DSP-chap7 2010cwlin/courses/dsp/notes/DSP... · 2010. 6. 11. · 2 Before we introduce the DFT, we consider the sampling of the Fourier transform of an aperiodic discrete-time sequence

1

The Discrete Fourier Transform: Its Properties and Applications

清大電機系林嘉文[email protected]

Chapter 7

� Frequency analysis of discrete-time signals is usually and most conveniently performed on a digital signal processor.

� To perform frequency analysis on a discrete-time signal, we convert the time-domain sequence to an equivalent frequency-domain representation.

2010/6/12 Introduction to Digital Signal Processing 2

Introduction

{ ( )} ( )Fx n X ω→

2

� Before we introduce the DFT, we consider the sampling of the Fourier transform of an aperiodic discrete-time sequence.

� Thus, we establish the relationship between the sampled Fourier transform and the DFT.


Frequency-Domain Sampling: The Discrete Fourier Transform

� We recall that aperiodic finite-energy signals have continuous spectra.

� Let us consider such an aperiodic discrete-time signal with Fourier transform


Frequency-Domain Sampling and Reconstruction of Discrete-Time Signals

( )x n

( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑

3

� Suppose that we sample periodically in frequency at a spacing of radius between successive samples.

� Since is periodic with period , only samples in the fundamental frequency range are necessary.



( )X ωδω

( )X ω 2π

� For convenience, we take N equidistant samples in the interval with spacing , as shown below:

� Frequency-domain sampling of the Fourier transform



2 / Nδω π=0 2ω π≤ <

4

� First, we consider the selection of N, the number of samples in the frequency domain.

� If we evaluate at

� We obtain



( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑ 2 /k Nω π=

2 /2( ) ( ) , 0,1,..., 1j kn N

n

X k x n e k NN

ππ ∞−

=−∞

= = −∑

� The summation above equation can be subdivided into an infinite number of summations, where each sum contains N terms. Thus



1 12 / 2 /

0

2 12 /

12 /

2( ) ... ( ) ( )

( ) ...

= ( )

Nj kn N j kn N

n N n

Nj kn N

n N

lN Nj kn N

l n lN

X k x n e x n eN

x n e

x n e

π π

π

π

π − −− −

=− =

−−

=

∞ + −−

=−∞ =

= + +

+ +

∑ ∑

∑

∑ ∑

5

� If we change the index in the inner summation from n to n - lN and interchange the order of the summation, we obtain the result



12 /

0

2( ) ( ) for 0,1,2,..., 1.

Nj kn N

n l

X k x n lN e k NN

ππ − ∞−

= =−∞

= − = −

∑ ∑

� The signal obtained by the

peoiodic repetition of every N samples, is clearly periodic with fundamental period N.

� It can be expanded in a Fourier series as



( ) ( )pl

x n x n lN∞

=−∞

= −∑

( )x n

12 /

0

( ) , 0,1,..., 1N

j kn Np k

k

x n c e n Nπ−

=

= = −∑

6

� With Fourier coefficients

� We can conclude that



12 /

0

1( ) , 0,1,..., 1

Nj kn N

k pn

c x n e k NN

π−

−

=

= = −∑

1 2( ), 0,1,..., 1kc X k k N

N N

π= = −

12 /

0

1 2( ) ( ) , 0,1,..., 1

Nj kn N

pk

x n X k e n NN N

ππ−

=

= = −∑

� The relationship in above equation provides the reconstruction of the periodic signal from the samples of the spectrum .

� However, it does not imply that we can recover or from the samples.

� To accomplish this, we need to consider the relationship between and .



( )px n( )X ω

( )X ω( )x n

( )px n ( )x n

7

� Since is the periodic extension of . It is clear that can be recovered from if there is no aliasing in the time domain, that is, if is time-limited to less than the period N of .



( )px n ( )x n( )x n ( )px n

( )x n( )px n

Original signal

No aliasing

Aliasing

� We conclude that the spectrum of an aperiodic discrete-time signal with finite duration L can be exactly recovered from its samples at frequencies

� The procedure is to compute

then

and finally, can be computed.



2 / , if k k N N Lω π= ≥

( ), 0,1,..., 1px n n N= −

( ), 0 1( )

0, elsewherepx n n N

x n≤ ≤ −

=

( )X ω

8

� As in the case of continuous-time signals, it is possible to express the spectrum directly in terms of its samples . To derive such an interpolation formula for , we assume that .

