∠−
++ +
I
Z
°
Finally
() ( )
*
1201.34226.6
7.23.6 VA
2
j
∠° ∠−°
==S +

11-18

P11.5-4
Before writing node equations, we can simplify the circuit using a source transformation:

The node equations are:

()
11 2
1250 1
22
j
j
10
−
−+ + =⇒ +−=
VV V
VV
() ()
21 2
11
1
04
28 0.80.4
j j
jj
−
++ =⇒ −++ =
+
VV V
V V
280V

Using Cramers’s rule
1
80
(16/3) 126.9 V
(4)8(1)jj j
== ∠
−− +
V °
then
()
12 1 1
11
1 10 2.66126.9 A
88
jj=− −=−−++= ∠ °IV V VV

Now the complex power can be calculated as

() ( )( )
1
* 2.66126.9(2/3)36.9
(1/8) 8
VA
292
j
°°
∠− − ∠
−
== =−
IV
S
Finally
88
0, VAR
9 9
PjQj P Q=+ = ⇒ = =S

11-19

P11.5-5

5012050120
2.583.13 A
16122036.87j
∠° ∠°
== =∠
+∠ °
I °

() ( ) 501202.583.13 125 36.8710075 VAj=∠ ° ∠− °= ∠ °=+
*
S=VI

P11.5-6

KVL:

( )
()
12
12
1020 502
1020 250
jj
jj
°
+= ∠−
⇒+ + =∠
II
II °

KCL:
12
60+=∠°II

Solving these equations using Cramer’s rule:

1020 2
1018
1 1
jj
j
+
∆= =+

1
5 215 12
= 0.63232 A 0.390.5 A
6 1 1018
j j
j
j
−
== ∠°=−−
∆+
I

21
= 6 63.9.56.39.56.414.47 Ajj−= ++= +=∠ °II

Now we are ready to calculate the powers. First, the powers delivered:

()() ()
[] () ()
50
60 2
Total 50 60
delivered
1
50 2.56.41(1804.47)16.01.1 VA
2
1
5 260 526.39.53 18.038.3 VA
2
2.037.2 VA
j
jj j j
j
°
∠°
°
∠°
∠° ∠°
=∠ −= ∠ − =−+
=− ∠=− + = −

=+ =−
*
2
SI
SI
SS S

11-20

Next, the powers absorbed:

()
() ()
2 2
10 1
2
j20 1
2 2
j2 2
Total
absorbed
11 0
10 .632.0 VA
22
20
4.0 VA
2
1
2 6.41 41.1 VA
2
2.037.1 VA
j
j
jj j
j
Ω
Ω
−Ω
== =
==
=− =− =−
=−
SI
SI
SI
S

To our numerical accuracy, the total complex power delivered is equal to the total complex
power absorbed.

P11.5-7

(a)

(b)

10020
430
2510
∠°
= =∠ °Ω
∠−°
V
Z=
I

()()cos 10025cos30
1082.5 W
22
P
θ °
== =
IV

(c) 0.25300.21650.125 Sj=∠ −°= −
1
Y=
Z
C 0.125 Sj=Y
. To cancel the phase angle we add a capacitor
having an admittance of . That requires =0.125 1.25 mFCCω ⇒= .

P11.5-8
Apply KCL at the top node to get

11
11
31 0
0 436.9 V
334
1
2
j
−
−−+ =⇒ =∠ °
−
VV
VV
Then
1
1 136.9 A
4
==∠ °
V
I
The complex power delivered by the source is calculated as

() ( )136.9*100
536.9 VA
2
∠° ∠°
== ∠−S °
Finally
( )cos36.9 .8 leadingpf=− °=
11-21

Section 11.6: Power Factor

P11.6-1
Heating: P = 30 kW

Motor:
()
1
150cos53.190 kWcos0.653.1
150sin53.1120 kVAR150 kVA
P
Q
θ
−
 =° === ° 
⇒
=° ==  
S

Total (plant): 3090120 kW
12012017045 VA
0120120 kVAR
P
j
Q
=+ = 
⇒= + =∠°
=+ = 
S

The power factor is cos 450.707pf= °= lagging.

The current required by the plant is
170 kVA
= 42.5 A
4 kV
==
S
I
V
.

P11.6-2
Load 1: () ()
()() () ( )
1
1
1
cos=12 kVA0.78.4 kW
Q sincos.7 12 kVAsin45.68.57 kVAR
P θ
−
==
== °=
S
S

Load 2: () ()
()() ()
2
1
2
10 kVA0.88 kW
10sincos0.810sin36.96.0 kVAR
P
Q
−
==
== °=

Total: ()8.488.576.0 16.414.5721.941.6 kVAPjQ j j=+ =++ + = + = ∠ °S

The power factor is ( )cos41.60.75pf= °= . The average power is P = 16.4 kW. The
apparent power is |S| = 21.9 kVA.

11-22

P11.6-3

The source current can be calculated from the apparent
power:
( )
-1
ss
s
250cos0.82
I 5 36.9A
s 2002

∗
∠
∗
= ⇒= = =∠°
∠°
VI S
S
V

s536.94j3 A=∠− °=−I
Next
s
1
200
253.11.21.6 A
681053.1
j
j
∠°
== =∠−°=−
+∠ °
V
I

2s 1431.21.62.81.4
3.1326.6 A
jj j=− =−−+ =−
=∠ −°
II I

Finally,
s
z
200
= 6.3926.6
3.1326.6
∠°
= =∠ °
∠− °
V
Z
I
Ω

P11.6-4
(Using all rms values.)
(a)
()()
2
22
= 50020 100 VrmsPR PR
R
=⇒ =⋅= ⇒ =
V
IV V

(b)
sL
10001000
55 5245
2020 20 20
j
jj
∠° ∠°
== + = + =−=∠−°
VV
II+I A

(c)

()()
s
2020
20 10245
2020
j
j
j
=− + = ∠−°Ω
+
Z

()
1
cos45 leading
2
pf
°
=− =

11-23

(d) No average power is dissipated in the capacitor or inductor. Therefore,

()
AVE AVE ss s
source 20
s
500 500
P P 500 W cos500 100 V
1cos
52
2
θ
θΩ
== ⇒ =⇒ = = =



VI V
I

P11.6-5
Load 1:
11
11 1
1
1
11
s
100160 V
2190 A1.970.348 A
23.2 W, 50 VAR
23.25055.1265.1 VA
cos65.10.422 lagging
55.1265.1
0.55194.9, so I0.55194.9 A
100160
j
PQ
PjQ j
pf
°
∗
=∠ °
=∠ °=−−
==
=+ = += ∠ °
==
∠°
== = ∠−° = ∠°
∠°
V
I
S
S
I
V

Load 2:
() ( )
21
22
1.970.3480.047.5492.12155 A
1001602.1215521245150150 VA
jj
j
°
∗ °° °
== −− + − = ∠−
== ∠ ∠ =∠−=−
II-I
SVI

()2
cos45 0.707 leadingpf
°
=− =

Total: () ( )
()
12 23.250150150173.210020030 VA
cos30 0.866 leading
jj j
pf
°
°
=+ = + + − = − =∠−
=− =
SS S

P11.6-6
refrig
120
14.12
8.5
==Z Ω

refrig14.12451010 j= ∠°=+ΩZ

()
2
2
lamp
120
144
100
R
P
== =
V
Ω

()
2
range
240
4.8
12,000
R== Ω

(a)
refrig lamp
1200 1200
8.545 Arms , 0.830 Arms
1010 144j
∠° ∠°
== ∠−° = =∠°
+
Ι I
and
11-24

range
2400
500 A
4.8
∠°
= =∠ °I
From KCL:
1r efrig range
2 lamp range
N1 2
566 = 56.36.1 A
50.83180 A
7.9249 A
j=+ =− ∠−°
=− − = ∠°
=−−= ∠−°
II I
II I
II I

(b)
22
refrig refrig refrig refrig refrig refrig 722.5 Wand 722.5 VARPR Q X== = =II

lamp lamp100 Wand 0PQ= =

total
total
72210012,00012.82 kW
12,822 72212.843.2 kVA
72200722 VAR
P
j
Q
=+ + = 
⇒= + = ∠°
=+ +=

S

The overall power factor is()cos3.20.998pf=° = ,

(c)

Mesh equations:
A
B
C
3010 20 1010 1200
20 164 144 1200
1010 144 158.810 0
jj
jj
+− −− ∠  
  
−− =
  
  −− − +
  
I
I
I
°
∠°
Solve to get:
A
B
C
54.31.5754.31.7 Arms
51.30.1951.30.5Arms
500500 Arms
j
j
j
= −= ∠−°
=− =∠−°
=+ =∠°
I
I
I

The voltage across the lamp is

lamp lampB CII1441.278.6183.2 VR=− = ∠−°=V

11-25

P11.6-7
(a) ()
1
VI=2207.61672 VA
P1317
pf .788
VI1672
=cospf38.0 Q=VIsin=1030VARθθ
−°
=
== =
=⇒

(b) To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing –Q, then
22
c
V (220)
1030 = = X = 47
11 C = = = 56.5F
X (377)(47)
ccXX
µ
ω
⇒Ω
∴

(c)
*
= VI coswhere = 0
then 1317 = 220I
I = 6.0A for corrected pf
Note I = 7.6A for uncorrected pf
P θθ
°
∴

P11.6-8
First load:

1
1
(1tan(cos(.6))) 500(1tan53.1)500677 kVAPjQP j j j
−
=+ =+ = + °=+S

Second load:
2
400600 kVAj=+S
Total:
12
9001277 kVAj=+S=S+S

1
desired
tan (cos(.90))900436 VAPjP j
−
=+ =+S

desiredFrom the vector diagram: Q=+S S. Therefore

9004369001277 841 VARjj Q Q j+= + +⇒ =−

22
2
*
*
(1000)
j841 1189 1189
j841 841 377
j
j j
j C
=− ⇒ = = = ⇒=− =−
−−
VV
ZZ
Z

Finally,
1
2.20 F
(1189)(377)
C µ==

11-26

P11.6-9
(a)
11
tan(cos)10001000tan(cos0.8)1000750 VAPjQPjP pf j j
−−
=+ =+ = + = +S

L
1000j750*
Let 1000 Vrms. Then I 10j7.5 107.5 A
L1000
j
+
=∠ ° = = =+ ⇒ =−
°
∠
S
VI
V

L
L
1000
836.96.44.8 V
12.536.9
j
∠°
== =∠°=+
∠− °
V
Z
I

LL
=[6.4(200)(.024))()(12.89.6)(107.5)2000 Vjj j++ = + − =VZ I ∠°
(b)
For maximum power transfer, we require ()
Ln ew
Ln ew
1*
6.44.8 ||
+
j+= =ZZ
YY
.
Ln ew new
11 1
0.15 S
(6.4j4.8) 6.44.86.44.8
j
jj
=⇒ = − =
−− +
Y+Y Y

newThen 6.67 so we need a capacitor given byj=− ΩZ

11
6.67 0.075 F
(6.67)(200)
C
C
µ
ω
=⇒ = =

Section 11-7: The Power Superposition Principle

P11.7-1

Use superposition since we have two different
frequency sources. First consider the dc source
(ω = 0):
1
22
11
12
14 12 A
122
(12)(2)288 WPR

==

+
== =
I
I

Next, consider the ac source (ω = 20 rad/s):

11-27

After a source transformation, current division gives

2
60
25(125)
9.166 116.6 A
60 5
24
(125)
j
j
j
j
j
−

−
=− =∠ °
−
 ++
−
I

Then
2
2
2
(125)(2)
(2) 125 W
22
P== =
I

Now using power superposition

12
288125413 WPPP=+ =+=

P11.7-2
Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s
source:

Current division yields

1
8
52
5 63.4 A
8 5
8
2
j
j


−
== ∠

+
−
I °
Then
2
1
1
8
20 W
2
P==
I

Next consider ω = 8000 rad/s source.

Current division yields
11-28

2
8
57
5 171.9 A
8 50
8
7
j
j
j


=− = ∠− °

+


I
Then
2
2
2
8
2 W
2
P==
I

Now using power superposition
12 22 WPPP=+=

P11.7-3
Use superposition since we have two different frequency. First consider the dc source (ω = 0):

21
1
()0 and ()10 1 A
10
it it

== 

=

1
2
22
R1 1
R
1(10)10 W
0 W
Pi R
P
== =
=

Next consider ω = 5 rad/s sources.

Apply KCL at the top node to get

()
1
21 2
(101040)
64 30
10
I
j
0
−∠°
−+ +−∠−°+ =II I
Apply KVL to get
12
10(52)0j−+− =II
Solving these equations gives

12
0.56 64.3 A and 1.0442.5 A=− ∠−° =−∠−°II
Then
11-29

() ()
1 2
2 222
11 22
RR
0.56(10) 1.04(5)
1.57 W and 2.7 W
222 2
RR
PP== = = = =
II

Now using power superposition

2RR
1
101.57 11.57 W and 02.72.7 WPP=+ = =+=

P11.7-4
Use superposition since we have two different frequency. First consider the ω = 10 rad/s source:

1
1
1
1
R
1
C
40
0.280.7 A
25
22(0.28.7)0.561.41.51 68.2 V
53.77 21.8 V
j
j
jj
jI
∠°
== +
−
== + = + = ∠ °
=− = ∠−°
1V
I=
Z
VI
V

Next consider ω = 5 rad/s source.

2
2
690
0.5770.12 A
210
j
j
∠−°
== = −
−
V
I
Z

2
R2
22(.5770.12)1.150.24
1.1711.8 V
jj== − =−
=∠ −°
VI

2
2
C
10=5.9258.3 Vj=− ∠ °VI

Now using superposition

R
C
(t)1.51cos(1068.2)1.17cos(511.8) V
(t)3.77cos(1021.8)5.9cos(5258.3) V
vt t
vt t
=+ °+ −
=− °+ −
°
°

Then
22
2
Reff Reff
22
2
Ceff Ceff
1.511.17
1.82 1.35 V
22
3.775.9
24.52 4.95 V
22
VV
VV
 
=+ = ⇒ =
 
 
 
=+ = ⇒ =
 
 

11-30

11-31

Section 11-8: Maximum Power Transfer Theorem

P11.8-1
t
*

Lt
4000||20008001600
800 1600
800
10008001600
1.6 H
jj
j
R
Rj L j
L
=− =− Ω
== + Ω
=Ω
+= + ⇒
=
Z
ZZ

P11.8-2
t
*
Lt
25,000||50,000 20,00010,000
=20,00010,000
20 k
20,00010,00010010,000
100 H
jj
j
R
RjL j L
L
ω
=− = −
=+ Ω
Ω
=Ω

+= + ⇒ =

=

Z
ZZ
After selecting these values of R and L,

()
2
2
3
max
0.1410
1.4 mA and 201019.5 mW
2
P
−
×
== ×=


I

Since , yes, we can deliver 12 mW to the load.
max12 mWP>

P11.8-3
2
tL 2
8001600 and
1( )
j
R
RjRCC
j
j RC
R
C
ωω
ω
ω
−

−
=+ Ω = =
+
−
ZZ
2
*
Lt 2
8001600
1( )
j
R
RjRCC
j
j RC
R
C
ωω
ω
ω
−

−
=⇒ = = −
+
−
ZZ Ω
Equating the real parts gives

22
4000
800 0.1 F
1()1[(5000)(4000)]
R
C
RC C
µ
ω
== ⇒ =
++

11-32

P11.8-4
tL400800 and 2000 || 1000400800 jj=+ Ω = − =−ZZ jΩ
t

Since the average power delivered to the load is maximum and cannot be increased by
adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is
*
L =ZZ

63.4L
R
tL
5 2.55 5.59 V
j
je
−
== −=
+
Z
V
ZZ

So
2
5.591
7.8 mW
20002
P

==



is the average power delivered to the 2000 Ω resistor.

P11.8-5
Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to
the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average
power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose
R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power
provided by source is delivered to the load.

Section 11-9: Mutual Inductance

P11-9-1

s1 2
s
12
0
(2 )
jL jM jL jM
jL LM
ωω ω ω
ω
++ + − =
⇒+ − =
VI I I I
V
I

Therefore
12 2
abLL L=+ −M

11-33

P11.9-2

KCL:
12 s+=II I

The coil voltages are given by:

11 2
jL jMω ω= +VI I

22 1
jL jMω ω= +VI I
Then
()
1
2
1

s
jL
jM L
ω
ω
−
=
−
VI
I
and
( )
22 2

s
jL jMω ω=+ −VI II
Then
()
()
2
12
12
11
()

2
s
s
s
jL jLM LLM
jM j
2

jML LLM
ωω
ωω
ω
−−   −
=+ ⇒ =
 
−+ − 
VI V
VI
I

Finally
2
12
ab
12
=
2
LLM
L
LL M
−
+−

P11.9-3

Mesh equations:
11 2
22 1 2
141.402 40 60 0
l200 60 60 0 I (0.2351)
jj
jj
−∠ °++ − =
+− =⇒= ∠°
II I
II I
1
I

Solving yields
12
4.1768 A and 0.9617A=∠ −° = ∠−°II
Finally
() ()
12 4.2cos(10068) A and 1.0cos(10017) Ait t it t=− ° = −°

11-34

P11.9-4

Mesh equations:
()
()
12
12
105 5010
50400500 0
jj
jj
+− =
−+ + =
II
II

Solving the mesh equations using Cramer’s rule:

() ()( )()
() ( )( )
2 2
1050 5010
0.06229.7 A
105400500 50
jj
jj j
+− −
==
++ −−
I ∠ °
Then
()
22
2
1
400
40400.06229.72.529.7
100
°
== = ∠ °=∠
∠
VI
I
V
°

P11.9-5

Mesh equations:
11 2
2 2 122
1005 9 30
28 6 39 3 0
jj j
jj j j
−∠°− + + =
+ ++ − =
II I
II I I I

Solving the mesh equations yields

12
0.25161 A and 2.5586 A=∠ ° = ∠−°II
then
( )
12
9( )92.6 81 23 9 Vjj=− = ∠−°=∠°VI I
Finally
()23cos(309) Vvt t= +°

11-35

P11.9-6
(a)

21 10 I10 0 A (0)10 Ai=⇒= ∠°⇒ =I

2
2
11
(0)(0.3) (10)
15 J
22
Li
w== =

(b)

Mesh equations:

21 1 2
1 2
6 - 3 0 2
100 A I= 50 A
jj =⇒=
=∠° ⇒ ∠°
II I I
I

Then
22 22
11 21 1 2
11 1 1
(0) (0) (0)(0)(0.3)(10)(1.2)(5)(0.6)(10)(5) 0
22 2 2
wL i Li Mii=+ − = + − =

(c)

21
(76) 3 0jj+ −=II

2
3.25 49.4 A= ∠°I

2()3.25cos(549.4) Ait t= +°

2(0)2.12 Ai=
Finally
2211
(0.3) (10)(1.2) (2.12)(0.6) (10) (2.12)5.0 J
22
w=+ − =

P11.9-7

Mesh equations:
T1 1 2 1 1 2 1
22 1 1
8 5( )6 6( )5
3 6( ) 5 0
jj j j j
jj
−+ + −− + −+ =
+− − =
VI II I II I
II I I
0

11-36

Solving yields
21
12
(1.64 27)
(18)(11) jj
°
=∠
+− =
II
II
TV

Then
2T
1
8.2 8.4 14 j== += ∠°
V
Z
I
Ω

P11.9-8

The coil voltages are given by

()
()
11 1 2 2 1
21 2 1 2 1
32 1 1 2 1
= 6 2( )4 4 2
= 4 2 2 2 2
= 8 4 2 2 6
jj j j j
jj j j
2
2
2
j
j jj jj
−− − = −
−− + = −
−+ −=− +
VI II I I I
VI I I I I
VI I II I
I
I

The mesh equations are
11 1 2 2
22 1 2 3 2
56 ( ) = 100
6( )2 5 = 0j
°
++ −+ ∠
−+ −+ +−
IV IIV
VI I IV I

Combining and solving yields

2
11610
64 0 60 40
1.2 0.28A
11664 50 33
64 83
j
j j
jj j
jj
+
−− +
== =
+− − +
−− +
I ∠ °
t

Finally

2
5 6.0 89.72 A ()6sin(2 89.7) Vjv t=− = ∠− °⇒ = − °VI

11-37

Section 11-10: The Ideal Transformer

P11.10-1
2
(10075)
(23) 6
5
j
j
−
=++ =Z Ω

1
120120
= 2 A
6

∠°∠ °
==I
Z

211
10075 10075
( 2) 1036.9V
25
jj
n
−− 
== =∠− 
 
VI °
°

21
= 5 (1036.9) = 50 36.9 Vn=∠ −° ∠−VV

1
2
2
A
5n
==
I
I

P11.10-2
(a)
3
0
02
11
(510)(10,000) 50 V
50
5
10
N
n
N
−
=× =
== = =
V
V
V

(b)
ab 22
11 3
(1010)400
25
RR
n
== × = Ω

(c)
s
ab
1010
0.025 A25 mA
400R
== = =I

11-38

P11.10-3

12 22
1
99 (58)4572
n
jj== =−=−ZZ Z Ω

Using voltage division, the voltage across is
1Z

()
1
4572
8050 74.473.3 V
45723020
j
jj
 −
=∠−° = ∠− °

−+ +
V
then
21
74.4 73.3
24.8 73.3 V
3
n
∠−°
== = ∠−°VV

Using voltage division again yields

()
c2
88 90
24.873.3 21.0105.3 V
58 8958
j
j
−∠ −°
== ∠−° = ∠−
 
− ∠−°
VV °

P11.10-4

()
()
11 2 12
200 8
5, 8 500400 V 2000 V
825
nn== =Ω⇒ = ∠°=∠°⇒ = =∠°
+
ZV V V

11-39

P11.10-5

22
320jL
nn
ω
=+Z

Maximum power transfer requires

22
320
160 kand 80
jL
j
nn
ω
=Ω =

so n = 2. Then 640 kLω=Ω so

3
5
64010
6.4H
10
L
×
==

P11.10-6

()
2
1
26 2
2
= += ΩZ

()
oc
62
21 60120
62 22
  
V= ∠°=∠°
  
++  
V

()
2
11
2
22
= =ΩZ

21
11 160
3.250 A
122
2
2
sc

∠°
=−= = =∠°
+

II I

Then
t
120
3.750
3.20
∠°
= =∠ °
∠°
Z Ω

11-40

P11.10-7

21
12
3
21
23
1
12
1
T3 1
1
T
1
2
24
66
1
21
6
6
=
−
==
=− =
=− =−
=− =
==
VV
VV V
I
VV
II
V
II
V
II I
V
Z
I
1
2
t

P11.10-8
Maximum power transfer requires . First
*
L
=ZZ

C1 L1 222
22
11 21
, 5
105
XX n
nn

== = ⇒


=

then
L 122 2
112
1 1 1 100 3
1 100
31
Rn
nnn
 
+= Ω⇒ = ⇒ = 

 
0

11-41

P11.10-9

L 2
120(17.54)8.123
0.30.13
520107.54 25
j
j
j
+∠ °
== =

++
Z + Ω

()
()
22 2
L2
L
2L
230
88 kW/home
22 20.3
P
RR
== = =
VV

Therefore, 529 kW are required for six homes.

P11.10-10

(a)

Coil voltages:
22
16
12
j
j
=
=
11VI
VI

Mesh equations:
85 45
12 0
0+−∠°=
−− =
11
22
IV
IV

Substitute the coil voltages into the mesh equations and do some algebra:

11-42

8 16545 0.2818.4
12 12 0 0
j
j
+= ∠°⇒ =∠−
+= ⇒ =
11 1
22 2
II I
II I
°

12 0=−=
22
VI

(b)

Coil voltages:
2
22
16 8
12 8
j j
j j
= +
=+
11
1
VI
VI
I
I
0

Mesh equations:
85 45
12 0
+−∠°=
−− =
11
22
IV
IV

Substitute the coil voltages into the mesh equations and do some algebra:

( )
2
81 6 8 54
12(12 8)0
jj
jj
5+ += ∠
++ =
11
22 1
II I
II I
°

()
1212 3
1
82
j
j
j
+
=− = −
1 2II
2I

() ()
22
3
816 18 545 0.138141
2
jj j
 
+− + =∠°⇒ = ∠−


II °

2
121.65639=−= ∠
2
VI °
11-43

(c )

Coil voltages and currents:
2
2
10
8.66
8.66
10
=
=−
1
1
VV
II

Mesh equations:
85 45
12 0
0+−∠°=
−− =
11
22
IV
IV

Substitute into the second mesh equation and do some algebra:

2
10 8.66 10
12 12
8.66 10 8.66
 
−− = ⇒ =
 
 
11 1IV V
1I

2
10
812 545 0.20845
8.66

+ =∠° ⇒ = ∠°


11 1II I

()121010
12 12 0.208452.8845
8.66 8.66

=− =− − = ∠°= ∠°


22 1
VI I
11-44

PSpice Problems

SP 11-1
The coupling coefficient is
3
0.94868
25
==
×
k .

11-45

SP 11-2
Here is the circuit with printers inserted to measure the coil voltages and currents:

Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the
current of coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the
voltage of coil 1 as 4.292∠-58.74°:

FREQ VM(N00984) VP(N00984)
7.958E-01 2.498E+00 1.072E+02

FREQ
IM(V_PRINT1)IP(V_PRINT1)
7.958E-01 4.484E-01 -9.457E+01

FREQ
IM(V_PRINT2)IP(V_PRINT2)
7.958E-01 6.245E-01 -7.277E+01

FREQ VM(N00959) VP(N00959)
7.958E-01 4.292E+00 -5.874E+01

The power received by the coupled inductors is

() ( )
()()
( )( )
()()
4.2920.4484 2.4980.6245
cos58.7494.57 cos107.272.77
22
0.78016.780000
p=− −− + −
=− ≈
−

11-46

SP 11-3

The inductance are selected so that
2 2
11
3
2
LN
LN
==and the impedance of these inductors
are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open
circuit while providing a connected circuit.

Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the
current of coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage
of coil 1 as 16.00∠114.1°:

FREQ VM(N00984) VP(N00984)
6.366E-01 2.400E+01 1.141E+02

FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01 4.000E+00 1.140E+02

FREQ
IM(V_PRINT2)IP(V_PRINT2)
6.366E-01 2.667E+00 -6.590E+01

FREQ VM(N00959) VP(N00959)
6.366E-01 1.600E+01 1.141E+02

The power received by the transformer is

()()
()
()( )
()()
164 242.667
cos114114 cos11466
22
3232.0040
p=− +
=− ≈
−−

11-47

SP 11-4

The inductance are selected so that
2 2
11
2
5
LN
LN
==and the impedance of these inductors
are much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open
circuit while providing a connected circuit.

FREQ VM(N00921) VP(N00921) VR(N00921) VI(N00921)
6.366E-01 1.011E+02 7.844E+01 2.025E+01 9.903E+01

The printer output gives the voltage across the current source as

20.2599.03101.178.44 Vj+ =∠ °
The input impedance is

20.2599.03
20.2599.03 101.178.44
1
t
j
j
+
== + Ω= ∠Z °Ω

(We expected ()()()
2
2
5
8 24420.5100
2
t j

=+ + = + Ω

Z j . That’s about 1% error.)
11-48

Verification Problems

VP 11-1
The average power supplied by the source is

()( )
()()
s
122.327
cos30 25.227.96 W
2
P=° −− °=

Capacitors and inductors receive zero average power, so the sum of the average powers
received by the other circuit elements is equal to the sum of the average powers received
by the resistors:

() ()
22
R
2.327 1.129
4 210.831.2712.10 W
22
P=+ = + =

The average power supplied by the voltage source is equal to the sum of the average
powers received by the other circuit elements. The mesh currents cannot be correct.

(What went wrong? It appears that the resistances of the two resistors were interchanged
when the data was entered for the computer analysis. Notice that

() ()
22
R
2.327 1.129
2 45.412.557.96 W
22
P=+ = + =

The mesh currents would be correct if the resistances of the two resistors were
interchanged. The computer was used to analyze the wrong circuit.)

VP 11-2
The average complex supplied by the source is

() ( )( )( )
s
12301.64717.92*12301.64717.92
9.8847.926.627.33 W
22
j
∠° ∠− ° ∠° ∠ °
== =∠ °=S +

The complex power received by the 4 Ω resistor is

( )( )
4
41.64717.921.64717.92*
5.430 VA
2
j
Ω
×∠ − ° ∠− °
==S +

The complex power received by the 2 Ω resistor is

( )( )
2
21.09413.151.09413.15*
1.200 VA
2
j
Ω
×∠ − ° ∠− °
==S +
11-49

The current in the 2 H inductor is

() ( )1.64717.921.09413.150.564027.19∠− °− ∠− °= ∠− °

The complex power received by the 2 H inductor is

() ( )
2H
80.564027.190.564027.19*
01.27 VA
2
j
j
×∠ − ° ∠− °
==S +

The complex power received by the 4 H inductor is

() ( )
4H
161.09413.151.09413.15*
09.57 VA
2
j
j
×∠ − ° ∠− °
==S +

() ( )( )( )
422 H 4H s
5.4301.20001.2709.576.6310.84jj j j j
ΩΩ
++ += ++ +++ ++ = + ≠SSS S S

The complex power supplied by the voltage source is equal to the sum of the complex
powers received by the other circuit elements. The mesh currents cannot be correct.

(Suppose the inductances of the inductors were interchanged. Then the complex power
received by the 4 H inductor would be

() ( )
4H
160.564027.190.564027.19*
02.54 VA
2
j
j
×∠ − ° ∠− °
==S +

The complex power received by the 2 H inductor would be

() ( )
2H
81.09413.151.09413.15*
04.79 VA
2
j
j
×∠ − ° ∠− °
==S +

() ( )( )( )
422 H 4H s
5.4301.20002.5404.796.637.33jj j j j
ΩΩ
+++= ++ ++ + ++ = + ≈SSS S S

The mesh currents would be correct if the inductances of the two inductors were
interchanged. The computer was used to analyze the wrong circuit.)

VP 11-3
The voltage across the right coil must be equal to the voltage source voltage. Notice that
the mesh currents both enter the undotted ends of the coils. In the frequency domain, the
voltage across the right coil is
11-50

()( )()( )
() (
161.00147.01 120.42431516.01642.995.09275
11.71510.9231.3184.918
13.03315.841
20.51350.55
jj
jj
j
∠− °+ ∠−°= ∠ °+ ∠°
=+ + +
=+
)
= ∠°

This isn’t equal to the voltage source voltage so the computer analysis isn’t correct.

What happened? A data entry error was made while doing the computer analysis. Both
coils were described as having the dotted end at the top. If both coils had the dot at the
top, the equation for the voltage across the right coil would be

()( )()( )
() (
161.00147.01 120.42431516.01642.995.09275
11.71510.9231.3184.918
10.3976.005
12.00730.01
jj
jj
j
∠− °− ∠−°= ∠ °− ∠°
=+ − +
=+
)
= ∠°

This is equal to the voltage source voltage. The computer was used to analyze the wrong
circuit.

VP 11-4
First check the ratio of the voltages across the coils.

()( )
1
2
1230 2
2.5
750.06430 5
n
n
∠°
=≠=
∠°

The transformer voltages don’t satisfy the equations describing the ideal transformer. The
given mesh currents are not correct.

That’ enough but let’s also check the ratio of coil currents. (Notice that the reference
direction of the i2(t) is different from the reference direction that we used when
discussing transformers.)
1
2
0.06430 2
2.5
0.025630 5
n
n
∠°
=≠=
∠°

The transformer currents don’t satisfy the equations describing the ideal transformer.

In both case, we calculated
1
2
n
n
to be 2.5 instead of
1
2.5
=0.4 . This suggests that a data
entry error was made while doing the computer analysis. The numbers of turns for the
two coils was interchanged.
11-51

11-52

11-53

Design Problems

DP 11. 1
1
100
125 kVA100 W
0.8
0.8
sin(cos0.8)125sin(36.9)75 kVAR
P
P
pf
pf
Q
−

== ==  
⇒
= 
== °=

S
S

(a) Now pf = 0.95 so
1
100
105.3 kVA
0.95
sin(cos0.95)105.3sin(18.2)32.9 kVAR
P
pf
Q
−
== =
== °=
S
S

so an additional 125 − 105.3 = 19.7 kVA is available.

(b) Now pf = 1 so
1
100
100 kVA
1
sin(cos1)0
P
pf
Q
−
== =
= =
S
S

and an additional 125-100= 25 kVA is available.

(c) In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In
part (b), the capacitors are required to reduce Q by 75 – 0 = 0 kVAR.

(d)

Corrected power factor 0.95 1.0
Additional available
apparent power
19.7 kVA 25 kVA
Reduction in reactive power 42.1 kVAR 75 kVAR

11-54

DP 11-2
This example demonstrates that loads can be specified either by kW or kVA. The
procedure is as follows:

First load: ()()
111
1
11
500.945 W50 VA
sin(cos0.9)50sin(25.8)21.8 kVAR0.9
Pp f
Qpf
−
= = == 
⇒
== °==  
SS
S

Second load:
2
22
1
22
45
49.45 kVA45 W
0.91
0.91
sin(cos0.91)49.45sin(24.5)20.5 kVAR
P
P
pf
pf
Q
−

== ==  
⇒
= 
== °=

S
S
Total load: (4545)j(21.8+20.5)9042.3 kVAj== ++ =+
L1 2
SS +S

Specified load:
s
ss
1
ss
90
92.8 kVA90 W
0.97
0.97
sin(cos0.97)92.8sin(14.1)22.6 kVAR
P
P
pf
pf
Q
−

== ==  
⇒
= 
== °=

S
S

The compensating capacitive load is
c42.322.619.7 kVARQ= −= .

The required capacitor is calculated as

2
32
c
c 3
c
(7.210) 1
2626 1.01 F
19.710 377 (2626)
XC
Q
µ
×
== = Ω⇒= =
×
V

11-55

DP 11-3
Find the open circuit voltage:

oc105100.5 0j−++ − =II V

and
oc10

5
−
=
V
I
so

oc836.96.44.8 Vj=∠° =+V

Find the short circuit current:

The short circuit forces the controlling voltage to be
zero. Then the controlled voltage is also zero.
Consequently the dependent source has been replaced
by a short circuit.

sc
100
20 A
5
∠°
= =∠°I

The the Thevenin impedance is:

oc
t
sc
3.2 2.4 j==+
V
Z
I
Ω

(a) Maximum power transfer requires
Lt* 3.2 2.4 j= =−ZZ Ω.

(c) ZL can be implemented as the series combination of a resistor and a capacitor with

()
1
3.2 and 4.17 mF
100(2.4)
RC=Ω = = .

(b)
()
2
oc
max
|| 64
2.5W
88 3.2
P
R
== =
V

11-56

DP 11-4

Using an equation from section 11.8, the power is given as

2
s
2
22
22
2
3
34
R
n
P
R
nn



=
 
++ +
 
 
V




When R = 4 Ω,
2
2
s
4 2
254825
nR
P
nn
=
+ +
V

42 2 3
2
s 42 2
54
2(254825)(10096)
0R
(254825)
50500 1 1
dP nn n n n n
dn n n
nn n n
 ++ − +
==

++
⇒− +=⇒ =⇒ =
V

When R = 8 Ω, a similar calculation gives n = 1.31.

