Hossein Sameti
Department of Computer Engineering
Sharif University of Technology

The DTFT is defined using an infinite sum over a
discrete time signal and yields a continuous
function X(ω)
◦ not very useful because the outcome cannot be stored on
a PC.


Now introduce the Discrete Fourier Transform
(DFT), which is discrete and can be stored on a
PC.
We will show that the DFT yields a sampled
version of the DTFT.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
2
C.
C.

X ()  CTFT {x(t )}   x(t )e  jt dt
x(t )
D.
D.
D.
periodic
D.
n  
Complex
x(n)

X ( z )  z{x(n)}   x(n) z  n
n  
Int.
~
~
~
N 1 ~
X (k )  DFS {x(n)}   x(n)e
x ( n)

N 1
X (k )  DFT {x(n)}   x(n)e
finite
j 2kn
N
n 0
periodic
x(n)
finite

X ( )  DTFT {x(n)}   x(n)e  jn
x(n)
Inf. or
Finite

C.

j 2kn
N
n 0
Int.
3

X ()  CTFT {x(t )}   x(t )e  jt dt
x(t )


x(n)
X ( )  DTFT {x(n)}   x(n)e  jn
n  

x(n)
~
X ( z )  z{x(n)}   x(n) z  n
n  
~
x ( n)
x(n)
~
N 1 ~
X (k )  DFS {x(n)}   x(n)e

j 2kn
N
n 0
N 1
X (k )  DFT {x(n)}   x(n)e

j 2kn
N
n 0
4
~
~
x(n)  x(n  kN) N : period
~
• Decompose x(n) in terms of complex exponentials that are
periodic with period
N.
j 2kn
~
N

1
1
x ( n) 
 X ( k )e N
N k 0
• How many exponentials? N
ek ( n )  e
j 2kn
N
k  0,1,..., N  1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
5
ek ( n )  e
j 2kn
N
k  0,1,..., N  1
Exponentials that are
periodic with period N.
ek (n)  ek rN (n)
arbitrary integer
1
* Proof:
ek  rN (n)  e
j 2 ( k  rN ) n
N
ek  rN (n)  e
ek  rN (n)  e
j 2kn
N
j 2kn j 2rnN
N .e N
 ek (n)
e0 (n)  eN (n)  e2 N (n)  ....
e1 (n)  eN 1 (n)  e2 N 1 (n)  ....
6
~
x ( n) 
N 1
1
 X ( k )e
N k 0
Answer:
j 2kn
N
How to find X(k)?
N 1 ~
X ( k )   x ( n )e

j 2kn
N
n 0
Proof : substitute X(k) in the first equation.
•It can also easily be shown that X(k) is periodic with period N:
X (k )  X (k  rN )
arbitrary integer
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
7
~
Analysis:
N 1 ~
X ( k )   x ( n )e

j 2kn
N
n 0
1 N 1 ~
x ( n) 
 X ( k )e
N k 0
~
Synthesis:
Periodic N pt.
seq. in time domain
j 2kn
N
Periodic N pt.
seq. in freq. domain
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
8

~
x(n)    (n  rN )
r  
1
-N
~
N 1 ~
X ( k )   x ( n )e

j 2kn
N
N 1
   ( n )e
n 0

1
1
0
N
j 2kn
N
n 0
1
-N
…
1
0
e

…
n
j 2k ( 0 )
N
1
1
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
n
9

~
x(n)    (n  rN )
(eq.1)
r  
N 1 ~
~
X ( k )   x ( n )e

j 2kn
N
n 0
1 N 1 ~
x ( n) 
 X ( k )e
N k 0
~
(eq.1) & (eq.2)
N 1
   ( n )e

j 2kn
N
n 0
j 2kn
N
N 1
1

 e
N k 0
j 2kn
N
e

j 2k ( 0 )
N
(eq.2)

  (n  rN )  1  e
r 
1
N 1
N
j 2kn
N
k 0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
10

Shift property:
~
~
x ( n)
X (k )
~
x(n  n0 )

e
j
2kn0
N
~
X (k )
Periodic convolution:
~
~
~
N 1 ~
~
x1  x 2  x 3   x1 (m) x 2 (n  m)
m0
~
Period N
~
~
X 1 (k ) X 2 (k )  X 3 (k )
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
11
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
12