� Since



N L≥

( ) ( ) for 0 1px n x n n N= ≤ ≤ −

( )X ω(2 / ), 0,1,..., 1X k N k Nπ = −

( )X ω

12 /

0

1 2( ) ( ) , 0 1

Nj kn N

k

x n X k e n NN N

ππ−

=

= ≤ ≤ −∑

� If we use and substitute for , we obtain



( )x n( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑

1 12 /

0 0

1 1( 2 / )

0 0

1 2( ) ( )

2 1 ( )

N Nj kn N j n

n k

N Nj k N n

k n

X X k e eN N

X k eN N

π ω

ω π

πω

π

− −−

= =

− −− −

= =

=

=

∑ ∑

∑ ∑

9

� The inner summation term in the brackets of above represents the basic interpolation function shifted by in frequency. Indeed, if we define



1

0

( 1)/2

1 1 1( )

1

sin( / 2)

sin( / 2)

j NNj n

jn

j N

eP e

N N e

Ne

N

ωω

ω

ω

ω

ωω

−−−

−=

− −

−= =

−

=

∑

2 /k Nπ

� Then, we can obtain

� The interpolation function is not the familiar but instead, it is a periodic counterpart of it, and it is due to the periodic nature of .



1

0

2 2( ) ( ) ( ),

N

k

X X k P k N LN N

π πω ω

−

=

= − ≥∑

( )P ω (sin ) /θ θ

( )X ω

10

� The phase shift reflects the fact that the signal is a causal, finite-duration sequence of length N. The function

is plotted for N = 5.



(sin / 2) / ( sin( / 2))N Nω ω

( )x n

� We observe that the function has the property

� Consequently, the interpolation formula gives exactly the sample values . At all other frequencies, the formula provides a properly weighted linear combination of the original spectral samples.



( )P ω

1, 02( )

0, 1,2,..., 1

kP k

k NN

π ==

= −

(2 / ) for 2 /X k N k Nπ ω π=

11

� Example : Consider the signal the spectrum of this signal is sampled at frequencies

� Determine the reconstructed spectra for when N=5 and N=50.



( ) ( ), 0 1nx n a u n a= < <

2 / for 0,1,..., 1.k k N k Nω π= = −0.8a =

� Solution : The Fourier transform of the sequence is ( )x n

0

1( )

1n j n

jn

X a eae

ωωω

∞−

−=

= =−∑

� Suppose that we sample at N equidistant frequencies Thus we obtain the spectral samples

� The periodic sequence , corresponding to the frequency samples can be obtained.



( )X ω2 / , 0,1,..., 1.k k N k Nω π= = −

2 /

2 1( ) ( ) , 0,1,..., 1

1 j k N

kX k X k N

N ae π

πω

−≡ = = −

−

( )px n(2 / ), 0,1,..., 1X k N k Nπ = −

12

� Hence

where the factor represents the effect of aliasing. Since , the aliasing error tends toward zero as .



0

0

( ) ( )

, 0 11

n lNp

l l

nn lN

Nl

x n x n lN a

aa a n N

a

∞−

=−∞ =−∞

∞

=

= − =

= = ≤ ≤ −−

∑ ∑

∑1/ (1 )Na−

0 1a< <N → ∞

� For , the sequence and its spectrum are shown below (a)(b):



( )X ω( )x n0.8a =

13

� The aliased sequences and the corresponding spectral samples are shown below (c)(d).



( ) for 5 and 50px n N N= =

� We note that the aliasing effects are negligible for N=50.

� If we define the aliased finite-duration sequence

� Then its Fourier transform is



( ) asx n

( ), 0 1ˆ( )

0, otherwisepx n n N

x n≤ ≤ −

=

1 1

0 0

1 1ˆ ˆ( ) ( ) ( )1 1

N j nN Nj n j N

p N jn n

a eX x n e x n e

a ae

ωω ω

ωω−− −

− −−

= =

−= = = ⋅

− −∑ ∑

14

� Note that although , the sample values atare identical. That is,



2 /k k Nω π=

ˆ ( ) ( )X Xω ω≠

2

2 1 1 2ˆ ( ) ( )1 1

N

N j kN

aX k X k

N a ae Nπ

π π−

−= ⋅ =

− −

� The development in the preceding section is concerned with the frequency-domain sampling of an aperiodic finite-energy sequence .

� In general, the equally spaced frequency samples

do not uniquely represent the original sequence when has infinite duration.


The Discrete Fourier Transform (DFT)

( )x n

(2 / ), 0,1,..., 1,X k N k Nπ = −

( )x n( )x n

15

� Instead, the frequency samples

correspond to a periodic sequence of period N, where is an aliased version of , as indicated by the relation in the preceding equation, that is,



(2 / ), 0,1,..., 1,X k N k Nπ = −

( )px n( )x n( )px n

( ) ( ).pl

x n x n lN∞

=−∞

= −∑

� When the sequence has a finite duration of length , then is simply a periodic repetition of ,

where over a single period is given as

� Consequently, the frequency samples

uniquely represent the finite-duration sequence .



