The Fourier Series for DiscreteTime Signals
Suppose that we are given a periodic sequence
with
period
N.
The
Fourier
series
representation for x[n] consists of N
harmonically related exponential functions
ej2kn/N, k = 0, 1,2,…….,N-1
and is expressed as
N 1
x[ n ] 

cke
j 2  kn / N
k 0
where the coefficients ck can be computed as:
ck 
1
N


x [ n ]e
 j 2  kn / N
n0
1
Example 2: Determine the spectra of the
following signals:
(a) x[n] = [1, 1, 0, 0], x[n] is periodic with period 4
(b) x[n] = cosn/3
(c) x[n] = cos(2)n
Solution: (a) x[n] = [1, 1, 0, 0]
 ck 
1
N
Now
c1 
1
N 1
 x[n ]e
c0 
1
4
x [ n ]e


4
1  1 cos
x[ n ] 
1
4
n0
 j2 n / 4
3
 x[n ]e

1
1
4
2
x[ 0 ]  x[1 ]  x[ 2 ]  x[ 3 ]  1  1  0  0  
1
3

4
x [ n ]e
 j sin

2
 j n / 2

1
4
n0

2
 j 2  kn / N
4 n0
3
n0
1

n0
3

4
 j 2  kn / N
1
x[ 0 ]  x[1 ]e
 j / 2
00

  1  0  j   1  j 
1
1
4
4
2
c2 
1
4
3

x [ n ]e
j2  2 n / 4

1
4
n0

3

c3 
3
x[ n ]e

4
 j2 n 3 / 4

1
4
n0

1
1
4
n0
4
1
x [ n ]e
j n
1  cos
1  1 .e 
j
  j sin    0
1
1
4
4
1  cos 3  / 2   j sin( 3  / 2 )   1  0  j  1  j
The magnitude spectra are:
c0 
1
2
c1 
2
4
c2  0
c3 
2
4
and the phase spectra are:
0  0
1 

4
 2  undefined
3 

4
3
(b) x[n] = cosn/3
Solution: In this case, f0 = 1/6 and hence x[n] is
periodic with fundamental period N = 6.
Now
ck 
5
1

6
x [ n ]e
 j 2  kn / N

n0

e

6
2
1
5
1
1
5

6
cos
n
e
 j n / 3
e
5
1

6
cos
n
 j  kn / 3


e

12
1
5
e
 j  kn / 3
3
n0
n0
j 3n 1  k 
e
 j 3n 1  k 
n0
 c0 
1
5
2 cos

12
n0


3
n0
j n / 3
e
 j 2  kn / 6
1
6
cos 0  cos

3
n
3
 cos

1
5
n
n0
3
cos

6
2
3
 cos
3
3
 cos
4
3
Similarly, c2 = c3 = c4 = 0, c1 = c5 = ½.
 cos
5
3
 0
4

(c) Cos(2)n
Solution: The frequency f0 of the signal is
1/2 Hz. Since f0 is not a rational number,
the signal is not periodic. Cosequently, this
signal cannot be expanded in a Fourier
series.
5
Power density Spectrum of Periodic Signals
The average power of a discrete time periodic
signal with period N is
1
Px 
N
N 1

x(n )
2
n0
The above relation may also be written as
N 1

*
*  2  kn / N
Px 
x
[
n
]
x
[
n
]

x
[
n
]
c



  ke
N n0
N n0
 n0
1
or P
N 1
1
N 1
x


c
n0

k 0
 1

N
2
N 1

*
k
ck

1
N
N 1
N 1

x [ n ]e
 j 2  kn / N
n0
N 1

x[ n ]






2
n0
This is Parseval’s Theorem for Discrete-Time Power
Signals.
6
Similarly, for discrete time energy signals, the
Parseval’s Theorem may be stated as follows:
2
N 1
Ex 

n0
x[ n ]
N 1
2
 N  ck
k 0
If the signal x[n] is real, [i.e. x*[n] = x[n]], then we
can easily show that
|c-k| = |ck|
-c-k = ck
|ck| = |cN-k|
ck = cN-k
(even symmetry)
(odd symmetry)
(Periodicity)
(periodicity)
7
More specifically, we have
|c0| = |cN|
c0 = - cN
|c1| = |cN-1|
c1 = - cN-1
|cN/2| = |cN/2|
cN/2 = 0 if N is even
|c(N-1)/2| = |c(N+1)/2| c(N-1)/2 = (N+1)/2 if N
is odd
8
Example: Determine the Fourier Series
Coefficients and the Power Density Spectrum of
the following periodic signal.
X[n]
A
n
-N
N
L
Solution:
ck 
1
N 1