11-57

DP 11-5

Maximum power transfer requires

()
12
106.28
10.628*
j
j
n
+
== +Z
Equating real parts gives
2
10
13 n
n
=⇒ =.16

Equating imaginary parts requires

2
0.628 6.28
3.16
jX
jX=− ⇒ =−

This reactance can be realized by adding a capacitance C in series with the resistor and
inductor that comprise Z2. Then

() () ()
55
11
6.28 6.28 0.1267
210 21012.56
X CF
C
µ
ππ
−= =− + ⇒ = =
××

11-58

DP 11-6

Maximum power transfer requires

()
76
7
7
1
||1001010*
10
10010
110
Rj
jC
R
j
jRC
−
=+ ×
=−
+

() ( )
78 9
(10010)110 10010 10 10Rj jRC RCjRC=− + =+ + −

Equating real and imaginary parts yields

89
10010 and10 100RR C RC=+ −=
then
88
88 10 10
10 10010 99 0.101 nF
99
RC R R C
R
−−
− 
=⇒ =+ =Ω⇒ = =


11-59

Chapter 12: Three-Phase Circuits
Exercises

Ex. 12.3-1
CA B

120240 so 1200and 120120
°°
=∠ − =∠ =∠−VV V
°

()
bc
3120 90
°
=∠ −V

Ex. 12.4-1
Four-wire Y-to-Y Circuit

Mathcad analysis (12v4_1.mcd):

180
π
argIcC()⋅ 134.036=
180
π
argIbB()⋅ 165−=
180
π
argIaA()⋅ 32.005−=
IcC1.164=IbB1.061=IaA1.272=
IcC0.809− 0.837i+=IbB1.025− 0.275i−=IaA1.0790.674i−=
IcC
Vc
ZC
:=IbB
Vb
ZB
:=IaA
Va
ZA
:=Calculate the line currents:
ZC100j25⋅−:=ZB80j80⋅+:=ZA80j50⋅+:=Describe the three-phase load:
VcVae
j
π
180
⋅120⋅
⋅:=VbVae
j
π
180
⋅ 120−⋅
⋅:=VaVpe
j
π
180
⋅0⋅
⋅:=
Vp120:=Describe the three-phase source:

12-1

SaSb+ Sc+ 354.968137.017i+=Total power delivered by the source:
Sc135.52933.882i−=Sb9090i+=Sa129.43880.899i+=
ScIcC

Vc⋅:=SbIbB

Vb⋅:=SaIaA

Va⋅:=
Calculate the power supplied by the source:
SASB+ SC+ 354.968137.017i+=Total power delivered to the load:
SC135.52933.882i−=SB9090i+=SA129.43880.899i+=
SCIcC

IcC⋅ZC⋅:=SBIbB

IbB⋅ZB⋅:=SAIaA

IaA⋅ZA⋅:=
Calculate the power delivered to the load:
INn0.755− 0.112i−=INnIaAIbB+ IcC+:=Calculate the current in the neutral wire:

Ex. 12.4-2
Four-wire Y-to-Y Circuit

Mathcad analysis (12x4_2.mcd):

180
π
argIcC()⋅ 83.13=
180
π
argIbB()⋅ 156.87−=
180
π
argIaA()⋅ 36.87−=
IcC2.4=IbB2.4=IaA2.4=
IcC0.2872.383i+=IbB2.207− 0.943i−=IaA1.921.44i−=
IcC
Vc
ZC
:=IbB
Vb
ZB
:=IaA
Va
ZA
:=Calculate the line currents:
ZCZA:=ZBZA:=ZA40j30⋅+:=Describe the three-phase load:
VcVae
j
π
180
⋅120⋅
⋅:=VbVae
j
π
180
⋅ 120−⋅
⋅:=VaVpe
j
π
180
⋅0⋅
⋅:=
Vp120:=Describe the three-phase source:

12-2

SaSb+ Sc+ 691.2518.4i+=Total power delivered by the source:
Sc230.4172.8i+=Sb230.4172.8i+=Sa230.4172.8i+=
ScIcC

Vc⋅:=SbIbB

Vb⋅:=SaIaA

Va⋅:=
Calculate the power supplied by the source:
SASB+ SC+ 691.2518.4i+=Total power delivered to the load:
SC230.4172.8i+=SB230.4172.8i+=SA230.4172.8i+=
SCIcC

IcC⋅ZC⋅:=SBIbB

IbB⋅ZB⋅:=SAIaA

IaA⋅ZA⋅:=
Calculate the power delivered to the load:
INn0=INnIaAIbB+ IcC+:=Calculate the current in the neutral wire:

Ex. 12.4-3
Three-wire unbalanced Y-to-Y Circuit with line impedances

Mathcad analysis (12x4_3.mcd):

Describe the three-phase source:Vp120:=
VaVpe
j
π
180
⋅0⋅
⋅:= VbVae
j
π
180
⋅ 120−⋅
⋅:= VcVae
j
π
180
⋅120⋅
⋅:=
Describe the three-phase load:ZA80j50⋅+:= ZB80j80⋅+:= ZC100j25⋅−:=
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZAZC⋅e
j
4
3
⋅π⋅
⋅ ZAZB⋅e
j
2
3
⋅π⋅
⋅+ ZBZC⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
Vp⋅:=
VnN 25.137− 14.236i−= VnN28.888=
180
π
argVnN()⋅ 150.475−=

12-3

SASB+ SC+ 391.88141.639i+=Total power delivered to the load:
SC142.84335.711i−=SB57.8757.87i+=SA191.168119.48i+=
SCIcC

IcC⋅ZC⋅:=SBIbB

IbB⋅ZB⋅:=SAIaA

IaA⋅ZA⋅:=
Calculate the power delivered to the load:
IaAIbB+ IcC+ 0=Check:
180
π
argIcC()⋅ 120.475=
180
π
argIbB()⋅ 156.242−=
180
π
argIaA()⋅ 26.403−=
IcC1.195=IbB0.851=IaA1.546=
IcC0.606− 1.03i+=IbB0.778− 0.343i−=IaA1.3850.687i−=
IcC
VcVnN−
ZC
:=IbB
VbVnN−
ZB
:=IaA
VaVnN−
ZA
:=Calculate the line currents:

Ex. 12.4-4
Three-wire balanced Y-to-Y Circuit with line impedances

Mathcad analysis (12x4_4.mcd):

Describe the three-phase source:Vp120:=
VaVpe
j
π
180
⋅ 0⋅
⋅:= VbVae
j
π
180
⋅ 120−⋅
⋅:= VcVae
j
π
180
⋅ 120⋅
⋅:=
Describe the three-phase load:ZA40j30⋅+:= ZBZA:= ZCZA:=
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZAZC⋅e
j
4
3
⋅π⋅
⋅ ZAZB⋅e
j
2
3
⋅π⋅
⋅+ ZBZC⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
Vp⋅:=

12-4

SASB+ SC+ 691.2518.4i+=Total power delivered to the load:
SC230.4172.8i+=SB230.4172.8i+=SA230.4172.8i+=
SCIcC

IcC⋅ZC⋅:=SBIbB

IbB⋅ZB⋅:=SAIaA

IaA⋅ZA⋅:=
Calculate the power delivered to the load:
IaAIbB+ IcC+ 1.05510
15−
× 2.22i10
15−
×−=Check:
180
π
argIcC()⋅ 83.13=
180
π
argIbB()⋅ 156.87−=
180
π
argIaA()⋅ 36.87−=
IcC2.4=IbB2.4=IaA2.4=
IcC0.2872.383i+=IbB2.207− 0.943i−=IaA1.921.44i−=
IcC
VcVnN−
ZC
:=IbB
VbVnN−
ZB
:=IaA
VaVnN−
ZA
:=Calculate the line currents:
180
π
argVnN()⋅ 124.695=VnN2.30110
14−
×=VnN 1.31− 10
14−
× 1.892i10
14−
×+=

Ex. 12.6-1

Balanced delta
load:

(See Table 12.5-1)
18045
°
∆=∠ −Z

phase currents:
3600
245A
18045
360120
275A
1845
360120
2165A
18045
°
°
∆
°
°
°
∆
°
°
°
∆
∠
== =∠
∠−
∠−
== =∠−
∠−
∠
== =∠
∠−
AB
AB
BC
BC
CA
CA
V
I
Z
V
I
Z
V
I
Z
D

line currents:
24521652315 A
23105 A
23135 A
°°
°
°
=− =∠−∠ =∠
=∠ −
=∠
AA B CA
B
C
II I
I
I
°

12-5

Ex. 12.7-1
Three-wire Y-to-Delta Circuit with line impedances

Mathcad analysis (12x4_4.mcd):

ZcCZaA:=ZbBZaA:=ZaA10j25⋅+:=Describe the three-phase line:
ZC5090i+=ZB5090i+=ZA5090i+=
ZC
Z1Z2⋅
Z1Z2+ Z3+
:=ZB
Z2Z3⋅
Z1Z2+ Z3+
:=ZA
Z1Z3⋅
Z1Z2+ Z3+
:=
Convert the delta connected load to the equivalent Y connected load:
Z3Z1:=Z2Z1:=Z1150j270⋅+:=Describe the delta connected load:
VcVae
j
π
180
⋅120⋅
⋅:=VbVae
j
π
180
⋅ 120−⋅
⋅:=VaVpe
j
π
180
⋅0⋅
⋅:=
Vp110:=Describe the three-phase source:

12-6

Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZaAZA+( )ZcCZC+()⋅ e
j
4
3
⋅π⋅
⋅ ZaAZA+( )ZbBZB+()⋅ e
j
2
3
⋅π⋅
⋅+ ZbBZB+( )ZcCZC+()⋅+
ZaAZA+( )ZcCZC+()⋅ ZaAZA+( )ZbBZB+()⋅+ ZbBZB+( )ZcCZC+()⋅+
Vp⋅:=
VnN 1.172− 10
14−
× 1.784i10
14−
×+= VnN2.13510
14−
×=
180
π
argVnN()⋅ 123.304=

180
π
argVBN()⋅ 121.502−=
180
π
argVCN()⋅ 118.498=
Calculate the line-to-line voltages at the load:
VABVANVBN−:= VBCVBNVCN−:= VCAVCNVAN−:=
VAB 151.227= VBC151.227= VCA 151.227=
180
π
argVAB()⋅ 28.498=
180
π
argVBC()⋅ 91.502−=
180
π
argVCA()⋅ 148.498=
Calculate the phase currents of the ∆-connected load:
IAB
VAB
Z3
:= IBC
VBC
Z1
:= ICA
VCA
Z2
:=
IAB0.49= IBC0.49= ICA0.49=
180
π
argIAB()⋅ 32.447−=
180
π
argIBC()⋅ 152.447−=
180
π
argICA()⋅ 87.553=
Calculate the line currents:IaA
VaVnN−
ZAZaA+
:= IbB
VbVnN−
ZBZbB+
:= IcC
VcVnN−
ZCZcC+
:=
IaA0.3920.752i−= IbB0.847− 0.036i+= IcC0.4550.716i+=
IaA0.848= IbB0.848= IcC0.848=
180
π
argIaA()⋅ 62.447−=
180
π
argIbB()⋅ 177.553=
180
π
argIcC()⋅ 57.553=
Check:IaAIbB+ IcC+ 0=
Calculate the phase voltages of the Y-connected load:
VANIaAZA⋅:= VBNIbBZB⋅:= VCNIcCZC⋅:=
VAN 87.311= VBN 87.311= VCN 87.311=
180
π
argVAN()⋅ 1.502−=

12-7

Ex. 12.8-1
Continuing Ex. 12.8-1:

Calculate the power delivered to the load:
SAIaA

IaA⋅ZA⋅:= SBIbB

IbB⋅ZB⋅:= SCIcC

IcC⋅ZC⋅:=
SA35.95864.725i+= SB35.95864.725i+= SC35.95864.725i+=
Total power delivered to the load:SASB+ SC+ 107.875194.175i+=

Ex. 12.9-1
1 2
1
12
P cos(+30) + cos(30)= PP
.4 61.97
450(24) cos 91.97+ cos 31.97 = 8791 W
P = 371 W P = 9162 W
ABA CBC
T
pflagging
SoP
θθ
θ
°°
°
°°
=−
=⇒ =
=

∴−
VI VI +

Ex. 12.9-2
Consider Fig. 12.9-1 with .
12
60 kW 40 kWP P==

(a.)
12
100 kWPP P=+ =

(b.) use equation 12.9-7 to get

21
L2
4060
tan 3 3 .346 19.11
PP 100
PP
θθ
−−
== =− ⇒ =−
+
°
then
cos (19.110) = 0.945 leadingpf=− °

12-8

Problems

Section 12-3: Three Phase Voltages

P12.3-1
()
()
C
A
B
Given 277 45 and an phase sequence:
2774512027775
277 45+120 277 165
abc
°
=∠ °
=∠ −°= ∠−°
=∠ °= ∠°
V
V
V

() ( )
() ( )
AB A B
27775277 165
71.69267.56267.5671.69
339.25339.25479.774548045
jj
j
=− = ∠−°− ∠°
=− −− +
=− = ∠−° ∠−
VV V
°

Similarly:

BC CA
480 165and 480 75
°°
=∠ − = ∠VV

P12.3-2
AB
AB A A330
330
=× ∠°⇒ =
∠°
V
VV V
In our case: ( )
AB BA 12470 35 12470 145 V=− =− ∠−°= ∠°VV
So
A
12470 145
7200115
330
∠°
==
∠°
V ∠°
°

Then, for an abc phase sequence:

()
()
C
B
7200 11512072002357200125
7200 11512072005V
=∠ + °= ∠°= ∠−
=∠ − °= ∠−°
V
V

P12.3-3
ab
ab a a330
330
=× ∠°⇒ =
∠°
V
VV V
In our case, the line-to-line voltage is
ab1500 30 V= ∠°V
So the phase voltage is
a
1500 30
8660 V
330
∠°
= =∠ °
∠°
V

12-9

Section 12-4: The Y-to-Y Circuit

P12.4-1
Balanced, three-wire, Y-Y circuit:

where
AB C123010.46j=== ∠= +ZZ Z

MathCAD analysis (12p4_1.mcd):

IaAIbB+ IcC+ 4.69610
15−
× 1.066i10
14−
×−=Check:
180
π
argIcC()⋅ 90.018=
180
π
argIbB()⋅ 149.982−=
180
π
argIaA()⋅ 29.982−=
IcC10.002=IbB10.002=IaA10.002=
IcC3.205− 10
3−
× 10.002i+=IbB8.66− 5.004i−=IaA8.6634.998i−=
IcC
VcVnN−
ZC
:=IbB
VbVnN−
ZB
:=IaA
VaVnN−
ZA
:=Calculate the line currents:
VnN2.76210
14−
×=VnN
ZAZC⋅e
j
4
3
⋅π⋅
⋅ ZAZB⋅e
j
2
3
⋅π⋅
⋅+ ZBZC⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
Vp⋅:=
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
ZCZB:=ZBZA:=ZA10.4j6⋅+:=Describe the balanced three-phase load:
VcVae
j
π
180
⋅120⋅
⋅:=VbVae
j
π
180
⋅ 120−⋅
⋅:=VaVpe
j
π
180
⋅0⋅
⋅:=
Vp
208
3
:=Describe the three-phase source:

12-10

Calculate the power delivered to the load:
SAIaA

IaA⋅ZA⋅:= SBIbB

IbB⋅ZB⋅:= SCIcC

IcC⋅ZC⋅:=
SA1.0410
3
× 600.222i+= SB1.0410
3
× 600.222i+= SC1.0410
3
× 600.222i+=
Total power delivered to the load:SASB+ SC+ 3.12110
3
× 1.801i10
3
×+=

Consequently:

(a) The phase voltages are

ab c
208
01200 V rms, 120120 V rms and 120120 V rms
3
= ∠°=∠° =∠ −° =∠ °VV V

(b) The currents are equal the line currents

aa A b bB1030 A rms, 10150 A rms == ∠−° ==∠−°II II
and
cc C1090 A rms== ∠°II
(c)

(d) The power delivered to the load is 3.1211.801 kVAj= +S .

P12.4-2
Balanced, three-wire, Y-Y circuit:

where
ab c1200 Vrms, 120120 Vrms and 120120 Vrms=∠ ° =∠−° =∠°VV V
( )( )
3
AB C102 60100101037.7jjπ
−
=== +×× × =+ZZ Z Ω
and
aA bB cC2===ZZ Z Ω

Mathcad Analysis (12p4_2.mcd):
12-11

IaA0.922.89i−= IbB2.963− 0.648i+= IcC2.0432.242i+=
IaA3.033= IbB3.033= IcC3.033=
180
π
argIaA()⋅ 72.344−=
180
π
argIbB()⋅ 167.656=
180
π
argIcC()⋅ 47.656=
Check: IaAIbB+ IcC+ 1.332− 10
15−
× 3.109i10
15−
×−=
Calculate the phase voltages at the load:VAZAIaA⋅:= VBZBIbB⋅:= VCZCIcC⋅:=
VA118.301= VB118.301= VC118.301=
180
π
argVA()⋅ 2.801=
180
π
argVB()⋅ 117.199−=
180
π
argVC()⋅ 122.801=
Describe the three-phase source:Vp120:=
VaVpe
j
π
180
⋅0⋅
⋅:= VbVae
j
π
180
⋅ 120−⋅
⋅:= VcVae
j
π
180
⋅120⋅
⋅:=
Describe the three-phase load:ZA10j37.7⋅+:= ZBZA:= ZCZB:=
Describe the three-phase line:ZaA2:= ZbBZaA:= ZcCZaA:=
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZaAZA+( )ZcCZC+()⋅ e
j
4
3
⋅π⋅
⋅ ZaAZA+( )ZbBZB+()⋅ e
j
2
3
⋅π⋅
⋅+ ZbBZB+( )ZcCZC+()⋅+
ZaAZA+( )ZcCZC+()⋅ ZaAZA+( )ZbBZB+()⋅+ ZbBZB+( )ZcCZC+()⋅+
Vp⋅:=
VnN 8.693− 10
15−
× 2.232i10
14−
×+= VnN2.39610
14−
×=
180
π
argVnN()⋅ 111.277=
Calculate the line currents:IaA
VaVnN−
ZAZaA+
:= IbB
VbVnN−
ZBZbB+
:= IcC
VcVnN−
ZCZcC+
:=

Consequently, the line-to-line voltages at the source are:

ab a
330120033020830 Vrms, =×∠°=∠°×∠°=∠°VV
bc ca208120 Vrms and 208120 Vrms=∠ −° =∠ °VV

The line-to-line voltages at the load are:

AB A330118.3333020533 Vrms, =×∠°= ∠°×∠°=∠°VV
bc ca205117 Vrms and 205123 Vrms=∠ −° =∠ °VV
and the phase currents are

aa A b bB c cC1072 A rms, 3168 A rms and 348 A rms==∠−° ==∠ ° ==∠°II II II
12-12

P12.4-3
Balanced, three-wire, Y-Y circuit:

where

ab c100 V7.070 V rms, 7.07120 V rms and 7.07120 V rms=∠°= ∠° = ∠−° = ∠°VV V
and
()()
AB C
121611216jj=== + =+ZZ Z Ω

MathCAD analysis (12p4_3.mcd):
IbB0.351− 0.042i−= IcC0.1390.325i+=
IaA0.354= IbB0.354= IcC0.354=
180
π
argIaA()⋅ 53.13−=
180
π
argIbB()⋅ 173.13−=
180
π
argIcC()⋅ 66.87=
Calculate the power delivered to the load:
SAIaA

IaA⋅ZA⋅:= SBIbB

IbB⋅ZB⋅:= SCIcC

IcC⋅ZC⋅:=
SA1.52i+= SB1.52i+= SC1.52i+=
Total power delivered to the load:SASB+ SC+ 4.56i+=
Describe the three-phase source:Vp
10
2
:=
VaVpe
j
π
180
⋅0⋅
⋅:= VbVae
j
π
180
⋅ 120−⋅
⋅:= VcVae
j
π
180
⋅120⋅
⋅:=
Describe the balanced three-phase load:ZA12j16⋅+:= ZBZA:= ZCZB:=
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN
ZAZC⋅e
j
4
3
⋅π⋅
⋅ ZAZB⋅e
j
2
3
⋅π⋅
⋅+ ZBZC⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
Vp⋅:= VnN1.67510
15−
×=
Calculate the line currents:IaA
VaVnN−
ZA
:= IbB
VbVnN−
ZB
:= IcC
VcVnN−
ZC
:=
IaA0.2120.283i−=

12-13

Consequently

(a) The rms value of ia(t) is 0.354 A rms.

(b) The average power delivered to the load is {} { }Re Re4.564.5 WPj== +=S

P12.4-4
Unbalanced, three-wire, Y-Y circuit:

where
ab c1000 V70.70 V rms, 70.7120 V rms and 7.07120 V rms=∠ °= ∠° = ∠−° = ∠°VV V
()() ()( )
33
AB203776010 2022.6, 40 3774010 4015.1jj j
−−
=+ × =+ Ω =+ × =+ZZ j Ω
()( )
3
C603772010 607.54jj
−
=+ × =+ ΩZ
and ()( )
3
aA bB cC10377510101.89jj
−
== =+ × =+ZZ Z Ω

Mathcad Analysis (12p4_4.mcd):

Describe the three-phase source:Vp100:=
VaVpe
j
π
180
⋅0⋅
⋅:= VbVae
j
π
180
⋅120⋅
⋅:= VcVae
j
π
180
⋅ 120−⋅
⋅:=
Enter the frequency of the 3-phase source:ω377:=
Describe the three-phase load:ZA20jω⋅0.06⋅+:= ZB40jω⋅0.04⋅+:= ZC60jω⋅0.02⋅+:=
Describe the three-phase line:ZaA10jω⋅0.005⋅+:= ZbBZaA:= ZcCZaA:=
12-14

Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZaAZA+( )ZcCZC+()⋅ e
j
4
3
⋅π⋅
⋅ ZaAZA+( )ZbBZB+()⋅ e
j
2
3
⋅π⋅
⋅+ ZbBZB+( )ZcCZC+()⋅+
ZaAZA+( )ZcCZC+()⋅ ZaAZA+( )ZbBZB+()⋅+ ZbBZB+( )ZcCZC+()⋅+
Vp⋅:=
VnN12.20924.552i−= VnN27.42=
180
π
argVnN()⋅ 63.561−=

Calculate the line currents:IaA
VaVnN−
ZAZaA+
:= IbB
VbVnN−
ZBZbB+
:= IcC
VcVnN−
ZCZcC+
:=
IaA2.1560.943i−= IbB0.439− 2.372i+= IcC0.99− 0.753i−=
IaA2.353= IbB2.412= IcC1.244=
180
π
argIaA()⋅ 23.619−=
180
π
argIbB()⋅ 100.492=
180
π
argIcC()⋅ 142.741−=

Calculate the power delivered to the load:
SA
IaA

IaA⋅()
2
ZA⋅:= SB
IbB

IbB⋅()
2
ZB⋅:= SC
IcC

IcC⋅()
2
ZC⋅:=
SA55.38262.637i+= SB116.40243.884i+= SC46.4255.834i+=
Total power delivered to the load:SASB+ SC+ 218.209112.355i+=

The average power delivered to the load is {} { }Re Re218.2112.4218.2 WPj== + =S

P12.4-5
Balanced, three-wire, Y-Y circuit:

where
ab c1000 V70.70 V rms, 70.7120 V rms and 7.07120 V rms=∠ °= ∠° = ∠−° = ∠°VV V
()( )
3
AB C203776010 2022.6jj
−
=== + × =+ZZ Z Ω
and ()( )
3
aA bB cC10377510101.89jj
−
== =+ × =+ZZ Z Ω

12-15

Mathcad Analysis (12p4_5.mcd):
180
π
argIcC()⋅ 159.243−=
180
π
argIbB()⋅ 80.757=
180
π
argIaA()⋅ 39.243−=
IcC2.582=IbB2.582=IaA2.582=
IcC2.414− 0.915i−=IbB0.4152.548i+=IaA1.9991.633i−=
IcC
VcVnN−
ZCZcC+
:=IbB
VbVnN−
ZBZbB+
:=IaA
VaVnN−
ZAZaA+
:=Calculate the line currents:
180
π
argVnN()⋅ 115.55=VnN2.08310
14−
×=VnN 8.982− 10
15−
× 1.879i10
14−
×+=
Describe the three-phase source:Vp100:=
VaVpe
j
π
180
⋅0⋅
⋅:= VbVae
j
π
180
⋅120⋅
⋅:= VcVae
j
π
180
⋅ 120−⋅
⋅:=
Enter the frequency of the 3-phase source:ω377:=
Describe the three-phase load:ZA20jω⋅0.06⋅+:= ZBZA:= ZCZA:=
Describe the three-phase line:ZaA10jω⋅0.005⋅+:= ZbBZaA:= ZcCZaA:=
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN
ZaAZA+( )ZcCZC+()⋅ e
j
4
3
⋅π⋅
⋅ ZaAZA+( )ZbBZB+()⋅ e
j
2
3
⋅π⋅
⋅+ ZbBZB+( )ZcCZC+()⋅+
ZaAZA+( )ZcCZC+()⋅ ZaAZA+( )ZbBZB+()⋅+ ZbBZB+( )ZcCZC+()⋅+
Vp⋅:=

Calculate the power delivered to the load:
SA
IaA

IaA⋅()
2
ZA⋅:= SB
IbB

IbB⋅()
2
ZB⋅:= SC
IcC

IcC⋅()
2
ZC⋅:=
SA66.64575.375i+= SB66.64575.375i+= SC66.64575.375i+=
Total power delivered to the load:SASB+ SC+ 199.934226.125i+=

The average power delivered to the load is {} { }Re Re200226200 WPj== + =S

12-16

P12.4-6
Unbalanced, three-wire, Y-Y circuit:

where
ab c1090 V7.0790 V rms, 7.07150 V rms and 7.0730 V rms=∠−° = ∠−° = ∠° = ∠°VV V
and
()() ()()( )(
AB C
44 144, 24 228 and 44 248jj j j j=+ =+ Ω =+ =+Ω =+ =+ ΩZZ Z ) j

Mathcad Analysis (12p4_6.mcd):

180
π
argIcC()⋅ 24.011−=
180
π
argIbB()⋅ 74.116=
180
π
argIaA()⋅ 144.495−=
IcC1.032=IbB1.426=IaA1.638=
IcC0.9430.42i−=IbB0.391.371i+=IaA1.333− 0.951i−=
IcC
VcVnN−
ZC
:=IbB
VbVnN−
ZB
:=IaA
VaVnN−
ZA
:=Calculate the line currents:
180
π
argVnN()⋅ 29.466−=VnN1.755=VnN1.5280.863i−=
VnN
ZAZC⋅Vb⋅ ZAZB⋅Vc⋅+ ZBZC⋅Va⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
:=
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
ZC4jω⋅2⋅+:=ZB2jω⋅2⋅+:=ZA4jω⋅1⋅+:=Describe the three-phase load:
ω4:=Enter the frequency of the 3-phase source:
VcVpe
j
π
180
⋅30⋅
⋅:=VbVpe
j
π
180
⋅150⋅
⋅:=VaVpe
j
π
180
⋅ 90−⋅
⋅:=
Vp10:=Describe the three-phase source:
12-17

Calculate the power delivered to the load:
SA
IaA

IaA⋅()
2
ZA⋅:= SB
IbB

IbB⋅()
2
ZB⋅:= SC
IcC

IcC⋅()
2
ZC⋅:=
SA5.3635.363i+= SB2.0328.128i+= SC2.1314.262i+=
Total power delivered to the load:SASB+ SC+ 9.52717.754i+=
{} { }Re Re9.52717.7549.527Pj== + =S
ab1090 V7.0790 V rms, 7.07150 V rms and 7.0730=∠−° = ∠−° = ∠° = ∠VV
()()
AB C
44 248jj=== + =+ΩZZ Z
Describe the three-phase source:Vp10:=
VaVpe
j
π
180
⋅ 90−⋅
⋅:= VbVpe
j
π
180
⋅150⋅
⋅:= VcVpe
j
π
180
⋅30⋅
⋅:=
Enter the frequency of the 3-phase source:ω4:=
Describe the three-phase load:ZA4jω⋅2⋅+:= ZBZA:= ZCZA:=
The voltage at the neutral of the load with respect to the neutral of the source should be zero:

The average power delivered to the load is W

P12.4-7
Unbalanced, three-wire, Y-Y circuit:

where
c V rms°V
and

Mathcad Analysis (12p4_7.mcd):

VnN
ZAZC⋅Vb⋅ ZAZB⋅Vc⋅+ ZBZC⋅Va⋅+
ZAZC⋅ ZAZB⋅+ ZBZC⋅+
:=
VnN1.51710
15−
×=

12-18

SASB+ SC+ 7.515i+=Total power delivered to the load:
SC2.55i+=SB2.55i+=SA2.55i+=
SC
IcC

IcC⋅()
2
ZC⋅:=SB
IbB

IbB⋅()
2
ZB⋅:=SA
IaA

IaA⋅()
2
ZA⋅:=
Calculate the power delivered to the load:
180
π
argIcC()⋅ 33.435−=
180
π
argIbB()⋅ 86.565=
180
π
argIaA()⋅ 153.435−=
IcC1.118=IbB1.118=IaA1.118=
IcC0.9330.616i−=IbB0.0671.116i+=IaA1−0.5i−=
IcC
VcVnN−
ZC
:=IbB
VbVnN−
ZB
:=IaA
VaVnN−
ZA
:=Calculate the line currents:

The average power delivered to the load is {} { }Re Re7.5157.5 WPj== + =S

Section 12-6: The ∆- Connected Source and Load

P12.5-1
Given I and assuming the abc phase sequence
we have
B
5040A rms=∠ −°

AC 5080A rms and 50200A rms=∠ ° =∠°II

From Eqn 12.6-4
A
AA B AB 33 0
330
=× ∠−°⇒ =
∠−°
I
II I
so
AB
BC CA
5080
I 28.9110A rms
330
I 28.910A rms and I 28.9130A rms
°
°°
∠°
== ∠
∠−°
=∠ − = ∠−

12-19

P12.5-2

The two delta loads connected in parallel are equivalent to a single delta
load with
5||204
∆
==Z Ω
The magnitude of phase current is
p
480
120 A rms
4
I==
The magnitude of line current is
Lp3 208 A rmsII==

ection 12-6:S The Y- to ∆- Circuit
12.6-1
e have a delta load with °. One phase current is

P

W 1230=∠Z

AB A A
AB
208 208
30 150
208033
17.3130 A rms
1230 1230
  
∠−°− ∠−°
  
−∠ °  
== = = = ∠−°
∠° ∠°
VV V
I
ZZ

he other phase currents are

One line currents is
T

BC CA17.31150 A rms and 17.3190 A rms=∠ −° =∠ °II

() ( )AA B 33017.3130 330300 A rms=×∠−°= ∠−°×∠−°=∠°II

The other line currents are

he power delivered to the load is

BC30120 A rms and 30120 A rms=∠−° =∠ °II
T

()
208
3()(30)cos030 9360 W
3
P=− °=

12-20

P12.6-2

The balanced delta load with 3940
∆
=∠−°ΩZ is
d with equivalent to a balanced Y loa

1340 9.968.36
3
Y
j
°∆
==∠−= − Ω
Z
Z

TY
A
413.968.3616.330.9
480
30
3
then 170.9 A rms
16.330.9
j
°
°
°
=+= − =∠− Ω
∠−
== ∠
∠−
I

ZZ

12.6-3

P
ab
ab a a330
330
=× ∠°⇒ =
∠°
V
VV V
In our case, the given line-to-line voltage is
So one phase voltage is

ab 380 30 V rms=∠ °V
a
380 30
2000 V rms
330
∠°
== ∠°
∠°
V
So

One phase current is
AB A
BC B
CA C
V 38030 V rms V 2200 V rms
V 380-90 V rms V 220120 V rms
V 380150 V rms V 220120 V rms
=∠ ° =∠°
=∠ ° =∠−°
=∠ ° =∠°
A
A
2200
4453.1 A rms
3j4
∠°
== ∠−°
+
V
I
Z

The other phase currents are

BC 44173.1 A rms amd 4466.9 A rms=∠− ° =∠ °II

12-21

P12.6-4
ab
ab a a330
330
=× ∠°⇒ =
∠°
V
VV V
In our case, the given line-to-line voltage is

ab380 0 V rms=∠°V

So one phase voltage is
a
380 0
20030 V rms
330
∠°
== ∠−°
∠°
V
So
One phase current is
ab a
bc b
ca c
V 3800 V rms V 22030 V rms
V 380-120 V rms V 220150 V rms
V 380120 V rms V 22090 V rms
=∠ ° = ∠−°
=∠ ° =∠−°
=∠ ° =∠°

a
A
22030
14.6783.1 A rms
9j12
∠−°
== = ∠−°
+
V
I
Z

The other phase currents are

BC 14.67203.1 A rms and 14.6736.9 A rms=∠ − ° =∠ °II

12-22

12-23

Section 12-7: Balanced Three-Phase Circuits

P12.7-1
3
a
25
100 Vrms
3
=× ∠°V
3
a
A
25
100
3
9625A rm
15025
°
×∠ °
== =∠−
∠°
V
I
Z
s

()
3
aA I
25
3 cos- 3 1096cos(025) 3.77 mWv
3
P θθ

== × −°=


VI

P12.7-2
Convert the delta load to an equivalent Y connected load:

Y
ˆ50
50
3
∆∆=Ω ⇒ =ZZ Z Ω
To get the per-phase equivalent circuit shown to the right:
The phase voltage of the source is

3
a
4510
0260 kV rm
3
×
=∠ °=∠°V s

The equivalent impedance of the load together with the line is

()
eq
50
1020
3
21251322.6
50
1020
3
j
j
j
+
=+ =+=∠
++
Z °Ω
The line current is
3
a
aA
eq
26100
200022.6 A rms
1322.6
×∠ °
== = ∠−°
∠°
V
Ι
Z

The power delivered to the parallel loads (per phase) is
()
2
6
Loads aA
50
1020
3
Re 4101040 MW
50
1020
3
j
P
j

+

=× =××=
++

I

The power lost in the line (per phase) is
12-24

{}
2
6
Line aA LineRe 41028 MWP=× =××=IZ

The percentage of the total power lost in the line is

Line
Load Line
8
100% 100% 16.7%
408
P
PP
×= ×=
++

P12.7-3
a
aa
T
530
0.523 A 0.5 A
68j
∠°
== =∠−°∴=
+
V
II
Z

{}
2
a
Load Load
3 Re 30.12541.5 W
2
P== × ×
I
Z =

also (but not required) :

{}
Source
2
a
line Line
(5)(0.5)
3 cos(3023) 2.25 W
2
3 Re 30.12520.75 W
2
P
P
=− − =
== × ×=
I
Z

12-25

Section 12-8: Power in a Balanced Load

P12.8-1
Assuming the abc phase sequence:

CB BC AB 20815 V rms 208195 V rms 208315 V rms=∠°⇒ =∠°⇒ =∠°VV V

Then
AB
A
208315208
285 V rms
330 330 3
∠°
== = ∠°
∠° ∠°
V
V
also
BA
3110 A rms 3230 A rms=∠ ° ⇒ =∠ °II
Finally
()
AB A V I
208
3 cos 3()(3)cos(285230)620 W
3
P θθ=− = °−°=VI

P12.8-2
Assuming a lagging power factor:

cos 0.8 36.9pfθ θ==⇒ = °

The power supplied by the three-phase source is given by

()
out
in
20745.7
17.55 kW where1 hp745.7 W
0.85
P
P
η
== = =

()
in
in A A A
A
3
17.5510
3 26.4 A rms
4803
30 .8
3
P
Pp f
pf
×
=⇒ = = =



IV I
V

AA

480
26.436.9 A rms when 0 V rms
3
°°
=∠ − = ∠IV

12-26

P12.8-3
(a) For a ∆-connected load, Eqn 12.8-5 gives

T
TP L L
1500
34
2203PL 3()(.8)
3
P
Pp f
pf
=⇒ = = =VI I
VI
.92 A rms
The phase current in the ∆-connected load is given by

LL
PP
4.92
2.84 A rms
33 3
=⇒ == =
II
II
The phase impedance is determined as:

()
LL 11L
VI
PP P
220
cos cos0.877.4436.9
2.84
pfθθ
−−
== ∠−= ∠ = ∠ = ∠ °
VVV
Z
II I
Ω

(b) For a ∆-connected load, Eqn 12.8-4 gives
T
TP L L
1500
34
2203PL 3()(.8)
3
P
Pp f
pf
=⇒ = = =VI I
VI
.92 A rms
The phase impedance is determined as:
()
PP 11P
VI
PP P
220
3
cos cos0.825.836.9
4.92
pfθθ
−−
== ∠−=∠ = ∠ = ∠°
VVV
Z
II I
Ω

P12.8-4
12
12
P
LP P L P
LL
Parallel loads
(4030) (5060)
31.2 8.7
4030 5060
600
, = 19.2 A rms, 3 33.3 A rms
Z 31.2
So = 3 = 3 (600) (33.3) cos (8.7) = 34Pp f
°°
°
∆ °°
∆
°
∆
∠∠ −
== = ∠−Ω
+∠ +∠−
== = = =
−
ZZ
Z
ΖΖ
V
VV Ι I Ι
VI .2 kW

12-27

P12.8-5
We will use
( )
1
cos sin sincospf pfθθ θ
−
=∠ = + = +SS S S S S
In our case:
()( )
()()
1
1
1
2
3
31 2
39(0.7)39sincos0.727.3 27.85 kVA
15
15 sincos0.2115 69.84 kVA
0.21
42.3 42.0 kVA 14.1 14.0 kVA
3
jj
j
jj
φ
φφ
−
−
=+ = +
=+ =−
=+ = − ⇒ == −
S
S
S
SS S S

The line current is
*
*
pL L
p
(14100 14000)
117.5 116.7 A rms = 167 45 A rms
208
3
j
j
 +
=⇒ = = = + ∠°


S
SVI I
V

The phase voltage at the load is required to be
208
01200 V rms.
3
∠°=∠° The source must
provide this voltage plus the voltage dropped across the line, therefore

S 1200(0.038 0.072)(117.5116.7)115.9 12.9116.6 6.4 V rmsjj j
φ=∠ °+ + + = + = ∠°V

Finally
S116.6 V rms
φ=V

P12.8-6
The required phase voltage at the load is
P
4.16
02.4020 kVrms
3
=∠ °= ∠°V .

