Chapter3 - Fourier Series Representation of Periodic Signals

241-306 Fourier Series Representation of Periodic Signals
1
Chapter 3
Fourier Series
Representation of
Periodic Signals

2
Outline
1 The Response of LTI Systems to Complex
Exponentials
2 Fourier Series Representation of Continuous-Time
Periodic Signals
3 Convergence of The Fourier Series
4 Properties of Continuous-Time Fourier Series
5 Fourier Series Representation of Discrete-Time
Periodic Signals

3
6 Properties of Discrete-Time Fourier Series
7 Fourier Series and LTI Systems

4
In Chapter 2, We discuss about a representation of
signals as linear combination of a set of basic
signals. Two properties of the basic signals that we
desire :
3.1 The Response of LTI Systems to Complex
Exponentials
1 The set of basic signals can be used to
construct a broad and useful class of signals.
2 The response of an LTI system to each signal
should be simple enough in structure to provide
us with a convenient representation for the
response of the system to any signal constructed
as a linear combination of the basic signals.

5
Eigenfunction and Eigenvalue of the system
LTI CT
e
st
H se
st
Eigenvalue Eigenfunction
yt=∫−∞
∞
hxt−d 
=∫−∞
∞
he
st−
d 
=e
st
∫−∞
∞
he
s
d 
H(s)

6
LTI DT
z
n
H z z
n
Eigenvalue Eigenfunction
y[n]= ∑
k=−∞
∞
h[k ] x[n−k ]
= ∑
k=−∞
∞
h[k ] z
n−k
=z
n
∑
k=−∞
∞
h[k ]z
−k
H(z)

7
If the input to a continuous-time LTI system is
represented as a linear combination of complex
exponentials
xt=∑
k
ak e
sk t
then the output will be
yt=∑
k
ak H sk e
sk t

8
For discrete-time LTI system,it the input is
represented as a linear combination of complex
exponentials
x[n]=∑
k
ak zk
n
then the output will be
y[n]=∑
k
ak H zk zk
n

9
Example 3.1
yt=xt−3
yt=e
j2t−3
=e
− j6
e
j2t
H s=∫−∞
∞
−3e−s 
d =e−3s
Consider the LTI system
If the input to this system is x(t) = e j2t
then
We would expect, since e j2t
is eigenfunction. The
associated eigenvalue is H(j2) = e -j6
. The impulse
response of the system is h(t) = δ(t-3)

10
yt=cos4t−3cos7t−3
xt=
1
2
e
j4t

1
2
e
− j4t

1
2
e
j7t

1
2
e
− j7t
xt=cos4tcos7t
Consider the input signal
The output is
By using Euler's relation

11
yt=
1
2
e
− j12
e
j4t

1
2
e
j12
e
− j4t

1
2
e
− j21
e
j7t

1
2
e
j21
e
− j7t
yt=
1
2
e
j4t−3

1
2
e
− j4t−3

1
2
e
j7 t−3

1
2
e
j7t−3
=cos4t−3cos7t−3
then
or

12
Problem section 3.1
1 Consider three continuous-time systems S1
, S2
and
S3
whose response to a complex exponential input ej5t
are specified as
S1 :e
j5t
te
j5t
S2 :e
j5t
te
j5t−1
S3 :e
j5t
cos5t
For each system, determine determine whether the
given information is sufficient to conclude that the
system is definitely not LTI.

13
2 Consider three discrete-time systems S1
, S2
and S3
whose response to a complex exponential input ejπn/2
are specified as
S1 :e jn/2
e j n/2
u[n]
S2 :e
j n/2
e
j3n/2
S3 :e
j n/2
2e
j5 n/2
For each system, determine determine whether the
given information is sufficient to conclude that the
system is definitely not LTI.
Problem section 3.1Problem section 3.1 (Cont.)