In the list of properties:
𝑊𝑁 = 𝑒
𝑗2𝜋
𝑁
𝑊𝑁 𝑘𝑛 =𝑒𝑘 (𝑛)
Where:
ek (n)  e
j 2kn
N
k  0,1,..., N  1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
13
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
14
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
15
x(n)
X (k )
N pt.
N pt.
N pt.
DFT
N pt.
DTFT
DFT
DFS
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
16


1) Start with a finite-length seq. x(n) with N points
(n=0,1,…, N-1).
~
2) Make x(n) periodic with period N to get x(n)
~
x ( n)  x ( n ) R N ( n )
~
Extracts one period of x(n)
1 n  0,1,..., N  1
R N ( n)  
otherwise
0
~

x(n)   x(n  rN )
r  
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
17
~


~
3) Take DFS of x(n)
X (k )
~
4) Take one period of X (k ) to get DFT of x(n): X (k )
~
X (k )  X (k ) RN (k )
x(n)
N pt.
~
~
x ( n)
X (k )
X (k )
N pt.
periodic
N pt.
N pt.
periodic
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
18
x(n)
3
2
1
1
0
~
3
3
x ( n)
3
2
2
1
1
~


2
1
1 1
0
X (k )
n
2 3
1



 
0

0



k
2 3
1

X (k )
n
2 3
1


1


1
2 3
k
19

Definition of DFT:
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
0k  N
otherwise
N pt. DFT of x(n)
2kn
 1 N 1
j
  X ( k )e N
0n N
x ( n)   N k  0
 0
otherwise
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
20
1 0 n 5

x ( n)  
0 otherwise

2kn
 4
j
 x ( n )e 5
0k 5
X (k )   n
0
0
otherwise
k 0
 5
X (k )  
0 otherwise
X (k )  ?

1  e  j 2k
0k 5

 j 2k
X (k )  
5
1 e
0
otherwise

21
Mehrdad Fatourechi, Electrical and
Computer Engineering, University of
British Columbia, Summer 2011
22
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0

X ( )   x(n)e
n  
 jn
0k  N
otherwise
N 1
  x(n)e  jn
n 0
 X ( )   2k
X (k )  
N
0
0k  N
otherwise
DFT thus consists of equally-spaced samples of DTFT.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
23
X ( )

3



8 pt. sequence
2k

8
2
3

8 pt. DFT
k  0,1,..., 7
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
24
So far we calculated the N pt. DFT of a seq. x(n) with N non-zero
values:
 X ( )   2k
X (k )  
N
0
0k  N
otherwise
Suppose we pad this N pt. seq. with (M-N) zeros to get a sequence
with length M.
3
x(n)
We can now take an M-pt.
2
DFT of the signal x(n)
1
1
0
1
2 3
n
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
25
2kn
M 1
j
  x ( n)e M
X (k )   n  0
 0
N pt.
DFT
0k M
otherwise
M pt.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
26
x(n)
3
2
1
0
1
1
n
2 3
2kn
 3
j
  x ( n )e 4
4 pt. DFT:
X (k )   n  0
0
2kn
 5
j
  x ( n )e 6
6 pt. DFT:
X (k )   n  0
0
2kn
 7
j
  x ( n )e 8
8 pt. DFT:
X (k )   n  0
0
2kn
 99
j
  x(n)e 100
100 pt. DFT: X (k )  n  0
 0
0k 4
otherwise
0k 6
otherwise
How are
these related
to each other?
0k 8
otherwise
0  k  100
otherwise
27
X ( )
Going from N pt. to
2N pt. DFT
0
2

Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
28
N pt. seq.
N pt.
N pt.
DFT
2N pt.
N pt. seq.
padded with N
zeros
2N pt.
DFT
What is the minimum number of N
needed to recover x(n)?
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
29