Let I1 be the line current required by the ∆-connected load. The apparent power per phase
required by the ∆-connected load is
1
500 kVA
167 kVA
3
= =S . Then

() ()
11
11 1cos 167cos0.8516731.8 kVApfθ
−−
=∠ =∠ =∠ =∠°SS S
and
()
()
*
*
* 3
1
1P 1 1 3
P
1671031.8
69.631.85936.56 A rms
2.402100
j
×∠ °
=⇒ = = = ∠−°=−
 ×∠ °

S
SVI I
V

12-28

Let I2 be the line current required by the first Y-connected load. The apparent power per phase
required by this load is
2
75 kVA
25 kVA
3
= =S . Then, noticing the leading power factor,

() ()
11
22 2cos 25cos02590 kVApfθ
−−
=∠ =∠ =∠ =∠−°SS S
and
()
()
*
*
* 3
2
2P 2 2 3
P
2510 90
10.49010.4 A rms
2.402100
j
×∠ −°
=⇒ = = =∠°=
 ×∠ °

S
SV I I
V

Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3
to be
3
2402024020
16 10.7 A rms
150 225
j
j
∠° ∠°
=+ =−I
The line current is
L1 2 3 + + = 75 36.8 A rmsj=−II II

The phase voltage at the load is required to be
P
4.16
02.4020 kVrms
3
=∠ °= ∠°V .The source
must provide this voltage plus the voltage dropped across the line, therefore

S 24020 (8.45 3.9) (75 36.8) 3179 0.3 Vrmsjj
φ
°
=∠ + + − = ∠−°V

Finally
SL = 3 (3179) = 5506 VrmsV

P12.8-7
The required phase voltage at the load is
P
4.16
02.4020 kVrms
3
=∠ °= ∠°V .

Let I1 be the line current required by the ∆-connected load. The apparent power per phase
required by the ∆-connected load is
1
1.5 MVA
0.5 MVA
3
= =S . Then

() ()
11
11 1cos 0.5cos0.750.541.4 MVApfθ
−−
=∠ =∠ =∠ =∠°SS S
and
()
()
*
*
* 6
1
1P 1 1 3
P
0.51041.4
2081.641.41561.41376.6 A rms
2.402100
j
×∠ °
=⇒ = = = ∠−°= −
 ×∠ °

S
SVI I
V

12-29

Let I2 be the line current required by the first Y-connected load. The complex power, per phase,
is
()()
1
2
0.67
0.67 sincos0.80.67 0.5 MVA
0.8
j
−
=+ = +S

()
()
( )
()
**
* 66
2
2 33
P
0.83310 36.90.670.510
2.402100 2.402100
346.936.9277.4208.3 A rms
j
j
  ×∠ −°+×
  == =
  ×∠ ° ×∠°
  
=∠ −°= −
S
I
V
The line current is
L1 2
433.7345.9554.738.6 A rmsj=+ = − = ∠−II I

The phase voltage at the load is required to be
P
4.16
02.4020 kVrms
3
=∠ °= ∠°V .The source
must provide this voltage plus the voltage dropped across the line, therefore

S 24020(0.4 0.8) (433.7 345.9)2859.6 38.6 Vrmsjj
φ
°
=∠ + + − = ∠−°V

Finally
SL = 3 (2859.6) = 4953 VrmsV

The power supplied by the source is

S 3 (4953) (554.7) cos (4.2 38.6) = 3.49 MWP
°°
=+

The power lost in the line is

() { }
2
Line 3 554.7Re0.40.8 = 0.369 MWPj=× × +

The percentage of the power consumed by the loads is

3.490.369
100%89.4%
3.49
−
×=

12-30

P12.8–8
The required phase voltage at the load is
P
600
0346.40 Vrms
3
=∠ °= ∠°V .

θ = cos
= 37
1−
°
(.)08

Let I be the line current required by the load. The complex power, per phase, is

()()
1160
160 sincos0.8160 120 kVA
0.8
j
−
=+ =+S

The line current is
()
**
3
P
16012010
461.9346.4 A rms
346.40
j
j
+×
== = −
∠°
 
S
I
V

The phase voltage at the load is required to be
P
600
0346.40 Vrms
3
=∠ °= ∠°V .The source
must provide this voltage plus the voltage dropped across the line, therefore

S 346.40(0.005 0.025) (461.9 346.4)357.5 1.6 Vrmsjj
φ
°
=∠ + + − = ∠°V

Finally
SL = 3 (357.5) = 619.2 VrmsV

The power factor of the source is

VI = cos ( ) = cos (1.6 (37)) = 0.78pf θθ− −−

12-31

Section 12-9: Two-Wattmeter Power Measurement

P12.9-1
()
out
out
in
3
in
in LL
LL
-1
W
20hp746 14920 W
hp
14920
= 20 kW
0.746
20 10
= 3 cos cos 0.50
3 3 (440) (52.5)
cos0.560
P
P
P
P
P
η
θθ
θ
=× =
==
×
⇒= = =
⇒= °
VI
VI

The powers read by the two wattmeters are

() ( )
1L L
cos 30(440) (52.5)cos 60300P θ=+ °= °+VI °=
and
() ( )
2L L
cos 30(440) (52.5)cos 603020 kWP θ= −°= °−°=VI

P12.9-2

PL
P
PL
1L L
2L
= = 4000 V rms = 40 + j 30 = 50 36.9
4000
= = = 80 A rms = 3 = 138.6 A rms
50
= cos = cos (36.9) = 0.80
P = VI cos ( + 30) = 4000 (138.6) cos 66.9 = 217.5 kW
P = V
L
pf
I
θ
θ
∆
°
°°
°
∠
∆
VV Z
V
I Ι I
Z
P
T1 2
TL L
cos ( 30) = 4000 (138.6) cos 6.9 = 550.4 kW
P = P + P = 767.9 kW
Check : P =3 cos = 3 (4000) (138.6) cos 36.9
= 768 kW which checks
θ
θ
°°
°
−
ΙV

12-32

P12.9–3
pp
200
= 115.47 Vrms
3
V==V

AB
=115.470 V rms, = 115.47120 V rms∠° ∠−°VV
C and = 115.47120 V rms∠°V

A
A
V 115.470
I= = = 1.63345 A rms
Z 70.745
∠°
∠−°
∠°

BC = 1.633 165 A rms and = 1.633 75 A rms∠−° ∠°II

TL L
BA C A 1
CB C B 2
= 3 cos = 3 (200) (1.633) cos 45 = 400 W
= cos = 200 (1.633) cos (45 30) = 315.47 W
= cos = 200 (1.633) cos (45+ 30) = 84.53 W
P
P
P
θ
θ
θ
°
°°
−
°°
VI
VI
VI

P12.9-4

() ( )
ˆˆ
YY
eq
1030 and 1530
Convert to 530
3
1030 530 500
then Z 3.7810.9
103053013.228 10.9
Y ∆
∆
∆
=∠−°Ω =∠°Ω
→= =∠°Ω
∠−°∠° ∠°
== =
∠−°+∠° ∠− °
ZZ
Z
ZZ Z
∠°Ω

()( )()
pp
AA
B
C
LL
1L L
2L L
208
120 V rms
3
1200
1200 V rms 31.7510.9
3.7810.9
31.75130.9
31.75109.1
3 cos 320831.75cos10.911.23
cos(30)6.24 kW
cos(30)4.99 kW
T
V
PV I k
WV I
WV I
θ
θ
θ
== =
∠°
=∠ ° ⇒= = ∠−°
∠°
=∠ − °
=∠ °
== =
=− °=
=+ °=
V
VI
I
I
W

12-33

P12.9-5

()
()
AC
AC
AC
9204601380W
460
tan 3 3 0.577 30
1380
1380
3 cos 7.67A rms
3cos 2120cos30
120
4.43A rms 27.1 27.130
4.433
T
T
TL L L
L
L
P
PP P
PP
PP
P
PV I soI
V
I
Ir
θθ
θ
θ
ο
°
°
∆∆
=+ =+ =
−−
== =− ⇒ =−
+
== = =
×× −
== ∴ = =Ω =∠ −ZZ

P12.9-6

()()()
()()()
1
2
12
0.8684.924580 80
380
380 V rms, 219.4 V rms
3
43.9 A rms
38043.9cos3010,723 W
38043.9cos30 5706 W
5017 W
LP
P
LP P
T
j
VV
V
IIandI
P
P
PP P
θ
θ
θ
°
°
=+ =∠°⇒ =
== =
== =
=− =
=+ =−
=+ =
Z
Z
°

12-34

PSpice Problems

SP 12-1

FREQ IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
6.000E+01 3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01

FREQ IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
6.000E+01 3.142E+00 -4.443E+01 2.244E+00 -2.200E+00

FREQ VM(N01496) VP(N01496) VR(N01496) VI(N01496)
6.000E+01 2.045E-14 2.211E+01 1.895E-14 7.698E-15

FREQ IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
6.000E+01 3.142E+00 7.557E+01 7.829E-01 3.043E+00

2
3.142
3.14243.43 Aand 20 2098.7 W
2
AA ARP=∠ − ° =Ω⇒ = =I
2
3.142
3.14275.57 Aand 20 2098.7 W
2
BB BRP=∠ ° =Ω⇒ = =I
2
3.142
3.142164.4 Aand 20 2098.7 W
2
CC CRP=∠ − ° =Ω⇒ = =I

()398.7696.1 WP==

12-35

SP 12-2

FREQ IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
6.000E+01 1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00

FREQ IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
6.000E+01 2.537E+00 -3.748E+01 2.013E+00 -1.544E+00

FREQ VM(N01496) VP(N01496) VR(N01496) VI(N01496)
6.000E+01 1.215E+01 -1.439E+01 1.177E+01 -3.018E+00

FREQ IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
6.000E+01 2.858E+00 1.084E+02 -9.023E-01 2.712E+00

2
2.537
2.53737.48 Aand 20 2064.4 W
2
AA ARP=∠ − ° =Ω⇒ = =I
2
2.858
2.858108.4 Aand 30 30122.5 W
2
BB BRP=∠ ° =Ω⇒ = =I
2
1.612
1.612133.6 Aand 600 6078 W
2
CC CRP=∠ − ° = Ω⇒ = =I

64.4122.578264.7 VP=+ +=
12-36

Verification Problems

VP 12-1

A
A
A
416
= = 240 V = V
3
= 10 + j4 = 10.77 21.8
240
= = = 22.28 A rms38.63 A rms
10.77
°
∠Ω
≠
AV
Z
V
I
Z

The report is not correct. (Notice that
38.63
22.3
3
= . It appears that the line-to-line voltage was
mistakenly used in place of the phase voltage.)

VP 12-2

LP
P
P
= = 2400 Vrms
40 30 50 36.9
2400
= = = 4.8 36.9 A rms
5036.9
j
°
°
∠°
=+ =∠ °Ω
∠°
∠−
∠
VV
Z
V
I
Z

The result is correct.

Design Problems

DP 12-1
()
-1
LL
L
L
= 400 W per phase,
0.94 cos = cos0.94=20
208
400 0.94 = 3.5 A rms
3
= = 2.04 A rms
3
208
= = = 101.8
2.04
= 101.8 20
P
pf θθ
∆
∆
°
== ⇒
=⇒
Ω
∠Ω
II
I
I
V
Z
I
Z
°

12-37

DP 12-2
L
AL L
CL L
LL
P
P

=
=
P
= = LP
= 240 V rms
cos (30+ ) = 1440 W
cos (30 ) = 0 W 30 = 90 or = 60
then 1440 = 240 cos (30) = 6.93 A rms
240
3
II = = 20
6.93
Finally, = 2
P
P
θ
θθ
⇒
°
°°
−⇒ −
°
−⇒
=Ω
V
VI
VI
II
VV
Z
ZI
Z
θ
°
−
0 60 ∠−°Ω

DP 12-3
inout
in
W
100 hp (746 )
hp
= = = 93.2 kW, 31.07 kW
0.8 3
PP
PP
φη
×
==

L
= 480 V rms, 0.9 and 0.75. Vp fc pf= = We need the impedance of the load so that we can use
Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor.

()
-1
0.75 cos = cos0.7541.4pf θθ== ⇒ =°

PP
480
31070 0.75 = 149.5 A rms
3
=⇒ II

P
P
480
3
= = = 1.85
149.5
= 1.85 41.4 1.3881.223 j
Ω
∠ °Ω= + Ω
V
Z
I
Z

The capacitance required to correct the power factor is given by

11
22
tan (cos0.75) tan (cos0.9)1.365
= = 434 F
1.3651.204 377
C µ
−−
 −

×
+

(Checked using LNAPAC 6/12/03)

12-38

DP 12-4

L
2
22 L
1
3
L
L
L
12
= 40 kV rms
25
Try = 25 then = =400001000 kVrms
1
4100
= 30 kA rms
4
3
30000
The line current in 2.5 is 1200 A rms
25
Thus = ( ) +
(2.5 40)
n
n
n
RjX
j
∠°
∠°=∠°
×∠ °
== ∠°
∠°
Ω= =∠°
+
=+
V
VV
V
I
Z
I
VI V
3
1
22
33
loss
(1200) 10010 100.4 2.7 kV
100.4 kV
Step need : = 5.02 5
20 kV
= 120 (2.5) 36 kW, (410) (3 10) 12 MW
12 .036
100% 99.7 % of the power supplied by
12
n
PR P
η
∠°+×= ∠°
=≅
== =× × =
−
∴= × =
I
the source
is delivered to the load.

12-39

Chapter 13: Frequency Response
Exercises
Ex. 13.3-1

o
s
2
1
() 1
()
() 1
1

1 ( )
= tan
jCR
gain
CR
phaseshift CR
ω
ω
ωω
ω
ω
−
==
+
=
+
−
V
H
V

46 o1
When 10, = 100, and 10, then = =0.707 and = 45
2
R C gain phaseshiftω
−
== −

Ex. 13.3-2

o
s
22
()
()
()

()
R
RjL
R
gain
RL
ω
ω
ω ω
ω
==
+
=
+
V
H
V

2
2
22
30
30
.630
0.6 20 rads
230(2)
ω
ω

−


=⇒ = =
+

Ex. 13.3-3

s
22
1
() 1
()
()
1

()
tan
RjL
gain
RL
L
phaseshift
R
ω
ω
ω ω
ω
ω
−
==
+
=
+
=−
I
H
V

When 30 , 2 H, and 20 rad/s, thenRL ω=Ω = =

1
22
1A 40
0.02 and tan 53.1
V330 40
gain phaseshift
−
= = =− =− °
+ 0

13-1

Ex. 13.3-4

o
s
2
1
() 1
()
()1
1

1 ( )
ta n
jCR
gain
CR
phaseshift CR
ω
ω
ωω
ω
ω
−
==
+
=
+
=−
V
H
V

16 3
6
tan (45)
45 tan(2010) 5010
2010
RR
°
−−
−
−°=− ⋅⋅⇒ = =⋅Ω
⋅

Ex. 13.3-5

o
s
2
() 1
()
()1
1

1 ( )
jCR
gain
CR
ω
ω
ωω
ω
==
+
=
+
V
H
V

,, and are all positive, or at least nonnegative, so gain 1. These specifications cannot be met.CRω ≤

Ex. 13.4-1
(a) dB = 20 log (.5) = 6.02 dB−

(b) dB = 20 log 2 = 6.02 dB

Ex. 13.4-2

-2
2
2
21 2 1
1
21
1
20log 20log 20 log () 40log
20 log ()20log () 40log 40log -40 log
dB
let 10 to consider 1 decade, then 40log10 40
decade
slope
slope
ωω
ω
ω
ωω ω ω
ω
ωω

== =−



=− = + = 

== −
H
HH
=−

13-2

Ex. 13.4-3
10 10
10 10
When , ()
()() in dB 20log() 20log
()() does not depend on so 0
When , ()
() in dB = 20 log () = 20 log
jA A
CB
jCC
A
d
C
b slope
jA A
CB j
BB
ω
ωω
ω
ωω
ωω
ω
ωω ω
ωω ω
>> − =

==


=

<< − =


H
HH
H
H
HH

10
10
+20 log
dB
()The slope is the coefficient of 20 log, that is, = 20
decade
()The break frequency is the frequency at which , that is,
A
B
c slope
B
aC
C
ω
ωω



B= =

Ex. 13.4-4

1
oc
2
1
s
2
o 1
s2
()1 ()
1
1(
1
() 1
() 1
() 1
R
R
R
Rj CR
R
)
R jCR
ωω
ω
ω
ω
ω
ωω

=+

 
=+ 
+
 
== + 
+
VV
V
V
H
V

1
2
When 0.1 and 3,
4
then ()
1
10
R
RC
R
j
ω
ω
==
=
+
H

13-3

Ex. 13.4-5
a)

o2
2
oo 1
s1 o
12
2
1
2
2
12
1
1
1

1R +
1
1
where = 16.7 rad/s
1
and = 5.56 rad/s
( )
R
jC
jR
jC
RRj
jC
RC
RR C
ω
ω
ωω
ω
ω ω
ω
ω
=+
++
== =
++ +
=
=
+
Z
VZ
VZ

()
()
()
ss
o
s
t10cos20 or 100
20
1
16.7

20
1
5.56
1 1.20
0.417 24.3
1 3.60
vt
j
j
j
j
==
+
∠°
∴ =
+
+
== ∠−
+
V
V
V
°

b) So

o
o
4.17 24.3
() 4.17cos(2024.3) Vvt t
=∠ −°
=−
V
°

13-4

Ex. 13.4-6

Ex. 13.5-1

a)
7
o 3
2.5 10
8000 20
40 10
C
QRCR
L
ω
−
−
×
== =
×

b)
37
11
500 rads
20(4010) (2.510)
o
BW
Q QLC
ω
−−
== = =
××

Ex. 13.5-2
7
0
5
o 27 2 12
o
10
= = = 50
2 10
11 1
Now = 1 mH
(10) (10 10)
Q
BW
L
LC C
ω
ω
ω
−
×
⇒= = =
×

Ex. 13.5-3

4
o 1
35 2
4
o
43
o
1
1
= 10 rads
(10)(10)
10
= = 100
2 (15.9)
(10)(10)
= = = 0.1
100
LC
Q
BW
L
R
Q
ω
ω
π
ω
−−
−
==


=
Ω

13-5

Ex. 13.5-4
a)
o 62
3
o
26 2 10
o
11 100 pF
(10) (0.01)
10
1 = = = 10
(10)(10)
o
C
LC
BW
QR
BW RC
C
ω
ω
ω
ω
−
=⇒ = =
⇒= = Ω

b) 6
o
3
66
o
64
o
10
1000
10
11

1.0510 10
1 11000
10 1.0510
1

1 97.6
Q
BW
jQ j
j
ω
ωω
ωω
== =
==
 ×
+− +−

 
×   
=
+
H
H

13-6

Problems

Section 13-3: Gain, Phase Shift, and the Network Function

P13.3-1

2
2
2
1
||
1
R
R
jCj Cωω
=
+ R

()
()
()
2
2o
2i
1
2
2
12
1
1
1
p
R
jCR
R
R
jCR
R
RR
jCR
ωω
ω
ω
ω
ω
+
==
+
+
+
=
+
V
H
V
where Rp = R1 || R2.
When R1 = 40 W, R2 = 10 W and C = 0.5 F
()
0.2
14j
ω
ω
=
+
H
(checked using ELab on 8/6/02)

P13.3-2
()
()
()
()
2
o
i
12
2
12
1
1
1
1
R
jC
RR
jC
jCR
jCR R
ω ω
ω
ω
ω
ω
ω
+
==
++
+
=
++
V
H
V

When R1 = 40 kW, R2 = 160 kW and C = 0.025 µF

()
( )
()
10.004
10.005
j
j
ω
ω
ω
+
=
+
H

(checked using ELab on 8/6/02)

13-7

P13.3-3

()
()
()
2o
i1 2
2
12
12
1
R
RRj
R
RR
L
j
L
RR
ω
ω
ω ω
ω
==
++
+
=
+
+
V
H
V

When R1 = 4 W, R2 = 6 W and L = 8 H
()
()
0.6
10.8j
ω
ω
=
+
H
(checked using ELab on 8/6/02)

P13.3-4

()
()
()
2o
i2
22
2
2
1
1
Rj L
RR jL
L
j
RR
LRR
j
RR
ωω
ω
ωω
ω
ω
+
==
++

+


=
 +
 +

+

V
H
V

Comparing the given and derived network functions, we require

()
2
2
22 2
2
2 2
0.6
1
1
12
0.6 12
11
20
20
R
RR
L
j
j
RR R
LRR L
jj
RR RR
L
ωω
ω
ω

=
+
 
+
 + 


 =⇒ 
 +
 ++

+ +

=


=
Since R2 = 60 W, we have
60
5 H
12
L== , then ()()2056040R= −= Ω.
(checked using ELab on 8/6/02)
13-8

P13.3-5

2
2
2
1
||
1
R
R
jCj Cωω
=
+ R

()
()
()
2
2o
2i
2
2
2
1
1
1
p
R
jCR
R
R
jCR
R
RR
jCR
ωω
ω
ω
ω
ω
+
==
+
+
+
=
+
V
H
V

where Rp = R || R2.

Comparing the given and derived network functions, we require

2
2
2
2
0.2
0.2
11 4
4
p
p
R
R
RR
RR
jCR j
CR
ωω

=+ 
+=⇒ 
++

=


Since R2 = 2 W, we have
2
0.2 8
2
R
R
=⇒ =
+
Ω. Then
()()28
1.6
28
pR= =Ω
+
.
Finally,
4
2.5 F
1.6
==C .
(checked using ELab on 8/6/02)

P13.3-6

()
()
() ()()
()
()
()
i
a
o
i
oa
1
1
A
Rj L CR
L
jjA
RjC
ω
ω
ωω
ω
ωωωω
ω

= 
+ 
⇒=

 += 
 

V
I
V
V
VI

13-9

When R = 20 W, L = 4 H, A = 3 A/A and C = 0.25 F

()
() ()()
0.6
10.2jj
ω
ω ω
=
+
H

(checked using LNAP on 12/29/02)

P13.3-7

In the frequency domain, use voltage division on the left side of the circuit to get:
() () ()
1
1
1
1
1 1
Ci
jC
VV
jCR
R
jC
ω
iVω ωω
ω
ω
==
+
+

Next, use voltage division on the right side of the circuit to get:
() () () ()
3
23 1
2
2 3
31
oC C
A
R
VA V AV
RR jCR
iVω ωω
ω
== =
++
ω

Compare the specified network function to the calculated network function:

1
22
42 33
4 and 2000
1 1 2000 3 100
1
100
AA
AC
jCR jC
j
ω ωω
== ⇒ = =
++
+
1

Thus, C = 5 µF and A = 6 V/V.
(checked using ELab on 8/6/02)

13-10

P13.3-8
2
o
i1
2
1
2
1

()
()
()
=
1
R
jC
R
R
R
jCR
ωω
ω
ω
ω
== −

−


+
V
H
V

()
12
2
2
1
When 10 k, = 50 k, and = 2 F, then
R1
5 and = so
R1 0
1
10
RR C
RC
j
µ
ω
ω
=Ω Ω
−
==
+
H
5

P13.3-9

2
2o
i
1
1
2
22
1
11
1
()
() =
() 1

1

1
R
jC
R
jC
R
jCR
R
jCR
ωω
ω
ω
ω
ω
ω
=−
+
=−
+
V
H
V

21
12
1
()
1
Rj CR
Rj CR
ω
ω
ω
 +
=− 
+
 
H
1
2

12 1 2
2
11 22
1
When 10 k, 50k , 4 F and 2F,
11
then 5 , and
25 10
so
RR C C
R
CR CR
R
µµ=Ω =Ω = =
== =
1
25
() = 5
1
10
j
j
ω
ω
ω

+

−
+

H
13-11

()
2
2
1
625
()5
1
100
gain
ω
ω
ω
+
==
+
H

11
( )180tan tan
25 10
phaseshift
ω ω
ω
−− 
=∠Η=+ −
 
 




P13.3-10

3
1
3
3
3
1
1

1 1
R
RjC
R
jC jCR
R
jC
ω
ωω
ω
==
+
+

3
2
32 3 2
11 1

1
()

R
R
jCR RRjRRC
3
3R RjRRC
ωω
ω
ω
+
+ ++
=− =−
+
H

23
0
1
2
21

1
31 2
5 lim ()
2 = lim () 2 20 k
then 5 30 k
RR
R
R
RR
R
RR R
ω
ω
ω
ω
→
→∞
+
==
=⇒= =
=− = Ω
H
H Ω

P13.3-11
2
2
11
1
2
1
22
2 37
22
1
1
1
1
()
() 180tan 90
() 135 tan = 45 1
1
10 k
1010
10 = lim () 1 k
10
R
jCRjC
Rj CR
CR
CR CR
R
RR
R
R
ω
ωω
ω
ω
ωω
ωω ω
ω
−
−
−
→∞
+
+
=− =−
∠= °+ −°
∠= °⇒ °⇒ =
⇒= =
=⇒ ==Ω
H
H
H
H
Ω

13-12

P13.3-12
()
22
1
2
21
1
1
1
2

1 1
10 lim() 10
1
() = 18090tan
tan (270 ()) 44
10tan(270 ()) 10 10 k
= 100 k
Rj CR
jCR
jC
R
RR
R
CR
R
C
R
ω
ω
ω
ω
ω
ω
ωω
ω
ω
ω
−
== −
+
== ⇒ =
→∞
−
∠° +°−
°−∠
⇒= = ⋅ °−∠ = =
⇒Ω
H
H
H
H
H Ω

P13.3-13

()
2
2o
s
1
1
12
11 22
1
()
() =
1()
(1 ) (1 )
R
jC
R
jC
CRj
jRC jRC
ωω
ω
ω
ω
ω
ωω
=−
+
−
=
++
V
H
V

()
11
22
When 5 k, 1 F,
10 k and 0.1 F,
then
0.01
()
1 1
200 1000
RC
RC
j
jj
µ
µ
ω
ω
ωω
=Ω=
=Ω =
−
=
 
++ 
 
H




so
() ()
00 90
5001.66175
25000.74116
ω ωω∠
−°
°
°
HH

Then

() () ()
112 2
(0) 50 (1.66)30cos(500115175)(0.74)20cos(250030116)
o
=49.8cos(50070)14.8cos(2500146) mV
When =5 k, =1 F, 10 k and 0.01 F, then
vt t t
tt
RC R Cµµ
= + +°+° − +°+°
−°− +°
Ω= Ω =

13-13

()=0.01
1 1
200 10,000
j
jj
ω
ω
ωω
−

++
 
 
H
So
() ()
00 90
5001.855161
25001.934170
ω ωω∠
−°
−°
°
HH

Then

() () ()()(0)50(1.855)30cos(500115161)(1.934)20cos(250030170)
o
55.65 cos(50046)38.68cos(2500190) mV
vt t t
tt
=+ +°−°− +°+
=− °− +°
°

P13.3-14
a)
s
o
o
s
2 V
(8 div)
div
= = 8
2
2 V
(6.2 div)
div
= = 6.2 V
2
6.2
gain == 0.775
8






=
V
V
V
V
V

b)
o
s
2
22 2
1
() 1
()
1() 1
11
Let =() then 1
1
jC
jCR
R
jC
gC
RgCR
ω ω
ω
ωω
ω
ω
ωω
== =
+
+

==

+ 
V
H
V
H
1
−

In this case = 25003142 rads,
()0.775 and 1000 so 0.26 F.RC
ωπ
ω µ
⋅=
== Ω =H

c)
1 tan( ())
()=tan so RC
RC
ω
ωω ω
− −∠
∠− =
H
H
Recalling that R = 1000 Ω and C=0.26µF, we calculate

13-14

() ()
2(200)0.95 18
2(2000)0.26 73
ω ωω
π
π
∠
−°
−°
HH

( )
()()
tan 45
()=45 requires = 3846 rad
6
1000.2610
tan ((135))
()135 requires = =3846 rads
6
(1000)(0.2610)
sωω
ωω
°
−−

∠− ° =
−
×
−−°
∠= −° −
−
×
H
Η

A negative frequency is not acceptable. We conclude that this circuit cannot
produce a phase shift equal to −135
°.

d)
tan ((60))
0.55F
(2500) (1000)tan (())

tan ((300))
0.55F
(2500 ) (1000)
C
C
R
C
µ
πω
ω
µ
π
°
°
 −
==

⋅−∠ 
=⇒ 
−−
==
 ⋅
H
−

A negative value of capacitance is not acceptable and indicates that this
circuit cannot be designed to produce a phase shift at −300
° at a frequency of
500 Hz.

e) tan((120))
0.55 F
(2500)(100)
C µ
π
°
−−
== −
⋅

This circuit cannot be designed to produce a phase shift of −120
° at 500 Hz.