14
3.2 Fourier Series Representation of Continuous-
Time Periodic Signals
Linear Combination of Harmonically Related
Complex Exponentials
xt=xtT
xt=cos0 t
xt=e
j
0
t
Definition of periodic signal
two basic periodic signals
ω0
– fundamental frequency
T = 2π/ω0
– fundamental period

15
Set of harmonically related complex exponentials
k=e
jk 0 t
=e
jk 2/T t
k=0,±1,±2,...
xt= ∑
k=−∞
∞
ak e
jk 0 t
= ∑
k=−∞
∞
ak e
jk 2/T t
A linear combination of harmonically related
exponentials of the from

16
Example 3.2
xt= ∑
k=−3
3
ak e
jk2t
Consider the signal x(t), with fundamental
frequency 2π.
where
a0=1 a1=a−1=
1
4
a2=a−2=
1
2 a3=a−3=
1
3

17
xt=1
1
4
e
j2t
e
− j2t

1
2
e
j4t
e
− j4t


1
3
e
j6 t
e
− j6t

xt=1
1
2
cos2tcos4t
2
3
cos6t
Rewrite the x(t)
Using Euler's relation, we can write x(t) in the from
We illustrate graphically how the signal x(t) is built
up from its harmonic component

18

19

20

21
xt= ∑
k=−∞
∞
ak
∗
e
− jk 0 t
Suppose that x(t) is real and can be represented in the
from
Fourier series of real periodic signals
xt= ∑
k=−∞
∞
ak e
jk 0 t
since x*(t) = x(t), we obtain
(*)

22
xt= ∑
k=−∞
∞
a−k
∗
e
jk 0 t
Replacing k by -k :
by comparision with (*) in slide 19
ak=a−k
∗
or
ak
∗
=a−k

23
xt=a0∑
k=1
∞
[ak e
jk 0 t
a−k e
− jk 0 t
]
xt=a0∑
k=1
∞
[ak e
jk 0 t
ak
∗
e
− jk 0 t
]
Alternative forms of the Fourier series
Suppose that x(t) is real and can be represented in the
from
xt= ∑
k=−∞
∞
ak e
jk 0 t
(*)
rearrange the summation
substituting ak
* for a-k
:

24
xt=a0∑
k=1
∞
2ℜe {ak e
jk 0 t
}
ak=Ak e
jk
xt=a0∑
k=1
∞
2ℜe {Ak e
jk 0 t−k 
}
since the two term inside the summation are
complex conjugates of each other.
If ak
is express in polar form
That is
xt=a02∑
k=1
∞
Ak cosk 0 t−k 

25
ak=Bk jCk
xt=a02∑
k=1
∞
[Bk cosk 0 t−Ck sink 0 t]
But if ak
is express in rectangular form
where Bk
and Ck
are both real. We can write :

26
Determination of the Fourier Series Representation
of a Continuous-Time Periodic Signal
Assuming that a given periodic signal can be
represented with the series
xt= ∑
k=−∞
∞
ak e
jk 0 t
We need the procedure for determining the
coefficients ak
.

27
Multiplying both sides by e
− jn
0
t
xte
− jn0 t
= ∑
k=−∞
∞
ak e
jk 0 t
e
− jn0 t
Integrating both sides from 0 to T=2π/ω0
:
∫
0
T
xte
− jn0 t
dt=∫
0
T
∑
k=−∞
∞
ak e
jk 0 t
e
− jn0 t
dt
or
∫
0
T
xte
− jn0 t
dt= ∑
k=−∞
∞
ak
[∫
0
T
e
jk−n0 t
dt
] (*)

28
By using Euler's relation
∫
0
T
e
jk−n0 t
dt =∫
0
T
cosk−n0 tdt
 j∫
0
T
sink−n0 tdt
For k ≠ n, the integral over one period of
sinusoidal signal is zero. For k=n, the integral
equal to T
∫
0
T
e
jk−n0 t
dt =
{T , k=n
0, k≠n

