The fourier series signals and systems by R ismail

Fourier series and signals and systems

1. The Fourier Series
BEE2113 – Signals & Systems
R. M. Taufika R. Ismail
FKEE, UMP

2. A Fourier series is an expansion of a
periodic function f (t) in terms of an infinite sum
of cosines and sines
Introduction




1
0
)sincos()(
n
nn
tnbtnaatf 

3. In other words, any periodic function can be
resolved as a summation of
constant value and cosine and sine functions:




1
0 )sincos()(
n
nn tnbtnaatf 
)sincos(
11
tbta 
0a
)2sin2cos(
22
tbta 
)3sin3cos(
33
tbta  
dc ac

4. The computation and study of Fourier series is
known as harmonic analysis and is extremely
useful as a way to break up an arbitrary
periodic function into a set of simple terms that
can be plugged in, solved individually, and
then recombined to obtain the solution to the
original problem or an approximation to it to
whatever accuracy is desired or practical.

5. =
+ +
+ + + …
Periodic Function
0a
tacos
1
ta2cos
2
tbsin
1
tb2sin
2
f(t)
t

6. 



1
000 )sincos()(
n
nn tnbtnaatf 
where frequency lFundementa
2
0

T





Tt
t
n dttntf
T
a
0
0
0cos)(
2




Tt
t
dttf
T
a
0
0
)(
1
0



Tt
t
n
dttntf
T
b
0
0
0
sin)(
2


7. Notice that
component dc
valueAverage
period aover graph below Area
)(
1 0
0
0





T
dttf
T
a
Tt
t
[*The right sign of the area below graph must be
obeyed. +ve sign for area above x-axis & −ve sign
for area below x-axis]

8. Symmetry Considerations
Symmetry functions:
(i)even symmetry
(ii)odd symmetry

9. Even symmetry
Any function f (t) is even if its plot is
symmetrical about the vertical axis, i.e.
)()( tftf

10. Even symmetry (cont.)
The examples of even functions are:
2
)(ttf
t t
t
||)(ttf
ttf cos)(

11. Even symmetry (cont.)
The integral of an even function from −A to
+A is twice the integral from 0 to +A
t 



AA
A
dttfdttf
0
eveneven
)(2)(
−A +A
)(
even
tf

12. Odd symmetry
Any function f (t) is odd if its plot is
antisymmetrical about the vertical axis, i.e.
)()( tftf 

13. Odd symmetry (cont.)
The examples of odd functions are:
3
)(ttf
t
t
t
ttf)(
ttf sin)(

14. Odd symmetry (cont.)
The integral of an odd function from −A to
+A is zero
t 0)(
odd 


A
A
dttf
−A +A
)(
odd
tf

15. Even and odd functions
(even) × (even) = (even)
(odd) × (odd) = (even)
(even) × (odd) = (odd)
(odd) × (even) = (odd)
The product properties of even and odd
functions are:

16. Symmetry consideration
From the properties of even and odd
functions, we can show that:
for even periodic function;


2/
0
cos)(
4
T
n tdtntf
T
a  0
nb
for odd periodic function;


2/
0
sin)(
4
T
n tdtntf
T
b 0
0 
naa

17. How?? [Even function]
2
T

2
T



2/
0
2/
2/
cos)(
4
cos)(
2
TT
T
n tdtntf
T
tdtntf
T
a 
(even) × (even)
| |
(even)
0sin)(
2
2/
2/


T
T
n tdtntf
T
b 
(even) × (odd)
| |
(odd)
)(tf
t

18. How?? [Odd function]
2
T

2
T



2/
0
2/
2/
sin)(
4
sin)(
2
TT
T
n tdtntf
T
tdtntf
T
b 
(odd) × (odd)
| |
(even)
0cos)(
2
2/
2/


T
T
n
tdtntf
T
a 
(odd) × (even)
| |
(odd)
)(tf
t
0)(
2
2/
2/
0


T
T
dttf
T
a
(odd)