( )px n( )x n

L N≤ ( )x n( )px n

( ), 0 1( )

0, 1p

x n n Lx n

L n N

≤ ≤ −=

≤ ≤ −

(2 / ), 0,1,..., 1,X k N k Nπ = −

( )x n

16

� Since over a single period (padded by N-L zeros), the original finite-duration sequence can be obtained from the frequency samples by means of the formula



( ) ( )px n x n≡( )x n

{ (2 / )}X k Nπ

12 /

0

1 2( ) ( ) , 0,1,..., 1.

Nj kn N

pk

x n X k e n NN N

ππ−

=

= = −∑

� Note that zero padding does not provide any additional information about the spectrum of the sequence

.

� In summary, a finite-duration sequence of length Lhas a Fourier transform

� Where the upper and lower indices in the summation reflect the fact that outside the range



{ ( )}x n( )X ω

( )x n

1

0

( ) ( ) , 0 2L

j n

n

X x n e ωω ω π−

−

=

= ≤ ≤∑

( ) 0x n = 0 1.n L≤ ≤ −

17

� When we sample at equally spaced frequencies

the resultant samples are

where for convenience, the upper index in the sum has been increased from L-1 to N-1 since x(n)=0 for .



2 / , 0,1,2,..., 1, where N Lk k N k Nω π= = − ≥

( )X ω

12 /

0

12 /

0

2( ) ( ) ( )

( ) ( ) , 0,1,2,..., 1

Lj kn N

n

Nj kn N

n

kX k X x n e

N

X k x n e k N

π

π

π −−

=

−−

=

≡ =

= = −

∑

∑

n L≥

� The relation

is called the discrete Fourier transform (DFT) of x(n).



12 /

0

12 /

0

2( ) ( ) ( )

( ) ( ) , 0,1,2,..., 1

Lj kn N

n

Nj kn N

n

kX k X x n e

N

X k x n e k N

π

π

π −−

=

−−

=

≡ =

= = −

∑

∑

18

� To summarize, the formulas for the DFT and IDFT are



12 /

0

12 /

0

( ) ( ) , 0,1,2,..., 1

1( ) ( ) , n 0,1,2,..., 1

Nj kn N

n

Nj kn N

k

X k x n e k N

x n X k e NN

π

π

−−

=

−

=

= = −

= = −

∑

∑

DFT:

IDFT:

� Example : A finite-duration sequence of length L is given

as

Determine the N-point DFT of this sequence for .


N L≥� Solution : The Fourier transform of this sequence is

1, 0 1( )

0, otherwise

n Lx n

≤ ≤ −=


1 1

0 0

( 1)/2

( ) ( )

1 sin( / 2)

1 sin( / 2)

L Lj n j n

n n

j Lj L

j

X x n e e

e Le

e

ω ω

ωω

ω

ω

ωω

− −− −

= =

−− −

−

= =

−= =

−

∑ ∑

19

� The magnitude and phase of are illustrated in the below for L=10.



( )X ω

Magnitude

Phase

� The N-point DFT of x(n) is simply evaluated at the set of N equally spaced frequencies

� Hence



( )X ω

2 / , 0,1,..., 1.k k N k Nω π= = −

2 /

2 /

( 1)/

1( ) , 0,1,..., 1

1sin( / )

sin( / )

j kL N

j k N

j k L N

eX k k N

ekL N

ek N

π

π

πππ

−

−

− −

−= = −

−

=

20

� If N is selected such that N=L, then the DFT becomes



, 0( )

0, 1,2,..., 1

L kX k

k L

==

= −



N=50 N=100

21

� The formulas for the DFT and IDFT may be expressed as

which is an Nth root of unity.


The DFT as a Linear Transformation

1

0

1

0

2 /

( ) ( ) , 0,1,2,..., 1

1( ) ( ) , n 0,1,2,..., 1

where

Nkn

Nn

Nkn

Nk

j NN

X k x n W k N

x n X k W NN

W e π

−

=

−−

=

−

= = −

= = −

=

∑

∑

� We note that the computation of each point of the DFT can be accomplished by N complex multiplications and (N-1) complex additions.

� Hence the N-point DFT values can be computed in a total of N2 complex multiplications and N(N-1) complex additions.

� It is instructive to view the DFT and IDFT as linear transformations on sequences and , respectively.



