Fourier Analysis of Discrete Time Signals
For a discrete time sequence we define two classes of Fourier Transforms:
• the DTFT (Discrete Time FT) for sequences having infinite duration,
• the DFT (Discrete FT) for sequences having finite duration.
The Discrete Time Fourier Transform (DTFT)
Given a sequence x(n) having infinite duration, we define the DTFT as follows:
X ( )  DTFT  x (n) 

 j n
x
(
n
)
e

n 
1
x (n)  IDTFT  X ( ) 
2


X ( )e j n d

X ( )
x ( n)
…..
…..
N 1
n


continuous frequency
discrete time

Observations:
• The DTFT X ( ) is periodic with period 2
• The frequency

;
is the digital frequency and therefore it is limited to the interval
     
Recall that the digital frequency  is a normalized frequency relative to the sampling frequency,
defined as
  2
F
Fs
one period of
 Fs
 2
 Fs / 2

0
0
X ( )
Fs / 2

Fs
2
F

Example:
DTFT
x[n]
1
0
n
N 1
since
N 1
X ( )   e
n 0
 j n
1  e  j N

1  e  j
 e  j ( N 1) / 2
sin  N / 2
sin  / 2
Example:
x[n]  A cos( 0n   )
X ( )  A e j  (   0) 
 A e  j  (   0)
Discrete Fourier Transform (DFT)
Definition (Discrete Fourier Transform): Given a finite sequence
x  [ x(0), x(1),...,x( N  1)]
its Discrete Fourier Transform (DFT) is a finite sequence
X  DFT( x)  [ X (0), X (1),...,X ( N  1)]
where
N 1
X ( k )   x(n) wN kn , wN  e  j 2 / N
n0
x
DFT
X
Definition (Inverse Discrete Fourier Transform): Given a sequence
X  [ X (0), X (1),...,X ( N  1)]
its Inverse Discrete Fourier Transform (IDFT) is a finite sequence
x  IDFT( X )  [ x(0), x(1),...,x( N  1)]
where
1
x ( n) 
N
N 1
 X (k )w
k 0
X
 kn
N
, wN  e  j 2  / N
IDFT
x
Observations:
• The DFT and the IDFT form a transform pair.
back to the same signal !
x
DFT
x
IDFT
X
X
• The DFT is a numerical algorithm, and it can be computed by a digital computer.
DFT as a Vector Operation
Let
 x[0] 
 x[1] 
,
x
  


x
[
N

1
]


 X [0] 
 X [1] 

X  DFT{x}  





X
[
N

1
]


 1 
 wk 
ek   N ,
  
  k ( N 1) 
 wN

Then:
x
X [k ]  ek*T x
x
1
 X [0]e0  X [1]e1  ...  X [ N  1]eN 1 
N
ek
1
X [ k ]ek
N
1

1
  x[0] 
 X [0]  1
  x[1] 
 X [1]  1 w
N 1
w

N
N



X  DFT{x}  
  
 




 

( N 1)( N 1)  
N 1
w
w
1
]
1

N
[
x
]
1

N
[
X

N
N
 

 

WN

X  WN x
1 *T
x  WN X
N 

WN1
Periodicity: From the IDFT expression, notice that the sequence x(n) can be interpreted as one
period of a periodic sequence x p (n) :
1
x p (n) 
N
N 1
 X (k )w
k 0
 kn
N
1

N
N 1
 X (k )w
k 0
 kn
N
wN
 kN
1

N
x ( n)
k 0
 k ( n N )
N
 x p (n  N )
n
x p (n)
N
 X (k )w
original sequence
N 1
 2N
N 1
periodic repetition
N
2N
n
This has a consequence when we define a time shift of the sequence.
For example see what we mean with x (n  1) . Start with the periodic extension x p (n)
x p (n)
A
B
D
N
N
C
n
x p (n 1)
A
D
B
N
N
C
n
If we look at just one period we can define the circular shift
x ( n)
x(n 1) N
A
A
B
D
B
D
C
n
C
A
B
C
D
D
A
B
C
0
1
2
3
0
1
2
3
D
Properties of the DFT:
• one to one
• time shift
where
x(n)  X ( k )


with no ambiguity;
DFT x(n  m) N  wN km X ( k )
x(n  m) N is a circular shift
periodic repetition
x ( n)
x ( N  m)  x ( N  1)
x(0) x(1)  x( N  m) x( N  m  1)  x( N  1)
x(n  m) N
x ( N  m)  x ( N  1) x (0) x (1)  x ( N  m  1)

X
(
k
)

X
( N  k)
• real sequences
| X ( k )|  | X ( N  k )|
• circular convolution
y (n)  x1 (n)  x2 (n)
N 1
  x1 ( k ) x2 (n  k ) N

k 0
circular shift
where both sequences x1 , x2
length N. Then:


must have the same
DFT x1 (n)  x2 (n)  X1 ( k ) X 2 ( k ), k  0,..., N  1
Extension to General Intervals of Definition
Take the case of a sequence defined on a different interval:
x[n]
n0
n0  N 1
How do we compute the DFT, without reinventing a new formula?
First see the periodic extension, which looks like this:
x[n]

n0  N 1
n0
Then look at the period

n
0  n  N 1
x[n]

n0
n0  N 1

N 1
n
Example: determine the DFT of the finite sequence
x[n]  0.8|n| if  3  n  3
x[n]
Then take the DFT of the vector
3
3
x  x[0], x[1],...,x[3], x[3],...,x[1]
n
x[n]
3
n