19. The amplitude-phase form

is known as the sine-cosine form
We can also express the Fourier series in the
cosine form only, that is
This form is called as the amplitude-phase form




1
000
)sincos()(
n
nn
tnbtnaatf 




1
00 )cos()(
n
nn tnAatf 

20. From this form, we can plot the amplitude spectrum,
vs. n and the phase spectrum, vs. n.
It can be shown that the combination of cosine and
sine function can be expressed as a cosine function
only:
Comparing both sides of eqn:
n
A
n

xAxA
xxA
xAxbxa
sin)sin(cos)cos(
)sinsincos(cos
)cos(sincos






.....(1) cosaA  .....(2) sinbA 

21. 22222
22222
222222
)sin(cos
sincos
baAbaA
baA
baAA

















a
b
a
b
a
b
A
A
1
tantan
cos
sin



:)2()1(
22

:
)1(
)2(

22. Hence
)cos(sincos
000 nnnn tnAtnbtna  
where
22
amplitude,
nnn baA 










n
n
n
a
b
1
tan phase,
Or in phasor/complex form:










n
n
nnnnnn
a
b
bajbaA
122
tan
00aA

23. Example 1
Determine the Fourier series of the following
waveform. Obtain the amplitude and phase
spectra.

24. Solution
First, determine the period & describe the one period
of the function:
T = 2






21,0
10,1
)(
t
t
tf )()2( tftf 


 
T
2
0We find that

25. Then, obtain the coefficients a
0
, a
n
and b
n
:
2
1
)01(
2
1
01
2
1
)(
2
1
)(
1
2
1
1
0
2
0
0
0










dtdtdttf
dttf
T
a
T
Or
2
1
2
11graph below Area
0



T
a

26. 





n
n
n
tn
dttdtn
tdtntf
T
a
n
sinsin
0cos1
cos)(
2
1
0
2
1
1
0
2
0
0











Notice that n is integer which leads ,
since
0sinn
03sin2sinsin  
Therefore, .0
na

27. 





n
n
n
tn
dttdtn
tdtntf
T
b
n
cos1cos
0sin1
sin)(
2
1
0
2
1
1
0
2
0
0












15cos3coscos  
16cos4cos2cos  
Notice that
Therefore,






even ,0
odd ,/2)1(1
n
nn
n
b
n
n


or
n
n )1(cos 

28. 
























ttt
tn
n
tn
n
tnbtnaatf
n
n
n
n
n
nn











5sin
5
2
3sin
3
2
sin
2
2
1
sin
2
2
1
sin
)1(1
2
1
)sincos()(
odd
1
1
1
000
Finally,

29. 





















)905cos(
5
2
)903cos(
3
2
)90cos(
2
2
1
)90cos(
2
2
1
sin
2
2
1
)(
odd
1
odd
1
t
tt
tn
n
tn
n
tf
n
n
n
n










In amplitude-phase form,

30. Amplitude spectrum: Phase spectrum:

31. Some helpful identities
For n integers,
n
n )1(cos 0sinn
02sinn 12cosn
xxsin)sin(
xxcos)cos(
)90cos(sin xx
)90cos(sin  xx
)180cos(cos  xx

32. The sum of the Fourier series terms can
evolve (progress) into the original
waveform
From Example 1, we obtain
 ttttf 





5sin
5
2
3sin
3
2
sin
2
2
1
)(
It can be demonstrated that the sum will
lead to the square wave:
Notes:

33. tttt 







7sin
7
2
5sin
5
2
3sin
3
2
sin
2
ttt 





5sin
5
2
3sin
3
2
sin
2

tt 



3sin
3
2
sin
2
t

sin
2
(a) (b)
(c) (d)

34. ttttt 









9sin
9
2
7sin
7
2
5sin
5
2
3sin
3
2
sin
2

ttt 





23sin
23
2
3sin
3
2
sin
2
2
1
 
(e)
(f)

36. Example 2
Given ,)(ttf 11t
)()2( tftf 
Sketch the graph of f (t) such that .33t
Then compute the Fourier series expansion of f (t).
Plot the amplitude and phase spectra until the forth
harmonic.