Assume y(n) is a signal of finite or infinite extent.
Y ( )  DTFT { y(n)}
• Sample Y ( ) at N equally-spaced points.
Y ( )   2k  X (k )
N pt. sequence.
N
IDFT { X (k )}  x(n)
N pt. sequence.
What is the relationship between x(n) and y(n)?
What happens if N is larger , equal or less than the length of y(n)?
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
30

We start with x(n) and find its relationship with y(n):
2kn
 1 N 1
j

X ( k )e N
0n N
x(n)   N k
0
 0
otherwise
Y ( )   2k  X (k )
N
Change the order
of summation:
2km
2kn
 1 N 1 
j
j

(  y ( m)e N )e N
0n N
x(n)   N k
 0 m  
 0
otherwise
2k ( m  n )
N 1  j
 
1
N

y ( m)  e
0n N
x ( n)   m
 
N k 0
0
otherwise
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
31
2k ( m  n )
N 1  j
 
1
N

y ( m)  e
x ( n)   m
 
N k 0
0
0n N
otherwise
However, we have shown that:
1 N 1 j
x ( n) 
 1.e
N k 0
~
2kn
N

   (n  rN )
r  
~
X (k )  1 0  k  N  1


f (n)    (n  rN )

f (n  m)    (n  m  rN )
r  


  y ( m) f ( n  m )
x(n)  m  

 0
r  
0n N
otherwise
Convolution with
train of delta
functions
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
32
  y (n  rN )

x(n)   r  
0
0n N
otherwise
One period of the replicated version of y(n)
• Examples
If we sample Y ( ) at a sampling rate that is higher than the
number of points in y(n), we should be able to recover y(n).
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
33

Shift property:
0k  N
x(n) DFT

 X (k )
N pt. seq.
x(n  n0 ) DFT

 X (k )e
j
2n0 k
N
0k  N
The above relationship is not correct, because of the
definition of DFT.
The signal should only be non-zero for the first N points.
~
x(n  n0 ) RN (n) DFT

 X (k )e
j
2n0 k
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
34

In the list of properties:
𝑊𝑁 = 𝑒
where:
𝑗2𝜋
𝑁
,
𝑊𝑁 𝑘𝑛 =𝑒𝑘 (𝑛)
ek (n)  e
j 2kn
N
k  0,1,..., N  1
~
and
where:
x[((n  n0 )) N ]  x(n  n0 ) RN (n)
1 n  0,1,..., N  1
RN ( n )  
otherwise
0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
35
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
36
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
37
Using DFT to calculate linear
convolution
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
38




We are familiar with, “linear convolution”.
Question: Can DFT be used for calculating the linear
convolution?
The answer is: NO! (at least not in its current format)
We now examine how DFT can be applied in order to
calculate linear convolution.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
39
• Linear convolution:
x3 (n)   x1 (k ) x2 (n  k )
x1 (n)  x2 (n)  x3 (n)
k
Application in the analysis of LTI systems
• Periodic convolution:
~
~
~
N 1 ~
~
x1  x 2  x 3   x1 (m) x 2 (n  m)
m0
A seq. with period N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
40
• Circular convolution:
x1  x 2  x 3
N pt. seq.
• Circular convolution is closely related to periodic convolution.
 ~
~
x 3 (n)  x1 (n)  x 2 (n) [ x1 (n)  x 2 (n)]. RN (n)
X 3 (k )  X 1 (k ). X 2 (k )
N pt. DFT of x3
N pt. DFT of x1
N pt. DFT of x2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
41
x1 (n)   (n  1)
x2 (n) : N
Circular convolution?
pt. seq.
Make
x1 (n) an N pt. seq.
0 0  n 1

x1 (n)  1
n 1
0 1  n  N

Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
42
43
 ~
~
x 3 (n)  x1 (n)  x 2 (n) [ x1 (n)  x 2 (n)]. RN (n)
We know from DFS properties:
~
~
~
~
x1  x 2  x 3
~
~
~
~
X 1 (k ) X 2 (k )  X 3 (k )
X 1 (k )  DFS {x1 (n)}
~
~
X 2 (k )  DFS {x 2 (n)}
~
~
X 3 (k )  DFS {x 3 (n)}
~
~
~

