Transcript Document

The Discrete Fourier Transform
Quote of the Day
Mathematics may be defined as the subject in which
we never know what we are talking about, nor
whether what we are saying is true.
Bertrand Russell
Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, ©1999-2000 Prentice Hall
Inc.
Sampling the Fourier Transform
• Consider an aperiodic sequence with a Fourier transform
 
DTFT
x[n] 
 X ej
• 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

•
•
•

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
Copyright (C) 2005 Güner Arslan
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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  m  rN
N k 0
r  
~
x[n]  xn 
Copyright (C) 2005 Güner Arslan


r  
r  
 n  rN   xn  rN
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Sampling the Fourier Transform Cont’d
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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
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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   Xk mod N  Xk N 
 0
else
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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]
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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
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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
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Properties of DFT
• Linearity
x1 n
DFT


X1 k 
x2 n
DFT


X2 k 
ax1 n  bx2 n DFT

 aX1 k   bX2 k 
• Duality
DFT
xn 


Xk 
DFT
Xn 

 Nx k N 
• Circular Shift of a Sequence
xn
DFT



Xk 
DFT
xn  mN  0  n  N - 1 

 Xk e j2k / Nm
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Example: Duality
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Symmetry Properties
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Circular Convolution
• Circular convolution of of two finite
length sequences
x3 n 
x3 n 
N 1
 x mx n  m 
m 0
1
2
N
N 1
 x mx n  m 
m 0
2
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N
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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
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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

1  e N
X3 k   
2 k
j
1e N

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



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2
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