29
Consequently, the equation
∫
0
T
xte
− jn0 t
dt= ∑
k=−∞
∞
ak
[∫
0
T
e
jk−n0 t
dt
]
reduce to
∫
0
T
xte
− jn0 t
dt=an T
or
an=
1
T
∫
0
T
xte
− jn0 t
dt=
1
T
∫T
xte
− jn0 t
dt

30
xt= ∑
k=−∞
∞
ak e
jk 0 t
= ∑
k=−∞
∞
ak e jk 2/T t
ak=
1
T
∫T
xte
− jk 0 t
dt=
1
T
∫T
xte− jk 2/T t
dt
a0=
1
T
∫T
xtdt
Equations of Fourier series of a periodic
continuous-time signal
For k=0

31
Example 3.3
Consider the signal
xt=sin0 t
Determine the Fourier series coefficients for this
signal

32
Solution
We can express sin ω0
t as
sin0 t=
1
2j
e
j 
0
t
−
1
2 j
e
− j 
0
t
Comparing to the equation of Fourier series
a1=
1
2j
a−1=−
1
2j
ak=0 k≠1 o r −1

33
∣ak∣
∢ak
10-1
●● ●
10
-1
●● ●
k
k
Graph of the magnitude and phase of ak

34
Example 3.4
xt=1sin 0 t2cos0 tcos20 t

4 
Determine the Fourier series coefficients for the
signal

35
Solution
Express this signal in the term of complex
exponentials
xt=1
1
2j
[e
j
0
t
−e
− j
0
t
][e
j
0
t
e
− j
0
t
]

1
2
[e
j2
0
t/4
e
− j2
0
t/4
]

36
Collecting terms :
xt=11
1
2je
j 
0
t
1−
1
2je
− j
0
t
1
2
e
j/4
e
j 2
0
t
1
2
e
− j/4
e
− j 2
0
t
The Fourier series coefficients are :
a0=1 a1=1
1
2j=1−
1
2
j

37
a2=
1
2
e
j/4
=2
4
1 j
a−2=
1
2
e
− j/4
=
2
4
1− j
ak=0, ∣k∣2
a−1=1−
1
2j=1
1
2
j

38
Magnitude and Phase of ak

39
Example 3.5
Determine the Fourier series coefficients for the
signal
xt=
{1, ∣t∣T1
0, T1∣t∣T /2

40
Solution
For k = 0 :
a0=
1
T
∫
−T 1
T 1
dt=
2T1
T
For k ≠ 0 :
ak=
1
T
∫
−T 1
T 1
e
− jk 0 t
dt=−
1
jk 0 T
e
− jk 0 t
∣−T1
T 1

41
ak=
2
k 0 T [e
jk 
0
T
1
−e
− jk 
0
T
1
2j ]
ak=
2sink T1
k T
=
sink T1
k 
, k≠0

42
Plot of the scaled Fourier series coefficients Tak
with T = 4T1

43
with T = 8T1
with T = 16T1

44
3.4 Convergence of the Fourier Series
The question of the validity of Fourier series
representation.
xt= ∑
k=−∞
∞
ak e
jk 0 t
Is the FS can approximate the periodic signal x(t)?
We can prove the validity by using the difference
in energy of x(t) and its FS

45
et=xt−∑−∞
∞
ak e
− jk 0 t
If x(t) and its FS are the same, there is no energy in
the difference.
∫T
∣et∣
2
dt=0
One class of periodic signals is the signal that have
finite energy over a signal period.
∫T
∣xt∣
2
dt∞

46
Condition 1. x(t) must be absolutely integrable over
one period.
The set of conditions (The DiriChlet condition)
guarantees that x(t) equal its Fourier series
representation
∫T
∣xt∣dt∞
xt=
1
T
violates the first
condition

47
Condition 2. In a finite time interval, x(t) has a finite
number of maxima and minima.
xt=sin2
t , 0t≤1 meets condition 1
but not condition 2