{ ( )}x n { ( )}X k

22

� Let us define an N-point vector xN of the signal sequence x(n), n=0,1,…,N-1, an N-point vector XN of frequency samples, and an N×N matrix WN as

� With these definitions, the N-point DFT may be expressed in matrix forms as .

� Where is the matrix of the linear transformation.



2 1

2 4 2( 1)

1 2( 1) ( 1)( 1)

1 1 1 1(0) X(0)

1(1) X(1)

x , X , W

( 1) X( 1)1

NN N N

NN N N N N N

N N N NN N N

xW W W

xW W W

x N NW W W

−

−

− − − −

= = = − −

�

�

� ��

� � � � �

�

X W xN N N=

WN

� We observe that is a symmetric matrix.

� If we assume that the inverse of exists, then

� But this is just an expression for the IDFT.

� In fact, the IDFT can be expressed in matrix form as



WN

WN

1x W XN N N−=

*1x W X

NN NN=

23

� Where denotes the complex conjugate of the matrix . Then we can conclude that

� Which, in turn, implies that

� Where is an N×N identity matrix.



WN

*WN

1 *1W W

N NN− =

*W W IN N NN=

I N

� Therefore, the matrix in the transformation is an orthogonal (unitary) matrix.

� Furthermore, its inverse exists and is given as .

� Of course, the existence of the inverse of was established previously from our derivation of the IDFT.



WN

*W /N

N

WN

24

� The DFT and IDFT are computational tools that play a very important role in many digital signal processing applications, such as frequency analysis (spectrum analysis) of signals, power spectrum estimation, and linear filtering.

� The importance of the DFT and IDFT in such practical applications is due to a large extent to the existence of computationally efficient algorithms, known collectively as fast Fourier transform (FFT) algorithms, for computing the DFT and IDFT.



� Relationship to the Fourier series coefficients of a periodic sequence

� A periodic sequence with fundamental period Ncan be represented in a Fourier series of the form

� Where is the Fourier series coefficients.


Relationship of the DFT to Other Transforms

{ ( )}px n

12 /

0

12 /

0

( ) ,

1( ) , 0,1,.., 1

Nj kn N

p kk

Nj kn N

k pn

x n c e n

c x n e k NN

π

π

−

=

−−

=

= −∞ < < ∞

= = −

∑

∑kc

25

� Relationship to the Fourier transform of an aperiod icsequence

� If is an aperiodic finite energy sequence with Fourier transform , which is sampled at N equally spaced frequencies the spectral components

are the DFT coefficients of the periodic sequence of period N.



2 /2 /( ) ( ) | ( ) , 0,1,.., 1j kn N

k Nn

X k X x n e k Nπω πω

∞−

==−∞

= = = −∑

2 / , 0,1,..., 1,k k N k Nω π= = −

( )x n( )X ω


� Thus is determined by aliasing over the interval . The finite-duration sequence



{ ( )}x n0 1n N≤ ≤ −

( ) ( )pl

x n x n lN∞

=−∞

= −∑

( )px n

( ), 0 1ˆ( )


x n≤ ≤ −

=

26


� Bears no resemblance to the original sequence , unless is of finite duration and length , in which case

� Only in this case will the IDFT of yield the original sequence .



{ ( )}x n( )x n

ˆ( ) ( ), 0 1x n x n n N= ≤ ≤ −L N≤

{ ( )}X k{ ( )}x n


� Bears no resemblance to the original sequence , unless is of finite duration and length , in which case

� Only in this case will the IDFT of yield the original sequence .



{ ( )}x n( )x n

ˆ( ) ( ), 0 1x n x n n N= ≤ ≤ −L N≤

{ ( )}X k{ ( )}x n

27

� Relationship to the z-transform

� Consider a sequence having the z-transform

with an ROC that includes the unit circle.



( )x n

( ) ( ) n

n

X z x n z∞

−

=−∞

= ∑

� Relationship to the z-transform

� If is sampled at the N equally spaced points on the unit circle , we obtain



( )X z2 / , 0,1,2,..., 1j k N

kz e k Nπ= = −

2 /

2 /

( ) ( ) | , 0,1,.., 1

( )

j nk Nz e

j nk N

n

X k X z k N

x n e

π

π

=

∞−

=−∞

≡ = −

= ∑

28

� Relationship to the Fourier series coefficients of a continuous-time signal

� Suppose that is a continuous-time periodic signal with fundamental period . The signal can be expressed in a Fourier series

� Where are the Fourier coefficients.