13-15

Section 13-4: Bode Plots

P13.4-1
520
201
205
()= 5 505
1
50
20
5 50
200
50
j
j
j
j
j
ω
ω
ω
ωω
ω
ω
ω
ω



<




+ 
 
≈ <<

+




<
 =




H

P13.4-2
()
12
11
55
() 10
11
50 50
j j
j j
ω ω
ωω
ω ω
++
==
++
HH

Both H1(ω) and H2(ω) have a pole at ω = 50 rad/s and a zero at ω = 5 rad/s. The slopes
of both magnitude Bode plots increase by 20 dB/decade at ω = 5 rad/s and decrease by 20
dB/decade at ω = 50rad/s. The difference is that for ω < 5rad/s

12
()10 dBand ()1020 dBωω==HH

13-16

P13.4-3
2
22
12
11 22
1
1
1
()
1 (1 )(1 )
R
jjCR
CR
jRC jRC
R
jC
ωω
ω
ωω
ω
+
=− =−
++
+
H

This network function has poles at

12
11 22
11
2000rads and 1000radspp
RC RC
== = =
so
12 1
2

12 1 2
11 1
12 2
11 22 21
()
() () 2
1
()
() ( )
CRj p
jR
CR p p
jCRR
j
CR p
jCRjCRjCR
ωω
ω
ωω
ω
ω
ω
ωω ω


<


==<


<
= >

H

13-17

P13.4-4
2
21 1 222
1 12 2 1 11
11
(1 ) 1 11
() so , and
(1 )
1
R
Rj CR RjCR
Kz p
R
22RjCR R CR CR
jCR
ωω
ω
ω
ω
++
=− =− =− = =
+
+
H

When z < p

When z > p

P13.4-5

Using voltage division twice gives:

()
() () ()
2
222
2 112 1 2 1 2
1
2
12
1
i
jLR
Rj L jLR Lj
jLR RRRj LRR LRR
R
j
Rj L
RR
ω
ωωω ω
ωω ω
ω
ω
+
== =
++ +
+
+
+
V
V

and
()
()
44
34 4 3 4
4323 4 34
3
34 3
1
o
RA
Rj CR AR RR
A
4
4
R
R CRRRR jCRR
Rj
RjCR RR
ωω
ωω
ω
ω
++
== =
++
++
++
V
V

Combining these equations gives

13-18

()
()
() () ()
4
13 4 12 34
12 3 4
11
o
i
ALR j
RR R LR R CRR
jj
RRR
ω ω
ω
ω
ωω
==
+ + 
++ 
 +

V
H
V
R

The Bode plot corresponds to the network function:

()
12
1111
200 20000
kj kj
jjjj
pp
ω ω
ω
ωωωω
==
   
++++    
 
  
H




()
1
1 1
1
12
2
12
11
1
kj
kj p
kj
kp p p
j
p
kj kpp
p
jj j
pp
ω
ωω
ω
ωω
ω
ω
ω
ωω ω




=≤

⋅


≈= ≤
 ⋅


 =≥

⋅


H
2
≤

This equation indicates that |H(ω)|=k p1 when p1 ≤ ω ≤ p2. The Bode plot indicates that
|H(ω)|=20 dB = 10 when p1 ≤ ω ≤ p2. Consequently

1
1010
0.05
200
k
p
== =
Finally,
()
0.05
11
200 20000
j
jj
ω
ω
ωω
=
 
++ 
 
H




Comparing the equation for H(ω) obtained from the circuit to the equation for
H(ω)obtained from the Bode plot gives:

()
4
13 4
0.05
ALR
RRR
=
+
,
()
12
12
200
RR
LRR
=
+
and
34
34
20000
RR
CRR
+
=

Pick L = 1 H, and R1 = R2 , then R1 = R2 = 400 Ω. Let C = 0.1 µF and R3 = R4 , then R3 =
R4 = 1000 Ω. Finally, A=40. (Checked using ELab 3/5/01)

13-19

P13.4-6
From Table 13.4-2:
()
2 3
2
1
32 dB40 401010400 k
R
kR
R
== = = × = Ω
()()
2
3
22
11
400 rad/s 6.25 nF
40040010
pC
CR
== ⇒ = =
×

()()
1
3
11
11
4000 rad/s 25 nF
40001010
zC
CR
== ⇒ = =
×

P13.4-7

()
()
()
2o
i2
22
2
2
1
1
Rj L
RR jL
L
j
RR
LRR
j
RR
ωω
ω
ωω
ω
ω
+
==
++

+


=
 +
 +

+

V
H
V

()
() ()()
()
0.2
0.210.25 1
4
10.05 0.25
1
20
0.05
k
j
z
j
p
ω
ω
ω

 =

+ 
= ⇒= =
+


==


H

P13.4-8
• The slope is 40dB/decade for low frequencies, so the numerator will include the
factor (jω)
2
.
• The slope decreases by 40 dB/decade at ω = 0.7rad/sec. So there is a second order
pole at ω 0 = 0.7 rad/s. The damping factor of this pole cannot be determined from the
asymptotic Bode plot; call it δ 1. The denominator of the network function will
contain the factor
2
112
0.70.7
j
ωω
δ

+− 


• The slope increases by 20 dB/decade at ω = 10 rad/s, indicating a zero at 10 rad/s.
13-20

• The slope decreases by 20 dB/decade at ω = 100 rad/s, indicating a pole at 100 rad/s.
• The slope decreases by 40 dB/decade at ω = 600 rad/s, indicating a second order pole
at ω 0 = 600rad/s. The damping factor of this pole cannot be determined from an
asymptotic Bode plot; call it δ 2. The denominator of the network function will
contain the factor
2
212
600600
j
ωω
δ

+− 


2
22
12
(1)()
10
()
12 12 1
0.70.7 600600 100
Kj j
jj j
ω
ω
ω
ωωω ω
δδ
+
=
  
 
+− + − +   
  
 
  
H
ω



()To determine , notice that 0 dB=1 when 0.7 < < 10. That isK ωω=H

2
2
2
(1)()
1( 0.7)
(1)(1)
0.7
Kj
KK
ω
ω
== ⇒

−


2=

P13.4-9
(a)
1
()
2
() 1
2
() 20 log 110
2
20 log 20log 20 log 110 10 10
Kj
z
j
K
z
K
dB
z
K
z
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω

+


=

=+



=+ 


=− + +


H
H
H

10 10
10
Let () 20 log 20log
and () 20 log
()
_Then () ~
()
L
H
L
H
dB K
K
dB
z
dB z
dB
dB z
ω ω
ω
ωω
ω
ωω
=−
=
 <<


>>
H
H
H
H
H

13-21

LHSo () dB and () dB are the required low and high-frequency asymptotes.ωωHH

The Bode plot will be within 1% of |H(ω)| dB both for ω << z and for ω >> z. The range
when ω << z is characterized by

()
L() 0.99 (gains not in dB)ωω=HH
or equivalently

()
10 L
2
10 10 10
2
10 10
2
20 log0.99 = () dB() dB (gains in dB)
20log 20log 20log 1
1
20log1 20log
1
K
K
z
z
z
ωω
ω
ω
ω
ω
ω
−

=− − +


=− + =



+

HH

Therefore
2
2
11
0.99 1 0.14
.99 7
1
z
zz
z
ω
ω

=⇒ = −=



+


−

The range when ω >> z is characterized by

H
().99() (gains not in dB)ωω=HH
or equivalently

13-22

10 H
2
10 10 10
2
10 10
2
20 log0.99 = () dB() dB (gains in dB)
20 log 20 log20 log 1
1
20 log 1 20 log
1
K
Kz
z
z
z
z
ωω
ω
ω
ω
ω
ω
−

=− − +



=− + =



+


HH

Therefore
2
2
1
1
.99 0.14
1
1
.99
zz
zω
ω

=− ⇒ = =



−

7
z
−

z
The error is less than 1% when < and when >7z.
7
ωω

P13.4-10
tt0
1s
tt1
1
t1 t
1t 1t 1t 1t
1t
()
()
() 1
1
(1 ) 1

1
RR
R
RRR
jCRjC
RjCR R jCR
RR jCRR RR CRR
j
1
RR
ω
ω
ω
ωω
ωω
ω
ω
== =
++
+
++
== 
++ + 

+
+

V
H
V

When
1t1 k, 1 F and 5 kRC Rµ=Ω = =Ω
5
1000
6
1
551000
() () 10001200
6 61000
1
1200 1 1200
j
j
j
ω
ω
ω
ωω
ω
ω

<


+
 
=⇒ ≅ < 
+
  <


HH ω<

13-23

P13.4-11

Mesh equations:

in 1 1 2 2
o2 2
()() [( )( )]
() () [( )]
RjLjM jMjL R
jMjLR
ωωω ω ωω
ωω ωω
=+ − +− + +
=− + +
VI
VI

Solving yields:
0 22
in 1 2 1 2
V() ()
()
V() ( 2)
Rj LM
RRj LLM
ω ω
ω
ωω
+ −
==
++ +−
H

Comparing to the given Bode plot yields:

2 2
12
0
12 1 2
21 2
21 2
lim lim
|()| 0.75 and |()| 0.2
2
333 rads and 1250 rads
2
LM R
KK
LL M RR
RR R
zp
LM LLM
ω
ω
ωω
→∞
→
−
== = = =
+− +
+
== = =
−+ −
HH =

13-24

P13.4-12
11 112
12 12
1
1
1 1(1
()
1
1
)jRC jRCjC
jRC RC j
R
jC
ωωω
ω
ωω
ω
++
=− =− =−H
()
12 11
1
11
12 2 11
11 1

()
11

RCj RC
C
RC
RCC
ω
ω
ω
ω
 
−< 
 
−

−= − >


H
RC

With the given values:
1
112
11
6 dB, 4000 rad/s
2
C
RCC
==− =

13-25

P13.4-13
Pick the appropriate circuit from Table 13.4-2.

We require
1
11 22 2
11
200 , 500 and 14 dB5
pC
zp k
CR CR zC
== == == =
12 1 2Pick 1 F, then 0.2 F, 5 k and 10 k.CC R Rµ µ== =Ω =Ω

P13.4-14
Pick the appropriate circuit from Table 13.4-2.

We require
2
21
1
500 and 34 dB50
R
p
CR R
== ==
21Pick 0.1 F, then 20 k and 400 .CR Rµ== Ω = Ω

13-26

P13.4-15
Pick the appropriate circuit from Table 13.4-2.

We require
1
11 22 2
11
500 , 200 and 14 dB5
pC
zp k
CR CR zC
== == == =
12 1 2Pick 0.1 F, then 0.05 F, 20 k and 100 k.CC R Rµ µ== = Ω = Ω
13-27

P13.4-16

Pick the appropriate circuit from Table 13.4-2.

We require
12
11 22
11
200 , 200 and 34 dB50
12p pk
CR CR
== == ===CR
12 1 2Pick 1 F, then 0.04 F, 5 k and 50 k.CC R Rµ µ== =Ω = Ω

P13.4-17
10(1 50)
()
(1 2)(1 20)(1 80)
j
jj j
ω
ω
ωω ω
+
=
++ +
H

13-28

() () () ( ) ()( )
11 1 1
tan 50tan 2tan 20tan 80ϕω ω ω ω ω
−− − −
=∠ = − + +H

P13.4-18

(a)

(b)

(c)
21
2
()
()
() 1
10

1
10,000
o
s
RR
jRC
j
ω
ω
ωω
ω
== −
+
=−
+
V
H
V
10 = 20 dB

10,000 rad/s

13-29

P13.4-19

2
oa
o1 1 22 11o
2
as
1a o
1
1
() ()
1
(1 )(1 )
() ()
0 (()())
jC
R
jCR jCR jCR
jC
jC
R
ω
ωω
ωω ω
ω
ωω
ωω ω


=

+
⇒+ + = +

−
=+ −


VV
VV
VV
VV
sV

()
o
22
s2 2 1212
() 11
()
10CRj CCRR j.81
ω
ω
ω ωω ω ω
== =
+ −− +
V
T
V +

This is a second order transfer function with 0 and 0.4
oω δ= =.

13-30

Section 13-5: Resonant Circuits

P13.5-1
0
6
11
60 kradsec
11
10
12030
LC
x
ω
−
== =
 
 
 

61
10
30
10 ,000 20
1
120
C
QR
L
−
×
== =
22
22
00 0 0
10 2 058.52krads and 61.52krads
22 22QQ QQ
ωω ωω
ωω ω ω
 
=− + += =+ +=
 
 

()
6
11
3 krads
1
10000 10
30
BW
RC −
== =

×



Notice that
0
21BW
Q
ω
ωω=− =.

P13.5-2
2
2 0
0
()
1
k
Q
ω
ωω
ωω
=

+−

H
so
00 3
8
() 400 and 1000 rads
2010
Rk ωω
−
== = = Ω =
⋅
H

3
4
At 897.6rads, () 200, so
20.10
ωω
−
== H =

2
2
400
200 8
897.61000
1
1000897.6
Q
Q
=⇒

+−

=
Then
0
1
1000
20 F

50 mH
400 8
LC C
L
C
Q
L
ω
µ

==

 =
⇒
=

==



13-31

P13.5-3
54
0
11
10rads, 10, 10rads
LR
QB W
RC LLC
ω== = = ==

P13.5-4
43
0
11
10rads, 10, 10rads
LR
QB W
RC LLC
ω== = = ==

P13.5-5
()
0
100 Rω= =ΩZ
1
500 20 F
100
BW C
C
µ== ⇒ =
()
0
6
1
2500 8 mH
2010
L
L
ω
−
== ⇒ =
⋅

P13.5-6
()
0
1
100 R
ω
= =Ω
Y

100
500 0.2 HBW L
L
== ⇒ =
()
0
1
2500 0.8 F
0.2
C
C
ω µ== ⇒=

13-32

P13.5-7

()
() ()
()
() () ()
12
2
12 2 12 1
121
22 2
11 22 12 1 12 1 22
22
21
11

()

()
jC
RjLR
RR CLRjLCRRRjL
RjLRR jL
RRR CLR LLCRRjRLCRRjLRR CLR
RR L
ωω
ω
ωω ω
ωω
ωω ω ω ω
ω
=+ +
+
+− + + −
=×
−+
+− − + + − +−
=
−
Y

()
0
is the frequency at which the imaginarypart of is zero:ωω ω= Y

() ()
2
21 22
11 2 1 2 0 2 0 2
2
0 12.9Mradsec
LRCRR
RLCRRLRR CLR
CLR
ωω
−
−+ − =⇒ = =

13-33

P13.5-8

(a) Using voltage division yields

()
()( )
()( )
()
o
100100
100100
10000
100100
100
100100
5
10
1002135 2135
10000 100090 V
1002135100502135
j
j
j
j
j
j
−
−°
=∠
−
+
−
∠−°° ∠−°
=∠ = = ∠°
∠−°+ ∠−°
V

|| 1000 V
o
∴ =V

(b) Do a source transformation to obtain

This is a resonant circuit with
0
1 400LCω= =rad/s. That’s also the frequency of the input, so
this circuit is being operated at resonance. At resonance the impedances of the capacitor and
inductor cancel each other, leaving the impedance of the resistor. Increasing the resistance by a
factor of 10 will increase the voltage Vo by a factor of 10. This increased voltage will cause
increased currents in both the inductance and the capacitance, causing the sparks and smoke.

13-34

P13.5-9
Let
2
2
1
G
R
=. Then
() ()
1
2
2
12 2 1
2
1
1
Rj L
Gj C
RGL CjLG C
Gj C
ω
ω
ωω ω
ω
=+ +
+
+− + +
=
+
Z
R

At resonance, ∠= so 0°Z

()
11 21
2
212
tan tan
1
LGC R
GRG LC
Cωωω
ω
−− +
=
+−

so
()
2
21 222
222
212
and
1
LG CR CLGC
CG L
GL CRG LC
ωω ω
ω
ω
+−
=⇒ = >
+−

12 0With 1 and 100 radsRR ω== Ω = ,
2 4
0 2
10
CL
LC
ω
−
== . Then choose C and calculate L:
10 mF 5 mHCL= ⇒=
Since , we are done.
2
2CG L>

13-35

P13.5-10

(a)
()
2
in

1 1
R
RRLCjLjC
jL
jRC
R
jC
ω ωω
ω
ω
ω
−+
=+ =
+
+
Z

Consequently,
() ()
()
2 2
2
2
||
1
in
RRLC L
RC
ωω
ω
−+
=
+
Z

(b) (c)

2
11
||
1
in
LC CR C
LL
ω=⇒ =

+


Z

P13.5-11

() ()
2
Let 0 and ABω ω=∠ =∠V V θ. Then

()
()
()
() ()
()
2
cos sinAB AB jB
R
AR AR
ωω
ω θ θθ
ω
ωω
−
−∠ − −
== = =
VV
I
Y
VV

()
() ( )
22
cos sin
||
AB B
AR
θ θ
ω
−+
=Y
13-36

PSpice Problems

SP13.1

Here are the magnitude and phase frequency response plots:

From the magnitude plot, the low frequency gain is k = 200m = 0.2.

From the phase plot, the angle is -45° at ( )239.891251 rad/spπ== .
13-37

SP13-2
Here is the magnitude frequency response plot:

The low frequency gain is ()
0
0.6lim k k
ω
ω
→
== ⇒H0.6 . =
The high frequency gain is () ()lim
p
kz
zω
ω
→∞
== ⇒ =H10 .6p
At , ()22.815717.69 rad/sωπ= =

()
()
2
2
22
22
2
17.69
1
0.6 16 869
0.80.6
9 31317.69
1
16
313 869
9
0.77778 312.56
20 rad/s
12 rad/s
p p
p
p
pp
p
p
z

+

+
=⇒ =
+
+


⇒+ =+
⇒=
⇒=
⇒=

13-38

SP13-3

From the magnitude plot, the low frequency gain is k = 4.0.

From the phase plot, the angle is -45° at ( )215.998100.5 rad/spπ= = .
13-39

SP13-4

From the magnitude plot, the low frequency gain is k = 5.0.
From the phase plot, the angle is 180°-45°=135° at ( )21.58499.958 rad/spπ== .

13-40

SP13-5
()
()
44
14
4
2
4
10 10
() tan 10
11 0
11 0
RR
C
jC
C
ωω
ω
ω
−
=− = ∠−
+
+
H

When ω = 200 rad/sec = 31.83 Hertz

()
()
4
14
2
4
10
1.8565158 tan 10
11 0
R
C
C
ω
ω
−
∠°= ∠−
+

Equating phase shifts gives

4
43
4
10
1010 tan(22)0.404 0.2F
10
CR
CC
R
ω µ== °= ⇒ =
+

Equating gains gives

() ()
44
22
4
10 10
1.8565 5k
10.40411 0
RR
R
Cω
== ⇒
++
=Ω

SP13-6
4
44
444
2 1
44 4
2
4
44
4
10 10 10
1 1010 10
() tan
10 10 10
101
1
11 0 10
10
jCR CRRR
CR R
CRRj
jC R
R
ω
ωω
ω
ω
ω
−
+ ++
== = ∠− 
+++
+
++
+
H

When ω = 1000 rad/sec = 159.1 Hertz

4
44
1
4
2
4
4
10
1010
0.17140859 tan
10
10
1
10
CRR
R
CR
R
ω
ω
−+
∠−°= ∠− 
+
+

+

Equating phase shifts gives

44
3
44
10 10
10 tan(59)1.665
10 10
CR CR
RR
ω == °=
++

Equating gains gives
13-41

()
44
44
22
4
4
10 10
10 10
0.171408 20k
11.66510
1
10
RR
R
CR
R
ω
++
== ⇒
+
+

+
= Ω

Substitute this value of R into the equation for phase shift to get:

( )
()
34
4
33
4 34
20101010
1.66510 10 0.25F
10 201010
CCR
C
R
µ
×
== ⇒ =
+ ×+

Verification Problems

VP13-1
When ω < 6300 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 200 and 400 rad/s.

When ω > 6300 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 12600, 25000, 50000 and 100000 rad/s.

At ω = 6300 rad/s, we expect | H(ω)| = −3 dB = 0.707. This agrees with the tabulated value of |
H(ω)| corresponding to ω = 6310 rad/s.

At ω = 630 rad/s, we expect | H(ω)| = −20 dB = 0.14. This agrees with the tabulated values of |
H(ω)| corresponding to ω = 400 and 795 rad/s.

This data does seem reasonable.

VP13-2
010,000
143 71.4 rads
70
BW
Q
ω
== = ≠ . Consequently, this report is not correct.

VP13-3
0
11
10 krads1.59 kHz, 20 and 500 rads79.6 Hz
LR
QB W
RC LLC
ω== = = = == =
The reported results are correct.

13-42

VP13-4
The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode
plot indicates a pole at 200 rad/s and a zero at 800 rad/s. Consequently, the Bode plot and
network function don’t correspond to each other.

Design Problems

DP13-1
Pick the appropriate circuit from Table 13.4-2.

We require

2
1
11 22 21
11
21000 , 210000 , 2 and 5
R pC
zp k
CR CR zCR
ππ×< = × >= == = =k

1
Try 22000. Pick 0.05 F.zCπ µ=× = Then

11
12 1 2
1
1
1.592 k, 2 3.183 k and 0.01 F
2
CC
RR R C
pCz
k
z
µ== Ω = = Ω = ==
22
1
Check: 31.42 krads210,000 rads.p
CR
π== <⋅

13-43

DP13-2

o
2
s
11
||
() 1
()
11() 1
||
1
R
R
jC jCR LC
R
jL jjL R
jCR RCLCjC
ω ωω
ω
ω
ωω ωω
ωω
+
== = =

+− ++

+

H
V
V
+

Pick
3
0
1
2(10010) rads
LC
ωπ== ⋅ . When
0ωω=
0
1
()
11 1
LC
j
LC RCLCLC
ω=
−+ +
H
1

So
0
()
C
R
L
ω=H . We require
0
3 dB 0.707 () 1000
CC
R
LL
ω−= = = =H

Finally
31
2(10010)
1.13 nF
2.26 mH
0.7071000
LC C
L
C
L
π

=⋅

=
⇒
=

=



13-44

DP13-3

1

2

3

4

5

6

1

2
10 k
866 k
8.06 k
1 M
2.37 M
499 k
0.47
0.1
R
R
R
R
R
R
CF
CF
µ
µ
=Ω
=Ω
=Ω
=Ω
=Ω
=Ω
=
=

Circuit A
33
ac s 1c
21
RR
RR
=− − =− −VV V HVH
2sV

Circuit B
5
4
oa
151
R
R
jCRω
=− =−
+
VV
3aHV

Circuit C
co
26
1
jCRω
=− =−VV
4oHV
a

Then
c3 4=VH HV
2
a2 s 134a a
13 41
s
−
=− − ⇒ =
+
H
VH VHHHV V
HHH
V
23
o3 a
13 41
=− =
+
HH
VH V
HHH
sV
After some algebra
3
14 1
os
3 2
246 12 51
R
j
RRC
R
j
RRR CC RC
ω
ω
ω
=
−+
VV

This MATLAB program plots the Bode plot:

R1=10; % units: kOhms and mF so RC has units of sec
R2=866;
R3=8.060;
R4=1000;
13-45

R5=2370;
R6=449;
C1=0.00047;
C2=0.0001;

pi=3.14159;
fmin=5*10^5;
fmax=2*10^6;
f=logspace(log10(fmin),log10(fmax),200);
w=2*pi*f;

b1=R3/R1/R4/C1;
a0=R3/R2/R4/R6/C1/C2;
a1=R5/C1;

for k=1:length(w)
H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1);
gain(k)=abs(H(k));
phase(k)=angle(H(k));
end

subplot(2,1,1), semilogx(f, 20*log10(gain))
xlabel('Frequency, Hz'), ylabel('Gain, dB')
title('Bode Plot')
subplot(2,1,2), semilogx(f, phase*180/pi)
xlabel('Frequency, Hz'), ylabel('Phase, deg')

13-46

DP13-4
Pick the appropriate circuits from Table 13.4-2.

We require
4
12 21 1 2
31 1
11
10 , 200 and 500
R
kkRC p p
24
R RC CR
=− = == ==
11
11
1
Pick 1 F. Then 5 k.CR
pC
µ== =Ω
24
22
1
Pick 0.1 µF. Then 20 k.CR
pC
= == Ω
Next
22 63
33
10 (10)(2010) 500
RR
RR
−
=⋅ ⇒ =
Ω

23Let 500 k and 1 k.RR=Ω =

13-47

DP13-5
Pick the appropriate circuits from Table 13.4-2.

We require
4
12 21 1 2
31 1
11
20 dB10 , 0.1 and 100
R
kkRC p p
24
R RC CR
==−= == ==
11
11
1
Pick 20 F. Then 500 k.CR
pC
µ== = Ω
24
22
1
Pick 1 µF. Then 10 k.CR
pC
= == Ω
Next
22 63
33
10 (2010)(1010) 50
RR
RR
−
=⋅ ⋅ ⇒ =
23Let 200 k and 4 k.RR=Ω =Ω

13-48

DP13-6
The network function of this circuit is
2
3
1
1
()
1
R
R
jRC
ω
ω
+
=
+
H
The phase shift of this network function is
1
1tanRCθω
−
=−
The gain of this network function is ()
()
33
22
22
1
11
1( ) 1tan
RR
RR
G
RC
ω
ω θ
++
== =
+ +
H

Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is
calculated from
1
tan()
R
C
θ
ω
−
= . Next pick R2 = 10kΩ (a convenient value) and calculated R3 using
2
32
(1 (tan)1)RGR θ=⋅ + −⋅. Finally

12 345 deg, 2, 1000 rads 10 k, 10 k, 18.284 k, 0.1 µFGR R R Cθ ω=− = = ⇒=Ω =Ω = Ω=

DP13-7
From Table 13.4-2 and the Bode plot:
16
1
1
800 2.5 k
(0.510)
zR
R
−
== ⇒= Ω
×

2

2
1
32 dB 40 100 k
R
R
R
==⇒ = Ω
3
2
1 1
200 0.05F
(200)(10010)
pC
RC
µ== ⇒ = =
×

(Check:
66
6
0.510 0.510
dB 10
0.0510
p
k
zC
−−
20
−
××
== = =
×
)

DP13-8
22
1
()
1 1
1
R jCR
jCR
jC
ω
ω
ω
ω
−
== −
+
+
H
1
11 6
tan(270195)
19518090tan 37.3 k
(1000)(0.110)
CR Rω
−
−
°−°
°=+− ⇒ = = Ω
×
2
21
1
10 () 10373 klim
R
RR
Rω
ω
→∞
== ⇒ = = ΩH

13-49

Chapter 14: The Laplace Transform

Exercises

Ex. 14.3-1
()
1
cos and
2
jt jt
atee
ft t e
sa
ωω
ω
+−
+
== =

−
L

()
22
11 1
[cos]
2
s
Fs t
sj sj s
ω
ωωω

== + =

− ++
L

Ex. 14.3-2
2
22
2 2
311
() [sin] [sin]
21 (2)(1
tt ss
Fs e t e t
ss ss
−− ++
=+ = + = + =

++ ++
LL L
)

Ex. 14.4-1
44 23
() [2()3()]2 [()]3 [()]
4
tt
Fs uteut ut eut
ss
−−
=+ = + =+
+
LL L

Ex. 14.4-2
22
2
1
() [sin(2)(2)] [sin]
1
ss
Fs tut e te
s
−− 
=− −= = 
+
LL

Ex. 14.4-3
1 22
1
11
()[][ ]
(1)

t
ss
ss
Fs te t
ss
−
→+
→+
== = =
+
LL

Ex. 14.4-4

() () () (
55
54 .2
33
ft tut t ut
  
=− + −−− −
  
  
)4.2

()
()
4.2
4.2
22
155 155 5
33 3
s
s
se
Fs e
ss s s
−
−
+−  
=− +− − =  
  
2

14-1

Ex. 14.4-5
2
2
2
00
0
3 3(1 )
() () 3
st s
st st e e
Fs ftedt edt
ss
− −
∞
−− −
== = =
−
∫∫

Ex. 14.4-6

()
52 0 2
0otherwi
tt
ft
<<
=
 se

() ()() () () ()()()
() ()
22
22
22 2
55 5 5
2 2 2 22(2
22 2 2
51 2 51
12
22
ss
ss
ft tutut tuttut tuttut ut
ee
Fs ft e se
ss s s
−−
−−
=− −= − −= −− −− 
 


)−

∴ == −− = −−
 

L

Ex. 14.5-1
()
() [] () () () () (
22 1
22
,t an
cos sin cos cos
jj
at at at
cjd cjd me me
d
Fs wheremcd
csajs ajsajs aj
)ftec td tute cd tutme tut
θθ
θ
ωω ωω
ω ω ωθ ωθ
−
−− −
+−
=+ =+ =+ =
+− ++ +− ++

∴ =− = + + = +


Ex. 14.5-2
(a)
()
()
()
()()
()()
() ( )()
22
22
1
2
28383 1
4132 29
382
2, 8, 3 & 3 6.33
3
6.33
tan 38.4, 86.3310.2
8
10.2cos338.40
t
ss
Fs
ss s
ac cad d
m
ft e t ut
ωω
θ
−°
−
−−
== ×
++ ++
−
∴== = −=−⇒= =
−

∴== = + =


⇒= +

14-2

(b)
() ()
()()
()
()()
() () () ()()
1 222
1
11 1
2333 1
Given , first consider .
217 2172 116
Identify 1, 0, 4 3 34. Then ||34, tan34090
So ()(34)sin4 . Next, 1. Finally
s
ts
e
Fs Fs
ss ss s
ac andd d md
ft e tut FseFsftft
ωω θ
−
−°
−−
== =×
++ ++ ++
=== −=⇒=− == = − =−
== ⇒ =−

() ()()
(1)
(34) sin41 1
t
fte tut
−−
∴ =− 

−

Ex. 14.5-3
(a)
()
() ()
2
22
5
111
sA B C
Fs
ssss s
−
== ++
++ +

where
() () ()
2
01
51
| 5 and 1 | 4
11
ss
AsFs CsFs
==
5
−
− −
== =− =+ =
−
=
6B

Multiply both sides by ()
2
1ss+

() ()
2
2
55 1 14ss Bss s−=−++ ++ ⇒ =
Then
()
()
2
56 4
1 1
Fs
ss s
−
=+ +
+ +

Finally
()( )()56 4
t t
fte te
−−
=−+ + ut

(b)
()
() () () ()
2
32
4
333
sA B C
Fs
sss
== + +
+++
3
3s+

Where
() () () ()
2
23
2
33
1
34 , 3
2 ss
dd
As Fs B sFs
ds ds=− =−
  =+ = = + =
  
24−
6

and
() ()
3
3
33
s
Cs Fs
=−
=+=
Then
()
() () ()
23
42 4 36
3 33
Fs
s ss
−
=+ +
+ ++

Finally
()( ) ()
23
42418
t
ftt te
−
=− + ut

14-3

Ex. 14.6-1
(a) ()
2
65
21
s
Fs
ss
+
=
++
() ()
( )
() ()
()
2
2
65
0lim l im
21
65
lim li m 0
2100
ss
fs Fs
ssss
ss
fs Fs
ssss
+
==

++→∞ →∞
+
∞= = =

++→→ 
6=

(b) ()
2
6
21
Fs
ss
=
−+
()
()
2
2
0
6
0lim 0
21
6
lim undefined no final value
21
s
s
s
f
ss
s
f
ss
→∞
→

==

−+

∞= = ⇒

−+

Ex. 14.7-1

KCL:
1 6
7
5
t
v
ie
−
+=
KVL:
1 143 0 4
di di
iv v i
dt dt
3+−= ⇒ = +
Then
66
43
35
72
54
tt
di
i
didt
ie i e
dt
− −
+
+= ⇒ +=

Taking the Laplace transform of the differential equation:

351 35 1
()(0)2() ()
46 4(2)(6
sIsi Is Is
ss
−+ = ⇒ =
)s+ ++

Where we have used . Next, we perform partial fraction expansion. (0)0i=

2
11 1
where and
(2)(6) 2 6 6 4 2 4
ss
AB
AB
ss s s s s
=− =−
=+ = = = =−
++ + + + +
6
1 1

Then
2635 35351351
() ()
162166 16 16
tt
Isi t
ss
e e
− −
=− ⇒ = −
++

14-4

Ex. 14.7-2

Apply KCL at node a to get

12 1 1
12
1
22
48 24
dvvv dv
vv
dt dt
−
=⇒ + =

Apply KCL at node b to get

22 1 2 2 2
12
50cos2 1
03 60
20 243024
vt vvv dv dv
vv
dt dt
−−
++ + =⇒−++ =cos2t

Take the Laplace transforms of these equations, using
1 2
(0)10 V and (0)25 Vv v= = , to get

() ()
2
12 1 2 2
2560100
2 ()2()10and ()3 ()
4
ss
sVsVs Vs sVs
s
++
+− = − ++ =
+

Solve these equations using Cramer’s rule to get

()
()
()
()()
()()()
( )
()()()
2
22
2
2 2
32
2
2560100
21 0
2256010010 44
2(3)2 41 4
25120220240
41 4
ss
s
ss s ss
Vs
ss ss s
ss s
ss s
++
++

+ ++ + ++
==
++ − ++ +
++ +
=
++ +

Next, partial fraction expansion gives

()
*
2
V
22 1 4
A AB C
s
sj sjss
=+ ++
+ −+ +

where
()()( )
() ()
() ()
32
2
*
32
2
1
32
2
4
25120220240 240240
66
14 2 40
66
25120220240 11523

15344
25120220240 32016

60341
sj
s
s
ss s j
Aj
ss sj
Aj
ss s
B
ss
ss s
C
ss
=−
=−
=−
++ + −−
==
++ − −
=−
++ +
== =
++
++ + −
== =
−++
=+

Then
14-5

()
2
66 66233163
22 1
jj
Vs
sj sjs s4
+ −
=+ + +
+ −+ +

Finally
4
2
23 16
() 12cos212sin2 V 0
33
tt
vt t te e t
−−
=+ + + ≥

Ex. 14.7-3
Taking Laplace Transform of the differential equation:

() () () ()() ()
2' 10
00 5 06
3
sFssf f sFsf Fs
s
=− + − + =

+

Using the given initial conditions

2
2 10 21640
(5 6) () 210
33
ss
ss Fs s
ss
++
++ = ++=
++

()
()()() () () ()
2
2
21640
323 3 23
ss A B C
Fs
ss s s ss
++
== +
++ + + ++
+

where . Then 10, 14,and 16AB C=− =− =

()
33 2
2
10 1416
() ()10 14 16 for 0
323
tt t
Fs ft te e e t
sss
−− −−−
=+ + ⇒ =− − +
+++
≥

Ex. 14.8-1

KCL at top node:

()
()
C
C
2
2
32
Vs s
Vs
s
+ =+

()
C
62
2
3
Vs
s
s
=−
+

()
()
( )
2/3
C
62 () V
t
vt e ut
−
=−
()
()
()
()C2
CC
2
23
2(
22 3
3
tVs
/3
) AIsi t
s
s
−
=− = ⇒ =
+
e ut

14-6

Ex. 14.8-2

Mesh Equations:
() () ()() () ()
41 4 1
60 6
22
CC C 6IsI sIs IsI
ss s s

−− − − =⇒−=+ +


s
() ()() () () () ()
10
63 4 0
9
CC CIsI s IsIs Is Is−+ + =⇒ =−
Solving for Ic(s):
() ()
42 1 6
332
4
CC
Is Is
ss
s

−=−+ ⇒ =


−

So Vo(s) is
() ()
24
4
3
4
oCVs Is
s
==
−

Back in the time domain:
()
0.75
24 ()
t
o
vt eut= V

Ex. 14.8-3

KVL:
()
82 0
48 4
LsIs
ss

+= ++



so
()
()
()
22
112
25 14
L
ss
Is
ss s
+++
==
++ ++

Taking the inverse Laplace transform:

() ()
1
cos2 sin2 A
2
tt
ite te tut
−−
=+



14-7

Ex. 14.9-1
(a) ()
1551 0
51 0 (
51 0
tt
response e eut
ss
−−
=− = −

++
L
10
)
−
impulse
(b) ()
11 011
()
10 5
tt
epresponse e eut
ss
−−
=− = −

++
L
5−
st

Ex. 14.9-2
() ()()
()
()
2
2 22
54 20
5sin4
4224
t
Hs e tut
sss
−
== =

0++++
L

()
-1 -1 2
2
14 1
(1 (cos4sin4))()
420 2
t
Hs s
stepresponse e t tut
ss ss
−
  +
== − =− − 
++
LL

Ex. 14.10-1

Voltage division yields

()
()
1
2
8
2
8
()
()
2()
8
2
2
8
16
16 1

4161620 1.25
16
c
s
Vs s
Hs
Vs
s
s
s
ss
ss



+
==



+
+
== =
++
++

so
() () ()
-1 1.25
1.25
t
ht Hs eut
−
==

L

14-8

Ex. 14.10-2
() ()
()() ()
()() ()()
()
()
2
11 1
1

2
1

11 2121
22 2
t
t
hte Hs
s
ftut Fs
s
htft HsFs eut
ss ss
−
−− −
=⇒ =
+
=⇒ =
 −  
∗= = = + =−   
++   
LL L
2−

Ex. 14.11-1
The poles of the transfer function are
() ()
2
1,2
33
2
kk
p
8−−± −−
= .
a.) When k = 2 V/V, the poles are
1,2
1
2
p
7−±−
= so the circuit is stable. The transfer function is

()
()
()
o
2
i
2
2
Vs s
Hs
Vs ss
==
++

The circuit is stable when k =2 V/V so we can determine the network function from the transfer
function by letting s = jω.

()
()
() ()
()
o
2 2
i
22
2 2
sj
sj
sj
Hs
ss j
ω
ω
ω ω
ω
ω ωω
=
=
== = =
++ −+
V
H
V

The input is vt . The phasor of the steady state response is determine by
multiplying the phasor of the input by the network function evaluated at ω = 2 rad/s.
()
i
5cos2 Vt=

() () ()
()
() ()
oi
22
2
24
50 507.0745
222
jj
jj
ω
ω
ω
ωω ω
ωω
=
=


=× = ∠°= ∠°= ∠−

 −+−+ 

VH V °

The steady state response is () ( )
o
7.07cos245 Vt=−vt . °
b. When 322k=− , the poles are
1,2
22 0
2,2
2
p
−±
= =−−so the circuit is stable. The
transfer function is
()
() () ()
22
0.17 0.170.172
222
s
Hs
sss
== −
+++

The impulse response is
14-9

() () ( )()
-1 2
0.17 12
t
ht Hs e tut
−
== −

L

We see that when 322k=− the circuit is stable and lim()0
t
ht
→∞
=.
c. When, 322=+k the poles are
1,2
22 0
2,2
2
p
±
= = so the circuit is not stable. The
transfer function is
()
() () ()
22
5.83 5.835.832
222
s
Hs
sss
== +
−−−

The impulse response is
() () ( )()
-1 2
5.8312
t
ht Hs e tut== +

L

We see that when 322k=+ the circuit is unstable and lim()
t
ht
→∞
=∞.

