NAS-Ch4-Application of Laplace Transform

This chapter provides complete solution of different circuits using Laplace transform method and also provides information about applications of Laplace transforms.

1. Network Analysis and Synthesis
Chapter 4: Application of Laplace Transform
By
Dr. K Hussain
Associate Professor and Head
Dept. of EE, SITCOE

2. Contents
•Solution of differential equation using Laplace
transform, Unit step, Impulse & ramp
functions
•Laplace transform of singular & shifted
function, Convolution integral
•Concept of complex frequency
•Transform impedance & transform
admittance, Series & parallel combination of
these transform networks.

3. Laplace Transform
An integral transform mapping functions from the time
domain to the Laplace domain or s‐domain
f

Time‐domain functions are functions of time, t


Laplace‐domain functions are functions of s


s is a complex variable

4. Laplace Transforms – Motivation
We’ll use Laplace transforms to solve differential
equations
Differential equations in the time domain
difficult to solve
Apply the Laplace transform
Transform to the s‐domain
Differential equations become algebraic equations
easy to solve
Transform the s‐domain solution back to the time domain
Transforming back and forth requires extra effort, but
the solution is greatly simplified

5. The Laplace Transform (LT) method is much superior to
the classical method due to the following reasons.

1.Laplace transformation transforms exponential and
trigonometric functions into algebraic functions.
2.Laplace transformation transforms differentiation and
integration into multiplication and division
respectively.
3.It transforms integro-differential equations into
algebraic equations which are much simpler to handle.
4.The arbitrary constants need not be determined
separately. Complete solution will be obtained directly.

6. Laplace Transform
The Laplace transform provides a useful method of solving certain
types of differential equations when certain initial conditions are
given, especially when the initial values are zero.
It is also very useful in the area of circuit analysis. It is often easier
to analyze the circuit in its Laplace form, than to form differential
equations.
The techniques of Laplace transform are not only used in circuit
analysis, but also in
•Proportional-Integral-Derivative (PID) controllers
•DC motor speed control systems
•DC motor position control systems
•Second order systems of differential equations (underdamped,
overdamped and critically damped)

7. Definition of Laplace Transform
The Laplace transform ℒ, of a function f(t) for t > 0 is defined by
the following integral over 0 to ∞:
The resulting expression is a function of s, which we write as F(s).
In words we say
"The Laplace Transform of f(t) equals function F of s".
ℒ {f(t)}=F(s)
ℒ {g(t)}=G(s)
Similarly, the Laplace transform of a function g(t) would be written:
and write:
ℒ {f(t)}=∫
0
∞
​e
−st
f(t)dt
Unilateral or one‐sided transform
Lower limit of integration is
Assumed that the time domain function is zero for all negative
time, i.e., g(t)<0

8. •The two singularity functions of interest are
1.unit step function, u(t)
and its construct: the gate function
2.delta or unit impulse function, (t)
and its construct: the sampling function
Laplace Transforms of Singularity Functions
•Singularity functions are either not finite or
don't have finite derivatives everywhere

9. Step Function
A useful and common way of characterizing a linear
system is with its step response
 The system’s response (output) to a unit step input
The unit step function or Heaviside step function: 





01
00
)(
t
t
tu

10. Extensions of Unit Step Function
•A more general unit step
function is u(t-a)







at
at
atu
1
0
)(
1
t 0 a
1
t 0  +T
•The gate function can be
constructed from u(t)
–a rectangular pulse that
starts at t= and ends at
t=+T : u(t-) – u(t--T)
–like an on/off switch

11. Unit Step Function – Laplace Transform
Using the definition of the Laplace transform
The Laplace transform of the unit step is
Note that the unilateral Laplace transform assumes that
the signal being transformed is zero for t<0
Equivalent to multiplying any signal by a unit step
£{f(t)}=1/s

12. Impulse Function
Another common way to describe a dynamic system
is with its impulse response
 System output in response to an impulse function input
Impulse function: If a unit step function u(t) is differentiated
with respect to t, the derivative is zero for time t greater than
zero, and is infinite for time t equal to zero.
An infinitely tall, infinitely narrow pulse
and
where the symbol δ(t) (delta) is used
to represent the unit impulse.
Mathematically, the function is defined as