48
Condition 3. In a finite time interval, x(t) has only a
finite number of discontinuities.

49
3.5 Properties of Continuous-Time Fourier Series
The notation
FS
xt ↔ ak
mean ak
are the Fourier series coefficients of the
periodic signal signal x(t) which have a period T
and fundamental frequency ω0
=2π/T

50
Linearity
FS
xt ↔ ak
FS
xt ↔ bk
FS
zt=AxtByt ↔ ck=AakBbk
Let x(t) and y(t) are periodic signals with period T

51
Time Shifting
FS
xt ↔ ak
FS
xt−t0 ↔ e
− jk 0 t0
ak=e
− jk2/T t0
ak
Let x(t) is a periodic signal with period T

52
Time Reversal
FS
xt ↔ ak
FS
x−t ↔ a−k
Time Scaling
The Fourier series coefficient dose not changed
but the Fourier series has changed because of the
change in the fundamental frequency.

53
Multiplication
FS
xt ↔ ak
FS
yt ↔ bk
xt yt ↔ hk= ∑
l=−∞
∞
al bk−l
FS

54
Conjugation and Conjugate Symmetry
FS
xt ↔ ak
FS
x∗
t ↔ a−k
∗
If x(t) is real, x(t) = x*(t), the Fourier series
coefficients will be conjugate symmetric.
a−k = ak
∗

55
Parseval's Relation for Continuous-Time Periodic
Signals
1
T
∫T
∣xt∣
2
dt= ∑
k=−∞
∞
∣ak∣
2
1
T
∫T
∣ak e
jk 0t
∣
2
dt=
1
T
∫T
∣ak∣
2
dt=∣ak∣
2
Parseval 's relation for continuous-time periodic
signals is
In one period of the periodic signal x(t)
The total average power in a periodic signal equal
the sum of the average powers in all of its harmonic
component.

56

57
Example 3.6
Determine the Fourier Series representation of g(t)
with a fundamental period of 4.

58
Solution
From the example 3.5 with T = 4 and T1
= 1 we
can write the function g(t) in the form of x(t).(add
dc offset -1/2 and shift the signal to right)
g t=xt−1−1/2
From shift property, the Fourier coefficients of
x(t-1) may be express as :
bk=ak e− jk /2
ak
is FS coefficient of x(t)

59
The Fourier coefficient of the dc offset is given by
ck=
{
0, for k≠0
−
1
2
, for k=0
By using linearity property, the coefficients for g(t)
may be expressed as
d k=
{
ak e
− jk /2
, for k≠0
a0−
1
2
, for k=0

60
by replacing ak
from example 3.5
d k=
{
sin k /2
k 
e
− jk /2
, for k≠0
0, for k=0

61
Example 3.7
Consider the triangular signal x(t) with period T = 4
and fundamental frequency ω0
= π/2. Determine
the Fourier coefficient from g(t) of example 3.6

62
Solution
The signal g(t) is the derivative of x(t) or x(t) is
the integration of signal g(t).
dk= jk /2ek
dk
is the Fourier coefficient of g(t) and ek
is the
Fourier coefficient of x(t)
Solution
ek=
2dk
jk 
=
2sin k /2
jk 2 e− jk /2
, k≠0
for k=0, e0
is determine by finding the area under
one period of x(t) and dividing by the period. e0
=1/2

63
3.6 Fourier Series Representation of Discrete-Time
Periodic Signals
Linear Combination of Harmonically Related
Complex Exponentials
The definition of discrete-time periodic signals
with period N
x[n]=x[nN ]
The fundamental period is the smallest positive
integer of N for which the equation above holds and
ω0
=2π/N

64
Set of harmonically related complex exponentials
k=e
jk 0 n
=e
jk 2/N n
k=0,±1,±2,...
There are N distinct signals in the set. The
consequence is the discrete-time complex
exponential which differ in frequency by multiple
of 2π are identical.
k [n]=krN [n]