( )ax t

01/pT F=

02( ) j ktFa k

k

x t c e π∞

=−∞

= ∑{ }kc


� If we sample at a uniform rate , we obtain the discrete-time sequence



( )ax t / 1/s pF N T T= =

02 2 /

12 /

0

( ) ( )

j kF nT j kn Na k k

k k

Nj kn N

k lNk l

x n x nt c e c e

c e

π π

π

∞ ∞

=−∞ =−∞

− ∞

−= =−∞

≡ = =

=

∑ ∑

∑ ∑

29


� It is clear the above equation is in the form of an IDFT formula, where

� Thus the sequence is an aliased version of the sequence .



( ) and k lN k k k lNl l

X k N c Nc c c∞ ∞

− −=−∞ =−∞

= ≡ =∑ ∑� �

{ }kc�{ }kc

� The DFT as a set of N samples of the Fourier transform for a finite-duration sequence of length .

� The sampling of occurs at the N equally spaced frequencies


Properties of the DFT

{ ( )}X k( )X ω { ( )}x n

L N≤

( )X ω2 / , 0,1,2,..., 1.k k N k Nω π= = −

1

0

1

0

2 /

( ) ( ) , 0,1,2,..., 1

1( ) ( ) , n 0,1,2,..., 1

where

Nkn

Nn

Nkn

Nk

j NN

X k x n W k N

x n X k W NN

W e π

−

=

−−

=

−

= = −

= = −

=

∑

∑

DFT:

IDFT:( ) ( )DFT

Nx n X k←→

30

� Periodicity . If and are an N-point DFT pair, then

� Linearity . If

then for any real-valued or complex-valued constants and ,


Periodicity, Linearity, and Symmetry Properties

( )X k( )x n

( ) ( )

( ) ( )

x n N x n n

X k N X k k

+ = ∀

+ = ∀

1 1 2 2( ) ( ) and ( ) ( )DFT DFTN N

x n X k x n X k←→ ←→

1a2a

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )DFTN

a x n a x n a X k a X k+ ←→ +

� Circular Symmetries of a Sequence . As we have seen, the N-point DFT of a finite duration sequence , of length , is equivalent to the N-point DFT of a periodic sequence , of period N, which is obtained by periodically extending , that is



( )x nL N≤

( )px n( )x n

( ) ( )pl

x n x n lN∞

=−∞

= −∑

31

� Circular Symmetries of a Sequence . Suppose that we shift the periodic sequence by k units to the right. Thus we obtain another periodic sequence

� The finite-duration sequence

is related to the original sequence by a circular shift.



( )px n

' ( ) ( ) ( )p pl

x n x n k x n k lN∞

=−∞

= − = − −∑

' ( ), 0 1'( )


x n ≤ ≤ −

=

( )x n

� Circular Symmetries of a Sequence . This relationship is illustrated as below for N=4.



32

� Circular Symmetries of a Sequence .

� In general, the circular shift of the sequence can be represented as the index modulo N. thus we can write

� Time reversal of N-point sequence



'( ) ( , modulo )

(( ))N

x n x n k N

x n k

= −

= −

(( )) ( ), 0 1Nx n x N n n N− = − ≤ ≤ −

� Circular Symmetries of a Sequence . An equivalent definition of even and odd sequences for the associated periodic sequence is given as follows

� If the periodic sequence is complex valued, we have



( )px n

( ) ( ) ( )

( ) ( ) ( )p p p

p p p

x n x n x N n

x n x n x N n

= − = −

= − − = − −

Even:

Odd:

*

*

( ) ( )

( ) ( )

p p

p p

x n x N n

x n x N n

= −

= − −

Conjugate even:

Conjugate odd:

33

� Circular Symmetries of a Sequence . These relationships suggest that we decompose the sequence as

where



( )px n

( ) ( ) ( )p pe pox n x n x n= +

*

*

1( ) [ ( ) ( )]

21

( ) [ ( ) ( )]2

pe p p

po p p

x n x n x N n

x n x n x N n

= + −

= − −

� Symmetry properties of the DFT . The sequences can be expressed as

� We can obtain



( ) ( ) ( ), 0 1

( ) ( ) ( ), 0 1R I

R I

x n x n jx n n N

X k X k jX k k N

= + ≤ ≤ −

= + ≤ ≤ −

1

0

1

0

2 2( ) ( )cos ( )sin

2 2( ) ( )sin ( )cos

N

R R In

N

I R In

kn knX k x n x n

N N

kn knX k x n x n

N N

π π

π π

−

=

−

=

= +

= − −

∑

∑

34

� Symmetry properties of the DFT .