13. Delta or Unit Impulse Function, (t)
•The delta or unit impulse function, (t)
–Mathematical definition


–Graphical illustration 





0
0
0
1
0
)(
tt
tt
tt
1
t 0
(t)
t
0

14. Extensions of the Delta Function
•An important property of the unit impulse
function is its sampling property
–Mathematical definition 





00
0
0
)(
0
)()(
tttf
tt
tttf
f(t)
t 0 t
0
f(t) (t-t
0)

15. The Laplace transform of the unit impulse function is

16. Unit Ramp Function – Laplace Transform
Could easily evaluate the transform integral
Requires integration by parts
Alternatively, recognize the relationship between the
unit ramp and the unit step
Unit ramp is the integral of the unit step
The unit ramp function is a useful input signal for
evaluating how well a system tracks a constantly‐
increasing input 





0
00
)(
tt
t
tu
The unit ramp function:

17. Apply the integration by parts,
Laplace transform of unit Ramp function is 1
!
)()()(


n
n
s
n
sFtuttf
In general 66
5
2
1
120!5
)()()(,5
!1
)()()(,1
1!0
)()()(,0
ss
sFtuttfn
s
sFttutfn
ss
sFtutfn




18. If L{f(t)}=F(s) when s>a then,
In words, the substitution (s−a) for s in the transform corresponds to
the multiplication of the original function by e
at
.
Proof:
L{e
at
f(t)}=F(s−a)
Laplace Transform of Shifted Function
First (frequency) Shifting Property:
F(s−a)=L{e
at
f(t)}
F(s)=∫
0
∞
e
−st
f(t)dt F(s-a)=∫
0
∞
e
−(s-a)t
f(t)dt
F(s-a)=∫
0
∞
e
−st+at
f(t)dt
F(s-a)=∫
0
∞
e
−st
e
at
f(t)dt F(s-a)= e
at
∫
0
∞
e
−st
f(t)dt
Therefore

19. Second (Time) Shifting Property
If L{f(t)}=F(s) and g(t)= f(t−a) t>a
0 t<a
then, L{g(t)}=e
−as
F(s)
Proof:
g(t)= f(t−a) t>a
0 t<a
L{g(t)}=∫
0
∞
e
−st
g(t)dt
L{g(t)}=∫
0
a
e
−st
(0)dt+ ∫
a
∞
e
−st
f(t-a)dt
L{g(t)}=∫
a
∞
e
−st
f(t-a)dt

20. Let z=t−a
t=z+a
dt=dz
L{g(t)}=e
−as
F(s) Therefore
L{g(t)}=∫
a
∞
e
−st
f(t-a)dt
L{g(t)}=∫
0
∞
e
−sz−sa
f(z)dz
L{g(t)}=∫
0
∞
e
−sz
e
−sa
f(z)dz L{g(t)}=∫
0
∞
e
−s(z+a)
f(z)dz
when t=a, z=0
when t=∞, z=∞
L{g(t)}=e
−sa
∫
0
∞
e
−sz
f(z)dz
L{g(t)}=e
−as
L{f(t−a)} L{g(t)}=e
−as
L{f(z)}

24. Convolution Theorem states that if we have two functions,
taking their convolution and then Laplace is the same as taking
the Laplace first (of the two functions separately) and then
multiplying the two Laplace Transforms.
Theorem: For any two functions f(t) and g(t) with Laplace
transforms F(s) and G(s) we have L(f ∗ g) = F(s) · G(s).
Proof:
We start by writing L(f ∗ g) as the convolution integral followed
by the Laplace integral
Convolution Integral

25. Next, we change the order of integration
Finally, change variables in the inner integral:
substitute v = t−u, dv = dt, (u =a constant)

26. Q. Find y(t) from the given functions using convolution integral.
Solution:
To evaluate the convolution integral we will use the convolution
property of the Laplace Transform:
We need the Laplace Transforms of f(t) and h(t), but we can look
them up in the tables:
So,
 (t) is step function