Ex. 14.12-1
For the poles to be in the left half of the s-plane, the s-term needs to be positive.

() ()
() ()
0 22 222
2
2
22
2 5 10 2 220
()0.1 2 0.1
10 + 125510 510
210125220
0.1
10125
250 25
0.1
10125 10125
ss
Vs
ss sss
ss ss
ss
ss ss

++ 
=− + = −
 +++ ++

++ −+
=
++
==
++ ++
s

These specifications are consistent.

14-10

Problems

Section 14-3: Laplace Transform

P14.3-1
() ()
() () ()
()
11
22
11 22
cos
AftAFs
As
Fs
s
sft t Fs
s
ωω
ω
=


⇒=
+=⇒ = 
+
L

P14.3-2
()
11 1
11 2
1!1
F
n
n
n
ts t
ss s
−−
++
 == =
 
LL
1
!
=

P14.3-3
() () () ()
() ()
() [] ()
()
11 22 11 22
12
3
11
22 2
2
Linearity: a a
Here 1
1
3
1
11
so
3
t
ftaft FsaFs
aa
ft e Fs
s
ft t Fs
s
Fs
ss
−
+= +

==
== =
 
+
== =

=+
+
L
LL
LL

P14.3-4
()() ()[ ] ()
()() ()() ()()()
() ()
()
()
11
12
23
1( )
11
11
,
11
bt
bt bt
ftAeut AftAFs
3ftA eututeutftft
Fs Fs
ss b
Ab
Fs A
ssbssb
−
−−
=− = =
=− = − = +
−
==
+

∴ =− =

++
L

14-11

Section 14-4: Impulse Function and Time Shift Property

P14.4-1
() ()()ftAututT=− − 
 

() () ()
( )1
sT
sT eAAe
Fs AutAutT A
ss s
−
− −
=− −=− =  
  
LL

P14.4-2
() ()() () ()()1
at at
ft ututTe Fs eututT =− − ⇒ = −− 
  
L
()()
() ()
()
()
()
1
1
saT
sT
e
ututT
e
s
Fs
sa
at
egtGsa
−
−
−
−− =
 −
⇒=
−

=−
 
L
L

P14.4-3
(a)
()
()
3
2
3
Fs
s
=
+

(b) ()() () () ()
sTs T
ftt TutT Fse teδδ
− −
=− − ⇒ = =

L

(c)
()
() ()()
22 22
555
848162545
Fs
sssss
== =
1++++ +++

P14.4-4
() ( ) ( ) ( )
((0.50.5)) 0.5(0.5)
0.5 0.5 0.5
tt t
gteut e ut ee ut
−− +− − −−
=− = −= −

() ( ) ()
0.50.50.50.5
0.5(0.5) 0.5 (0.5) 0.50.5
0.5 0.5
1 1
ss
tt s t eee
ee ut e e ut ee eut
s s
−−
−− − − −− −− −
    −= −= = =
    
+ +
LL L

P14.4-5
() () ()
2

sTs
sT eetT t
utT e ut tut
TT TT
−−
−−  
−− = − = − =
  
  
LL L
T
s

14-12

Section 14-5: Inverse Laplace Transform

P14.5-1
()()
32 22
33
()
36 4 1 211 3
ss A
Fs
ss s s ssss
++
== = +
++ + + ++++ +

4
BsC+

where
()
2
1
32
313 s
s
A
s =−
+
= =
++

Then
()
()()
()
2
22
2
3 243
3( )
12 4 3 3 312 4
s BsC
sB s BCs
ss sss s
+ +  
=+ ⇒+=+ +++ ++
 
++ +++ + 
8
C

Equating coefficient yields
2 22
s:0
33
42 1
s:1
33 3
BB
CC
=+⇒ =−
=−+ ⇒ =

Then
()
()
()
() ()
22
122 1 2 2
31
333 3 3 3
11 13 13 13
ss
Fs
ss ss s
−+ −+
=+ = + +
++
2
+++ + ++

Taking the inverse Laplace transform yields

()
22 1
cos3 sin3
33 3
tt t
ft e e t e
−− −
=− +

14-13

P14.5-2
()
()( )( )
22
32
21 2 1
34 2 11 1 1 1
ss ss a a b
Fs
sss ssjsj sjsj s
−+ −+
== = +
+++ ++− ++ +− ++ +
*
1
+
where
()
()( )
2
2
2
*
21
4
11 1
21 343
2
11 221
3
2
2
ss
b
s s
ss j
aj
ss jsj
aj
−+
==
++ =−
−+ −
==
++ + −=−+
=−−
=−+
Then
()
33
22
4
22
11
jj
Fs
sj sjs
−+ −−
=++
1+−+ + +

Next
() ()
22 152
-32 2 and tan 126.9
32
2
m θ
−


= += = =
−

°
From Equation 14.5-8
() ( ) ()5cos1274
tt
fte t eu
−−
=+ °+

t

P14.5-3
()() ()
22
51
()
1212 1
sA B
Fs
ssss s
C−
== + +
+ −+− +

where
()
2
12
51 51
2 and 1
2 1ss
ss
BC
s s=− =
−−
== =
− +
=
Then
() ()
()
2
2
1
1
9
11
2
s
s
d
As Fs
ds s
=−
=−
−
=+ = =

−
−
Finally
()
() ()
2
2
12 1
() 2 +
121
tt t
Fs ft e teeut
sss
−−−
=+ + ⇒ =−+

+−+

14-14

P14.5-4
()
()() ()() ()
222
11
112 2 1111 1
A BsC
Ys
sss s sss
+
== = +
+++ + ++++ +


where
2
1
1
1
22
s
A
ss
=−
= =
++

Next
()()
()
() ( )
2
22
2
11
12 2 (1)
12 212 2
11 2
BsC
ss BsCs
ss sss s
Bs BC sC2
+
=+ ⇒=++++ +
++ +++ +
⇒= + +++++

Equating coefficients:
2
s:0 1 1
:0 2 1
BB
sB C C
=+ ⇒ =−
=++ ⇒ =−

Finally
()
()
() ()
2
1
cos
1 11
tts
Ys ytee tut
s s
−−1+
=− ⇒ = −

+ ++

P14.5-5
()
()
()()
()
() ()
222
23 112
112 5 14 14
ss
Fs
sss s ss
+− +
== + +
+++ + +++ +

() () ()()cos2 sin2
tt t
fte e te tu
−− −
=− +

t

P14.5-6
()
()
()()
2s+3
ss+12 1 2
AB C
Fs
ss s s
== + +
+ ++

where
()
()
()()
() ()
()
()
01
1
0
23 23
3, 1 4
12 2
ss
s
s
ss
AsFs BsFs
ss ss
== −
=−
=
++
== = =+ =
++ +
=−
and
() ()
()
()
2
2
23
21
1
s
s
s
sF s C
ss
=−
=−
+
+= =
+
=
Finally
() ()() ()
234 1
34
12
tt
Fs ft eeut
ss s
−−−
=+ + ⇒ =− +
++

14-15

Section 14-6: Initial and Final Value Theorems

P14.6-1

(a)
() ()
22
22
23 42
0lim lim
32
ss s
fs Fs
ss sss
−+
== =
++→∞ →∞
2=

(b)
() ()
4
lim 2
20
fs Fs
s
∞= ==
→

P14.6-2
Initial value:
() ()
()
2
22
16 16
0lim lim lim 1
412 412
ss s
ss s s
vs Vs
ss ss
→∞ →∞ →∞
+ +
== =
++ ++
=

Final value:
()
2
22
00
16 16
lim lim 0
412 412
ss
ss
vs
ss ss
→→
++
∞= = =

++ ++
s

(Check: V(s) is stable because {}Re 0since 2 2.828
ii
pp<= −± j. We
expect the final value to exist.)

P14.6-3
Initial value:
2
32
10
(0)lim()lim 0
32 1
ss
ss
vs Vs
ss
→∞ →∞
+
==
++
=

Final value:
() ()
()
()
2
00
10
lim lim 10
32 1
ss
ss
vs Vs
ss s
→→
+
∞= = =
++

(Check: V(s) is stable because 0.3330.471p ii=−± . We expect the
final value to exist.)

P14.6-4
Initial value:
() ()
2
2
214
0lim lim
210
s
s
ss
fs Fs
ss
→∞
→∞
−−
==
−+
2=−

Final value: F(s) is not stable because{}
1
Re 0since 1 3
i
p p> =i±. No final value
exists.

14-16

Section 14-7: Solution of Differential Equations Describing a Circuit

P14.7-1

KVL:
4
210
500.001 2
tdi
iv
dt
−×
++ =e
The capacitor current and voltage are related
by

()
6
2.510
dv
i
dt
−
=×

4t
210
1
= 2e Vv
−×
, i (0)1 A, (0)8 Vv==

Taking the Laplace transforms of these equations yields

[]
() ()
4
6
2
50 ()0.001() (0)()
210
()2.510 () 0
Is sIsi Vs
s
Is sVsv
−
+− + =
+×
=× −


Solving for I(s) yields

()
() ( )( )
24 8
4444 4
1.4 101.610
10 210 41010 2 10 4 10
ss A B
Is
ss sss s
+× −×
== +
++ × +×++ × +×
4
C
+
where
() ()
() ()
() ()
() ( )
() ()
24 8 8
4
4 844
= 10
4
= 10
24 8 8
4
4 844
= 210
4
= 210
2
4
4
= 410
1.410 1.6 10 210 2
s10
3 10 32 10 4 10
s1.410s 1.6 10 .4 101
s210
2 105s10s410
s1.4
s410
s
s
s
s
s
ss
AI s
ss
BI s
CI s
−
−
−×
−×
−×
+× −× −× −
=+ = = =
×+× +×
+× −× ×
=+× = = =
×++ ×
+×
=+× =
() ( )
48 8
844
4
= 410
10s 1.6 10 8.8 1022

6 10 15s10 s210
s−×
−× ×
==
×++ ×

Then

() () ()
44 4
10t 2x10 4x10
44
123 15 2215
10e 3e 22e A
10 210 4 10 15
t t
Is it ut
ss s
−− −
=− + + ⇒ =− + +
++ × +×

14-17

P14.7-2

We are given ()160cos400t=vt .
The capacitor is initially uncharged, so
()
C
00 v =V. Then

()
( )160cos40000
0 160 A
1
i
×−
==
KCL yields
C C
3
10
100
dv v
i
dt
−
+ =
Apply Ohm’s law to the 1 Ω resistor to get
C
C
1
vv
iv vi
−
= ⇒ =−
Solving yields
()
4
10101600cos4006.410sin400
di
it
dt
+= −× t

Taking the Laplace transform yields
()
()
( )()
()
2
22
22
6.4104001600s
()(0)1010()
s400 400
sIsi Is
s
×
−+ = −
++

so
()
7
22
160 1600s2.510
()
1010 1010 (400)
Is
s ss
−×
=+
+  ++
 

Next
()
7*
22
1600s2.510

1010 400 4001010 (400)
AB B
ss j sjss
−×
=+ +
++ −++


where
()
7
2
2
= 1010
16002.510
23.1
400
s
s
A
s
−
−×
==
+
− ,
() ( )
77
*
5
= 400
16002.510 2.56101.4
11.527.2 and 11.527.2
1010 400
8.691068.4sj
sx x
Bj
ss j
x
°
°
−
−∠
== = −
+−
∠
B j= +
Then
()
136.911.527.211.527.2

1010 400 400
jj
Is
ss j sj
− +
=+ +
++ −

Finally
() () ()
1010
1010
136.9 211.5cos400227.2sin400for 0
136.9 23.0cos40054.4sin400for0
t
t
it e t t t
et t t
−
−
=+ −
=+ − >
>

14-18

P14.7-3

C(0)0v=
c
c
c
c
1510cos2
220cos1
30
vi t
dv
vtdv
dti
dt
+= 

⇒+ =
=


2

Taking the Laplace Transform yields:

() () ()
()()
*
cc c c2 2
20
02 ()20
42 24
ss A
sVsv Vs Vs
ss ss
−+ = ⇒ = = + +
2 2
B B
sjsj+ ++ −++

where
()( )
2
= 2
= 2
20 40 20 555 55*
5, and
48 2 2 1 22 2
s
sj
ss j
AB j
ss sj j
−
−
−
= = =− = = =+ =−
++ − + 2
B j
Then
() () ( )
2
c
55 55
5
22 22
55 cos2sin2
22 2
t
c
jj
Vs vt e t t
ss jsj
−
+−
−
=+ + ⇒ =− + +
++ −
V

P14.7-4
L
c
cL L122 8 and
di dv
vi iC
dt dt
++ =− =

Taking the Laplace transform yields

() () ()()
cL L L
8
12 2 0Vs Is sIsi
s
++ − =

−
() ()()
L c 0Is CsVsv=−
 c

cL(0)0, (0)0v i==

Solving yields
14-19

()
c
2
4

6
2
C
Vs
C
ss s
−
=

++



(a)
1
F
18
=C ()
() ()
2 2
72

33 3
c
cab
Vs
ssss s
−
== + +
++ +

()
()
2
2488
8, 8, and 24
3 3
cab c Vs
ss s
−
=− = = ⇒ =+ +
+ +

()
33
88 24
tt
c
vt e te
− −
=−+ +

(b)
1
F
10
=C ()
()()
40

15 1 5
c
cab
Vs
ss s sss
−
== +
++ + +
+

()
2810
8, 10, and 2
1 5
cab c Vs
ss s
−−
=− = =−⇒ =++
+ +

()
5
810 2
tt
c
vt e e
− −
=−+ −

P14.7-5
()cL(0) 10 V, 00 Avi
−−
==

()
6 c
c510 and 4001 0
dv di
ii
dt dt
−
=× ++=v

Taking Laplace transforms yields
()() ()()
() ()() ()
()
()
6
c
225 2
c
1
400
510 10 10 40
400210400 0 0 200400
Is sVs
Is
ssIs sIs Vs s
−

−
=× − − 
⇒= =
++ ×+− + = ++

so
() () ()
2001
sin400 A
40
t
it e tut
−
=−

14-20

Section 14-8: Circuit Analysis Using Impedance and Initial Conditions

P14.8-1

400
66
0.010 0.002
.003.005
()
52000(400) 400
2 mA 0
()
35 mA 0
L
L t
s
ss
Is
ss s ss
t
it
et
−
−−
== = =
++
<
=
−>
+

P14.8-2

()
4000
15
10 8
()
().015() 3
0(
40002000 4000 5
15
8
()0.15 0.0050.00215
() 0.015
4000400055
1515
()53 mA,0
L
LL
L
L
L
t
L
Vs
VsVss
Vs
s
s
Vs
Is
s
sss
it e t
−
−
−−
=+ + ⇒ =
+
+
== + = −

++


=− >
)

14-21

P14.8-3

6
1000
8
()
()0.006
0
102000
.5
6000 8
500()0.5() 0
81200012 4
()
(1000) 1000
()124 V,0
c
c
c c
c
t
c
Vs
Vs s
s
s
Vs sVs
ss
s
Vs
ss ss
Vt e t
−
−
−+ + =

−+ + −=


+
== −
++
=− >

P14.8-4

()
6
1500
6
()
()0.5 8
() 0
2000400010
68
500() 250()0.5() 0
600084 4
()
1500 1500
()44 V,0
c
c
c
cc c
c
t
c
Vs
Vs ss
Vs
s
Vs Vs sVs
ss
s
Vs
ss ss
vt e t
−
−
 
++ −=
 
 
  
−+ + −=
  
  
+
== +
++
=+ >

14-22

P14.8-5

Node equations:

() () ()
() ()
aC a
aC
16
66
Vs VsVs
Vs Vs
ss s
−
+= ⇒ = +
6
6s+ +

() () ()
()
CC
C
C
666
13663
0
44
Vs VsVs
ssss
Vs
ss

−+− 
+++
++ +
2
−=
)

After quite a bite of algebra:
()
()()(
2
C
656132
23
ss
Vs
ss s
++
=
5+ ++

Partial fraction expansion:
()()()
2
44 1
656132 933
()
c 325 2 3
ss
Vs
ss s s s s
++
== −
5
+
+++ + + +

Inverse Laplace transform:
23 5
()443 9 (13) V
c
tt t
vt e e e
−− −
=− +

14-23

P14.8–6

Write a node equation in the frequency domain:
() ()
()
22
11oo
o
12
2
2
10
10 51 0 510
5
11
1
1
R R
s
RCRVs Vs
s
CV s
RR s
s
ss
Cs RC
RC
+−
=− + ⇒ = = +

+
+


R

Inverse Laplace transform:
()
222 1000
o
11
10 510 105 V for 0
tRC t
RR
vt e e t
RR
− −

=+ − =−


>
14-24

P14.8-7
Here are the equations describing the coupled coils:

() ( )
() ( )
12
11 1 1 2 1 2
21
22 2 1 2 1 2
() ()3()2 ()33() ()9
() () ()22()3 ()2()8
di di
vtL M Vs sIs sIs sIssIs
dt dt
di di
vtL M VssIs sIs sIssIs
dt dt
=+ ⇒ = −+ −= +
=+ ⇒ = −+ −= +
−
−

Writing mesh equations:

() () ( )()
()
12 1 1 2 1 2 1 2
1 2 2 1 2 1 22 12
55
2()() 2()()3()()9 32 2 9
() ()1() 3() ()9 ()2()8() 2 11
IsIsV IsIs sIssIs sIsI
s s
VsVsIs sIssIs sIssIs Is sIsI
==+ + + + + −⇒ + ++ =+
=+ ⇒ + −= + −+ ⇒ −+=

Solving the mesh equations for I2(s):

()
() ( )
2 2
158 31.6 0.64 2.36
= +
0.261.54 + 0.26 + 1.5459 2
ss
Is
ss s sss
++
==
++++

Taking the inverse Laplace transform:

0.26 1.54
2
()0.64 2.36 A for 0
tt
it e e t
−−
=+ >

P14.8-8

t<0

time domain

frequency domain

Mesh equations in the frequency domain:

() () ()() () () ()
11 2 1 1 2
12 2 2
66 6 0
33
Is Is Is Is Is Is
ss
+− + +=⇒ = −
14-25

() () ()() () ()
21 2 2 1
26 2
60 6 6Is IsIs Is Is
ss s

−− − =⇒ + − =


6
s

Solving for I2(s):
() () ()
22 2
1
22 26
2
66
133
2
Is Is Is
ss s
s

+− − =⇒ =
  +

Calculate for Vo(s):
() ()
o2
1
16 1 6 2
2
1122
22
Vs Is
ss
ss

 −
=− = −=
++

4
s
−
Take the Inverse Laplace transform:

()( )
/2
42 V for 0
t
ovt e t
−
=−+ >

(Checked using LNAP, 12/29/02)

P14.8-9

t<0

time domain

frequency domain

Writing a mesh equation:
() () ()
2
6
12 335
45 30 0
44
55
s
sIs Is
ss
sss
 
−+



++ +=⇒ = =−+

++ 


Take the Inverse Laplace transform:

() ( )
0.8
31 A for 0
t
it e t
−
=−+ >

(Checked using LNAP, 12/29/02)
14-26

P14.8-10
Steady-state for t<0:

From the equation for vo(t):

()
()
o
2
612 6 Vve
−∞
∞=+ =

From the circuit:
() ()
o
3
18
3
v
R
∞=
+

Therefore:
()
3
61 8 6
3
R
R
=⇒ =
+
Ω

Steady-state for t>0:

() ()
1186 6
20
1
2
Is Is
Cs ss
s
C
 −
++ −=⇒ =
 +

() ()
o
11 81 6 1812121812
1 1
22
Vs Is
Cs sCs ss s s
ss s
CC

−−
=+ = += + +=
++

6
1
2C
+
+

Taking the inverse Laplace transform:

()
/2
o
612 V for 0
tC
vt e t
−
=+>
Comparing this to the given equation for vo(t), we see that
1
2 0.25 F
2
C
C
=⇒ = .

(Checked using LNAP, 12/29/02)
14-27

Section 14-9: Transfer Function and Impedance

P14.9-1
1
21 1
12
112 11 22
1
1
() where and
11
R
ZR Cs
Hs Z Z
R
2R
ZZR Cs
R
Cs
== =
RCs
=
+ ++
+

()
( )
() ( )
()
()
() ( )
21
11 1 2 22
12 1 2
21 2
22
12 1 2
Rs 1
Let and then Hs
Rs 1 s1R
1
When constant, as required.
1
s
RC RC
Rs R
Hs
RR RR
τ
ττ
ττ
τ
τττ
τ
+
== =
++ +
+
==⇒ = = =
++ +

11 22we require RCR C∴ =

P14.9-2
12
1
Let and then the input impedance isZR ZRLs
Cs
=+ =+
()
()
2
12
2
12
1
1

1 21
L
R RLs LCsRC s
ZZ Cs R
Zs R
ZZ LCsRCs
RR Ls
Cs
  
+ ++ +
  
  
== =
++ 
++ +


+
+

2
Now require : 2 then
L
RCR C LRC Z
R
+= ⇒ = =R

P14.9-3
The transfer function is
()
2
21
21
1
21
1
1
1
R
RCs RC
Hs
2
2
R RR
Rs
RCs RRC
+
==
+
++
+

Using
12
2, 8and 5 FRR C=Ω =Ω =
gives
()
0.1
0.125
Hs
s
=
+

The impulse response is () () ()
0.125
0.1 V
t
ht Hs e ut
-1
L
−
==

.

The step response is
14-28

()
()
() ()
0.1250.1 0.8 0.8
0.81 V
0.125 0.125
t
Hs
eu t
ss s ss
-1 -1 -1
L= L =L
−
  
 − = −
 ++   

(Checked using LNAP, 12/29/02)
P14.9-4
The transfer function is:

() ()
()
4
2 2
12 12
12
8164
t
Hs teut
sss
−
== =

+++
L
The Laplace transform of the step response is:
()
() ()
22
3
12 3
4
444
Hs k
ssss s
−
== + +
+++ s

The constant k is evaluated by multiplying both sides of the last equation by . ()
2
4ss+
() () ()
2
233
12 43 4 34 12
44
ss kss ks ks k

=+ −++ =+ ++ + ⇒ =− 
3
4

The step response is
()
()
14 33
3 V
44
t
Hs
et ut
s
L
−−
  
 =− +    


P14.9-5
The transfer function can also be calculated form the circuit itself. The circuit can be represented
in the frequency domain as

We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R
and the op amp comprise a non-inverting amplifier. Thus

() ()
ac 1
10000
R
Vs Vs

=+  

Now, writing node equations,
14-29

() ()
()
() () ()
ci o a o
c 0and 0
1000 5000
Vs Vs VsVsVs
CsVs
Ls
−−
+= + =

Solving these node equations gives
()
1 5000
1
1000 10000
1 5000
1000
R
CL
Hs
ss
CL

+  
=
 
++   



Comparing these two equations for the transfer function gives

() ()
11
2000or 5000
1000 1000
ss ss
CC
 
+= + + =+  

() ()
5000 5000
2000or 5000ss ss
LL
 
+= + + =+  

1
1000
1
10000
5000
1510
6
C
R
L
+





 =×

The solution isn’t unique, but there are only two possibilities. One of these possibilities is

()
1
2000 0.5F
1000
ss C
C
µ

+= + ⇒ = 

()s
L
sL+





=+ ⇒ =
5000
5000 1H

()
6
6
1 5000
11 510
10000110000.510
R
R

+= × ⇒ =Ω 
×
5k
(Checked using LNAP, 12/29/02)

14-30

P14.9-6
The transfer function of the circuit is
()
2
21
1
2
1
1
1
R
RCs RC
Hs
R
s
RC
+
=− =−
+

The give step response is () ( )()
250
41 V
t
ovt eut
−
=−− . The correspond transfer function is
calculated as

()
() (){}
()
()
250 4 4 1000 1000
41
250 250 250
t
Hs
eu t Hs
ss s ss
− −−
=− − =−− = ⇒ =

++
L
s+

Comparing these results gives

( )
2
6
2
11 1
250 40k
2502500.110
R
RC C
−
=⇒ = = =
×
Ω
( )
1
6
1
11 1
1000 10k
100010000.110
R
RC C
−
=⇒ = = =
×
Ω

(Checked using LNAP, 12/29/02)

P14.9-7

() () ()
ai
42
42 2
Vs Vs Vs
ss
 
== 
++ 
i

() () () () ()
ob b a
12
22 2 2
55
12 22 2 2
6
s
Vs Vs Vs Vs Vs
ss s s
s

    
== = =    
++ + +   +

i

The transfer function is:
()
()
()()
o
2
i
20
2
Vs
Hs
Vs s
==
+

14-31

The Laplace transform of the step response is:
()
() ()
o 22
20 55 10
222
Vs
ssss s
− −
== + +
+++

Taking the inverse Laplace transform:
() ( )()
2
o
55 12 V
t
vt e tut
−
=− +


(checked using LNAP 8/15/02)

P 14.9-8

From the circuit:
() () ()
11 4
4 6
144
6
6
CCs L
Hs k k
Ls
ss
Cs L C
 
 
==  
+  ++ + 
 
1

From the given step response:
()
() ()
()()
32 24 6 12
24 6
32 3
tt
Hs
ee ut
ss s s
−−
=+ − =+ − =

2ss s+ ++ +
L

so
()
()()
12
32
Hs
ss s
=
+ +

Comparing the two representations of the transfer functions let
11
3 F
61
C
C
=⇒ =
8
,
4
22 L H
L
=⇒ = and 23 . 12 2 V/Vkk××= ⇒ =

(Checked using LNAP, 12/29/02)

P 14.9.9
From the circuit:
14-32

()
()
()
o
i
1212
R
s
Vs RLs
L
Hs
RVs RLs
s
L
+
+
== =
+++
+

From the given step response:

()
() ()
()
()
4 0.50.5 2 2
0.51
44
t
Hs ss
eut Hs
ss s ss
−
4s
+ +
=+ =+ = ⇒ =

+ ++
L

Comparing these two forms of the transfer function gives:

2
122
4 6 H, 12
12
4
R
LL
LR
R L
L

=
 +
⇒= ⇒ = =
+

=

Ω
(Checked using LNAP, 12/29/02)

P14.9-10

Mesh equations:

() () ()
() ()
11
21
11 1

11
0
Vs R Is Is
CsCs Cs
RR Is Is
Cs Cs

=+ + −



=+ + −


2

Solving for I2(s): ()
()
2
1 2
1
21
2
()
Vs
Cs
RR
Cs Cs Cs



=
  
++ −  
  
1
Is
Then Vs gives () ()
o 2
RIs=

()
()
() [] [ ]
()
0
1
2 1
1 22
2
1
1
22 11
41
2
2
2
Vs RCs s
Hs
Vs RCs RCs
RCRC
RCs s
RRC
RRC
== =
++ −  
+
 ++
 
 

14-33

P14.9-11
Let
2
1
1
1 1
xx
R
RCs
Z
RCs
R
Cs
ZR Ls



==
+
+
=+

Then
()
()
22
2
11 2 x x x x
xx
2
xx21 x
xx
1

1
1

R
VZ RRCs
RVZ Z LRCsLRRCsR
RLs
RCs
V LC
LRRCV RR
ss
LRC LRC
+
== =
++ +
++
+
=
+ +
++
R++

P14.9-12

Node equations:
() ()
1 out
1i n 1 11 1 11in o
1
0
out 2 1 22 out
2

01
0
VV
VVsC RCsVRCsVV
R
V
VsC VRCsV
R
−
−+ =⇒ + = +
++ =⇒ =−
ut

Solving gives:
()
11 22out
2
2
in 1212 22
11 1212
1

11 1

s
RCs RCV
Hs
VRRCCsRCs
ss
RCR RCC
−
−
== =
++
++

14-34

P14.9-13

Node equations in the frequency domain:

11 12
12 3
0
iVV VVVV
RR R
0− −−
+ +=

0
1
12 3 3
11 1
i
VV
V
1
RRR RR

⇒+ + −=


21
20 1 220
2
0
VV
sCV V sCRV
R
−
−= ⇒ =−

After a little algebra:
()
03
223 213 21 2 1i
VR
Hs
VsCRRsCRRsCRRR
−
==
+ ++

P14.9-14

()
() 2
11
()
11
o
i
Vs
Cs LC
Hs
RVs
LsR s s
Cs LLC
== =
++ ++

L, H C, F
R, Ω H(s)

2

0.025

18
()()
2
20 20
920 4 5ss s s
=
+++ +

2

0.025

8
()
22 4
20 20
420 24ss s
=
++ ++

1

0.391

4
() ( )
2
2.56 2.56
42.56 0.83.2ss s s
=
++ + +

2

0.125

8
()
22
20 20
44 2ss s
=
++ +

14-35

a) ()
()()
20
45
Hs
ss
=
++
{} ()()
{}
()
45
54
2020
() () = 20 20 ()
45
() 20 154
step response
(4)(5) 45
step response = 14 5 ()
tt
tt
htHs ht e eut
ss
Hs
ss ss sss
ee ut
−−
−−
=− ⇒ = −
++
−
= == ++
++ ++
+−
L
L ⇒

b) ()
()
2
4
20
24
Hs
s
=
++
(){} ()
2
22
5(4)
(s ) 5sin4()
(2)4
t
ht H hte tut
s
−
== ⇒ =
++
L
{}
() ()( )
{}
()
()
()
()
12
22
22
12 1 2
12
2 22
2
() 20 1
step response
(420) 420
20 420 1 4 20
1, 4
1
4
21
2
step response
2424
1
step response1 cos4sin4()
2
t
Hs KsK
ss ss sss
ss sKsKsKsK
KK
s
sss
et tut
−
+
== =+
+++
=+ ++ +=++++
⇒= − =−
−
−+
=+ +
++++

=− +

L
L
+
)

c)
()
() (
2.56
0.83.2
Hs
ss
=
++

(){} () ()
{}
.8t 3.2t
3.2 .8
1.071.07
() 1.07eeu(t)
.83.2
41
() 2.56 1 33
stepresponse
( .8)(3.2) .83.2
14
stepresponse1 ()
33
tt
ht Hs ht
ss
Hs
ss s s ss s
ee ut
−−
−−
== − ⇒ = −
++
−
== =+ +
++ + +

=+ −


L
L

d) ()
()
2
20
2
Hs
s
=
+

()
()()
2
2
4( )
stepresponse112 ()
t
t
ht teut
teut
−
−
=
=−+

14-36

P14.9-15
For an impulse response, take Vs()
1
1 =. Then

()
()
() ( )
*
0
32
32 3 2 32 32
s AB B
Vs
ss jsj ssjsj
+
== +
+− ++ +− ++
+
Where

() ()
*
00 32
0
.462, (32) 0.47119.7 and 0.47 119.7
oo
sj
s
AsVs Bs jVs B
=−+
=
== =+− = ∠− = ∠

Then
()
0
0.4620.47 119.7 0.47 119.7
32 32
oo
Vs
ss j sj
∠− ∠
=+ +
+− ++

The impulse response is

( )()
3
0()0.4622(0.47)cos2119.7 V
to
vt e t ut
−
=+ −


Section 14-10: Convolution Theorem

P14.10-1
()()() () ()()
11
11
s s
ee
ftutut Fs utut
ss s
− −
−
=− −⇒ = −−=−=

L
()() ()
()() () ()(
2
2
12 1 1
2
11 2
*
21 1 2 2
ss s
ee e
ftft Fs
ss
tut tut tut
−− −
−− −

  −− +
== = 
  

)= −− −+− −
LL L

P14.10-2
() ()() ()
2
22
22
s
e
ft utut Fs
ss
−
=− − ⇒ =−


()() ()() () () ()
24
11
22 2
48 4
48 2 244
ss
ee
4ff FsFs tut tut tut
ss s
−−
−− 
∗= = − + = −− −+− −


LL

14-37

P14.10-3
() () ()
()
()
()
()() () ()()
() ()()
() ()
11 2
2
1
1
21 1
21 2
/
2
1
11
11
1
1
1
1, 0
tRC
vt tut Vs
s
Vs
Cs RC
Hs
Vs
Rs
Cs RC
vt htvt VsHs
RC
Vs VsHs
s
s
RC
vt tRCe t
−
−
=⇒ =
== =
++
 =∗ =
 


== 

+

=− − ≥
L

P14.10-4
()() ()() () ()
()()
()()
1
2
2
2
22
()
22
11
wher e and
11
So
Solvingthepartialfractionsyields:1,1,1
1
So , 0
at
ht ft HsFs Hs Fs
ss
AB C
HsFs
sa sssa
s
Aa B aC a
te
htft t
aa a
−
−
∗= = =

+

== ++
++
=− = =
−
∗= ++ ≥
L
a

14-38

Section 14-11: Stability

P14.11-1
a. From the given step response:

()
() ()
()
10037
1
4 100
t
Hs
eu t
ss
−
=− =

+
L
5
s

From the circuit:
()
()
55
R
HsR
L
Hs
RRL s s
ss
L
=⇒ =
+++ 
+


Comparing gives
75
15
5 0.2 H
100
R
RL
RL
L

=
 =Ω
⇒
+=

=


b. The impulse response is
() ()
-1 10075
75
100
t
ht eut
s
−
==

+
L
c.
()
100
75 3
45
10010042j
ω
ω
=
= =∠
+
H °
() ()
o
31 5
4550 45 V
42 42
ω

=∠ °∠°= ∠°


V
() ( )2.652cos10045 V
o
vt t=− °

(Checked using LNAP, 12/29/02)

P14.11-2
The transfer function of this circuit is given by

()
()() ()
() ()
()
()
2
22
55 10 20 20
55 12
2 22
t
Hs
et ut Hs
ss s ss ss
− −−
=− + =+ + = ⇒ =

+ ++ +
L
2
2

This transfer function is stable so we can determine the network function as

() ()
() ()
22
20 20
22
sj
sj
Hs
sj
ω
ω
ω
ω
=
=
== =
++
H

14-39

The phasor of the output is

()
()
()
()
()
o 22
20 20
545 54512.545 V
22 2245j
ω=∠ °= ∠°=∠−
+ ∠°
V °

The steady-state response is
() ( )
o 12.5cos245 Vvt t=− °

(Checked using LNAP, 12/29/02)

P 14.11-3
The transfer function of the circuit is ()
()
15
2
30
30 ()
5
t
Hs teut
s
−−
 = =
 
+
L . The circuit is stable
so we can determine the network function as

() ()
() ()
22
30 30
55
sj
sj
Hs
sj
ω
ω
ω
ω
=
=
== =
++
H

The phasor of the output is

()
()
()
()
()
o 22
30 30
100 1008.8262 V
53 5.8331j
ω=∠ °= ∠°=∠−
+∠ °
V °

The steady-state response is
() ( )
o 8.82cos362 Vvt t=− °

14-40

PSpice Problems

SP 14-1

14-41

SP 14-2

/
() for > 0
t
vtABe t
τ−
=+
0
7.2(0) 7.2
0.8 V
8.0() 8.0 V
vA Be AB
B
vA Be A
−∞
== + ⇒ =+
⇒= −
=∞=+ ⇒ = 

0.05/ 0.05 87.7728
7.7728(0.05)80.8 ln 1.25878
0.8
0.05
39.72 ms
1.25878
ve
τ
τ
τ
− −
== − ⇒− = =−


⇒= =

Therefore
/0.03972
()80.8 for > 0
t
vt e t
−
=−
14-42

SP 14-3

/
() for > 0
t
itABe t
τ−
=+
3
33
0
0(0) 0
410 A
410 () 410 A
iA Be AB
B
iA Be A
−
−− −∞
== + ⇒ =+ 
⇒= −×
×= ∞=+ ⇒ =× 

() ()
( )
()
36 3 3
36
3
6
6
510/
2.451410 (510)410 410
42.451410510
ln 0.94894
410
510
5.269 s
0.94894
ve
τ
τ
τµ
−− − −
−−
−
−
−
−×
×= × =× −×
−××
⇒− = =−
×

×
⇒= =

Therefore
6
/5.26910
()44 for > 0
t
it e t
−
−×
=−
14-43

SP 14-4

Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)

14-44

V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.