 X 1 (k ) X 2 (k ) RN (k )  X 3 (k ) RN (k )
X 1 (k ). X 2 (k )  X 3 (k )
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
44
 ~
~
x 3 (n)  x1 (n)  x 2 (n) [ x1 (n)  x 2 (n)]. RN (n)
X 1 (k ). X 2 (k )  X 3 (k )
If we multiply the DFTs of two N pt. sequences, we get the
DFT of their circular convolution and not the DFT of their
linear convolution.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
45
1 0  n  L

x1 (n)  x2 (n)  
0 otherwise

Calculate N pt. circular convolution of x1 and x2 for the
following two cases of N:
1) N=L
2) N=2L
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
46
2kn
 L 1
j
  x ( n )e L
X ( k )  n  0
 0
1 0  n  L  1

x1 (n)  
 0 otherwise

0  k  L 1
otherwise
N pt. DFT of x1
k 0
L
X 1 (k )  X 2 (k )  
0 otherwise
X 3 (k )  X 1 (k ). X 2 (k )
IDFT
 L2
k 0

0 otherwise
L 0  n  L  1
x3 ( n )  
otherwise
0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
47
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
48
• Pad each signal with L extra zeros to get an 2L pt. seq.:
N=2L pt. DFT of x1
2kn
2 L 1
j
  x ( n )e 2 L
X (k )   n  0
 0
0  k  2L  1
otherwise
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
49
Same as linear
convolution!!
50

Our hypothesis is that if we pad two DT signals with enough
zeros so that its length becomes N, we can use DFT to calculate
linear convolution.
x1 (n)
L pt. seq.
NL
x 2 ( n)
P pt. seq.
NP
Goal: calculate
x3 (n)  x1 (n) * x2 (n) Using DFT
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
51
x3 (n)  x1 (n) * x2 (n)
X 3 ( )  X 1 ( ) X 2 ( )

X ( )  DTFT {x(n)}   x(n)e  jn
n  
To get DFT, we have to sample the above DTFT at N equally-spaced points:
Y (k )  X 3 ( )   2k
N
IDFT {Y (k )} 
Solution to the
problem statement


  x3 (n  rN )
 r  

0
0n N
otherwise
(Eq.1)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
52
On the other hand, we know that:
Y (k )  X 3 ( )   2k  X 1 ( )   2k X 2 ( )   2k
N
N pt. DFT of x1
N
N
N pt. DFT of x2
IDFT {Y (k )}  x1 (n)  x 2 (n)
(Eq.1)
(Eq.2)
x1 ( n)  x 2 ( n)
Circular
convolution
(Eq.2)


  x (n  rN )
  r   3

0
0n N
otherwise
Replicated version of the linear convolution
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
53

In other words, the N pt. circular convolution of two
DT signals is the same as their linear convolution, if we
make the result of linear convolution periodic with
period N and extract one period.
x3 (n)  L  P  1
To avoid aliasing: N  L  P  1
We can thus use DFT in order to calculate the linear
convolution of two sequences!
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
54
1) Start with
x1 (n)
L pt. seq.
x 2 ( n)
P pt. seq.
2) Choose N  L  P  1
3) Pad x1 (n) with N-L zeros to get N points.
4) Pad x2 (n) with N-P zeros to get N points.
5) Calculate the N pt. DFTs of the above two sequences and
multiply them together.
6) Calculate IDFT of the resulting N pt. sequence.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
55
y(n)
x(n)
h(n)
A very long
sequence
An FIR filter with a limited
number of taps (P)
• Examples of this situation?
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
56
Solution to
the problem
Overlap - add
Main idea: using the following property:
[ x1(n)  x2 (n)] * h(n)  x1(n) * h(n)  x2 (n) * h(n)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
57



Segment the long sequence into non-overlapping
chunks of data with the length of L.
Convolve each chunk with h(n) to get (L+P-1) new
points.
Add the results of the convolution of all chunks to get
the final answer.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
58
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
59
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
60
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
61



Discussed DFS and DFT and examined the relationship
between DFT and DTFT
Showed how DFT can be used for calculating
convolution sum.
Next: Fast Fourier Transform
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
62