� Similarly,



1

0

1

0

1 2 2( ) ( )cos ( )sin

1 2 2( ) ( )sin ( )cos

N

R R Ik

N

I R Ik

kn knx n X k X k

N N N

kn knx k X k X k

N N N

π π

π π

−

=

−

=

= −

= +

∑

∑

� Real-valued sequences . If the sequence is real, it follows

� Consequently,



*( ) ( ) ( )X N k X k X k− = = −

( )x n

( ) ( )

( ) ( )

X N k X k

X N k X k

− =

∠ − = −∠

35

� Real and even sequences . If is real and even, that is

� And . Hence the DFT reduces to

� IDFT



( ) ( ), 0 1x n x N n n N= − ≤ ≤ −

( )x n

1

0

2( ) ( )cos , 0 1

N

n

knX k x n k N

N

π−

=

= ≤ ≤ −∑

( ) 0IX k =

1

0

1 2( ) ( )cos , 0 1

N

k

knx n X k n N

N N

π−

=

= ≤ ≤ −∑

� Real and odd sequences . If is real and odd, that is

� And . Hence the DFT reduces to

� IDFT



( ) ( ), 0 1x n x N n n N= − − ≤ ≤ −

( )x n

1

0

2( ) ( )sin , 0 1

N

n

knX k j x n k N

N

π−

=

= − ≤ ≤ −∑

( ) 0RX k =

1

0

1 2( ) ( )sin , 0 1

N

k

knx n j X k n N

N N

π−

=

= ≤ ≤ −∑

36

� Purely imaginary sequences .

� We observe that is odd and is even.



( ) ( )Ix n jx n=

1

0

1

0

2( ) ( )sin

2( ) ( )cos

N

R In

N

I In

knX k x n

N

knX k x n

N

π

π

−

=

−

=

=

=

∑

∑

( )RX k ( )IX k



37

� Suppose that we have two finite-duration sequences of length N, . Their respective N-point DFTs are

� If we multiply the two DFTs together, the result is a DFT, say , of a sequence of length N.


Multiplication of Two DFTs and Circular Convolution

1 2( ) and ( )x n x n

12 /

1 10

12 /

2 20

( ) ( ) , 0,1,..., 1

( ) ( ) , 0,1,..., 1

Nj nk N

n

Nj nk N

n

X k x n e k N

X k x n e k N

π

π

−−

=

−−

=

= = −

= = −

∑

∑

3( )X k 3( )x n

3 1 2( ) ( ) ( ), 0,1,..., 1X k X k X k k N= = −

� The IDFT of is

� Suppose that we substitute for using the DFTs, thus we obtain



1 12 / 2 /

3 3 1 20 0

1 1( ) ( ) ( ) ( )

N Nj km N j km N

k k

x m X k e X k X k eN N

π π− −

= =

= =∑ ∑

3{ ( )}X k

1 2( ) and ( )X k X k

1 1 12 / 2 / 2 /

3 1 20 0 0

1 1 12 ( )/

1 20 0 0

1( ) ( ) ( )

1 ( ) ( )

N N Nj kn N j kn N j km N

k n l

N N Nj k m n l N

n l k

x m x n e x l e eN

x n x l eN

π π π

π

− − −− −

= = =

− − −− −

= = =

=

=

∑ ∑ ∑

∑ ∑ ∑

38

� The inner sum in the brackets in the above equation has the form

where is defined as

� We observe that when is a multiple of N.



1

0

, 1

1, 1

1

Nk N

k

N a

a aa

a

−

=

=

= −≠ −

∑

a 2 ( )/j m n l Na e π − −=

1a = m n l− −

� On the other hand, for any value of .

� Consequently, the above equation reduces to

� Then we obtain the desired expression for in the form



1Na = 0a ≠

1

0

, (( )) , p an integer

0, otherwise

NNk

k

N l m n pN m na

−

=

= − + = −=

∑

3( )x m

1

3 1 20

( ) ( ) (( )) , 0,1,..., 1N

Nn

x m x n x m n m N−

=

= − = −∑

39

� Circular convolution . If

then

where denotes the circular convolution of the sequence .



1 1 2 2( ) ( ) and ( ) ( )DFT DFTN N


1 2 1 2( ) ( ) ( ) ( )DFTN

x n x n X k X k←→N

1 2( ) ( )x n x nN

1 2( ) and ( )x n x n

� Time reversal of a sequence . If

then

� Hence reversing the N-point sequence in time is equivalent to reversing the DFT values.