37. Solution
The function is described by the following graph:
T = 2


 
T
2
0We find that

38. 












nnn
n
n
n
n
tn
n
n
dt
n
tn
n
tnt
tdtnttdtntf
T
b
nn
n
1
22
1
0
22
1
0
1
0
1
0
1
0
0
)1(2
0
)1(2sin2cos2
sin
2
cos2
cos
2
cos
2
sin
2
4
sin)(
4

























Then we compute the coefficients:
0
0

n
aa since f(t) is an odd function.

39. 














)904cos(
2
1
)903cos(
3
2
)902cos(
1
)90cos(
2
4sin
2
1
3sin
3
2
2sin
1
sin
2
sin
)1(2
)sincos()(
1
1
1
000
tt
tt
tttt
tn
n
tnbtnaatf
n
n
n
nn



















Hence,

40. Amplitude spectrum: Phase spectrum:
0.64
0.32
0.21
0.16

41. Example 3
Given






42,0
20,2
)(
t
tt
tv
)()4( tvtv 
(i) Sketch the graph of v (t) such that .120t
(ii) Compute the trigonometric Fourier series of v (t).
(iii) Express the Fourier series of v (t) in the
amplitude-phase form. Then plot the amplitude
and phase spectra until the forth harmonic.

42. Solution
(i) The function is described by the following graph:
T = 4
2
2
0

 
T
We find that
0 2 4 6 810 12
t
v (t)
2

43. (ii) Then we compute the coefficients:
2
1
2
2
4
1
)2(
4
1
0)2(
4
1
)(
1
2
0
22
0
4
2
2
0
4
0
0

















t
tdtt
dtdttdttv
T
a
Or
2
1
4
22
2
1
0


a

44. 



























odd ,/4
even ,0])1(1[2)cos1(2
2
2cos1cos
2
1
0
sin
2
1sin)2(
2
1
0cos)2(
2
1
cos)(
2
222222
2
0
2
0
2
0
2
0
2
0
2
0 0
0
2
00
0
4
2
2
0
0
4
0
0
nn
n
nn
n
n
n
n
tn
dt
n
tn
n
tnt
tdtnttdtntv
T
a
n
n












45. 










nnn
n
n
n
tn
n
dt
n
tn
n
tnt
tdtnttdtntv
T
b
n
21
2
2sin1
sin
2
11
cos
2
1cos)2(
2
1
0sin)2(
2
1
sin)(
2
0
2
0
2
0
0
2
0
2
0
2
0
0
2
0 0
0
2
00
0
4
2
2
0
0
4
0
0



















since 0sin2sin
0
 nn

46. 





























1
00
1
22
1
000
)cos(
2
sin
2
2
cos
])1(1[2
2
1
)sincos()(
n
nn
n
n
n
nn
tnAa
tn
n
tn
n
tnbtnaatv






Hence,

47. 










































odd ,
2
tan1
42
even ,90
2
odd ,
22
even ,
2
1
22
n
n
nn
n
n
nj
nn
n
n
j
jbaA
nnnn






(iii) Where

48.  5.5775.0
11
A
 0.7822.0
33
A
 9032.0
22
A
 9016.0
44
A
5.0
00
aA
We obtained that
 
 
















902cos16.00.78
2
3
cos22.0
90cos32.05.57
2
cos75.05.0
)cos()(
1
00
t
t
t
t
tnAatv
n
nn






49. Amplitude spectrum: Phase spectrum:
0.5
0.75
0.32
0.22
0.16
0π/2π3π/22π
π/2π3π/22π
−57.5°
−78.0°
−90° −90°

50. Example 4
Given









21,1
11,
12,1
)(
t
tt
t
tf
)()4( tftf 
Sketch the graph of f (t) such that .66t
Then compute the Fourier series expansion of f (t).