65
x[n]=∑
k
ak e
jk 0 n
=∑
k
ak e
jk 2/N n
A linear combination of harmonically related
exponentials of the from
Since the sequence Φk
[n] are distinct only over a
range of N successive values of k. we can
express the limits of the summation as k = <N>.
x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e jk2/ N n
k could take on values, for example, k = 1,2,...,N-1
or k = 3,4,...,N+2

66
Determination of the Fourier Series
Representation of a Periodic Signal
x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e jk 2/ N n
The equation of the discrete-time Fourier series
ak
is the Fourier series coefficients
If we evaluate the equation above for N
successive values of n corresponding to one
period of x[n] :

67
x[0]= ∑
k=N 
ak
x[1]= ∑
k= N 
ak e
j2k/ N
x[N −1]= ∑
k=N 
ak e
j2 k N −1/ N

68
∑
n=〈 N 〉
e jk 2/ N n
=
{N , k=0,±N ,±2N ,...
0, otherwise
∑
n=〈 N 〉
x[n]e
− jr2/N n
= ∑
n=〈N 〉
∑
k=〈 N 〉
ak e
jk−r2/ N n
The basic result is
Multiplying both side of x[n] by e-jr(2π/N)n
and
summing over N terms

69
∑
n=〈 N 〉
x[n]e
− jr2/N n
= ∑
k=〈 N 〉
ak ∑
n=〈 N 〉
e
jk−r2/ N n
ar=
1
N
∑
n=〈 N 〉
x[n]e
− jr2/N n
Interchanging the order of summation on right-
hand side
N , k−r=0,±N ,±2N ,...
∑
n=〈 N 〉
x[n]e
− jr2/N n
=ar N

70
x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e
jk2/ N n
ak=
1
N
∑
n=〈N 〉
x[n]e
− jk 0 n
=
1
N
∑
n=〈N 〉
x[n]e
− jr2/N n
Discrete-time Fourier Series pair

71
Example 3.10
x[n]=sin 0 n
0=
2
N
Fine the Fourier series coefficient for the signal
when

72
x[n]=
1
2j
e
j2/N n
−
1
2j
e
− j2/N n
a1=
1
2j
a−1=−
1
2j
Expanding the signal as a sum of two complex
exponentials
We see by inspection that
Solution

73
These coefficient repeat with period N thus aN+1
also equal to 1/2j and aN-1
equals (-1/2j)
For example with N = 5

74
Consider 0=
2M
N
Assuming that M and N do not have any common
factors
x[n]=
1
2j
e jM 2/N n
−
1
2j
e− jM 2/N n
By inspection aM
= (1/2j), a-M
= (-1/2j) and the
remaining coefficients over one period of length
N are zero

75
For example with M = 3 and N = 5

76
Example 3.11
x[n]=1sin2
N n3cos2
N n
cos4
N
n

2 
Fine the Fourier series coefficient for the signal

77
x[n]=1
1
2j
[e
j2/N n
−e
− j2/N n
]

3
2
[e
j2/N n
e
− j2/N n
]

1
2
[e
j4n/ N /2
e
− j4n/N /2
]
Expanding the signal as a sum of two complex
exponentials
Solution

78
x[n]=13
2

1
2je
j2/ N n
3
2
−
1
2je
− j2/N n
1
2
e
j/2
e
j22/ N n
1
2
e
− j/2
e
− j22/N n
Collecting terms
The Fourier series coefficient are