Additional DFT Properties

( ) ( )DFTN

x n X k←→

(( )) ( ) (( )) ( )DFTN NN

x n x N n X k X N k− = − ←→ − = −

40

� Circular time shift of a sequence . If

then

� Circular frequency shift . If



( ) ( )DFTN

x n X k←→

2 /(( )) ( )DFT j kl NN N

x n l X k e π−− ←→

( ) ( )DFTN

x n X k←→

2 ln/( ) (( ))DFTj NNN

x n e X k lπ ←→ −

� Complex-conjugate properties . If

then

� The IDFT of is

therefore,



( ) ( )DFTN

x n X k←→

* * *( ) (( )) ( )DFTNN

x n X k X N k←→ − = −

1 1* 2 / 2 ( )/

0 0

1 1( ) ( )

N Nj kn N j k N n N

k k

X k e X k eN N

π π− −

−

= =

=

∑ ∑

* * *(( )) ( ) ( )DFTN N

x n x N n X k− = − ←→

* ( )X k

41

� Circular correlation . If

then

where is the (unnormalized) circular crosscorrelation sequence, defined as



( ) ( ) and ( ) ( )DFT DFTN N

x n X k y n Y k←→ ←→

*( ) ( ) ( ) ( )DFTxy xyN

r l R k X k Y k←→ =��

1*

0

( ) ( ) (( ))N

xy Nn

r l x n y n l−

=

= −∑�

( )xyr l�

� Multiplication of two sequences . If

then

� Parseval’s Theorem . If

then



1 1 2 2( ) ( ) and ( ) ( )DFT DFTN N


1 2 1 2

1( ) ( ) ( ) ( )DFT

Nx n x n X k X k

N←→ N

( ) ( ) and ( ) ( )DFT DFTN N

x n X k y n Y k←→ ←→1 1

* *

0 0

1( ) ( ) ( ) ( )

N N

n k

x n y n X k Y kN

− −

= =

=∑ ∑

42

� The DFT provides a discrete frequency representation of a finite-duration sequence in the frequency domain, it explore its use as a computational tool for linear system analysis and, especially, for linear filtering.

� We have already established that a system with frequency response , when excited with an input signal that has a spectrum , possesses an output spectrum .

� The output sequence is determined from its spectrum via the inverse Fourier transform.


Linear Filtering Methods based on the DFT

( )H ω( )X ω

( ) ( ) ( )Y X Hω ω ω=

( )y n

� Suppose that we have a finite-duration sequence of length L which excites an FIR filter of length M. Without loss of generality, let

where is the impulse response of the FIR filter.

� The output sequence


Use of the DFT in Linear Filtering

( )x n

( ) 0, 0 and

( ) 0, 0 and

x n n n L

h n n n M

= < ≥

= < ≥

( )h n

1

0

( ) ( ) ( )M

k

y n h k x n k−

=

= −∑

43

� The frequency-domain equivalent to the above is

� If

then

where are the N-point DFTs of the corresponding sequences .


Use of the DFT in Linear Filtering

( ) ( ) ( )Y X Hω ω ω=

2 /

2 /

( ) ( ) | , 0,1,..., 1

( ) ( ) | , 0,1,..., 1k N

k N

Y k Y k N

X H k Nω π

ω π

ω

ω ω=

=

≡ = −

= = −

( ) ( ) ( ), 0,1,..., 1Y k X k H k k N= = −

{ ( )} and { ( )}X k H k( ) and ( )x n h n

� In practical applications involving linear filtering of signals, the input sequence is often a very long sequence. This is especially true in some real-time signal processing applications concerned with signal monitoring and analysis.

� Since linear filtering performed via the DFT involves operations on a block of data, which by necessity must be limited in size due to limited memory of a digital computer, a long input signal sequence must be segmented to fixed-size blocks prior to processing.


Filtering of Long Data Sequences

( )x n

44

Overlap-Add Method

2010/6/12 87Introduction to Digital Signal Processing

Overlap-Add Method

� We first segment x(n), assumed to be a causal sequence here without any loss of generality, into a set of contiguous finite-length subsequences of length L each:

where

� Thus we can write

where

� Since h(n) is of length M and xm(n) is of length L, ym(n) is of length L + M −1


0

( ) ( )

( ), 0 1( )

0, otherwise

mm

m

x n x n mL

x n mL n Lx n

∞

=

= −

+ ≤ ≤ −=

∑

0

( ) ( ) ( ) ( )

( ) ( ) ( )

mm

m m

y n h n x n y n mL

y n h n x n

∞

=

= ∗ = −

= ∗

∑

45

Overlap-Add Method

� The desired linear convolution y(n) = h(n) ＊ x(n) is broken up into a sum of infinite number of short-length linear convolutions of length L + M −1 each: ym(n) = h(n) ＊ xm(n)