51. Solution
The function is described by the following graph:
T = 4
2
2
 
T
We find that
0−4−6 2 46
t
f (t)
−2
1
−1

52. Then we compute the coefficients. Since f (t) is
an odd function, then
0)(
2
2
2
0


dttf
T
a
0cos)(
2
2
2


tdtntf
T
a
n

and

53. 2222
1
0
22
2
1
1
0
1
0
2
1
1
0
2
0
2
2
)2/sin(4cos2sin2cos
cos2cossincos
coscoscos
sin1sin
4
4
sin)(
4
sin)(
2






















n
n
n
n
n
n
n
n
n
nn
n
tn
n
n
n
tn
dt
n
tn
n
tnt
tdtntdtnt
tdtntf
T
tdtntf
T
b
n




































since 0sin2sin  nn

54. 



















1
1
1
1
0
2
sin
)1(
2
2
sin
cos2
)sincos(
2
)(
n
n
n
n
nn
tn
n
tn
n
n
tnbtna
a
tf






Finally,
♣

55. Example 5
Compute the Fourier series expansion of f (t).

56. Solution
The function is described by
T = 3
3
22
0

 
T
and









32,1
21,2
10,1
)(
t
t
t
tf
)()3( tftf 
T = 3

57. Then we compute the coefficients.
3
8
1
2
3
2)01(
3
4
21
3
4
)(
4
)(
2
2/3
1
1
0
2/3
0
3
0
0 


















 
dtdtdttf
T
dttf
T
a
 
3
8
)23()12(2)01(
3
2
121
3
2
)(
2
3
2
2
1
1
0
3
0
0 





 
dtdtdtdttf
T
a
Or, since f (t) is an even function, then
Or, simply
3
8
4
3
2
period ain
graph below area Total2
)(
2
3
0
0










T
dttf
T
a

58. 3
2
sin
2
3
2
sinsin2
2
sin
2
3
sin2
3
4
sin
2
3
sin2sin
3
4
sin2
3
4sin
3
4
cos2cos1
3
4
cos)(
4
cos)(
2
2/3
1
1
0
2/3
1
1
0
2/3
0
3
0


















n
n
n
n
n
n
n
n
n
n
n
n
n
tn
n
tn
tdtntdtn
tdtntf
T
tdtntf
T
a
n



















































;
3
2


59. 























1
1
1
0
3
2
cos
3
2
sin
12
3
4
3
2
cos
3
2
sin
2
3
4
)sincos(
2
)(
n
n
n
nn
tnn
n
tnn
n
tnbtna
a
tf





Finally,
♣
and 0
n
b since f (t) is an even function.

60. Parseval’s Theorem
Parserval’s theorem states that the
average power in a periodic signal is equal
to the sum of the average power in its DC
component and the average powers in its
harmonics

61. =
+ +
+ + + …
tacos
1
ta2cos
2
tbsin
1
tb2sin
2
f(t)
t
P
avg
P
dc
P
a1 P
b1
P
a2
P
b2
0a

62. For sinusoidal (cosine or sine) signal,
R
V
R
V
R
V
P
2
peak
2
peak
2
rms
2
12








For simplicity, we often assume R = 1Ω,
which yields
2
peak
2
rms
2
1
VVP 

63. And since ,
Therefore, the total power of all harmonics is




2
2
2
2
2
1
2
1
2
0
dcavg
2
1
2
1
2
1
2
1
2211
babaa
PPPPPP
baba






1
22
0
1
222
0avg
2
1
)(
2
1
n
n
n
nn
AabaaP
2
rms
VP






1
22
0
1
222
0rms
2
1
)(
2
1
n
n
n
nn
AabaaV

64. Circuit applications
Steps for applying Fourier series:
1.Express the excitation as a Fourier series
2.Transform the circuit from the time domain to
the frequency domain
3.Find the response of the dc and ac
components in the Fourier series
4.Add the individual dc and ac responses using
the superposition principle