14-45

SP 14-5

Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)

14-46

V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.

14-47

Verification Problems

VP 14-1

() ()
2.1 15.9
36 2
tt
LL
d
vt it e e
dt
−−
== − −

() ()
2.1 15.91
0.092 0.575
75
tt
CC
d
it vt e e
dt
−−
== − −

() ()
2.1 15.9
1
12 126 2
tt
RL
vt vt e e
−−
=− =+ +

()
()()()
2.1 15.9
2
12
10.456 0.123
6
LC tt
R
vt vt
it e e
−−
−+
== + −

()
()
2.1 15.9
3 10.548 0.452
6
C tt
R
vt
it e e
−−
== + +
Thus,
() () ()()()
12
12 0 and +
LR R C R
vt vt ititit−+ + = =
3

as required. The analysis is correct.

14-48

VP 14-2

() ()
12
18 20
and
33
44
Is Is
ss
==
− −

KVL for left mesh:
12118 18 20
60
33 32
44 4
ss
ss s
  
  
++ −  
  −− −
  
= (ok)
KVL for right mesh:
18 20 20 18
63 4
33 3 3
44 4 4
ss s s
  
  
−− + −  
  −− − −
  
0


=


(ok)
The analysis is correct.

VP 14-3
Initial value of IL (s):
2
lim 2
1
5
s
s
s ss
+
=
→∞ ++
(ok)
Final value of IL (s):
2
lim 2
0
0 5
s
s
s ss
+
=
→ ++
(ok)
Initial value of VC (s):
()
()
2
lim 202
0
5
s
s
s sss
−+
=
→∞ ++
(not ok)
Final value of VC (s):
()
()
2
lim 202
8
0 5
s
s
s sss
−+
=−
→ ++
(not ok)
14-49

Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s)
was calculated as ()
20
LIs
s
− instead of ()
20 8
LIs
ss
− +. After correcting this error
()
2
20 2 8
5
C
s
Vs
ss s s
+
=− +

++
.
Initial value of VC (s):
()
()
2
lim 2028
8
5
s
s
s ssss

−+
 +=
→∞ ++

(ok)
Final value of VC (s):
()
()
2
lim 2028
0
0 5
s
s
s ssss

−+
 +=
→ ++

(ok)
14-50

Design Problems

DP 14-1

Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
()
()
2
2
5
1
4
o
kR
L
Vs
R
s
ss
LLC
==
+
++

Equating the poles:
2
1,2
4
40
2
RR
LL LC
s

−± −

== −±

Summarizing the results of these comparisons:

2
4, and 5
2
Rk
R
LL LC
R
= ==

Pick L = 1 H, then k = 0.625 V/V, R = 8 Ω and C = 0.0625 F.

DP 14-2

Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
14-51

()
()
2 2
2
10 10
1 8244
o
kR
L
Vs
R sss
ss
LLC
== =
0++++
++

Equating coefficients:
1
8, 20 , and 10
Rk
LL C L
== =
R

Pick L = 1 H, then k = 1.25 V/V, R = 8 Ω and C = 0.05 F.

DP 14-3

Equating the Laplace transform of the step response of the give circuit to the Laplace transform
of the given step response:
()
() ()
2
2
55 10
1 24 6
o
kR
L
Vs
R ss ss
ss
LLC
== − =
8+ ++
++
+

Equating coefficients:
1
6, 8 , and 10
Rk
LL C L
R
= ==

Pick L = 1 H, then k = 1.667 V/V, R = 6 Ω and C = 0.125 F.

DP 14-4

14-52

Comparing the Laplace transform of the step response of the give circuit to the Laplace
transform of the given step response:
()
() ()
2
2
5 5 1030
1 24 6
o
kR
s
L
Vs
R ss ss
ss
LLC
+
= ≠+=
8+ ++
++
+

These two functions can not be made equal by any choice of k, R, C and L because the
numerators have different forms.

DP 14-5
a) Use voltage division to get

()
2
22
121
11 22
111
1212
12 1 2

1()

()
11
1

o
R
sCRVs
RRVs
sCR sCR
s
CRC
RRCC
s
RRCC
+
=
+
++

+

=
++
+
 +


b) To make the natural response be zero, we eliminate the pole by causing it to cancel the zero.

112 2
11 121 2 12
1

()
RRR C
CR RRCC CR
+
−= − ⇒ =
+

c) Let
11
1
() () ()ut Vs
s
=⇒vt . Then =
1111
12
12 12
12 1 2
12 1 2
1
()
()()

o
s
RCCK
Vs
RRCC sRR
sss
2
K
RRCCRRCC

 +

== +
 ++  +
 ++
+ +

22 1
12
12 1 2 12
where and
RR C
KK
RRC CR
== −
++ R+

Then
()
22 1
12 1 2 1 2
()
t
o
RR C
vt eut
RR CCRR
τ
−
 
=+ −
++ +


14-53

12 1 2
12
()
where
RRCC
RR
τ
+
=
+
. To make the step response be proportional to the step in put, we
require
21
12 1 2
RC
CC RR
=
+ +

Then
()
2
12
()
o
R
vt ut
RR
=
+

DP 14-6
The initial conditions are v . Consider the circuit after t = 0. A
source transformation yields
c(0)0.4 V and (0) 0 Ai=− =

The mesh equations are
() ()
() ()
12
12
880
2
88
0
Is Is
ss s
Is Ls Is
ss

+− =



−+ + =


.4

Solving for ()
2
Isyields
()
2 2
1.6
(4 8
Is
sLsLs
=
)++

Therefore, the characteristic equation is
2 8
40ss
L
++=

We require complex roots with significant damping. Try L = 1 H. Then

22 2
1.6 0.20.2(4)0.2 0.2(2) 0.8
()
(48) 48 (2)4(2)4
ss
Is
ss s s ss s s s
2
− +− +
== + = + −
++ ++ ++ ++

Finally
()
22
()0.20.2cos20.4sin2 A
tt t
it e e tut
−−
=− −


14-54

Chapter 15: – Fourier Series
Exercises

Ex. 15.3-1
Notice that ()
12 1 2
() () () ()()ftTftTftTftftft−= −+−= + =

Therefore, f(t) is a periodic function having the same period, T. Next

() () ()
() ( )()
() ()()
() ( )( )( )(()
11 22
110 1 0 1 0
1
22 0 2 0 2 0
1
110 220 11 22 0 11 22 0
1
cos sin
cos sin
cos sin
nn
n
nn
n
nn n n
n
ftkftkft
ka a ntb nt
ka a ntb nt
kaka kaka ntkbkb nt
ωω
ωω
ωω
∞
=
∞
=
∞
=
=+

=+ +



++ +


=+ + + + +
∑
∑
∑ )

Ex. 15.3-1
f(t) = K is a Fourier Series. The coefficients are a0 = K; an = bn = 0 for n ≥ 1.

Ex. 15.3-2
f(t) = Acosw0t is a Fourier Series. a1 = A and all other coefficients are zero.

15-1

Ex. 15.4-1

4
82
T
ππ
= =

,
0
2
rad
4
sT
π
ω==

Set origin at t = 0, so have an odd function; then an = 0 for n = 0,1, . . . Also, f(t) has half wave
symmetry, so bn = 0 for n = even. For odd n, we have

() () ()
()
()
0
2 2
00 0
0
22
2
0
0
0
0
22 2
( ) sin sin sin
4
sin
44
(1cos ) 1, 3, 5, . . .
2
T T
n TT
T
b ft ntdt ntdt ntdt
TT T
ntdt
T
nT n
nfT n
ωω ω
ω
ω
ππ
−−
== − +
=
=− = =
∫∫ ∫
∫

Finally,
00
41
() sin ; odd and 4 rads
N
n
ft ntn
n
ωω
π
==∑

Ex. 15.4-2
0

0
4

0
0
2
, 2
0 , 0 for all
odd function with quarter wave symmety
0 = even
2
0
68
() sin where () = 6
2
64
Thus
n
n
n
n
T
T
aa
bn
t
t
bf t ntdt ft
t
b
π
n
π
πω
π
π
ω
π
ππ
== =
==
⇒
=

−
<<


= 

−≤ <

−
=
∫
22
22
1
241
sin
3
24 1
so () sin sin (2)
3
N
n
oddn
n
n
n
ft nt
n
π
π
π
π =



− 
= ∑ 


15-2

Ex. 15.4-3
a) is neither even nor odd. f(t) will contain both sine and cosine terms
b)
1
wave symmetry no even harmonics
4
⇒
c) average value of f(t) = 0 ⇒ a0 = 0

Ex. 15.5-1
0
2
2 s, rad/sT
T
π
ωπ== =

21 2
00 1
2
11 1
()
22 2
11
1 (1
2
2

0
jnt jnt
jnt
n
jn jn jn jn
n
C ftedt edt edt
ee e e
jn jn
nodd
jnC
neven
ππ
π
)
π ππ
ππ
π
−−
−
−− − −
== −
=− ++ − = −



=


∫∫ ∫
π

Finally,
35 3 521 1 1 1
() ... ...
35 3 5
jt jt jt jt jt jt
ft e e e e e e
j
ππ π π π π
π
−− −
=+ + +− − − −



Ex. 15.5-3
4
2/4
0
4
4
11 1

22
T
T
jntT jn jnjnt
n T
T
T
Ce dt e e e
TT jn jn
ππω
ππ
−−−
−
−

2 /2π
 == =

−
 
−−
∫

(1)
(1)
2
odd
0 even , 0
12 0
n
n
n
n
Cn
n
π
−
−




=≠

=



n

15-3

Ex. 15.6-1
Use the “stem plot” in Matlab to plot the required Fourier spectra:

% Fourier Spectrum of a Pulse Train

A = 8; % pulse amplitude
T = 4; % period
d = T/8; % pulse width
pi = 3.14159;
w0 = 2*pi/T; %fundamental frequency

N = 49;
n = linspace(-N,N,2*N+1);
x = n*w0*d/2;

% Eqn.15.6-3. Division by zero when n=0 causes Cn(N+1) to be NaN.
Cn = (A*d/T)*sin(x)./x;
Cn(N+1)=A*d/T; % Fix Cn(N+1); sin(0)/0 = 1

% Plot the spectrum using a stem plot
stem(n,Cn,'filled');
xlabel('n');
ylabel('|Cn|');
title('Fourier Spectrum of Pules Train with d = T/8');

15-4

Ex. 15.8-1

0= 4 rad/sω
From Example 15.4-1:

02
1
odd
11 1
()3.24 sinsin 3.24sin4sin12 sin20 sin28
2 9 25 49
N
s
n
n
n
vt nt t t t t
n
π
ω
=
 
== − + −∑ 
 
"
1 



The network function of the circuit is
()
()
()
o
s
1
11
1 11
jC
4jCR j
R
jC
ω ω
ω
ω ωω
ω
== = =
++
+
V
H
V

Evaluating the network function at the frequencies of the input series
()
1
4 1,3,5...
116
nn
jn
==
+
H
n H(n4)
1 0.062∠-86°
3 0.021∠-89°
5 0.012∠-89°
7 0.0009∠-89°
Using superposition
() ( ) () () ()
0.021 0.012 0.0009
()3.240.062sin486 sin1289 sin2089 sin2889
92 5 49
ovt t t t t
 
= −°− −°+ −°− −° 
 
"

() ( )( )( )
() ( )() ()
5
()0.2009sin486.00756sin1289
0.00156sin2089 5.9510sin2889
ovt t t
tt
−
=− °− −°
+ −°− × −°"

Discarding the terms that are smaller than 25 of the fundamental term leaves

() ( )( )( )()0.2009sin486.00756sin1289
ovt t t=− °− −°

15-5

Ex. 15.9-1

()
()
0
0
() ()
1
() ()
at
ajt
jt atjt
fteut
e
F ftedteedt
aj aj
ω
ωω
ω
ω ω
−
∞
−+
+∞ ∞
−
−∞
=
== = =
−+ +
∫∫

Ex. 15.10-1

() ()
() () ()
()
{}
Let
11
{}
jt
jaja
fat fatedt
at t
a
fat fed fe d F
aa aa
ω
ωτωτ
τ
τ
τ ω
ττ τ
∞
−
−∞
∞∞
−−
−∞ −∞
=
=⇒ =

== =


∫
∫∫
F
F

Ex. 15.10-2

() ()() ()()
0
0
11
22
22
jt
ftA edt
ω
πδω πδω
ππ
+
−
∞
−∞
==∫∫
AdtA=

Ex. 15.11-1
(){} ()
() ()
{} () ( )() () (()
0
0
00
00
00
0
0 00
Take the Fourier Transform of both sides to get:
11
22
2
cos 2 2
22 2

jtjt
jt
jt jt
jt jt
edt e
e
ee A A
At A e e
ωω
ω
ωω
ωω
δωω δωω
ππ
πδωω
ωπ
∞
−∞
−
−
−= − =
=−
+
== + = −+

∫
-1
F
F
FF F F
() ( )
00
AAπδωωπδωω=− + +
)δωωπδωω+

15-6

Ex. 15.12-1
a)
() ()
2
22
120 120 14400
24 24 576
in in
VV
j
ωω
2
ω ωω
=⇒ = =
++ +

b)
1
2
0
0
48
48
1
2
24
24
114400 140001
tan 300 J
576 24 24
1 14400 140001
tan 61.3 J
576 24 24
61.3
100% 100%20.5%
300
in
out
out
in
Wd
Wd
W
W
ω
ω
πω π
ω
ω
πω π
η
∞
∞
−
−
 
== =

+ 
 
==

+ 
=
∴=× = × =
∫
∫

Ex. 15.13-1

()
() ()
()
()
()
()
22
11
and
at
at at
ft te
ftte ftte
Fs Fs
sa sa
+−
−− −
+−
=
=⇒ −=−
−
∴ ==
++

() () ()
() ()
() () ()
22
22
22
11
Then
11 4
sj sj
sj sj
FF s Fs
sa sa
ja
aj aj a
ωω
ωω
ω
2
ω
ωω ω
+−
== −
== −
−
=+ = +
++
−
=− =
+− +

15-7

Problems

Section 15.3: The Fourier Series

P15.3-1
2
0
2
2 s = rad/s and () for 0 2
2
T ftt t
π
ωπ=⇒ = = ≤≤. The coefficients of the Fourier
series are given by:

()
2
2
0
0
2
2
2
0
2
2
0
1
4

3
2
24
cos
2
24
sin
2
n
n
at dt
at ntdt
n
bt ntdt
n
π
π
π
π
==
==
−
==
∫
∫
∫

()
22
11
44 1 41
cos sin
3
N
nn
ftn t
nn
ntπ π
ππ
∞
==
∴ =+ −∑∑

15-8

P15.3-2
42
0
4
4 2
0
4
22 2
cos 2 cos
12 2
sin 2sin
1
(sin 0)2(sin )sin
22
(1)
1
sin
2
TT
n T
T T
T
a ntdt ntdt
TT T
nt nt
nT T
nn
n
n
n
n
n
ππ
ππ
π
ππ
π
π
π
π
  
=+  
 


 
=+
 
 

  
=− + −  
 
+
−

=− =


∫∫
1
2
odd
0 even
n
n
n
π






( )
42
0
4
4 2
0
4
22 2
sin 2 sin
12 2
cos 2 cos
1
(2cos ()1)cos
2
3
is odd
2
2,6,10,
0 4,8,12,

TT
Tn
T T
T
bn tdt n
TTT
nt nt
nT T
n
n
n
n
n
n
n
n
ππ
ππ
π
π
π
π
π
π
 
=+ 



 
tdt
=− +
 
 


=− −−






=− = …

=…


∫∫

15-9

P15.3-3

()
0average value of
2
A
af t= =
() 1f or 0
t
ftA t
T

T= −≤


≤

() () ()
00 0
2
0
22
22 2 2 1 2
1 cos cos cos
22 2
cos sin
21
0
2
cos2 cos02sin2 0
2
0
TT T
n
T
tA
a A ntdt ntdt t ntdt
TT T T T T T
nt nt nt
A TT T
TT
n
T
A
nn n
n
ππ
ππ π
π
ππ π
π
π    
=− = −
    
    
 
  
 +
  
   
=−
 

 

 
 
−
=− + − 
 
=
∫∫ ∫

() ()() ()()
00 0
2
0
22
22 2 2 1 2
1 sin sin sin
22 2
sin cos
21
0
2
sin2 sin0 2cos2 0
2
TT T
n
T
tA
b A ntdt ntdt tntdt
TT T T T T T
nt nt nt
A TT T
TT
n
T
A
nn n
n
A
n
ππ π
ππ π
π
ππ
π
   
=− = −
   
   
π
 
  
 −
  
   
=−
 

 

 
 
−
 =− − −
 
=
∫∫ ∫
π

()
1
2
sin
2
n
AA
ftn
nT
t
π
π
∞
=

=+


∑
15-10

P15.3-4
0
2
2 s, rad/s
2
T
π
ωπ== = , ()
0
average value of 1af t= =,

() for 0 2ftt t= ≤≤

()
() () ()
()
() () ()
2
2
2
0
0
22
cos sin2
cos
2
1
cos2 cos02sin2 0
0
n
nt nt nt
at ntdt
n
nn n
n
ππ π
π
π
ππ
π
+
==
π= −+ − 
 
=
∫

()
() () ()
()
() ()() ()()
2
2
2
0
0
22
sin cos2
sin
2
1
sin2 sin0 2cos2 0
2
n
nt nt nt
bt ntdt
n
nn n
n
n
ππ π
π
π
ππ
π
π
−
==
π =− − −
 
−
=
∫

()
1
22
1s in
n
ftn
nT
t
π
π
∞
=

=−


∑

Use Matlab to check this answer:

% P15.3-4
pi=3.14159;
A=2; % input waveform parameters
T=2; % period

w0=2*pi/T; % fundamental frequency, rad/s
tf=2*T; % final time
dt=tf/200; % time increment
t=0:dt:tf; % time, s

a0=A/2; % avarage value of input
v1=0*t+a0; % initialize input as vector

for n=1:1:51 % for each term in the Fourier series ...
an=0; % specify coefficients of the input
series
bn=-A/pi/n;
cn=sqrt(an*an + bn*bn); % convert to magnitude and angle form
thetan=-atan2(bn,an);
v1=v1+cn*cos(n*w0*t+thetan); % add the next term of the input
Fourier series
end

15-11

plot(t, v1,'black') % plot the Fourier series

grid
xlabel('t, s')
ylabel('f(t)')
title('P15.3-4')

15-12

Section 15-4: Symmetry of the Function f(t)

15.4-1
o
2
4 s rad/s
42
T
ππ
ω=⇒ == .
The coefficients of the Fourier series are:

()
0
average value of 0
d
av== t

0 because v
na= d(t) is an odd function of t.

()
()() ()() ()()()
4
0
44
00
4
4
22
0
0
22
1
63sin
22
3
3sin sin
22 2
cos
312
3s in c
22 2 2
2 4
66
1cos2 sin2 02cos2
n
bt ntdt
ntdt tntdt
nt
nt nt nt
n
n
nn n
nn
π
ππ
π
ππ π
π π
ππ π
ππ

=−


 
=−
 
 

os
nπ
 
−
  
  
=− −   
    
   
=− + − −−
=
∫
∫∫
12
nπ

The Fourier series is:

()
1
12
sin
2
d
n
vt nt
n
π
π
∞
=

=


∑

P15.4-2
() () ()
11
12 12
16 6 sin 1 6 sin
22
cd
nn
vt vt nt ntn
nn 2
π ππ
ππ
∞∞
==
 
= −−=−+ −=−+ −
 
 
∑∑




15-13

P15.4-3
o
21000
6 ms 0.006 s rad/s krad/s
.0063 3
T
π π π
ω== ⇒ = = =
The coefficients of the Fourier series are:
()
0
32
1
2
average value of V
62
aa vt
×
== =
because v0
nb= a(t) is an even function of t.

()
()
0.001
0
0.001 1
6
00
26 2
2 1000
2 33000cos
0.006 3
1000 1000
2000 cos 210 cos
33
1000
sin
1000 1000 1000 10003
2000 cos sin
1000 10 33 3
3 9
nat n tdt
nt dt tn tdt
nt
nt nt nt
n
n
π
ππ
π
ππ π
π π
  
=−
  
  
 
=− ×
 
 


  
=− +
  
  
∫
∫∫
0.001
0
23 2
22
22
39
2000 sin 0 cos 1 sin 0
1000 3 10 3 3 3
61 8 6
sin cos 1 sin
33 3
18
nn n n
nn
nn n
nn n
n
ππ π
ππ
ππ π
ππ π
π





 


      
=− − −+      
     
   
=− −−
   
   
=−
π 
−


cos 1
3
n
π 
−
 
 

The Fourier series is

()
a 22
1
1 18 1000
1cos cos
23
n
n
vt n t
n 3
π π
π
∞
=
  
=+ −
 
 
∑




P15.4-4
() () ()
ba 22
1
22
1
11 8
21 1 1cos cos 2
23
1 18 1000 2
1cos cos
23 3 3
n
n
n
vt vt nt
n
n
nt n
n
ππ
π
ππ π
π
∞
=
∞
=
  
=− −=−++ − −
 
 
   
=−+ − −
  
  
∑
∑
3



15-14

P15.4-5

()ftt tππ=− <<
()
0
0
0

0
0

Choose t =
2
2, 1
2
average value: 0
2
sin
0 since have odd function
T
n
n
T
a
bf t ntdt
T
a
π
π
πω
π
ω
−
== =
=
=
=
∫

2
1
2
3
2
sin
2
1sin cos
1
2
11
1
23
nb tntdt
ntt nt
nn
b
b
b
π
π
π
π
π
π
ππ
π
−
−
=
 
=−
 
 

=+ =


=−
=
∫

P15.4-6

0
8 s, 4 rad/sT ωπ==
()0 because is an even functon
nbf t=
()
0
22 21
average 14
8
a
×−×
== =
()
2
0
0
12
01
4
cos
4
2 cos cos
448
2
3 sinsin
42

T
n
af t ntdt
T
ntdt ntdt
nn
n
ω
ππ
ππ
π
=
 
=−
 
 

=−


∫
∫∫
12 3.714, .955, .662aa a= ==

15-15

P15.4-7

()
()
()
()
() ()
()
() ()() ()
0
2
0
2
2
2
2
2
2
2,
2
cos
2
cos cos 2
sin21 sin212
221 221
sin21sin21
2
22
21 21
2
21 sin21 21 sin21
241
n
T
A
aA tdt
aA t ntdt
nt n tA
nn
nn
A
nn
A
nn n n
n
π
ω
π
ω
π
ω
π
ω
π
ω
π
ω
π
ωω
ω
ω
ω
ππ
ω
ωω
π
ωωω
πω ω
ππ
π
π
π
−
−
−
==
==
=
 −+
=+
−+


−+

=+

−+


=+ −−−
−
∫
∫
−
()
()
()
()
2
2
2
4
cos
41
41

41
0 due to symmetry
n
n
A
n
n
A
n
b
π
π
π
π



=−
−
−
=−
−
=

P15.4-8
0
2
0.4 s, 5 radsT
T
π
ωπ=⇒ ==

0
0
cos 0 .1
() 0 .1 .3
cos .3 .4
At t
ft t
At t
ω
ω
≤≤

= ≤<

≤≤


Choose period .1 .3 for integralt−≤ ≤

.1
00
.1
.1
00
.1
1
cos
2
cos cos
n
aA tA
T
aA t n
T
ωπ
ωω
−
−
==
=
∫
∫
tdt

15-16

[]
.1
2
10
.1
.1
00
.1
.1
.1
2
5c os
2
5cos cos
1
5c os 5(1)cos 5(1)
2
2cos (/2)
1
1
n
A
aA tdt
aA tntdt
A nt ntdt
An
n
n
ω
ωω
ππ
π
π
−
−
−
==
=
=+ +
=≠
−
∫
∫
∫
−

0 because the function is even.
n
b=

P15.4-9

00 because the average value is zero
0 because the function is odd
1
0 for even due to wave symmetry
4
n
n
a
a
b
=
=
=

Next:

()
22
4

0 22
4
22
8
8 sin 4 cos 1,5,9, ...
22
sin
8
3,7,11, ...

T
n
T
nn
n forn
n
bt ntdt
n
forn
n
ππ
π
π
ω
π
π
−
  
− =  
 
== =

−=

∫

Section 15.5: Exponential Form of the Fourier Series

P15.5-1
2
1
1
oT 2
π
ω π=⇒ ==, the coefficients of the complex Fourier series are given by:
()
() ()
( )
()
()
()
()
11
22
00
1
21 21
0
1
21 21
2
0
1
sin
12
2
2
22 1 21 (41)
jt jt
jnt jnt
n
jn t jnt
jn t jnt
ee
A ted tA ed t
j
A
ee dt
j
Ae e A
jj n jn n
ππ
ππ
ππ
ππ
π
ππ π
−
−−
−− −+
−− −+
−
== 


=−

−
=− =
−− −+ −

∫∫
∫
C

where we have used e and
2
1
jnπ±
=
j j
ee
π π−
= .

15-17

P15.5-2

22
11
200
1
jnt jnt
TT
n
A A
te dt te dt
TT T
ππ
−−



∫∫
==C
Recall the formula for integrating by parts:
22
2
1
11
tt t
ttt
udvuv vdu=−∫ ∫
. Take and ut=
2
jnt
T
dve dt
π
−
= . When , we get 0n≠

2
2
2 0
0
2
2
2
0
22
2
1
22
2 2
1
2 2
2
T
jnt
T jntT
T
n
T
jnt
jn T
jn jn
Ate
ed
T
jn jn
TT
ATe e
Tj n
jn
T
ATe e
Tj n
jn
T
A
j
n
π
π
π
π
ππ
ππ
π π
π π
π
−
−
−
−
−−


=+

−





=+
−







−

=+
−




=
∫
C t

Now for n = 0 we have
0
0
1
2
TAA
Ct dt
TT
= =∫

Finally,
()
2
0
1
22
n
jnt
T
n
n
AA
ft j e
n
π
π
=∞
=−∞
≠
=+ ∑

15-18

P15.5-3
/2
2
2
/2
/2
/2
22
2
sin
sin
d
jnt jnd jnd
TTjntd
T
n
d
d
2
T
jnd jnd
TT
AA e Ae e
ed t
TT T
jn jn jn
TT
Ae e
nj
An d
nT
nd
Ad T
ndT
T
ππ
π
ππ
ππ
π
π
π
π
π
−−
−
−
−
−
T
π
π
 

== = −
−

 

 −
=



=




 
=


∫
C


P15.5-4
m
()()
0
0
1
o
tT
jnt
nd
t
afttbe dt
T
ω
+
−
=− +∫
C
Let
dttτ=−, then
dttτ=+ .

m
()()
()
()()
()()
() () ()
0
0
0
0
0
0
0 0
0 0
1
1
11
d
od
d
d
oo d
d
od
d
o
d
d d
od o od o
d d
tTt
jn t
n
tt
tTt
jn jnt
tt
jnt
tTt
jn
tt
tTt tTt
jnt jn jnt jn
tt tt
af be d
T
af be e d
T
e
af be d
T
ae fe de be d
TT
ωτ
ωτω
ω
ωτ
ωω τ ω
ττ
ττ
ττ
ωτ
τ ττ
+−
−+
−
+−
−−
−
−
+−
−
−
+− +−
−− −
−−
=+
=+
=+
=+
∫
∫
∫
∫∫
C
−

But
0
0
0
0
00
0
d
o
d
o
d
d
tTt
jn
tTt
jn
tt
o
tt
ne
be db
bjn
ωτ
ωτ
τ
ω
+−
−
+−
−
−
−
 ≠
== 
=− 
∫
so

m
00CaCb= +
and
m
0
odjnt
nnae n
ω−
= ≠CC

15-19

P15.5-5
()
00
2221121
8 s, rad/s, average value
48
TC
T
ππ
ω
×− ×
== = = =
4
=

The coefficients of the exponential Fourier series are calculated as

()
11 2
44
n
21 1
11 2
44 4
21 1
42 44 2 4
1
12 1
8
1
12 1
8
44 4
2
2
nn
jt jt jt
nn n
jt jt jt
nn nn n n
jj jj j j
ed t e dt ed
ee e
nn n
jj j
j
e e ee ee
n
ππ
ππ π
ππ ππ π π
ππ π
π
−− −−
−−
−
−− −
−−
−−

= −×+ ×+ −×



=− × +× +−×

−− −


  −
=− − − + −  
  
∫∫ ∫
C
4
n
t
π
−



and

()
11 2
44
n
21 1
11 2
44 4
21 1
42 44 2 4
1
12 1
8
1
12 1
8
44 4
2
2
nn
j t jt jt
nn n
jt jt jt
nn n n n n
jj jj j j
ed t e dt ed
ee e
nn n
jj j
j
e e ee ee
n
ππ
ππ π
ππ ππ π π
ππ π
π
−−
−− −−
−
−−
−
−−
−− −

=− × + × +−×



=− × +× +−×



  
=− − − + −  
  
=−
∫∫ ∫
C
nC
4
n
t
π−




The function is represented as

()
00
0n n
11
jnt jnt
nn
ftC e e
ωω
∞− ∞
−
−
== −
=+ +∑∑CC

15-20

This result can be checked using MATLAB:

pi = 3.14159;
N=100;

T = 8; % period
t = linspace(0,2*T,200); % time
c0 = 1/4; % average value
w0 = 2*pi/T; % fundamental frequency

for n = 1: N
C(n) = -j*((exp(+j*n*pi/4)-exp(+j*n*pi/2))-2*(exp(-j*n*pi/4)-
exp(+j*n*pi/4))+(exp(-j*n*pi/2)-exp(-j*n*pi/4)))/(2*pi*n);
end

for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+C(n)*exp(-j*n*w0*t(i));
end
end

plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');

15-21

Alternately, this result can be checked using Mathcad:

N15:= n12,N..:= m12,N..:=
T8:= ω2
π
T
:=
d
T
200
:= i12,400..:= t
i
di⋅:=
C
n
2−
1−
t1−expj−n⋅ω⋅t⋅()⋅
⌠

⌡
d
1−
1
t2expj−n⋅ω⋅t⋅()⋅
⌠

⌡
d+
1
2
t1−expj−n⋅ω⋅t⋅()
⌠

⌡
d+
T
:=
C
m
2−
1−
t1−expjm⋅ω⋅t⋅()⋅
⌠

⌡
d
1−
1
t2expjm⋅ω⋅t⋅()⋅
⌠

⌡
d+
1
2
t1−expjm⋅ω⋅t⋅()
⌠

⌡
d+
T
:=
fi()
1
N
n
C
n
expjn⋅ω⋅t
i
⋅()
⋅∑
= 1
N
m
C
m
exp1−j⋅m⋅ω⋅t
i
⋅()
⋅∑
=
+:=
fi()
1.643
1.685
1.745
1.807
1.856
1.88
1.872
1.831
1.767
1.693
1.628
1.589
1.589
1.633
1.717
1.825
=
C
n
0.357
0.477
0.331
0
0.199
0.159
0.051
0
0.04
0.095
0.09
0
0.076
0.068
0.024
= C
m
0.357
0.477
0.331
0
0.199
0.159
0.051
0
0.04
0.095
0.09
0
0.076
0.068
0.024
=
100 200 300 400
2
0
2
fi()
i

15-22

P15.5-6
The function shown at right is related to the
given function by
() ()
1
16vt vt=− +−
(Multiply by –1 to flip v1 upside-down; subtract 6
to fix the average value; replace t by t+1 to shift
to the left by 1 s.)

