The Discrete Fourier Transform
Sampling the Fourier Transform
• Consider an aperiodic sequence with a Fourier transform
 
DT FT
x[n] 
 X e j
• Assume that a sequence is obtained by sampling the DTFT
~
Xk   X e j
 X e j2 / Nk
 
  2  / Nk


• Since the DTFT is periodic resulting sequence is also periodic
• We can also write it in terms of the z-transform
~
Xk   Xz z  e2  / Nk  X e j2 / Nk

•
•
•
DSP
The sampling points are shown in figure
~
Xk  could be the DFS of a sequence
Write the corresponding sequence
1 N1 ~
~
x[n]   Xk e j2  / Nkn
N k 0

Sampling the Fourier Transform Cont’d
• The only assumption made on the sequence is that DTFT exist
    xme
Xe
j

~
Xk   X e j2 / Nk

 jm
m  

1 N1 ~
~
x[n]   Xk e j2  / Nkn
N k 0
• Combine equation to get

1 N 1  
~
x[n]     xme  j2  / Nkm e j2  / Nkn
N k  0 m  


 1 N 1 j2  / Nk n  m  
~


  xm  e

x
m
p n  m


m  
N k 0
 m  
• Term in the parenthesis is

• So we get
1 N1 j2  / Nk n m
~
p n  m   e

N k 0
~
x[n]  xn 
DSP

 n  m  rN
r  


r  
r  
 n  rN   xn  rN
Sampling the Fourier Transform Cont’d
DSP
Sampling the Fourier Transform Cont’d
• Samples of the DTFT of an aperiodic sequence
– can be thought of as DFS coefficients
– of a periodic sequence
– obtained through summing periodic replicas of original sequence
• If the original sequence
– is of finite length
– and we take sufficient number of samples of its DTFT
– the original sequence can be recovered by
~
x n 0  n  N  1
xn  
else
 0
• It is not necessary to know the DTFT at all frequencies
– To recover the discrete-time sequence in time domain
• Discrete Fourier Transform
– Representing a finite length sequence by samples of DTFT
DSP
The Discrete Fourier Transform
• Consider a finite length sequence x[n] of length N
xn  0 outside of 0  n  N  1
• For given length-N sequence associate a periodic sequence
~
x n 

 xn  rN
r  
• The DFS coefficients of the periodic sequence are samples of
the DTFT of x[n]
• Since x[n] is of length N there is no overlap between terms of
x[n-rN] and we can write the periodic sequence as
~
x n  xn mod N  xnN 
• To maintain duality between time and frequency
~
– We choose one period of Xk  as the Fourier transform of x[n]
~
~
Xk  0  k  N  1
Xk   Xk mod N  Xk N 
Xk   
 0
else
DSP
The Discrete Fourier Transform Cont’d
• The DFS pair
~
Xk  
N 1
~
 x[n]e j2 / Nkn
n0
1 N1 ~
~
x[n]   Xk e j2  / Nkn
N k 0
• The equations involve only on period so we can write
N1 ~
 x[n]e j2 / Nkn 0  k  N  1
Xk   n0

0
else

 1 N1 ~
  Xk e j2 / Nkn 0  k  N  1
x[n]  N k 0

0
else

• The Discrete Fourier Transform
Xk  
N 1
 x[n]e
n0
 j2  / N kn
1 N 1
x[n]   Xk e j2  / Nkn
N k 0
• The DFT pair can also be written as
DFT
Xk  

 x[n]
DSP
Example
• The DFT of a rectangular pulse
• x[n] is of length 5
• We can consider x[n] of any
length greater than 5
• Let’s pick N=5
• Calculate the DFS of the periodic
form of x[n]
~
Xk  
4
 j2 k / 5 n
e

n0
1  e  j2 k

1  e  j2 k / 5 
5 k  0,5,10,...

else
0
DSP
Example Cont’d
• If we consider x[n] of length 10
• We get a different set of DFT
coefficients
• Still samples of the DTFT but in
different places
DSP
Properties of DFT
• Linearity
x1 n
DFT


x2 n
DFT


X1 k 
X2 k 
ax1 n  bx2 n DFT

 aX1 k   bX2 k 
• Duality
xn DFT


Xk 
Xn DFT

 Nx k N 
• Circular Shift of a Sequence
xn
DFT


Xk 
xn  mN  0  n  N - 1 DFT

 Xk e j2k / Nm
DSP
Example: Duality
DSP
Symmetry Properties
DSP
Circular Convolution
• Circular convolution of of two finite
length sequences
x3 n 
N 1
 x mx n  m 
1
2
N
m0
x3 n 
N 1
 x mx n  m 
2
m0
DSP
1
N
Example
• Circular convolution of two rectangular pulses L=N=6
1 0  n  L  1
x1 n  x2 n  
else
0
• DFT of each sequence
X1 k   X2 k  
N 1
e
j
2
kn
N
n0
N k  0

0 else
• Multiplication of DFTs
N2 k  0
X3 k   X1 k X2 k   
 0 else
• And the inverse DFT
N 0  n  N  1
x3 n  
else
0
DSP
Example
• We can augment zeros to
each sequence L=2N=12
• The DFT of each sequence
X1 k   X2 k  
1e
j
1e
2 Lk
N
j
2 k
N
• Multiplication of DFTs
2 Lk
j


N
1  e

X3 k   
2 k 
j
1e N 


DSP
2