79
a0=1
a1=
3
2

1
2j
=
3
2
−
1
2
j
a−1=
3
2
−
1
2j
=
3
2

1
2
j
a2=
1
2
e
j/2
=
1
2
j
a−2=
1
2
e
− j/2
=−
1
2
j

80
Fourier series coefficient for N = 10

81
Fourier series coefficient for N = 10

82
Example 3.12
Evaluate the Fourier series for the signal

83
ak=
1
N
∑
m=0
2N1
e
− jk 2/N m−N1
=
1
N
e
− jk 2/N N1
∑
m=0
2N1
e
− jk 2/N m
ak=
1
N
∑
n=−N 1
N 1
e− jk 2/ N n
For this signal, the Fourier coefficient become
Let m = n+N1

84
ak=
1
N
e
jk 2/ N N 1
1−e
− jk22N11/ N
1−e
− jk 2/N  
=
1
N
e− jk 2/2N
[e
jk2N 11/2/N
−e
− jk2N11/2/ N
]
e
− jk 2/2N
[e
jk 2/2N
−e
− jk 2/2N
]
=
1
N
sin[2 k N11/2/ N ]
sin k / N 
, k≠0,±N ,±2N ,...
ak=
2N11
N
, k=0,±N ,±2N ,...
and

85
Plot of Nak
for 2N1
+1 = 5 (a) N = 10; (b) N = 20;
(c) N = 40

86
Multiplication
FS
x[n] ↔ ak
FS
y[n] ↔ bk
x[n] y[n] ↔ dk= ∑
l=〈N 〉
al bk−l
FS
3.6 Properties of Discrete-Time Fourier Series

87
First Difference
FS
x[n] ↔ ak
FS
x[n]−x[n−1] ↔ 1−e− jk 2/N 
ak
The first-difference operation define as x[n] -x[n-1]

88
Parseval's Relation for Discrete-Time Periodic
Signals
1
N
∑
n=〈 N 〉
∣x[n]∣2
= ∑
k=〈 N 〉
∣ak∣2
Where the ak
are the Fourier series coefficients of
x[n] and N is the period

89

90
Example 3.13
Find the Fourier series coefficient of the x[n] below
solution
By linearity property, x[n] = x1
[n]+x2
[n], ak
= bk
+ck
when ak
FS coefficients of x[n], bk
FS coefficients
of x1
[n], ck
FS coefficients of x2
[n]

91
x[n] = x1
[n]+x2
[n]

92
bk=
{
1
5
sin3k /5
sink /5
, for k≠0,±5,±10,...
3
5
, for k=0,±5,±10,...
c0=
1
5
∑
n=0
4
x2 [n]=1
From example 3.12, with N1
= 1 and N = 5
The sequence x2[n] has only DC value, which is
captured by its zeroth Fourier series coefficient.

93
ak=
{
1
5
sin3k /5
sink /5
, for k≠0,±5,±10,...
8
5
, for k=0,±5,±10,...
The Fourier series coefficient of x[n] are :

94
Example 3.14
Suppose we given the following facts about a
sequence x[n].
1 x[n] is periodic with period N = 6
2
3
4 x[n] has the minimum power per period
among the set of signals satisfying the
preceding three conditions.
∑n=0
5
x[n]=2
∑n=2
7
−1
n
x[n]=1
Find the sequence x[n]

95
p=∑
k=0
5
∣ak∣
2
From fact 2, we conclude that a0
= 1/3
From fact 3, (-1)n
= e-jπn
= e-j(2π/6)3n
, we conclude
that a3
= 1/6
From fact 4, the average power in x[n] using
Parseval's relation is

96
x[n]=a0a3 e
jn
=
1
3

1
6
−1n
The value of P is minimized by choosing
a1
=a2
=a4
=a5
=0
We can sketch x[n] :

97
3.7 Fourier Series and LTI Systems
H s=∫−∞
∞
he
−s 
d 
H z= ∑
k=−∞
∞
h[k ]z
−k
From section 3.2, if x(t) = est
is the input to a
continuous-time LTI system, then the output is
given by y(t) = H(s)est
, where
If x[n] = zn
is the input to a continuous-time LTI
system, then the output is given by y[n]= H(z)zn
,
where