65. The signal in Figure 6.1 is applied to the circuit in
Figure 6.2.
(i)Find the Fourier series of i
s
(t).
(ii)Plot the amplitude spectrum for the dc
component and the first three non-zero harmonics
of i
s
(t).
(iii)Find the load current i
L
(t).
(iv)Find the average power dissipated in the
resistor R
L.
Example 6

66. )(ti
s
)(ti
s
H 1
 1
 1
LR
)(ti
L
Figure 6.2
Figure 6.1

67. Solution
(i)The current source is described by
and







0,
0,
,||)(
tt
tt
ttti
s



1,2
0
 T

68. Compute the Fourier series. Since i
s
(t) is an even
function,
or
22
1
2
2
)(
2
0
2
0
2/
0
0











 
t
tdtdtti
T
a
T
s
22
11
0over
graph below Area1
0




















t
a

69. 

























even ,0
odd ,/4]1)1[(2
)1(cos2
cos2sin2
sin2sin2
cos
2
4
cos)(
4
2
2
2
0
2
000
2/
0
0
n
nn
n
n
n
n
nt
n
n
dt
n
nt
n
ntt
ntdtt
tdtnti
T
a
n
T
sn












70. Therefore, the Fourier series of i
s
(t) is
0
nb






















)1805cos(
25
4
)1803cos(
9
4
)180cos(
4
2
)180cos(
4
2
cos
4
2
)(
odd
1
2
odd
1
2
t
tt
nt
n
nt
n
ti
n
n
n
n
s








71. nsI
n
sI
n n
1.571
1.273
0.141
0.051
(ii)Hence,
,5,3,1,180
4
2
 n
n
I
n
s
2
0


s
I

72. (iii)To compute i
L(t), separate to dc and ac analysis:
dc analysis (n = 0)
S/c the inductor, hence
0sI
 1
 1
42
2/
2
0
0


s
L
I
I

73. 
































180
2
tantan
4
14
180
4
2
1
,
)(
11
2
2
2
2
0
n
n
n
n
n
njn
jn
nnI
LjRR
LjR
I
ns
nL
n
L
nn





s
L
L I
LjRR
LjR
I
)(




ac analysis (n ≥ 1)
or

74. 


)49.1905cos(048.0)26.1953cos(124.0
)43.198cos(805.0785.0)(
tt
tti
L
Hence,



49.190048.0
,26.195124.0,43.198805.0
,785.0
5
31
0
L
LL
L
I
II
I

75. W949.0)048.0124.0805.0(
2
1
785.0
2
1
2222
1
22
2
)(
0












L
n
LL
LrmsL
RII
RIP
n
(iv)Average power absorbed by R
l is

76. The signal in Figure 7.1 is applied to the circuit in
Figure 7.2.
(i)Find the Fourier series of v
s
(t).
(ii)Find the output voltage v
o
(t).
(iii)Plot the amplitude spectrum for the dc
component and the first three non-zero harmonics
of the output voltage v
o(t).
(iv)Find the r.m.s value of v
o
(t) .
Example 7

77. )(tv
s
F 1
 1
)(tv
o
Figure 7.2
Figure 7.1
 1
)(tv
s



78. Solution
(i)The voltage source is described by
and






01,1
10,1
)(
t
t
tv
s

0
,2T
Compute the Fourier series. Since v
s(t) is an odd
function,
0
0

n
aa

79. 

















even ,0
odd ,/4])1(1[2
)cos1(2
cos
2sin
2
4
sin)(
4
1
0
1
0
2/
0
0
n
nn
n
n
n
n
tn
tdtn
tdtntv
T
b
n
T
sn









80. Therefore, the Fourier series of v
s(t) is














 




)905cos(
5
4
)903cos(
3
4
)90cos(
4
)90cos(
4
sin
4
)(
odd
1
odd
1
t
tt
tn
n
tn
n
tv
n
n
n
n
s









Hence,
,5,3,1,90
4
 n
n
V
n
s

0
0

sV

81. (ii)To compute v
o(t), separate to dc and ac analysis:
dc analysis (n = 0)
O/c the capacitor, hence
0
0

sV
 1
0
0

oV
 1


0I

82.  


