27. We can now write y(t) (which is implicitly zero for t<0)
We can look up both of these terms in the tables.

28. Solution of Differential Equation using
Laplace Transform

29. s-Domain Circuit Analysis
Time domain
(t domain)
Complex frequency
domain (s domain)
Linear
Circuit
Differential
equation
Classical
techniques
Response
waveform
Laplace Transform
Inverse Transform
Algebraic
equation
Algebraic
techniques
Response
transform
L
L
-1
Laplace Transform
L
Transformed
Circuit

30. Kirchhoff’s Laws in s-Domain
s domain
i
2 (t)
t domain

Kirchhoff’s current law (KCL)

i
1 (t)  i
2 (t)  i
3 (t)  i
4 (t)  0
Kirchhoff’s voltage law (KVL)


 v
1 (t)  v
2 (t)  v
3 (t)  0
  

V
1 (s)  V
2 (s)  V
3 (s)  0
i
1 (t) i
3 (t)
i
4 (t)
I
1 (s)  I
2 (s)  I
3 (s)  I
4 (s)  0

 v
2 (t)   v
4 (t) 
  
5
v
1 (t) v
3 (t) v (t)

31. Signal Sources in s Domain
L
L
t domain s domain
+
v(t)
_
v
S (t)
+
i(t)
Voltage Source:
v(t)  v
S (t)
i(t)  depends
on circuit
_
+
V (s)
+
I (s)
Voltage Source:
V
S (s) V (s)  V
S (s)
I (s)  depends
on circuit
_
v(t)
+
i
S (t)
i(t)
Current Source:
i(t)  i
S (t)
v(t)  depends
on circuit
I
S (s) V (s)
+
_
I (s)
Current Source:
I (s)  I
S (s)
V (s)  depends
on circuit
- -

32. Time and s-Domain Element Models
L
L
L
+
v
R (t)
_
R
v
R (t)  Ri
R (t)
Resistor:
+
V
R (s)
_
R
Impedance and Voltage Source for Initial Conditions
Time Domain s-Domain
V
R (s)  RI
R (s)
Resistor:
+
v (t)
L
_
L
i
L (t)
dt
L
L
di (t)
v (t)  L
Inductor:
_
+
+
V (s)
L
_
Ls
L
Li (0)
I
L (s)
L
Li (0)
V
L (s)  LsI
L (s) 
Inductor:
_
+
v (t)
C C
i
C (t)
0
C
t
C C
C
 v (0)
v (t) 
1
i ( 
 )d
Capacitor:
_
+
+
V (s)
C
_
1 Cs
C
s
v (0)
I
C (s)
Cs
C
s
C C
v ( 0 )
V (s) 
1
I (s) 
Capacitor:
I
R (s) i
R (t)

33. Impedance and Voltage Source for Initial
Conditions
R
R
R
 R
I (s)
V (s)
Z (s) 
L
L
L
L
 Ls with i (0)  0
I (s)
V (s)
Z (s) 
 with v (0)  0
I (s) Cs
V (s) 1
Z (s) 
C
C
C
C
•Impedance Z(s)
Z (s) 
voltage transform
current transform

with all initial conditions set to zero
•Impedance of the three passive elements

34. Time and s-Domain Element Models
L
L
L
+
v
R (t)
_
R
R
R R
i (t) 
1
v (t)
Resistor:
+
V
R (s)
_
R
Admittance and Current Source for Initial Conditions
Time Domain s-Domain
R
R R
I (s) 
1
V (s)
Resistor:
+
v (t)
L
_
L
i
L (t)
Inductor:
0
t
L L
L
 i
L (0)
i (t) 
1
v ( 
 )d
_
+
v
C (t) C
i
c(t)
dt
C
C
dv (t)
i (t)  C
Capacitor:
Inductor:
Ls
L L
i
L( 0 )
I (s) 
1
V (s) 
+
V (s)
L
_ Ls
s
i
L (0)
I
L (s)
C
Cv
C (0)
I (s)  CsV
C (s) 
s
Capacitor:
_
+
V
C (s)
1 Cs
C
Cv (0)
I (s) C
i
R (t) I
R (s)