98
Where s and z are complex number, H(s) and
H(z) are referred as the system functions of the
corresponding systems.
For continuous-time signals and systems, we
focus of the specific case in which Re{s} = 0, so
that s = jω. The system function is :
H  j =∫−∞
∞
hte
− j t
dt
This system function is referred to as the
frequency response of the system

99
H e
j
= ∑
n=−∞
∞
h[n]e
− jn
For discrete-time signals and systems, we focus of
the specific case in which |z| = 1, so that z = ejω
.
The system function is :
This system function is referred to as the
frequency response of the system

100
xt= ∑
k=−∞
∞
ak e
jk 0 t
yt= ∑
k=−∞
∞
ak H  jk 0e
jk 0 t
For continuous-time system, Let x(t) be a
periodic signal apply as the input to the
system:
The output of the system is

101
x[n]= ∑
k=〈N 〉
ak e jk2/ N n
For discrete-time system, Let x[n] be a periodic
signal apply as the input to the system:
y[n]= ∑
k=〈 N 〉
ak H e
jk2/ N 
e
jk2/ N n

102
Example 3.16
ht=e
−t
ut
Suppose that the periodic signal x(t) in example
3.2 is the input to LTI system with impulse
response
Find the Fourier series coefficient of output y(t)

103
H  j =∫
0
∞
e
−
e
− j
d 
=−
1
1 j 
e
−
e
− j 
∣0
∞
=
1
1 j 
solution
To calculate the Fourier series coefficients of
the output y(t), we first compute the frequency
response

104
= ∑
k=−3
3
bk e jk2t
yt= ∑
k=−∞
∞
ak H  jk 0e
jk 0 t
xt= ∑
k=−3
3
ak e
jk2t
The input of the system is

105
b0=1
b1=
1
4 1
1 j2, b−1=
1
4 1
1− j2
b2=
1
2 1
1 j4 , b−2=
1
2 1
1− j4
b3=
1
3  1
1 j6 , b−3=
1
3  1
1− j6 
With bk
= ak
H(jk2π), so that

106
yt=12∑
k=1
3
Dk cos jk2tk 
yt=12∑
k=1
3
[Ek cos2kt−Fk sin2kt]
bk=Dk e
j 
k
=Ek jFk , k=1,2,3,...
y(t) must be a real-valued signal, since it is the
convolution of x(t) and h(t) which are both real.
Therefore, y(t) can be expressed in the form
or
where

107
D1=∣b1∣=
1
414
2
, 1=∡b1=−tan
−1
2
E1=ℜe {b1}=
1
414
2

,
F1=ℑm{b1}=−

214
2

This coefficients can be evaluated from bk
, for
example

108
Example 3.17
x[n]=cos2n
N 
Find the Fourier series coefficient of output y(t)
Consider an LTI system with impulse response
h[n] = αn
u[n], -1< α <1, and with the input

109
x[n]=
1
2
e
j2/ N n

1
2
e
− j2/ N n
solution
x[n] can be written in Fourier series form as
Frequency response of system
H e
j
=∑
n=0
∞

n
e
− j n
=∑
n=0
∞
 e
− j 

n

110
H e
j
=
1
1−e
− j
y[n]=
1
2
H e j2/N
e j2/N n

1
2
H e− j2/N
e− j2/N n
The summation of this series is
We obtain the Fourier series for the output :

111
y[n]=
1
2 1
1−e
− j2/ N e
j2/N n

1
2 1
1− e
j2/ N e
− j2/N n
1
1−e− j2/ N
=r e
j
If we write
The output become
y[n]=r cos2
N
n

112
1
1−e
− j2/4
=
1
1 j
=
1
1
2
e
j−tan
−1

y[n]=
1
1
2
cosn
2
−tan
−1

For example, if N = 4,
and thus

113

Chapter3 - Fourier Series Representation of Periodic Signals

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