N
x[ n ]e
 j 2  kn / N
1
N 1

N

n0
Ae
 j 2  kn / N
n0
k = 0, 1, 2, …., N-1
ck 
L 1
e

N
A
n0
n
 j2 k / N


 A
N
AL
N
1 e
,
 j 2  kL / N
1 e
 j 2 k / N
k0
, k  1 , 2 ,..., N  1
9
But
1e
 j 2  kL / N
1e
e
 j2 k / N
 j k ( L  1 ) / N
 e  j kL / N
   j  k / N
 e
  e j kL / N  e  j  kL / N
  j k / N
 j k / N
 e

e





sin(  kL / N )
sin(  k / N )
Therefore,
ck
2
AL 2

N 

2

sin  kL / N 
A 2

 N  
 sin  k / N 

k 0
otherwise
10
The Fourier Transform of Discrete-Time
Aperiodic Signals
The Fourier Transform of a finite energy discrete
time signal x[n] is defined as

X(w ) 
 x[ n ]e
 jwn
n  
X(w) may be regarded as a decomposition of x[n]
into its Frequency components. It is not difficult to
Verify that X(w) is periodic with frequency 2.
The Inverse Fourier Transform of X(w) may be
defined as
x[ n ] 
1

2
2
X ( w )e
jwn
dw
11
Energy Density Spectrum of Aperiodic
Signals
Energy of a discrete time signal x[n] is defined as
2

Ex 

x[ n ]
n  
Let us now express the energy Ex in terms of the
spectral characteristic X(w). First we have


 1
E x   x [ n ]x [ n ]   x [ n ] 
 2
n  
n  
*




X ( w )e
 jwn

dw 

If we interchange the order of integration and
summation in the above equation, we obtain
 
 jwn
Ex 
X
(
w
)
x
[
n
]
e



2 
 n  
1



1
dw


2




2
X ( w ) dw
12
Therefore, the energy relation between x[n] and
X(w) is

Ex 

n  
2
x[ n ] 
1
2


2
X ( w ) dw

This is Parseval’s relation for discrete-time
aperiodic signals.
13
Example: Determine and sketch the energy
density spectrum of the signal x[n] = anu[n],
-1<a<1
Solution:

X(w ) 

x [ n ]e
 jwn
n  



n
a e
 jwn

 ae 

n0
 jw n

n0
1
1  ae
 jw
The energy density spectrum (ESD) is given by
2

1
S xx ( w )  X ( w )  X ( w ) X ( w ) 
1  ae 1  ae 
X(w)
1

 jw
jw
1  2 a cos w  a
a = 0.5
2
a= -0.5
w

0

14
Example: Determine the Fourier
Transform and the energy density
spectrum of the sequence
A,
x[ n ]  
 0,
0  n  L 1
otherwise
Solution:

X(w ) 
 x[ n ]e
 jwn
L 1

n  
 Ae
 jwn
A
0
1e
 jwL
1e
 jw
 Ae
 j ( w / 2 )( L  1 )
sin( wL / 2 )
sin( w / 2 )
The magnitude of x[n] is


X(w )  
A


w  0
A L,
sin( wL / 2 )
sin( w / 2 )
,
otherwise
and the phase spectrum is
 X (w )   A   (L  2)  
w
2
sin( wL / 2 )
sin( w / 2 )
The signal x[n] and its magnitude is plotted on the next slide. 15
The
Phase spectrum is left as an exercise.
x[n]
|X(w)|
16
Properties of Discrete Time Fourier
Transform (DTFT)
Symmetry Properties: Suppose that both the
signal x[n] and its transform X(w) are complex
valued. Then
x[n] = xR[n] + jxI[n]
(1)
X(w) = XR(w) + jXI[w]
(2)

 X(w ) 

x [ n ]e
 jwn
n  
Substitution of (1) and (2) gives

X R ( w )  jX I ( w ) 
 x
n  
R
[ n ]  x I [ n ]e
 jwn


 x
R
[ n ]  x I [ n ]cos wn  j sin wn
n  
Separating the real and imaginary parts, we have
17