35. Admittance and Current Source for Initial
Conditions
R
R
R
I (s) 1
V (s) R
 Y (s) 

L
L
L
L
with i (0)  0
V (s) Ls
I (s) 1
Y (s) 
 Cs with v (0)  0
V (s)
I (s)
Y (s) 
C
C
C
C
•Admittance Y(s)
Y (s) 
current transform

1
voltage transform Z (s)

with all initial conditions set to zero
•Admittance of the three passive elements

36. L
+
V
A(t)
i(t)
L
+
V (s)
L
_ L
Li (0)
 V (s) 
R
+ V
A
s
Example: Solve for Current Waveform i(t).
I (s)
Ls
_
+
s
R L
V
A
By KVL:  V (s)  V (s)  0
Resistor: V
R (s)  RI (s) Inductor: V
L (s)  LsI (s)  Li
L (0)
L  RI (s)  LsI (s)  Li (0)  0
s
V
A
V
A L



V
A R

V
A R i
L (0)
s s  R L s  R L
i
L (0)
s(s  R L) s  R L
I (s) 
V
L L
V
 i
L (0)e u(t
 A
e
 R R

R
t  
R
t
Inverse Transform: i(t)  
A
forced response natural response
- -
R R

37. Consider the RL circuit shown in Fig.(a). Assume that the switch is
closed at time t = 0 and assume that the current i at the time of
switching is zero.
The transform circuit in s domain is shown in Fig.(b).
From this,
Taking inverse LT
Thus, inductor current rises exponentially with time constant L/R.
Analysis of RL Circuit using Laplace Transform Method

38. Inductor voltage decreases exponentially with time constant L/R.
The current and voltage transients are shown in Fig.
Voltage across the inductor is given by
Taking inverse LT

39. Example: Initially relaxed series RL circuit with R = 100 Ω and L = 20
H has dc voltage of 200 V applied at time t = 0. Find (a) the equation
for current and voltages across different elements (b) the current at
time t = 0.5 s and 1.0 s and (c) the time at which the voltages across
the resistor and inductor are equal.
Solution: Transform circuit for time t > 0 is
shown.
Therefore, (a) current = i(t) = 2 (1-e
-5t
) A
Voltage=V
R = R i(t) = 200 (1-e
-5t
) V
Voltage V
L
(c) Let t
1 be the time at which V
R(t) = V
L(t). Then
This gives t
1 = 0.1386 s
200 (1- e
-5t1
)= 200 e
-5t1
i.e., e
-5t1
= 0.5
(b) i(0.5) = 2 (1- e
-2.5
) = 1.8358 A and
i(1.0) = 2 (1-e
-5
) = 1.9865 A

40. Use Partial fraction method and then
Apply inverse Laplace Transform
Example:
Solution:
This is series R-L circuit.

41. Voltage across the capacitor
Taking inverse LT, we get the capacitor voltage as
The circuit current and the voltage across the capacitor vary as shown in Fig.
The transform circuit for time t > 0 is shown in Fig.(b). From this
Taking inverse LT
Consider the RC circuit shown in Fig. (a). Assume that the switch is closed at time t = 0
and assume that the voltage across the capacitor at the time of switching is zero.
Analysis of RC Circuit using Laplace Transform Method

42. Example: In the RC circuit shown in fig (a)below, the capacitor has
an initial charge q
0 = 2500 μC. At time t = 0, the switch is closed.
Find the circuit current for time t > 0.
Solution:
Transform circuit for time t > 0 is shown in Fig. (b)
Taking inverse LT, current i(t) = 15 e
-2000t
A
Fig (a)
Fig (b)
From Fig (b)

43. Consider the RLC series circuit shown in Fig. (a). Assume that there is no initial
charge on the capacitor and there is no initial current through the inductor.
The switch is closed at time t = 0.
Transform circuit for time t > 0 is shown in
Fig. (b).
Using the transform circuit, expression for the
current is obtained as
The roots of the denominator polynomial are
Where and
Analysis of RLC Circuit using Laplace Transform Method