From Table 15.5-1

()
() ()
0 2
1
16 1
nn
jnt
jnt
nn
jA j
vt e e
nn
π
ω
ππ
∞∞
=−∞ =−∞
−−
==∑∑

Therefore
()
() () ()1
22
61 61
66
nn
jnt jn jnt
nn
jj
vt e e e
nn
2
π ππ
ππ
∞∞
+
=−∞ =−∞
−−
=−− =−−


∑∑

The coefficients of this series are:
()
2
0n
61
6 and
n
jnj
Ce
n
π
π
−
=− =−C
This result can be checked using Matlab:

pi = 3.14159;
N=100;
A = 6; % amplitude
T = 4; % period
t = linspace(0,2*T,200); % time
c0 = -6; % average value
w0 = 2*pi/T; % fundamental frequency

for n = 1: N
C(n) = (-j*A*(-1)^n/n/pi)*exp(+j*n*pi/2);
D(n) = (+j*A*(-1)^n/n/pi)*exp(-j*n*pi/2);
end

for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));
end
end

plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
title('p15.5-6')

15-23

15-24

P15.5-7
Represent the function as
()
()
5
51 5
10
12
t
t
et
ft
ee t
−
−− −
 1− ≤≤
=
− ≤≤

(Check: () () ()
55
00,11 1,2ff e f ee
−−
== − =−.
5
0
−
=)
00
21
2 s, , also average value
22
TC
π
ωπ== = = =

The coefficients of the exponential Fourier series are calculated as

()
()
( )
( )
()
( )
()
()
()
()
12
5155
n
01
11 2 2
555 5
00 1 1
121 2
55
55
0 101
1
1
2
1
2
1
25 5
ttj nt jnt
jntjnt t jnt jnt
jnt jntjnt jnt
ee dt e eedt
edteedtee dteedt
ee e e
ee
jn jn jn jn
ππ
πππ π
πππ π
ππ π π
−−−− − −
−+−− − − −
−+ −+− −
−

=− + −


=− + −

 
 =− + −
 −− + −+ −
 
∫∫
∫∫ ∫ ∫
C




()
() ()
()
() ()
52 552
55
55 2 2
5
11 1
25 5
11 1
25 5
jn jnjn jn jn jn
jn jn jn jn jn jn
ee e e e e e
ee
jn jn jn jn
e ee eee ee
e
jn jn jn jn
ππππ π
ππ π π π π
ππ π π
ππ π π
−+ −+−− − − −
−
− −− −−− −−
−




 −− − −
=− + − 
 
−− + −+ −
 
 −− − −
=− + − 
 
−− + −+ −
 
π








() ()
()
()
()
()
55
5
11 11 1 111
25 5
nn n n
ee
e
jn jn jn jnππ π π
−−
−


 
  −− −− −− −−
=− + −  
  −− + −+ −
  

The terms that include the factor e are small and can be ignored.
5
0.00674
−
=

()
()
()
()
()( )
n
11 111
25 5
11
odd
5
0e ven
5
odd
5
0e ven
nn
jn jn jn
n
jn jn
n
n
jn jn
n
ππ
ππ
ππ
 −− −−−
=− + 
 −− + −+
 

−
+=




+=


C
π





15-25

This result can be checked using Matlab:

pi = 3.14159;
N=101;
T = 2; % period
t = linspace(0,2*T,200); % time
c0 = 0.5; % average value
w0 = 2*pi/T; % fundamental frequency

for n = 1:2:N
if n == 2*(n/2)
C(n) = 5/((+j*pi*n)*(5+j*pi*n));
D(n) = 5/((-j*pi*n)*(5-j*pi*n));
else
C(n)=0;
D(n)=0
end
end

for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));
end
end

plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');
title('p15.5-7')

15-26

Section 15-6: The Fourier Spectrum

P15.6-1
0Average value 0 0a=⇒ =
()()
2
0
2
0
44 2 4
cos (cos()1)
22
halfwave symmetry ()
44 2 2
sin 1cos
T
T
AA
at ntdtn
TT T
n
ft
AA
bt ntdtn
TT T n
π
π
π
π
π
π
   
n
n
= −= −∫  
  


−⇒ 

  
=− =− −∫  
   
(
−

22 1
n
= a = tan
11.509 57.5
20 0
30.434 78.0
40 0
50.257 82.7
60 0
70.183 84.8
n
nn n
n
b
nC b
a
A
A
A
A
θ
−
°
°
°
°

+ 

⋅−
⋅−
⋅−
⋅−

15-27

P15.6-2
Mathcad spreadsheet (p15_6_2.mcd):
N100:= n12,N..:= T32:= ω02
π
T
:=
Calculate the coefficients of the exponential Fourier series:
C1
n
4
T
3
T
16
⋅
T
4
t16
t
T
⋅3−





expj−n⋅ω0⋅t⋅()
⌠



⌡
d⋅:= C2
n
4
T
T
4
T
2
tsin
2π⋅
T
t⋅





expj−n⋅ω0⋅t⋅()
⌠



⌡
d⋅:=
C4
n
4
T
3
T
4
⋅
T
tsin
2π⋅
T
t⋅





expj−n⋅ω0⋅t⋅()
⌠



⌡
d⋅:=
C3
n
4
T
11
T
16
⋅
3T⋅
4
t11
16t⋅
T
−





expj−n⋅ω0⋅t⋅()
⌠



⌡
d⋅:=
C
n
C1
n
C2
n
+ C3
n
+ C4
n
+:=

Check: Plot the function using it's exponential Fourier series:
d
T
200
:= i12,400..:= t
i
di⋅:= fi()
1
N
n
C
n
expjn⋅ω0⋅t
i
⋅()
⋅∑
= 1
N
n
C
n

expj−n⋅ω0⋅t
i
⋅()
⋅∑
=
+:=
10 20 30 40 50 60
5
0
5
fi()
t
i
15-28

Plot the magnitude spectrum:
123 45 67 89 10
0
0.5
1
1.5
C
n
n

That’s not a very nice plot. Here are the values of the coefficients:

C
n
1.385
0
0.589
0
0.195
0
0.139
0
0.082
0
0.039
0
0.027
0
1.226·10
-3
0
= argC
n()
180
π
⋅
-115.853
-90
22.197
-24.775
-113.34
106.837
66.392
-78.232
-69.062
-48.814
109.584
90.415
-25.598
78.14
63.432
163.724
=

15-29

P15.6-3
Use Euler’s formula to convert the trigonometric series of the input to an exponential
series:

()
i
10 10 100 100
100 10 10 100
10cos10cos1010cos100 V
10 10 10
22 2
55 5 55 5
tt t t t
tt t t t t
vt t t t
ee e e e e
ee e ee e
−− − − − −
−− −
=+ +
++ +
=+ +
=+ + ++ +
t

The corresponding Fourier spectrum is:

Evaluating the network function at the frequencies of the input:

ω, rad/s |H(ω)| ∠ H(ω), °
1 1.923 -23
10 0.400 -127
100 0.005 -174

Using superposition:

() ( ) ( ) ( )
o
19.23cos234.0cos101270.05cos100174 Vvt t t t=− °+ −°+ −°

Use Euler’s formula to convert the trigonometric series of the output to an exponential
series:

()
() () () () () ()23 23 10127 10127 100174 100174
o
174 127 10 23 23 12710 174
19.23 4.0 0.05 V
22 2
=19.23 4.0 19.23 19.23 4.0 19.23
jt jt jt jt jt jt
jj t j jt j jt j jt j jt j
ee e e e e
vt
ee ee ee ee ee ee
−° −−° −° − −° −° − −°
°− °− °− −° − ° − °
++ +
=+ +
+ +++
jt

15-30

P15.6-4
00
21
1 s, 2 rad/s,
2
TC
T
π
ωπ== = =
1 s

()1w hen 0ft t t=−≤ <

The coefficients of the exponential Fourier series are given by

()
11 1
22 2
00 0
1
1
1
jnt jnt jnt
n
te dte dtte dt
ππ−− −
=− = −∫∫ ∫
C
π

Evaluate the first integral as
1
22
1
2
0
0
1
0
22
jnt jn
jnt ee
ed t
jn jn
ππ
π
ππ
−−
− −
= ==
−−
∫

To evaluate the second integral, recall the formula for integrating by parts:

22
2
1
11
tt t
ttt
udvuv vdu=−∫∫
. Take ut=and
2jnt
dve dt
π−
= . Then
() ()
1
2
11
22
00
0
1
22 2 2
22
0
1
22
11
22 22
jnt
jnt jnt
jn jnt jn jn
te
te dt e dt
jn jn
ee e e
j
jn jnjn jn
π
ππ
ππ π π
ππ
2nπ ππππ
−
−−
−− − −
=+
−
−
=+ =+ =
−−
∫∫

Therefore
1
0
2
0
2
n
n
C
j
n
nπ

=


=
−
 ≠


To check these coefficients, represent the function by it’s Fourier series:

()
22
1
1
22 2
n
jnt jnt
n
jj
ft e e
nn
ππ
π π
=∞
−
=
−
=+ +

∑

Next, use Matlab to plot the function from its Fourier seris (p15_6_4check.m):

pi = 3.14159;
N=20;

T = 1; % period
t = linspace(0,2*T,200); % time
15-31

c0 = 1/2; % average value
w0 = 2*pi/T; % fundamental frequency

for n = 1: N
C(n) = -j/(2*pi*n);
end

for i=1:length(t)
f(i)=c0;
for n=1:length(C)
f(i)=f(i)+C(n)*exp(j*n*w0*t(i))-C(n)*exp(-j*n*w0*t(i));
end
end

plot(t,f,'black');
xlabel('t, sec');
ylabel('f(t)');

This plot agrees with the given function, so we are confident that the coefficients are
correct. The magnitudes of the coefficients of the exponential Fourier series are:

n
1
0
2
1
0
2
n
n
nπ

=


=
 ≠

C

Finally, use the “stemplot” in Matlab to plot the Fourier spectrum (p15_6_4spectrum.m):

pi = 3.14159;
N=20;
n = linspace(-N,N,2*N+1);

Cn = abs(1/(2*pi)./n); % Division by 0 when n=0 causes Cn(N+1)=
NaN.
15-32

Cn(N+1)=1/2; % Fix Cn(N+1); C0=1/2

% Plot the spectrum using a stem plot
stem(n,Cn,'-*k');
xlabel('n');
ylabel('|Cn|');

15-33

Section 15.8: Circuits and Fourier Series

P15.8-1
The network function of the circuit is:
()
()
()
o
i
100
1
100
10 1
10
j
j
j
ω ω
ω
ωω
ω
== =
++
V
H
V

Evaluating the network function at the harmonic frequencies:
1
22
12 0 20
tan
22 0 400
1
20
n
n
n jn n
j
20
π π
π π π
− 
== = ∠−
 
+ +
+
H
From problem 15.4-2, the Fourier series of the input voltage is

()
c
1
12
6s in
22
n
vt ntn
n
ππ
π
∞
=

=−+ −


∑

Using superposition, the Fourier series of the output voltage is

()
1
o
22
1
240
6 sin tan
22 20400n
n
vt ntn
nn
ππ π
ππ
∞
−
=
  
=−+ − + 
+ 
∑

P15.8-2
The network function of the circuit is:
()
()
() () ( )
()()
()()() ()()()
2
22 12o
i 11 22
1
1
6
66
1
1 11
102000
1 1010001 102000
500
11
1000 500
R
jCR jCR
jCR jCR
R
jC
j
jj
j
jj
ωωω
ω
ω ωω
ω
ω
ωω
ω
ωω
−
−−
+
== − =−
++
+
=−
++
=−
  
++  
  
V
H
V

Evaluating the network function at the harmonic frequencies:
15-34

2
1000 3
23
11
33
jn
n
jn jn
π
π
π π

=−

 
++
 
 
H




From problem 15.4-4, the Fourier series of the input voltage is

()
b 22
1
1 18 1000 2
1cos cos
23 3
n
n
vt n tn
n 3
π ππ
π
∞
=
  
=−+ − −
 
 
∑




Using superposition, the Fourier series of the output voltage is

()
b 22
1
1000
18
31 1000 2 1000
1cos cos
23 3 3
n
n
n
vt n tn n
n
π
3
π ππ π
π
∞
=

×

   
=−+ − − +∠
   
   
∑
H
H




P15.8-3
()
()
()
o
1
H
j
p
ω
ω
ωω
==
+
o
i
V
H
V

When ω = 0 (dc)
()
4
5 2 25 k
10
R
R−=− ⇒ = Ω
When ω = 100 rad/s

() ( ) ()()( )
1
135 180tan tan45 10025000
0.4µF
CR C
C
ωω
−
°=∠ =°− ⇒ °=
⇒=
H

()()()
()() ()
4
4
6
25000
10
54005 3.032
14000.41025000
c
j
−
== =
+×
H
() ( )
16
445 40045180tan4000.41025000149θ
−−
=°+∠ =°+− ××× =H °

15-35

P15.8-4
When ω = 0 (dc)
()
o o
52 2.5 VHH=⇒ = /V
When ω = 25 rad/s
() ()
1 25
45 tan tan45 25 rad/sp
pp
ω
ω
−
−°=∠ =− ⇒ °= ⇒ =


H
()()()
4
2.5
51005 3.03
100
1
25
c
j
==
+
H =
()
1
4
100
45 10045tan 31
25
θ
−
=°+∠ =°− =−


H °

P15.8-5
()
()
()
2
12 1
R
2RRj CRR
ω
ω
ωω
==
++
o
i
V
H
V

When ω = 0 (dc)
2
12
12
3.75 2.25 2.25
(500)300
6 3.75 3.75
R
RR
RR
 
=⇒ = = =
 
+  
Ω
When ω = 1000 rad/s

() () ()
()()121
12
300500
20.5 tan tan20.51000
800
2µF
RR
CC
RR
C
ωω
−
 
−°=∠ =− ⇒ °= 
 +

⇒=
H

()() ()()
()
33
6
500
5452.0763.4
8003000210500300
c
j
θ
−

∠= ∠°= ∠−
+×


15-36

P15.8-6

ˆRather than find the Fourier Series of () directly, consider the signal () shown above.
These two signals are related by
vt vt

ˆ() ( 1)6vtvt=−−

since () is delayed by 1 ms and shifted down by 6 V.vt

ˆThe Fourier series of ()isobtained asfollows:vt

0
2radians
4 ms rad/ms
4ms 2
T
π π
ω=⇒ = =

n
ˆ ˆa 0 because the averagevalueof() 0vt==
()
4
0
44
00
4
4
22
0
0
1
ˆ ˆ 63sin because ()is an odd function.
22
3
3sin sin
222
cos
312
3s in cos
22 2 2
2 4
nbt ntdt vt
ntdt tntdt
nt
n
nt t nt
n
n
π
ππ
π
ππ π
π π

=−


 
=−  
 
 
−
 
 
=− −  
 


∫
∫∫
()() ()() ()()()
22
66
1 cos2 sin2 02cos20nn n
nn
ππ π π
12
nπ ππ


=− + − −− − −=

Finally,
1
12
ˆ() sin
2n
vt nt
n
π
π
∞
=
=∑

ˆThe Fourier series of ()isobtainedfromtheFourier series of () asfollows:vt vt

()
1
12 12
() 6 sin 1 6 sin
22
1
n
vt nt ntn
nn
n
2
π ππ
ππ
∞
=
∞

=−+ −=−+ −∑∑ 

=

where t is in ms. Equivalently,

15-37

3
1
121
() 6 sin 10
22n
vt n tn
n
π π
π
∞
=

=−+ −∑ 


where t is in s.

Next, the transfer function of the circuit is
2
()
11
R
s
R
L
Hs
R
LsR ss
Cs LLC
==
++ ++
.
The network function of the circuit is
()
4
82 4
2
10
()
1 10 10
R
j
j
L
R j
j
LC L
ω
ω
ω
ω ω
ωω
= =
 −+
−+

H .
We see that H(0) = 0 and

()
()
()
1
22
20
90tan
3 400
0 22 2
22 22
20 1
10
2 400 20
400 400
n
j
n
jn
nn e
nj n
nn
π
π
ππ
ω
ππ
ππ
−
−

−
== =

−+
−+
HH

Finally, ()
()
31
22
0
2
1 22 22
20
sin 10 90tan
2 2 40012

400 400
n
n
nt n
n
nn n
ππ π
π
π
ππ
°−
∞
=
 
−+ −

−
=∑
−+
vt

15-38

P15.8-7
()
ˆRather than find the Fourier Series of () directly, consider the signal () shown below.
ˆThese two signals are related by () 21
vt vt
vt vt=− −

ˆLet's calculate the Fourier Series of (),taking advantage of its symmetry.vt

0
2rad
= 6 ms radms
6 ms3
T
π π
ω⇒= =

()
1
0
3.2
1
2
ˆ average value of () V
62
ˆ0because() is an even function
2
23 3cos
36
o
n
n
av t
bv t
at ntdt
π
==
=

=−


∫
=

11
00
1
22
0
22 22
2cos 2cos
33
sin
13
2c os sin
33 3
3 9
61 8 6 18
sin cos 1 sin cos1
33 3
nan tdttntdt
n
ntnt nt
n
n
nn n
nn n n 3
n
π π
π
πππ
ππ
ππ π
ππ π π
=−



=− + 



 
=− −+ =− 
 
∫∫
π
−


so

()()
22
22
11 8
ˆ() 1coscos
23 3
1
11 8
ˆ21 1coscos
23
1
vt n nt
n
n
n
vtvt ntn
n
n
ππ
π
2
3 3
π ππ
π
∞

=+ −∑ 

=
∞
 
=− −=−+ − −∑  
 



=

15-39

where t is in ms. Equivalently,

3
22
1
11 8 2
() 1coscos10
23 3n
n
vt n tn
n 3
π ππ
π
∞
=
 
=−+ − −∑  
 




where t is in s.

The network function of the circuit is:
() ( )
2
1222
11 22
1
1
1
()
1 11
R
jCRjCR
jRC jRC
R
jC
ωω
ω
ωω
ω
−
+
==
++
+
H
Evaluate the network function at the harmonic frequencies of the input to get.
()
3
0
3
10
23
11
33
jn
nn
jn jn
π
π
ω
π π
−

==

 
++ 
 
HH




The gain and phase shift are

()
() ( )
()
0
22 22 22 22
11
0
3

4 99 4
11
99
2
90tan tan
33
n
n
n
nn nn
nn n
π
π
ω
ππ π π
ππ
ω
°− −
==
   ++
++
  
  

∠= −− +

H
H

The output voltage is
()
() ( )
31
0
22 22 22
1
22
181coscos10 90tan tan
33 3 3

99 4
n
n
nt n n n
vt
nn n
1
3
π ππ π
ππ π
°− −
∞
=
 
−− −− − 
 
=∑
++
π



At t = 4 ms =0.004 s

()
() ( )
11
0
22 22 22
1
42 2
181coscos 90tan tan
33 3 3
.004
99 4
n
n
nn n n
v
nn n
3
π ππ π π
ππ π
°− −
∞
=
 
−− −− −
 
 
=∑
++




15-40

Section 15.9 The Fourier Transform

P15.9-1
Let . Notice that () () ()
at at
gteuteut
−
=− − () ()
0
lim .
a
ft g
→
= t Next

()
() ()
() ()
0
() ( )
0
0
0
22
11
00
ajt ajt
atjt atjt ee
G eedteedt
aj aj
2j
aj aj a
ωω
ωω
ω
ωω
ω
ω ωω
∞
−+ −
∞
−− −
−∞
−∞
=− = −
−+ −
  −
=− − −=  
  
−+ − +
  
∫∫

Finally () ()
22
00
22
lim lim
aa
j
FG
aj
ω
ωω
ωω
→→
−
== =
+

P15.9-2
() ()
() ()
()
0
0
0
ajt
at jt atjt Ae A A
FA eutedt Aeedt
aj aj aj
ω
ωω
ω
ω ωω
∞
−+
∞∞
−− −−
−∞
== = =−
−+ −+ +
∫∫
=

P15.9-3
First notice that

Then, from line 6 of Table 15.10-2: (){}
2
22
1
22 4 4 4
ATT T AT T
ft Sa Sa
ωω

−  
=− =
  
  

F




From line 7 of Table 15.10-2: (){} () (){}
2
2
11
44
dA
ft ftj ft j Sa
dt
ω
ωω
T T 
== =− 
 
FF F



This can be written as: (){}
2
sin
2
44 2
sin
244
4
T
AT A T
ft j
j
T
ω
ω
ω
ω
ω



=− =





F
15-41

P15.9-4
First notice that: (){ } ()
111 0
0022
jt
jt
ed e
ω
ω
δωω δωω ω
ππ
∞
−∞
−
−−
−= − =∫
F
Therefore {} (
0
02
jt
e
ω
)πδωω
−
= −F . Next, 10 .
50 50
cos505 5
jt jt
te e
−
=+
Therefore { }{ }{ }
50 50
10cos50 5 5 10(50)10(50)
jt jt
te e πδω πδω
−
=+ = −+FF F +.

P15.9-5

() () () (() )
() ()
2
2
2
1
1
22 2
2 cos2 sin2 cos sin
22
coscos2 sinsin2
jt
jt j je
Fe dt e e j j
jj j
j
ω
ωω ω
ω ωω ω
ωω ω
ωω ω ω
ω
ω ω
−
−− −−
=− = = − = − − −
−
=− + −
∫

P15.9-6

()
()
() ()
2 22
0
0
22
1
11
1
B
jt jB
B
jt
jB jB
AA e Ae
Ft edt jt jB
BB Bj
ABe e
Bj
ωω
ω
ωω
ωω
ωωω
ω
ωωω
−−
−
−−
  
== −− = −−  
−−  
 −
=+ −
 
 
∫

P15.9-7

() () ()
()
21
21
22
21
21
11
2
sin2sin
jt jt
jt jt j j j jee
Fe dtedt e e ee
jj j j
ωω
ω ωω ω
ω
ωω ω ω
ωω
ω ω
ω
−−
−− −
−−
−−
=− = − = − − −
−−
=−
∫∫
−

P15.12-1
() ()
()
()
()
()
() ()()
() () ()
4
40signum
28 0
40
1
4
18 02020
44
10signum 20
s
s
s
s
t
it t
I
jj
I
H
Ij
IH I
jjj j
it t eut
ω
ωω
ω
ω
ωω
ωω ω
ωωω
−
=

==


==
+
== ×=−
++ ω
∴ =−

15-42

P15.12-2
()
() ( )( )
()
()
()
()
() ( )
100cos3A
100 3 3
1
4
33
100
4
s
s
s
it t
I
I
H
Ij
I
j
ωπ δωδω
ω
ω
ωω
δω δω
ωπ
ω
=
=− +

==
+
−+ +
= 
+

+

()
() ( )
() ()
()
33
336.9 336.9
33100
50
24 434
10
10cos336.9
jt jt
jt
jt jt
ee
it ed
jj
ee
t
ω
δω δωπ
ω
πω
−
∞
−∞
−− −
−+ + 
==  
+− 
3j
+
+
 =+
 
=−
∫

P15.12-3
()
() ( )( )
()
() ()()
() ( )
10cos2
10 2 2
1
2
10 2 2
2
vt t
j
j
ωπ δω δω
ω
ω
ωω ω
πδω δω
ω
=
=+ + −

=
+
=
++ −

=
+
V
Y
IY V

()
() ( )
() ( )
()
22
245 245
2210
5
22 2222
55 cos2
jt jt
jt
jt jt
ee
it ed
jj j
ee t
ω
δω δωπ
ω
πω
−
∞
−∞
−− −
++ − 
== + 
+− +
=+ = −

∫
45A

15-43

P15.12-4
() ()()
(){} ()
()
(){} ()
() ()
0
1
0 1
11
1
11
1
t
jt
tt jt tjt
vteutut
e
eut eutedt eedt
j j
ut
j
V
jj
ω
ωω
ω ω
πδω
ω
ωπ δω
ωω
−
∞
−−
−∞ −∞
−∞
=− +
−= − = = =
−−
=+
∴ =+ +
−
∫∫
F
F

()
() ()
()
() ()
() () () ()
1
3
11 1
2 11 2
,
11 123
1
22
111
11 1 312 4
31 3 1 3
11
32 3 6
11 1 1
signum
12 4 3 6
o
jt
tt
o
j j
H
jj
jj
V
jj j j jj
ed
jj
vt euteut t
ω
ω ω
ω
ωω
ωω
j
πδω
ωπ δω
ωωω ω ωω
πδω πδω
ω
ωπ ω
∞
−−
−∞
−


+
== =
++
++
+
−

=+ + = + ++

+− + − +

==
++

ω
∴ =− + −+ +
∫
F

P15.12-5

() () ()
()() ( )
()
()
5
22
55
0
2 2
22
00
15
15 V
5
15 15 22.5J
1 1
11
=10 F. Try =10 k. Then
10 15
10 5
1 10 15 1 300 300
15J
10 5 25 100
t
s
tt
s
o
o
vt eut V
j
We utdt edt
jC RC
H
Rj
jC RC
CR
V
jj
Wd
jj
ω
ω
ω
ω
ω
ω
µ
ω
ωω
ωω
πω ω π ω ω
−
∞∞
−−
−∞
∞∞
=⇒ =
+
== =
==
++
Ω
=×
++
 
=× = − =
 
++ + +
∫∫
∫∫
d

15-44

P15.12-6

()
() ()(){} () ()
() () () ()
() ()
4
4
88
88 1 8 8
8
1s ince
4 8 88 88
1
44
j
s
jj
s
jj
o
H
j
Vu tut
jj
Ve e
j
Ve
jj jj jj4
e
e
ω
ωω
ωω
ω
ω
ωπ δω πδ
ωω
ωδ ω δω
ω
ω
ωω ωω ωω
−
−−
−−
=
+
 
=− −= + − +
 
 
=− =
 
=× − = − − −
 
++  
F ω
+

() ()
() () () () ()
() ()
() () () ()
()
()()
414
11
Next use to write
18 18
88
44
18 18
88
44
88 8 18 1
8(1
j
o
j
tt
o
jj
V e
jj jj
e
jj jj
vtut eutut eut
e
ω
ω
πδωπδω
ωω
ω πδωπδω πδωπδω
ωω ωω
πδω πδω
ωω ωω
−
−
−−−
−
=+ −
  
=+ − − − + − −  
++  
  
= + −− +−  
++  
=− − −− −
=− ()
()
()
414
)8 (1 )1V
tt
ut e ut
−−
−− −

15-45

PSpice Problems

SP 15-1

Vin 1 0 pulse (25 5 0 0 0 4 5)
R1 1 0 1

.tran 0.01 5
.four 0.2 v(1)
.probe
.end

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 8.960000E+00

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE(DEG)

1 2.000E-01 7.419E+00 1.000E+00 1.253E+02 0.000E+00
2 4.000E-01 6.030E+00 8.127E-01 1.606E+02 3.528E+01
3 6.000E-01 4.061E+00 5.473E-01 -1.642E+02 -2.894E+02
4 8.000E-01 1.935E+00 2.609E-01 -1.289E+02 -2.542E+02
5 1.000E+00 8.000E-02 1.078E-02 -9.360E+01 -2.189E+02
6 1.200E+00 1.182E+00 1.593E-01 1.217E+02 -3.600E+00
7 1.400E+00 1.704E+00 2.297E-01 1.570E+02 3.168E+01
8 1.600E+00 1.537E+00 2.072E-01 -1.678E+02 -2.930E+02
9 1.800E+00 8.954E−01 1.207E-01 -1.325E+02 -2.578E+02

SP 15-2

Vin 1 0 pulse (1 -1 -0.5 1 0 0 1)
R1 1 0 1

.tran 0.1 1
.four 1 v(1)
.probe
.end

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 1.299437E-02

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 1.000E+00 6.364E-01 1.000E+00 -1.777E+02 0.000E+00
2 2.000E+00 3.180E-01 4.996E-01 4.679E+00 1.823E+02
3 3.000E+00 2.117E-01 3.326E-01 -1.730E+02 4.682E+00
4 4.000E+00 1.585E-01 2.490E-01 9.366E+00 1.870E+02
5 5.000E+00 1.264E-01 1.987E-01 -1.683E+02 9.376E+00
6 6.000E+00 1.051E-01 1.651E-01 1.407E+01 1.917E+02
7 7.000E+00 8.972E-02 1.410E-01 -1.636E+02 1.409E+01
8 8.000E+00 7.817E-02 1.228E-01 1.880E+01 1.965E+02
9 9.000E+00 6.916E-02 1.087E-01 -1.588E+02 1.883E+01

Verification Problems

15-46

VP 15-1

01()2cos 2 , 1 and all other coefficients are zero.
2
The computer printout is correct.
t
ft a a=+ ⇒ = =

VP 15-2

0
Table 15.4-2 shows that the average value of a full wave rectified sinewave is
2 2(400)
where is the amplitude of the sinewave. In this case 255.
Unfortunately the report says, "half-wave rectif
A
Aa
ππ
==
ied." The report is not correct.

Design Problems
DP 15-1

0
/2
0
0
3
For sinusoidal analysis, shift horizontal axis to average, which is 6 V.
Now we have an odd function so 0
s , 2/ 2 rad/s
22
()sin
Need third harmonic :
4
sin 6
n
T
n
a
T
bf tntdt
T
bt dt
T
πω ππ
ω
=
== =
×
=
=
∫
() () ()
() ()()()
/2 /2
00
1 1
0
20 1 0 0
4
cos6 0.424
6
0.424sin60.424cos(690) V 0.42490
for third harmonic
6
16
transfer function is (3)
16
6
= 3 = (3) (3) (0.42490)
Ch

c
t
vt t t
jj
CC
j
C
π π
π
ω
ω
ω
ωω ω ω ω
°
=− =
== −° ⇒ =
−−
==
∠−°
∴ =
−
∠∠ −
∫
V
Z
H
VH V H H
() ()
()
()
20
0
2
oose = 1.36 so 3 3.2
11 6
This requires = F. Then 3 3.264.9
205 1634
third harmonic of 1.36sin(664.9) V
C
j
vt t
ωω
ω
°
⇒=
== ∠ °
−
∴ =+
VH
H

15-47

DP 15-2
Refer to Table 15.4-2.
24 1
() cos(2)
02
411
N
AA
vt nt
s
nn
ω
ππ

=− ∑ 
−=

In our case:
360 6401
() cos(2377)
2
411
N
vt nt
s
nn
ππ

=− ∑ 
−=

()
s0 s 0 o0 o
11
oo 0 o1
1
o0 s0
Let () () and () ()
We require ripple 0.04 dc output
max () 0.04 () 0.04
but because the inductor acts like a short at dc.
NN
sn n
nn
N
n
n
vtv vt vtv vt
vt v vt v
vv
==
=
=+ =+∑∑
≤⋅
≤⋅ ⇒ ≤∑
=
o0

Next, using the network function of the circuit gives
os
0
=
nn
n
R
Rj Lω


+

VV .

For n=1:
o1 s1 s1 s1 01
0
1 640 1 640
= = , but so
1377 (3) 13773
R
RjL jL jLω ππ

==

++ + 
VV V V V

o1 o0 o0 s0
22
360 1 640 360
We require 0.04 and = = . Then 0.04
31(377)
vv v
Lπ ππ

≤⋅

+
V ≤
Solving for yields > 1.54 mHLL

15-48

DP 15-3
From Table 15.5-1, the Fourier series can represent the input to the circuit as:

()
()
00
2
even 2
11
44 1
jt jt jnt
s
n
jj
vt e e e
n
ωω
π π
∞
=
=+ + +
−
∑
0
ω

The transfer function of the circuit is calculated as
p
o1 s1 p
Lp
= where =
+1
R
jRCω+
Z
VV Z
ZZ

So
o
2
s
1

11
()()
LC
jj
RCLC
ωω
==
++
V
V

The gain at dc, 0ω=, is 1 so
o0 s0
1
vv
π
==
For n = 1
o1 o0 s0
22
4
11 1 1/
20 20 20 20
1
LC
vv
RC LC
1
π π
ω
ω
== = ⇒ =
 
++
 
 
V

We are given =800 and =75 k. Choosing =0.1 mH yields =0.1 FRL Cωπ Ω

15-49

Chapter 16: Filter Circuits

Exercises
Ex. 16.3-1
1
()
1
1 1250
()
1250 1250
1
1250
n
n
Ts
s
s
Ts T
s s
=
+

== =

+
+

Problems
Section 16.3: Filters

P16.3-1
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s.
2
1
()
(1)( 1)
n
Hs
ss s
=
+ ++

Frequency scaling so that
c
= 2100=628 rad/sωπ :
3
22 22
1 628 247673152
()
(628)(628628)(628)(628394384)
11
628 628628
L
Hs
ss s s s sss s
== =
++ + + + +
 
++ + 

 


P16.3-2
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a
cutoff frequency equal to 1 rad/s and a dc gain equal to 1.

2
1
()
(1)( 1
n
Hs
ss s
=
)+ ++

Multiplying by 5 to change the dc gain to 5 and frequency scaling to change the cutoff
frequency to ωc = 100 rad/s:

3
22 22
5 5100 5000000
()
(100)(100100)(100)(10010000)
1 1
100 100100
L
Hs
ss s s s sss s
⋅
== =
++ + + ++
 
++ + 

 


16-1

P16.3-3
Use Table 16-3.2 to obtain the transfer function of a third-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

3
2
5
()
(1)( 1)
n
s
Hs
ss s
=
+ ++

Frequency scaling to change the cutoff frequency to
c
100 rad/sω=

3
33
22 22
5
55100
()
(100)(100100)(100)(10010000)
1 1
100 100100
H
s
ss
Hs
ss s s s sss s


⋅⋅
== =
++ + + ++
 
++ + 

 


P16.3-4
Use Table 16-3.2 to obtain the transfer function of a fourth-order Butterworth high-pass filter
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

()
() ( )
4
22
5

0.7651 1.8481
s
Hs
n
ss s s
⋅
=
+ ++ +

Frequency scaling can be used to adjust the cutoff frequency 500 hertz = 3142 rad/s:

()
() ( )
4
22
4
22 2 2
5
3142
0.765 1 1.848 1
3142 3142 3142 3142
5
2403.63142 5806.43142
H
s
Hs
ss s s
s
ss s s

⋅


=
 
   
++ +    
 
   
 
⋅
=
++ + +

+



16-2

P16.3-5
First, obtain the transfer function of a second-order Butterworth low-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:

()
2 2
2 8000000

28284000000
1.414 1
2000 2000
L
Hs
ssss
==
++ 
++
 
 

Next, obtain the transfer function of a second-order Butterworth high-pass filter having a
passband gain equal to 2 and a cutoff frequency equal to 100 rad/s:

()
2
2
2 2
2
2100

141.410000
1.414 1
100 100
H
s
s
Hs
ssss

⋅
⋅
==
++ 
++ 
 

Finally, the transfer function of the bandpass filter is

() () ()
() ( )
2
22
16000000

141.410000 2828 4000000
BL
s
Hs HsHs
H
ss s s
⋅
=⋅ =
++ + +

P16.3-6
()
()
2
2
2
2
22
250
250000
1
4
250
25062500250
1
B
s
s
Hs
ssss


==
++++


P16.3-7
First, obtain the transfer function of a second-order Butterworth high-pass filter having a dc gain
equal to 2 and a cutoff frequency equal to 2000 rad/s:

()
2
2
2 2
2
22000
28284000000
1.414 1
2000 2000
L
s
s
Hs
ssss



==
++ 
++ 
 

Next, obtain the transfer function of a second-order Butterworth low-pass filter having a pass-
band gain equal to 2 and a cutoff frequency equal to 100 rad/s:

16-3

()
2 2
2 20000
141.410000
1.414 1
100 100
HHs
ssss
==
++ 
++ 
 

Finally, the transfer function of the band-stop filter is

() () ()
( ) ( )
() ( )
() ( )
22 2
22
43 2 10
22
2 141.4100002000028284000000
141.410000 28284000000
2282.84000056560000810
141.410000 28284000000
NL H
ss s s s
Hs HsHs
ss s s
ss s s
ss s s
++ + + +
=+ =
++ + +
++ + +⋅
=
++ + +

P16.3-8
()
()
()
2
2
2
2
2
22
250
462500
1
44
250
25062500250
1
N
s
s
Hs
ssss

+
=− =
++++


P16.3-9
()
()
2
24
2
22
250 4250
4
250
2
250 25062500
1
LHs
ss ss

 ⋅
= =
++ ++


P16.3-10
()
()
2
24
2
2
22
4
4
250
25062500250
1
H
ss
Hs
ssss

 ⋅
==
++++


16-4

Section 16.4: Second-Order Filters

P16.4-1
The transfer function is
()
()
()
0
2 1
s
Vs
RC
Ts
sVss s
RCLC
==
++

so
2 0
00
11
1 , and
C
KQ
LC RCQ L
ω
ωω== =⇒= =RC R

2
0
1
Pick 1 F. Then 1 H and 1000
L
CL RQ
CC
µ
ω
== = = = Ω

P16.4-2
The transfer function is
0
2
1
()
()
1()
s
Is LC
Ts
sIs
s
RCLC
==
++

so
2 0
00
11
1 , and
C
KQ
LC RCQ L
ω
ωω== =⇒= =RC R

2
0
1
Pick 1F then 25 H and 3535
L
CL RQ
CC
µ
ω
== = = = Ω

16-5

P16.4-3
The transfer function is
2
1
2
22
1
1
()
11
2
RRC
Ts
R
ss
RCR RC
−
=

++ +


Pick 0.01 FC µ= , then
0
0
1
1
1
2000 5000050
1
2 8333 8.33
2
Rk
RC
RR
R k
QRC R Q
ω
ω
== ⇒= = Ω

=+ ⇒= = =
−
Ω

P16.4-4
12 3Pick 0.02 F. Then 40 k, 400 k and R=3.252 k.CR Rµ== Ω= Ω Ω

P16.4-5
12
Pick 1 FCC Cµ== = . Then
6
0
12
10

RR
ω=
and
110
2 2
12
1

R R
QR
RCQ R Q
ω
=⇒ = ⇒ =

6
21 1
10
In this case and 10001 k
1000
RR R== = =Ω

16-6

P16.4-6

The node equations are
() ()
()()
()()()
2
0a
2
2
0a
1a i
1
1
0
R
Vs Vs
R
Cs
VsVs
CsVsVs
R
=
+
−
− −=

The transfer function is:
()
()
()
2
0
2
i
22 1212
1
Vs s
Ts
sVs
s
RCR RCC
==
++

02 2
12 12
0
2112
11
Pick 1 F. Then and
R
CC C Q RQR
RC Q RCRR
ω
µω== = = =⇒= ⇒ =.
12 0
1
In this case and 1000 RR R R
CR
ω== =⇒= Ω.