902tantan
41
14
90
4
21
1
,
21
1
11
22
22
0









nn
n
n
n
nnj
jn
nnV
RCj
RCj
V
ns
n
n
o
nn
sso
V
RCj
RCj
V
CjRR
CjR
V




21
1
)]/1([
)/1(






ac analysis (n ≥ 1)
or

83. 

)82.915cos(128.0
)02.933cos(213.0)61.98cos(660.0)(
t
tttv
o
Hence,



82.91128.0
,02.93213.0,61.98660.0
,0
5
31
0
o
oo
o
V
VV
V

84. noV
n
oV
n
n
0.660
0.213
0.128
−98.61°
(iii)
−93.02°
−91.82°
(iv)
V 705.0)128.0213.0660.0(
2
1
2
1
222
1
22
)rms(
0

 

n
ooo
n
VVV

85. Exponential Fourier series
Recall that, from the Euler’s identity,
xjxe
jx
sincos

yields
2
cos
jxjx
ee
x



2
sin
j
ee
x
jxjx 

and

86. Then the Fourier series representation becomes






































































































11
0
1
0
1
0
1
0
1
0
222
222
222
222
)sincos(
2
)(
n
tjnnn
n
tjnnn
n
tjnnntjnnn
n
tjntjn
n
tjntjn
n
n
tjntjn
n
tjntjn
n
n
nn
e
jba
e
jbaa
e
jba
e
jbaa
ee
jb
ee
a
a
j
ee
b
ee
a
a
tnbtna
a
tf






87. Here, let we name



















11
0
222
)(
n
tjnnn
n
tjnnn
e
jba
e
jbaa
tf

2
nn
n
jba
c

 ,
2
nn
n
jba
c



Hence,






















n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
ececcec
ececc
ececc



1
0
1
11
0
11
0
and .
2
0
0
a
c
c
0
c
−nc
n

88. Then, the coefficient c
n can be derived from

















T
tjn
T
TT
TT
nn
n
dtetf
T
dttnjtntf
T
tdtntfjtdtntf
T
tdtntf
T
j
tdtntf
T
jba
c
0
0
00
00
)(
1
]sin)[cos(
1
sin)(cos)(
1
sin)(
2
2
cos)(
2
2
1
2





89. In fact, in many cases, the complex
Fourier series is easier to obtain rather
than the trigonometrical Fourier series
In summary, the relationship between the
complex and trigonometrical Fourier series
are:
nn
nn
n A
jba
c 


2
1
2 2
nn
n
jba
c





T
dttf
T
ac
0
00 )(
1



T
tjn
n dtetf
T
c
0
)(
1



nncc
or
22
22
nnn
n
Aba
c 



90. The average power and the rms value in
the term of Fourier complex coefficient are







1
2
2
0
2
1
2
n
n
n
n
cccP






1
2
2
0
2
rms 2
n
n
n
n cccF

91. Example 8
For the given function,
(i)Obtain the complex Fourier series
(ii)Plot the amplitude and the phase spectra
of the complex Fourier series for −5 ≤ n ≤ 5
(iii)Calculate the average power and the rms value
of the signal
2 44 2 0
f (t) =
e

t
2
e
1
)(tv
t

92. (i)Since , . Hence
Solution






2
1
2
1
2
1
)(
1
2
2
0
2
0
0
0






e
e
dte
dttv
T
c
t
t
T
1
0
2T

93. )1(2
1
)1(2
1
)1(2
1
12
1
2
1
2
1
)(
1
222)1(2
2
0
)1(
2
0
)1(
2
0
0
0
jn
e
jn
ee
jn
e
jn
e
dtedtee
dtetv
T
c
njjn
tjn
tjnjntt
T
tjn
n
































since 1012sin2cos
2




njne
nj

94. jnt
nn
tjn
n e
jn
e
ectv 



 


)1(2
1
)(
2
0



Therefore, the complex Fourier series of v(t) is
0
2
0
2
0
2
1
)1(2
1
c
e
jn
e
c
n
n
n 









*Notes: Even though c
0 can be found by substituting
c
n with n = 0, sometimes it doesn’t works (as shown
in the next example). Therefore, it is always better to
calculate c
0
alone.