44. Depending on whether , , or ,

the discriminant value will be positive, zero or negative and three
different cases of solutions are possible.
The value of R, for which the discriminant is zero, is called the
critical resistance, R
C.
Then
If the circuit resistance R > RC, then
Thus
If the circuit resistance R < RC, then

46. The plot of this current transient is
shown in Fig.
In this case, the current is said to be critically damped.

47. As seen in Equation, α will be a negative number.
Thus, for this under damped case, the current is oscillatory and at the same
time it decays.
For this case, the roots are complex conjugate, s
1=α+jβ and s
2=α-jβ
Taking inverse LT, we get
Case (iii):
Waveform shown is a exponentially decaying sinusoidal wave.

48. Example: For the RLC circuit shown, find the expression for the transient current when the
switch is closed at time t = 0. Assume initially relaxed circuit conditions.
Solution: The transform circuit is shown in Fig. (b)
The roots of the denominator polynomial are

49. This is an example for over-damped.
Taking inverse LT, we get current

50. Example: Taking the initial conditions as zero, find the transient
current in the circuit shown in Fig. (a) when the switch is closed at
time t = 0.
Solution
The transform circuit is shown in Fig (b).
The roots of the denominator polynomial are
Fig (b)
It can be seen that
Fig
(a)

51. This is an example for under-damped.
Taking inverse LT, we get

52. Network Functions
Input Signal Transform
Network function 
Zero - State Response Transform
I (s) Y (s)
V (s) 1
 Z (s) 
V (s)
•Transfer function relates an input and response at
different ports in the circuit
V (s)
V
T (s)  Voltage Transfer Function 
2
Circuit in
the
zero-state

V (s)
I (s)
Circuit in
the
zero-state
V
1 or I
1 V
2 or I
2
Input Output
_
+ 
V
2
T
V (s)

V
1

In

Out
I
1 (s)
1
I (s)
T
I (s)  Current Transfer Function 
2
2
V (s)
I (s)
Y
T (s)  Transfer Admittance 
I
1 (s)
1
V
2 (s)
T
Z (s)  Transfer Impedance  _
+
1
V
2
I
T
Y (s)
In Out
1
I

V
2
T
Z (s)
In

Out
I
1
T
I (s)
In Out
I
2
Driving-point function relates the
voltage and current at a given
pair of terminals called a port

53. Calculating Network Functions
Z
1
Z
2


V
2 (s) V
1 (s)
_
+
Y
1 Y
2 I
1 (s)
2
I (s)
1 1 2
2
Z
2 (s) V (s)
V
V (s) Z (s)  Z (s)
 T (s) 
•Driving-point impedance
Z
EQ (s)  Z
1 (s)  Z
2 (s)
•Voltage transfer function:
2
V (s)
Z
2 (s)
 1
2 


 1
Z (s)  Z (s)
V (s) 

1 1 2
2
Y
2 (s) I (s)
I
I (s) Y (s)  Y (s)
 T (s) 
•Driving-point admittance
Y
EQ (s)  Y
1 (s)  Y
2 (s)
•Current transfer function:
Y (s)  
Y
1 (s)  Y
2 (s) 
I
2 (s)  
2
 I
1 (s)

54. Circuit Analysis in s-Domain
The equivalent impedance Z
eq(s) of two impedances Z
1(s) and
Z
2(s) in series is
V(s) Z
2
Z
1
+
-
Same current flows
V (s)  Z1(s)I s Z2 (s)I (s)  Zeq (s)I s
The equivalent admittance Y
eq(s) of two admittances Y
1(s) and
Y
2(s) in parallel is
+
s) Y
1 Y
2
-
V(
Same voltage
I (s)  Y1(s)V (s)  Y2 (s)V (s)  Yeq (s)V (s)
I(s)
Basic Rules:
I
1(s)
Y
eq
(s)  Y
1
(s)  Y
2 (s)
Zeq (s)  Z1(s)  Z2 (s)

55. Limitation of Laplace Transform
•Only be used to solve differential equations
with known constants.
•An equation without the known constants,
then this method is useless.
Laplace Transform is powerful tool using in
differential areas of mathematics, physics, and
engineering. But

56. THANK YOU