16-7

P16.4-7

()
0
2
i
11
()
11()
Vs Cs LC
Ts
RVs
LsR ss
Cs LLC
== =
++ ++

2 6
When 25 , 10 H and 410 F, thenRL C
− −
=Ω = =× the transfer function is

()
6
26
2510
25002510
Ts
ss
×
=
++ ×

so
6
old25105000ω=× =
and
new
f
old
250
0.05
5000
k
ω
ω
== =
The scaled circuit is

P16.4-8

The transfer function of this circuit is

16-8

()
()
()
2
22 120
i 2
1
1
11 22 1212
1
1
1
11 1

R
s
RCs RCVs
Ts
Vs
R
ss
Cs
RCRC RRCC
+
== − =−

+
++ +



()
m
11 2 2
100 F 500 F
Pick 1000 so that the scaled capacitances will be 0.1 F and 0.5 F.
1000 1000
Before scaling 20 , 100 F, = 10 and 500
k
RC R C F
µ µ
µ µ
µµ
==
=Ω = Ω =
=

()
25
100s
Ts
s700s+10
−
=
+

()11 2 2After scaling 20000 20 k, 0.1 F, 10000 10 k, 0.5 FRC R Cµ µ=Ω =Ω= = Ω=Ω=

()
25
100
70010
s
Ts
ss
−
=
++

P16.4-9
This is the frequency response of a bandpass filter, so

()
0
202
0
Ks
Q
Ts
ss
Q
ω
ω
ω
=
++

From peak of the frequency response

66
0
2101062.810 rad/s and k=10 dB3.16pω=× ×= × =
Next
66 6 60
BW (10.1109.910) 2 (0.210)2 1.2610 rad/s
Q
ω
ππ== ×−× = × = ×

So the transfer function is

66
26 2 12 2 6
3.16(1.26)10 (3.98)10
()
(1.26)1062.8.10 (1.26)103.944.10
ss
Ts
ss ss
==
++ ++
15

16-9

P16.4-10

(a)
()
()
()
2
220

11 2
s1 1
2
2
1
= where and
1
R
jCj
R
C
R
jC
ωω
ω
ωω
ω
×
=− =− =
+
ZV
HZ Z
VZ

()
21
12
11 22
12
11
= where ,
1+ 1+
jRC
RCR
jj
ω
ωω
ωω
ωω
∴− = =


 
 
H
C
ω

(b)
12
11 22
11
,
RCR
ω ω==
C

(c)
()
()
21 221
12 1
1
1
1

=
0+ 1+0
RCRjRC
RC
Rj
ωω
ωω
ω
ω
ω
ω
−= =




H =

P16.4-11

Voltage division:
() () () ()
01 1
1
s, (1 )
1
Cs
Vs V Vs sRCVs
R
Cs
=⇒ =+
+
0
KCL:
()
1s 1 0
10+ + 0
VV VV
VVnCs
mR R
−−
− =
Combining these equations gives:

22
0
1

sVsC
Vs C snRC
mR m mR

++ + =



16-10

Therefore
()
() ()
0
222 2
00
V 11
() =
V1 1 s
1j
sm RCnmRCs
Q
ω
ω
ω
ω ω
ω ω
==
++ + 
−+


H
0
1
where and
1
mn
Q
mmnRC
ω==
+

P16.4-12

2
2
2 22 120
11S 2
1
11 11 22 1212
1 1
1()
() =
11() 11 1

R
R s
Cs RCs RCVs
Hs
RCsVs
R
ss
Cs Cs RCR C RRCC
−
+
=− = =
+ 
+
++ +



where
()
()
0
12 12
0
11 22
1
70.7 kradsec= 211.25 kHz
11
BW= 150 k rad =223.9 kHz
Q
RRCC
s
RC RC
ωπ
ω
π
==
=+ =

16-11

P16.4-13

()
a0
1a s
1 220
2
as
20
11 1212
2
1
= 0 s
()
()=
11()
ss 0
VV
CsVV
R RCVs
Hs
VV s
sCV
RCR RCCR
− 
−+ −


⇒=
 ++−− =



Comparing this transfer function to the standard form of the transfer function of a second order
bandpass filter gives:

4
0
12 12
3

11
0
1
10radsec
1
BW = 10rad/sec
10
BW
RRCC
RC
Q
ω
ω
==
=
==

16-12

P16.4-14

Node equations:

()()
()()
cs c 0
c
sc
s0
()() ()()
() 0
() 0
VsVs VsVs
aCsVs
RR
VsVsC
sVsVs
aR

− −
+ +=
−
−+ =

Solving these equations yields the transfer function:

()
()
()
()
2
2
0
2
s
2
21 1

()
21 1

sa s
aR CVs RC
Hs
Vs
ss
aRCRC

++ +


==

++


We require
51
10
RC
= . Pick 0.01F then 1000 C Rµ= = Ω. Next at
0sjω=
()
2
0
2
1
2 2
a
aa
a
ω
+
= =+H
The specifications require
()
2
0201 1 20
2
a
aω== + ⇒ =H

16-13

P16.4-15

Node equations:

()
aa 0
21
ab
a
bs
b0
=0
0
+
VV V
RR
VV
CsV
R
VV
CsVV 0
R
−
+
−
+=
−
− =

Solving the node equation yields:
()
()
1
22
20
s 12
22
2
1
1
11
2
R
RRCVs
Vs R
ss
RRC RC

+


=

+− +



() ( )
0 39
11
41.67 kradsec
1.210 2010RC
ω
−
== =
××

16-14

Section 16.5: High-Order Filters

P16.5-1
This filter is designed as a cascade connection of a Sallen-key low-pass filter designed as
described in Table 16.4-2 and a first-order low-pass filter designed as described in Table 16.5-2.

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_sklp.mcd)
A1.586=Calculate the dc gain.
RA 1−()⋅ 9.33110
3
×=R1.59210
4
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q0.707=ω0628=Q
ω0
b
:=ω0a:=Determine the Filter Specifications:
b628:=a628
2
:=Enter the transfer function coefficitents:
c
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

16-15

First-Order Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_1stlp.mcd)

The transfer function is of the form T(s) =
-k
-------- .
s + p
Enter the transfer function coefficitents:p628:= k0.5p:=
Pick a convenient value for the capacitance:C0.110
6−
⋅:=
Calculate resistance values:R2
1
Cp⋅
:= R1
1
Ck⋅
:= R13.18510
4
×= R21.59210
4
×=

P16.5-2
This filter is designed as a cascade connection of a Sallen-key high-pass filter, designed as
described in Table 16.4-2, and a first-order high-pass filter, designed as described in Table 16.5-
2.

The passband gain of the Sallen key stage is 2 and the passband gain of the first-order stage is
2.5 So the overall passband gain is 2 × 2.5 = 5

Sallen-Key High-Pass Filter:

16-16

MathCad Spreadsheet (p16_5_2_skhp.mcd)
A2=Calculate the passband gain.
RA 1−()⋅ 110
5
×=R1 10
5
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q1=ω0100=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b100:=a10000:=Enter the transfer function coefficitents:
A s^2
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

First-Order High-Pass Filter:

MathCad Spreadsheet (p16_5_2_1sthp.mcd)

The transfer function is of the form T(s) =
-ks
-------- .
s + p
Enter the transfer function coefficitents:p100:= k2.5:=
Pick a convenient value for the capacitance:C0.110
6−
⋅:=
Calculate resistance values:R1
1
Cp⋅
:= R2kR1⋅:= R1110
5
×= R22.510
5
×=

16-17

P16.5-3
This filter is designed as a cascade connection of a Sallen-key low-pass filter, a Sallen-key high-
pass filter and an inverting amplifier.

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_3_sklp.mcd)
A1.586=Calculate the dc gain.
RA 1−()⋅ 2.9310
3
×=R5 10
3
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q0.707=ω0210
3
×=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b2828:=a4000000:=Enter the transfer function coefficitents:
c
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

16-18

Sallen-Key High-Pass Filter:

MathCad Spreadsheet (p16_5_3_skhp.mcd)
A1.586=Calculate the passband gain.
RA 1−()⋅ 5.8610
4
×=R1 10
5
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q0.707=ω0100=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b141.4:=a10000:=Enter the transfer function coefficitents:
c s^2
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

Amplifier: The required passband gain is
6
1.610
4.00
141.42828
×
=
×
. An amplifier with a gain equal to
4.0
1.59
2.515
= is needed to achieve the specified gain.

16-19

P16.5-4
This filter is designed as the cascade connection of two identical Sallen-key bandpass filters:

Sallen-Key BandPass Filter:

MathCad Spreadsheet (p16_5_4_skbp.mcd)
AQ⋅2=Calculate the pass-band gain.
RA 1−()⋅ 410
4
×=2R⋅810
4
×=R4 10
4
×=
A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q1=ω0250=Q
ω0
b
:=ω0a:=Determine the Filter Specifications:
b250:=a62500:=Enter the transfer function coefficitents:
cs
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

16-20

P16.5-5
This filter is designed using this structure:

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_5_sklp.mcd)
A1.586=Calculate the dc gain.
RA 1−()⋅ 5.8610
4
×=R1 10
5
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q0.707=ω0100=Q
ω0
b
:=ω0a:=Determine the Filter Specifications:
b141.4:=a10000:=Enter the transfer function coefficitents:
c
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

16-21

Sallen-Key High-Pass Filter:

MathCad Spreadsheet (p16_5_5_skhp.mcd)
A1.586=Calculate the passband gain.
RA 1−()⋅ 2.9310
3
×=R5 10
3
×=A3
1
Q
−:=R
1
Cω0⋅
:=Calculate resistance values:
C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q0.707=ω0210
3
×=Q
ω0
b
:=ω0a:=Determine the Filter Specifications:
b2828:=a4000000:=Enter the transfer function coefficitents:
c s^2
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

Amplifier: The required gain is 2, but both Sallen-Key filters have passband gains equal to 1.586.
The amplifier has a gain of
2
1.26
1.586
= to make the passband gain of the entire filter equal to 2.

16-22

P16.5-6
This filter is designed as the cascade connection of two identical Sallen-key notch filters.

Sallen-Key Notch Filter:

MathCad Spreadsheet (p16_5_6_skn.mcd)
A1.5=Calculate the pass-band gain.
RA 1−()⋅ 210
4
×=
R
2
210
4
×=R4 10
4
×=
A2
1
2Q⋅
−:=R
1
Cω0⋅
:=Calculate resistance values:
2C⋅210
7−
×=C0.110
6−
⋅:=Pick a convenient value for the capacitance:
Q1=ω0250=Q
ω0
b
:=ω0a:=Determine the Filter Specifications:
b250:=a62500:=Enter the transfer function coefficitents:
c(s^2 + a)
----------------- .
s^2 + bs + a

The transfer function is of the form T(s) =

Amplifier: The required passband gain is 4. An amplifier having gain equal to
4
1.78
(1.5)(1.5)
=
is needed to achieve the required gain.

16-23

P16.5-7

(a)

Voltage division gives: ()
()
()
11 1
a
s1
1
11
Vs R RCs
Hs
Vs RCs
R
Cs
== =
+
+

(b)

Voltage division gives: ()
()
()
2
b
12
Vs Ls
Hs
Vs RLs
==
+

(c)

Voltage division gives:

Doing some algebra:
()
()
()
( )
()
122
c
s2
12
||
1
||
RR LsVs Ls
Hs
Vs RLs
RR Ls
Cs
+
== ×
+
++

()
()
()
( )
()
()
()
()
()
()
12
122
c
s2 12
12
2
12 1
2
12 1 1 2 2
12
2
211 2 1 2
2
1
2
11 2 1 2
1
RR Ls
RR LsVs Ls
Hs
Vs RLsRR Ls
CsRR Ls
RRCsRLCs Ls
RRCsRLCsRRLsRLs
RCsRLs Ls
RLsRLCsRRCLsRR
RLCs
RLCsRRCLsRR
×+
++
== ×
+×+
+
++
+
=×
++ ++ +
+
=×
+++ ++
=
++ ++

(d)
() () ()
ca b
Hs HsHs≠× because the
2,RLs voltage divider loads the
1
1
,R
Cs
voltage
divider.

16-24

P16.5-8
()
100 20
11 1 1
200 20,000 20 2000
2000
11 1 1
20 200 2000 20,000
Hs
ss s
ss s s
s
π ππ
ππ π π
=×
   
++ + +
   
   
=
   
++ + +
   
   
π







P16.5-9
(a) The transfer function of each stage is
()
2
22
22
21
i
11 1
1
11
||
1
1
R
Cs
RR
RR
RCs RCs Cs
Hs
2
R RR R
×
+
+
=− =− =− =−
+Cs

The specification that the dc gain is 0 db = 1 requires
21RR=.
The specification of a break frequency of 1000 rad/s requires
2
1
1000
RC
= .
Pick 0.1 FCµ= . Then
210 kR=Ω so
110 kR=Ω.

(b)
() ()
2
2
11 1 1
10,000 40.1 dB
101110
11
1000 1000
jj
ω
ωω
−−
=× ⇒ = ==−
+++
HH
16-25

PSpice Problems
SP 16-1

16-26

SP 16-2

16-27

SP 16-3

16-28

SP 16-4

16-29

SP 16-5

16-30

SP 16-6

16-31

SP 16-7

Vs 7 0 ac 1
R1 7 6 200
R2 7 1 100
R3 7 2 50
L1 6 5 10m
L2 1 3 10m
L3 2 4 10m
C1 5 0 1u
C2 3 0 1u
C3 4 0 1u

.ac dec 100 100 10k
.probe
.end

SP 16-8

Vs 1 0 ac 1
R1 1 2 100

16-32

C1 2 3 0.2u
R2 3 4 200k
C2 3 4 50p

Xoa5 3 0 4 FGOA

.subckt FGOA 1 2 4
*nodes listed in order - + o
Ri 1 2 500k
E 3 0 1 2 100k
Ro 4 3 1k
.ends FGOA

.ac dec 100 1k 100k
.probe
.end

SP 16-9

Vs 1 0 ac 1
L1 1 2 2.5m
Rw 2 0 8
C2 1 3 34.82u
L2 3 4 0.364m
Rmr 4 0 8
C3 1 5 5u
Rt 5 0 8

.ac dec 100 10 100k
.probe
.end

Bw=4.07k - 493 HZ ∼ 3600HZ

Verification Problems

16-33

VP 16.1
0
0
100
10000100 rad and 25 4 5
25
sQ
Q
ω
ω== = ⇒= =≠
This filter does not satisfy the specifications.

VP 16.2
0
0
100 75
10000 100 rad, 25 4 and 3
25 25
sQ k
Q
ω
ω== = ⇒= = ==

This filter does satisfy the specifications.

VP 16.3
0
0
20 600
40020 rads, 25 0.8 and 1.5
25 400
Qk
Q
ω
ω== =⇒== = =

This filter does satisfy the specifications.

VP 16.4
0
0
25 750
625 25 rads, 62.5 0.4 and 1.2
62.5 625
Qk
Q
ω
ω== = ⇒ = = = =

This filter does satisfy the specifications.

VP 16.5
0
0
12
144 12 rad/s and 30 0.4
30
Q
Q
ω
ω== = ⇒ = =

This filter does not satisfy the specifications.

16-34

Design Problems

DP 16.1
0
1
2
33
()

22()
s
Vs RC
Vs
ss
RCRRC
−
=
++

3 0
0 2
33
22 3
2(100.10) and 2(10.10) BW =
QRRC RC
ω
πω π== = =

3 12 3 22
30
22
100 pF is specified so 318 k and 1.6 k
(10010) (21010)
CR R
RCπω
−
== = Ω =
×× ×
= Ω

DP 16.2

16-35

DP 16.3
12 3 4
Choose 0.1 , 2 , 5 , 100 radsω ωω ω== = = . The corresponding Bode magnitude plot
is:

( )
() ()
22
13
22
24
11
()
11
ss
Hs
ss
ωω
ωω

++ 

=
++

min
Minimum gain is 46.2 dB at 0.505 Hzf−=

16-36

Chapter 17- Two-Port and Three Port Networks
Exercises

Ex. 17.4-1

1
25(100)
10
250
ac
ab c
RR
R
RR R
= ==
++
Ω
2
(125)(125)
12.5
250
bc
ab c
RR
R
RR R
= ==
++
Ω
3
100(125)
50
250
ab
ab c
RR
R
RR R
= ==
++
Ω

Ex. 17.5-1

12 21
1
21
YY−=−=
()
()
11 12 11
22 21 22
11
1

2142 42 42
1
1
10.5 1/7
2110.5
YY Y
YY Y
+= ⇒ = −− =
+= ⇒ = −−=
3

11

4121

11

217
 
−
 
=
 
−
 
 
Y

2
1
1
1
11 0
1
2
22 0
2
1
12 21 0
2
42 (2110.5)
18
4231.5
10.5(63)
9
73.5
6
I
I
I
V
Z
I
V
Z
I
V
ZZ
I
=
=
=
+
== =
+
== =Ω
== =Ω
Ω

21 2 2
18610.5 42(10.5)
Since , then 6 Z =
73.5 73.5 69
II V I I

== = ⇒



17-1

Ex. 17.6-1

1
11
1
2
21
1
1
6
1
.167
6
I
Y
V
I
Y
V
==
==−=−

1
12
2
2
22
2
0.0567
0.944
I
Y
V
I
Y
V
==
==

Ex. 17.7-1

22 1
6, (91) 10 , 1I iV i iV i= =+ = =

2
22
2
1
12
2
6
0.6 S
10
0.1
10
Ii
h
Vi
Vi
h
Vi
== =
== =

1
1
1
1
2
1
10

99
44
5
99
Vi
V
Iii
V
Iii
=
=+ =
=− =

Therefore
()
()
()
1
11
1
2
21
1
0.9
10
9
44
9
4.4
10
9
Vi
h
I i
i
I
h
I i
= ==
== =
Ω

17-2

Ex. 17.8-1
21 21

12 641 1155 55
an d S 30
12 755030 12 3 4

105 1015
−  

  =∆ =−= ⇒ =

  − 
  
YY Z =

Ex. 17.8-2
() ()
() ()
2/5 1

1/10 1/10 4 10
1/30 2/15 1/3 4/3

1/10 1/10

−−

−−  
==
 

 
−−

−−

T

Ex. 17.9-1
ab c
1 12 1 0 1 3
= , = and =
0 1 1/6 1 0 1
  
  
  
TT T







1 121 0 3 121 33 21
0 11/6 1 1/6 10 11/6 3/2
abc c
   
== =
   
   
TTT T

Problems

Section 17-4: T-to-T1 Transformations

P17.4-1

17-3

P17.4-2

P17.4-3

211 2 21
2i
22 L 1 22 L

(forward current gain)

zI I z
IA
zR I zR
−−
=⇒ = =
++

()
12i111 11122 1221
in 11 11
11 1 22 L

(input resistance)
zAIVz IzI zz
Rz z
II I zR
+
== =− =−
+

iL2
22 L iL1 1 in1 v
1i n
and (forward voltage gain)
ARV
VI RARI VRI A
VR
=− = = ⇒ ==
L2
pi v i
in

R
AA AA
R
∴==

17-4

P17.4-4
First, simplify the circuit using a ∆-Y transformation:

eq 1||5||204
3
R
RR= == Ω
Mesh equations:
2
2
30 1810
50 1020
II
II
=−
=−

Solving for the required current:
30 10
50 20 100
0.385 A
18(20) (10)10260
I
−
− −
== =
−− − −

P17.4-5

17-5

Section 17-5: Equations of Two-Port Networks

P17-5-1

12
11 12 11
22 21 22
6
12 18
3 9
Z
ZZ Z
ZZ Z

=Ω
−=Ω ⇒ =Ω
−=Ω ⇒ =Ω

2
1
1
1
11
1 0
12
12 21
220
22
22
220
1
S
14
6 1
S
(612) 21
/71
S
7
V
V
V
I
Y
V
II
Y
VV
IV
Y
VV
=
=
=
==
−
== =− =
+
== =
Υ

11
14 21
=
11

217
 −


−

Υ

P17.5-2

24 4

4 2
j j
j j
−− 
=
 
−+ 
Z

17-6

P17.5-3

2
2
12
11 21
11 00
and
VV
II
YY
VV
==
==

12 1 2 2
11
11 2

and
II II I
Vb
GG G
V
+ +
=+ =
1

so
11 21 1 2 21
((1)) 1and (1) 3

IGb GV V IbGV V=− − =− =− =
Finally
11 21Y 1 S and Y3 S=−=
Next

12
22
32
and
2III
VV
GG
+ −
==

1
1
1
12 2
2
0
2
22 2 3
2
0
1 S
4 S
V
V
I
YG
V
I
YG G
V
=
=
== −=−
== +=

P17.5-4

Using Fig. 17.5-2 as shown:

12 21 12 21
11 12
22 21
0.1 S or 0.1 S
0.2 0.3 S
0.05 0.15 S
YY YY
YY
YY
−=−= ==−
=− =
=− =

P17.5-5

12 21
11 12
11
10 S
13.33 S
23.33 S
YY
YY
Y
µ
µ
µ
=−=
+=
=

22 21 2220 S 30 YY Y Sµ µ+=⇒ =
17-7

P17.5-6

2
2
1
11
1 0
21
21
11 0
332 (3)
2
2
I
I
V
Zj j
I
Vj I
Zj
II
=
=
j= =+ −=+Ω
−
== =− Ω

1
1
1
12
2 0
2
22
2 0
2
2
I
I
V
Zj
I
V
Zj
I
=
=
== −
== −
Ω
Ω

P17.5-7

11 21
11
21 12
2
22 21 22
Z 4
41
1 4
1Z
s
21
1 2 2
Z
s
Z
ssZ
s
ZZ s Z s
s s
−= 
+
⇒= +=
−=

+
−= ⇒ =+=

P17.5-8
Given:
1
1
=
1 1
s
s
s
+
−


−+
Y

Try a π circuit as shown at the right.

12
11 11 12
22 21
1
11
1
(1)1
Ys
ss
YY Y
ss
YY s s
=−
++
=⇒ += −
+= +−=
1
s
=

17-8

P17.5-9
Given:
2
22
2
22
22 1

11
=
11

11
ss
ss ss
s
ss ss
++

++ ++

+

++ ++
Z
Try :

From the circuit, we calculate:
()
()
2
2
11 2 12
11 1 1 22
22
2
1
1 11
RLs
LCRsRRCLsRRRLsCs
zR R
RCsLCs LCsRCs
RLs
Cs
+
++ +++
=+ =+ =
++ ++
++
2

Comparing to the given yields:
11z
1
2
2
1
12
12
1
1
1
1
1
1 H
2
1 F
2
LC
R
RC
R
LCR
L
RRCL
C
RR
= 
=Ω
=

=Ω
= ⇒
=

+=
 =
+= 


Then check . The are all okay. If they were not, we would have to try a different
circuit structure..
12 21 22, and zz z

P17.5-10
It is sufficient to require that the input resistance of each section of the circuit is equal to Ro, that
is

Then
22(2 )
44 (2) 3 (31)
3
o
oo
o
RR R
R RR R R R R
RR
R
+
=⇒ =± + =± =
+
−

17-9

Section 17-6: Z and Y Parameters

P17.6-1

11
21
11
12
12
12
( )
and

Vb
iI
RR
bR R
2
1
R
V
R
IIi V
RR
+
=− =−
++
=−−=



2 2
12 112
11 21
11 12 0
( )
and
VV
bR R bRII
YY
VV RR
==
++ +
== = =−
12 0
RR

21
2
22 2 2
2

II
V
VR I I
R
=−
=⇒ =

11
21
22 12
22 2 0
11
and
VV
II
YY
VR V
==
== == −
20
R
1i

P17.6-2

()
11
2
21
13 4
0
3
vi
i
vi
= + =
=⇒ 
=

therefore
11 214 and 3 zz=Ω =Ω

()
()
12 2
1
12 2
3
0
32
vi i
i
vi i
α
α
 =+

=⇒ 
=+ +

2i

therefore
()
12 21
31 and 53zz α α=+ =+
Finally,
4 3(1)
=
3 53
α
α
+ 
 
+ 
Z
17-10

P17.6-3
Treat the circuit as the parallel connection of two 2-port networks:

The admittance matrix of the entire network
can be obtained as the sum of the admittance
matrices of these two 2-port networks

1 02 12
2 1 2 2 12
ss ss
ss s s
−+ −  
=+ =
  
−− +  
Y

When ()()
1
:itut=
11 1
22
21
11
() () 2 21

2() ()
36 1
00
ss
Vs Vs ss
ss
Vs Vs
ss
−
+
  
  − +  
=⇒ = =
 
  
  ++
  
YY
1
0
s
so
2 2
(2) 161.257.25
()
(361)3 1.820.184
S
S
SS S S S
Vs
 −− −
== + +
 
++ + + 

Taking the inverse Laplace transform

1.82 0.184
2
1
(t) = 61.25 7.25 t0
3
tt
ve e
−−
−− + ≥


17-11

P17.6-4

KVL: ()11 2 2 1
1
20
2
iv vvv−+ +−=

KCL: iv
11 142vv−= +
2

11 1
11 11
111 2
11 2 12 2
12 21
1
5 2
36
2936
52 6 1
52
31
iv v
iv z
iiv v
iv v iv v
iv z
i
−

8
=−⇒ ==

=−  
⇒
=+ +
 
Ω
= +⇒ ==−


Ω

KVL:
21 1 1
11
1 5
22
vv vv=+ +=
1
3
v

KCL:
2
21 252 5
12
v
iv v=+ = +
1v

21 1 1 12
13 1
25 18
21
iv v v z

=+ = ⇒ =



8
Ω
and

22 2 1 22
2
25 2.769 0.361
13
iv v v z

=+ = ⇒ =


Ω

17-12

P17.6-5

KCL:
1
12
1
v
ii
R
+=
KVL:
22 1 100Ribv v−− +−=

Then
()
21
21 1 1
21 2 1
111 1 RR bbb
i vandi v v
RR R R
 +++ +
=− = + =


1
2R

so
()
21 2
21 11
12 1 12
11
and
ii Rb
y y
vR v RR
1Rb+ ++
==− ==

KVL:
21 2 1 2
2
1
0Riv i v
R
+= ⇒ =−

KCL: ()23 1 2 3 2 3
2
1
vR iiR vRi
R

=+ =− + 
 

2

Then
13 2
12 2 32 22
22 2 2 3
11
and 1
iR i
yv Ri y
vR R vR

==− + = ⇒ == +

 2
1
R

17-13

Section 17-7: Hybrid Transmission Parameters

17.7-1

2
11
21
2 0
34 5
6.8 since
52
V
VV
4||1034
B IV
I
=
== =Ω −= =
−+

2
1
21
2 0
104 10
1.4since
10 104
V
I
DI
I
=
+
== = =−
−+
I

2
1
21
2 0
12 10
1.2 since
10 102
I
V
AV
V
=
== = =
+
V

2
1
2 0
1
0.1S
10
I
I
C
V
=
== =

1)I
17.7-2
20V=
so
1i 1 2( ||VR RR=+
therefore

2
1
11 i 1 2
1
0
||600 k
V
V
hR RR
I
=
== + = Ω
KVL:
1i
21
12 o
RR
1I IA
RR R
+ =−
+
I
therefore
2
2i 1 6
21
1o 1 2
0
()
V
IR R
hA
IR RR
=
== − + =−
+
10

17-14

1i00Iv A=⇒ =⇒ =
i0v
so
2
2
o1 2()
V
I
RRR
=
+

therefore
1
2o 1 2 3
22
2o 1 2
0
10
()
I
IR RR
h
VR RR
−
=
++
== =
+

Next,
1
12
12
R
VV
RR
=
+

therefore
1
11
12
21 2
0
1
2
I
VR
h
VR R
=
==
+
=

P17.7-3

Compare :
21
12
VnV
InI
=
=−
to
11 11 12
22 11 22

Vh IhV
2
2
IhIhV
= +
=+

11 22 12 21
11Then 0, 0, and hh h h
nn
== = =
−

P17.7-4

()
23
11 2 3 1 11 1
23
22
21 21
23 23
|| +

RR
VR RRI hR
RR
RR
II h
RR RR
=+ ⇒ =
+
=− ⇒ =−
++

2
22 2
23 2 3
22
12 12
23 23
1

V
Ih
RR RR
RR
VV h
RRR
=⇒ =
++
=⇒ =
++ R

17-15

P17.7-5

21
21
0.1 and 950
so 95
I vv
II
I= =
=

2
2
1
11
1
0
2
21
1
0
509501000
95
V
V
V
h
I
I
h
I
=
=
= =+ = Ω
==

100Iv=⇒=

1
1
1
12
2
0
2 4
22
2
0
0
10 S
I
I
V
h
V
I
h
V
=
−
=
==
==

Section 17-8: Relationships between Two-Port Parameters

P17.8-1
Start with
11 11 1211 11 122
2 211 222 2 211 222

Y parameters: and H parameters:

Vh IhVIY VYV 2
IYVYV IhIhV
=+=+ 

=+ = + 

Solve the Y parameter equations for V to put them in the same form as the H parameter
equations.
1and
2I
I
1
11 1 12 1 1 11 12 1111 1 122
211 2 222 21 2 22 2 2 21 22 2
Y 0 1 0 1

Y 1 0 1 0
VY I V Y YYVIYV
YVIYV I YV I Y Y V
−
−− − − −−= +      
⇒= ⇒ =
     
−+ = −−     
−



12
1
11 12 11 12 11 11
21 22 21 22 21 1221
22
11 11 11
11
0
0 1 1
=
1 0 0
1
Y
YY Y Y Y Y
YY Y Y Y
Y
YY
−
 
−−
 
−− −  
 ∴==
  
−    
−−
 
 
H
YY
Y







17-16

P17.8-2
First ∆= Then (3)(6)(2)(2)14− =Z.
22 12
21 11
62

14 14
= =
23

1414
ZZ
ZZ
  
−−
  
∆∆
  
−  
−
   ∆ ∆
ZZ
Y
ZZ
.

P17.8-3
First ∆= . Then (0.1)(0.5)(0.4)(0.1).01S− =Y
12
11 11
21
11 11
1

10 1
=
4 0.1

Y
YY
Y
YY

−

− 
=
 
∆  


H
Y
.

P17.8-4
First ∆= . Then (0.5)(0.6)(0.4)(0.4)S−−−Y
12
11 11
21
11 11
1Y
-
YY 2 0.8

Y ∆ 0.8 0.28

YY


 
==
 
−  


H
Y

Section17-9: Interconnection of Two-Port Networks

P17.9-1

12 21
22 21
11 12 11
1 S
3
10 S
3
41 S S
3
YY
YY
YY Y
==−
=− =
+= ⇒ =

a
41

33
11

33
−
=

−

Y

12 21
11 12 11
2122 22
1 S
31S S
22
14 S
33
YY
YY Y
YY Y S
==−
+= ⇒ =
+= ⇒ =

b
3
1
2

4
1
3
−
=

−

Y

31 744
( )
32 3 6 3
5544

33 3
− +−
 
==
 −−
 
Y
4
3





17-17

P17.9-2

Admittance parameters:

106

4444
68

4444
− 
 
= 
−
 
  
Y
Transmission parameters:

844

66
110

66


=


T

p
2012

4444
1216

4444
−

=+ =
−


YY Y

C
108792

3636
'
18144

3636


=⋅=


TT T

P17.9-3
12 2
22
1

1

ss
GG G
s
GG G
ss
s

+−
+−
3
 
=+  
−+   −+

Y

17-18

Verification Problems
VP 17-1

12
75
50 15
17575
VI

==
+
2I

1
1
12
2 0
15
I
V
Z
I
=
= =Ω

11
11
0.028
50125
1I VV

=+ =


2
1
11
1 0
28 mS
V
V
Y
I
=
= =

1124 mS, so the report is not correct.Y≠

VP 17-2

1111
2121
20.20.2(10) (20.2)
0.1 (0.1)
Zs sVs I
ZsVs I
=+ = +=+ 
⇒
== 

22 1220.2and 0.1Z sZ s=+=

2
(20.2)(20.2)(0.1)(0.1)0.01(38040)ss ss s s∆= + + − = ++Z

17-19

2
11
21 21
22
21 21
2(10)0.1(38040)
==
1 2(10)
0.1
Z
ss s
ZZ
ss
Z s
s
ZZ s
∆  ++

+
 
  
 + 
  
 
Z
T

This is not the transmission matrix given in the report.

Design Problems

DP 17-1

We will need to find
1 and R Rby trial and error. A Mathcad spreadsheet will help with
the calculations. Given the restrictions
110 and 10 R R≤Ω ≤Ωwe will start with
110 and 10 RR=Ω =Ω:

R110:= R10:=
Ra
1420⋅
1420+ R+
:= Rb
14R⋅
1420+ R+
:= Rc
R20⋅
1420+ R+
:=
RinR1
Rb2+( )Rc20+()⋅
Rb2+ Rc+ 20+
+:= Rin14.279=

The specifications cannot be satisfied.
1and R R are at their maximum values but
inRneeds to be larger. Reducing either
1or RRwill reduce
inR.

17-20

DP 17-2

12
13 2 4
34
13 2 4
Need for balance
(1)
(2)
AB
VV
RV RV
RR RR
RV RV
RR RR
+
=
++
=
++

12
34
Dividing (1) by (2) yields: .
RR
RR
=

DP 17-3
11 11 12
22 11 22
Vh IhV
2
2IhIhV
=+
=+
and VI
22 L 2 211 22LR IhIhR
2I=−⇒ = −
Next
2L 2
21 i 21
12 2L 1 1 22
11
11
II I
hA h
LI hR I I hR
 
=⇒ ==−=− 
 ++
 

We require
L
L3
22L
17 9
7980 1 1 1.013 k1 k
18 08010
R
R
hR
 
=⇒ + =⇒ = Ω 
+× 
≅Ω
Next
2
2 211 222 222 L 211(1/)
L
V
I hIhV Vh R hI
R
=−= + ⇒ + =−
Substituting this expression into the second hybrid equation gives:

12 21
11 11
22
L
()
1()
hh
VhI I
h
R
1
−
=+
+

The input resistance is given by

in 11 12L21 22
L
1(since )Rh hRh h
R
≈− <<
Finally
43
in45(510)(10)(80)5 10 R
−
=− × =Ω<Ω

17-21

DP 17-4
11 22
4(12) 8(8)
25 and
412 88
ZZ=+ =Ω = =Ω
++
4

2
2
21 1 21
1 0
4
82 Z
412
I
V
VI I
I
=

2 = =⇒ = =

+
Ω

12Similarly 2 Z=Ω

22
L
Thèvenin: 4 so for maximum
power transfer, use =4
T
ZZ
R
== Ω
Ω

2
s
RL s
2
89.3 WV37.8 V
4
V
P



== ⇒=

17-22

DP 17-5
The circuit consists of 4 cascaded stages. Represent each stage by a transmission matrix
using:

1 ()
0 1
Zs 
=
 
 
T

1 0
() 1Ys

=


T

1a
1
1
0 1
Cs


=



T

1
12b
1
1
0 1
Ls
LCs


+=



T

3
23
1 0
1
1
Cs
LCs


=

 +

T

d
L
1 0
1
1
R


=



T

11 3 11
22
1211 L23 L 1211
ab cd
3
L2 3 L
1
1
LCs Cs LCs
LCCsCsRLCsRLCCsCs
Cs
RLCsR

+×

++

==


+
TTTTT
+

17-23

Solution manual for introduction to electric circuits

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