95. n
n
n
n
e
c
nn
1
2
1
2
2
tan
1
85
1
tan0
12
1
























(ii)
The complex frequency spectra are

96. kW 19.20
)7.166.209.26381.60(285
222222
2
1






n
ncP
(ii)The average power is (assume R = 1 Ω)
and since
2
rms 1VP

V 09.142k19.20
rms
V

97. Example 9
Obtain the complex Fourier series of the function in
Example 1. Then plot the complex frequency spectra
for −4 ≤ n ≤4 and calculate the rms value of the signal.

98. Solution
2
1
1
2
1
)(
1
1
00
0  
dtdttf
T
c
T

0
)1(
22
1
01
2
1
)(
1
1
0
2
1
1
00


















jn
tjn
tjn
T
tjn
n
e
n
j
jn
e
dtedtetf
T
c

99. )1(
2



jn
n e
n
j
c
But
njn
nnjne )1(cossincos 



Thus,




even ,0
odd ,/
]1)1[(
2 n
nnj
n
j
n


Therefore, 












odd
0
2
1
)(
0
n
n
n
tjn
n
tjn
n e
n
j
ectf


*Here notice that .0
0
cc
n
n


100. 2
1
0
c
 0,0
nn
cFor n even,
For n odd,






0,90
0,90
,
1
n
n
n
c
nn


0.5

101. 692.0)11.032.0(25.0
222
2
rms



n
n
cF
The average rms value is

102. Summary
Sine-cosine form
Amplitude-phase form
Exponential (or complex) form




1
000 )sincos()(
n
nn tnbtnaatf 
,cos)(
2
0
0

T
n
tdtntf
T
a  

T
n
tdtntf
T
b
0
0
sin)(
2
,)(
1
0
0

T
dttf
T
a,
2
0
T






1
00 )cos()(
n
nn tnAatf 




n
tjn
nectf
0
)(

nnnn
jbaA 
,
00
ac ,
2
nn
n
jba
c


*
nn
cc


103. Prob.17.37, pg.803
If the sawtooth waveform in Fig.17.9 is the
voltage source v
s(t) in the circuit of
Fig.17.22, find the response v
o(t).

104. Prob.17.37, pg.803
If the periodic current waveform in
Fig.17.73(a) is applied to the circuit in
Fig.17.73(b), find v
o.


3
)(ti
s
0 1 2 3
−1
3
t
(a) (b)

105. Problem


1 2 30 4
1
5
2

3
6
 1
o
R
 1
s
R F 1C


)(tv
o
)(tv
s
)(tv
s
t
(i) Compute the output voltage v
o
(t).
(ii) Plot the amplitude and phase spectra of v
o (t)
for the first three nonzero harmonics.
(ii) Calculate the average power dissipated in
the resistor R
o
.

106. QUIZ 1
)(tv
s


)(tv
o
H1
H1
1
1
Given the Fourier series expansion of the
voltage signal v
s(t) is
Find v
o
(t) in the amplitude-phase form.










1
sin
10
5)(
n
s nt
n
tv


107. Quiz1
Determine the Fourier series of the following
waveform.
)(tv
4

108. QUIZ 2
Find i(t) if v(t) is the signal as given in QUIZ 1.


















odd
1
odd
1
)90cos(
2
2
1
sin
2
2
1
)(
n
n
n
n
tn
n
tn
n
tv 




109. QUIZ 3
Find the output voltage v
o(t) if the current
source i
s(t) is given by










1
)90cos(
2
1)(
n
s nt
n
ti

)(ti
s


)(tv
o5 F5.0
3

110. QUIZ 4
Using time differentiation technique, find the
Fourier transform of f(t).