� Consider implementing the following convolutions using the DFT-based method, where now the DFTs (and the IDFT) are computed on the basis of (L + M −1) points

� The first convolution in the above sum, y0(n) = h(n) ＊ x0(n), is of length L + M −1 and is defined for 0 ≤ n ≤ L + M − 2

� The second short convolution y1[n] = h(n) ＊x1(n), is also of length L + M −1 but is defined for L ≤ n ≤ 3L + M − 2

� There is an overlap of samples between these two short linear convolutions


0

( ) ( )mm

y n y n mL∞

=

= −∑

Overlap-Add Method

� In general, there will be an overlap of M −1 samples between the samples of the short convolutions h(n) ＊ xr-1(n) and h(n) ＊ xm(n) for (r −1)L ≤ n ≤ rL + M − 2


46

Overlap-Add Method

� Therefore, y(n) obtained by a linear convolution of x(n) and h(n) is given by


0

1 1

1

( ) ( ), 0 6

( ) ( ) ( 7), 7 10

( ) ( 7), 11 13

y n y n n

y n y n y n n

y n y n n

= ≤ ≤

= + − ≤ ≤

= − ≤ ≤

�

Overlap-Add Method

� The above procedure is called the overlap-add method since the results of the short linear convolutions overlap and the overlapped portions are added to get the correct final result

� The MATLAB function fftfilt can be used to implement the above method

� The following illustrates an example of filtering of a noise-corrupted signal using a length-3 moving average filter:


47

Overlap-Save Method


Overlap-Save Method

� In implementing the overlap-add method using the DFT, we need to compute two (L + M −1)-point DFTs and one (L + M−1)-point IDFT for each short linear convolution

� It is possible to implement the overall linear convolution by performing instead circular convolution of length shorter than (L + M −1)

� To this end, it is necessary to segment x(n) into overlapping blocks xm(n), keep the terms of the circular convolution of h(n) with that corresponds to the terms obtained by a linear convolution of h(n) and xm(n), and throw away the other parts of the circular convolution


48

Overlap-Save Method

� To understand the correspondence between the linear and circular convolutions, consider a length-4 sequence x(n) and a length-3 sequence h(n)

� Let yL(n) denote the result of a linear convolution of x(n) with h(n)

� The six samples of yL(n) are given byyL(0) = h(0)x(0)yL(1) = h(0)x(1) + h(1)x(0)yL(2) = h(0)x(2) + h(1)x(1) + h(2)x(0)yL(3) = h(0)x(3) + h(1)x(2) + h(2)x(1)yL(0) = h(1)x(3) + h(2)x(2)yL(5) = h(2)x(3)


Overlap-Save Method

� If we append h(n) with a single zero-valued sample and convert it into a length-4 sequence he(n), the 4-point circular convolution Yc(n) of he(n) and x(n) is given by

yC(0) = h(0)x(0) + h(1)x(3) + h(2)x(2)yC(1) = h(0)x(1) + h(1)x(0) + h(2)x(3)yC(2) = h(0)x(2) + h(1)x(1) + h(2)x(0)yC(3) = h(0)x(3) + h(1)x(2) + h(2)x(1)

� If we compare the expressions for the samples yL(n) of with those of yC(n), we observe that the first 2 terms of yC(n) do not correspond to the first 2 terms of yL(n), whereas the last 2 terms of yC(n) are precisely the same as the 3rd and 4th terms of yL(n), i.e.

yL(0) ≠ yC(0), yL(1) ≠ yC(1), yL(2) = yC(2), yL(3) = yC(3)


49

Overlap-Save Method

� General case : N-point circular convolution of a length-M sequence h(n) with a length-L sequence x(n) with N > M

� First M − 1 samples of the circular convolution are incorrect and are rejected

� Remaining L − M + 1 samples correspond to the correct samples of the linear convolution of h(n) with x(n)

� Now, consider an infinitely long or very long sequence x(n)� Break it up as a collection of smaller length (length-4)

overlapping sequences xm(n) as xm(n) = x(n + 2m), 0 ≤ n ≤ 3, 0 ≤ m ≤ ∞

� Next, form


( ) ( ) ( )m mw n h n x n= 4

Overlap-Save Method

� Or, equivalently, wm(0) = h(0)xm(0) + h(1)xm(3) + h(2)xm(2)wm(1) = h(0)xm(1) + h(1)xm(0) + h(2)xm(3)wm(2) = h(0)xm(2) + h(1)xm(1) + h(2)xm(0)wm(3) = h(0)xm(3) + h(1)xm(2) + h(2)xm(1)

� Computing the above for m = 0, 1, 2, 3, . . . , and substituting the values of xm[n] we arrive at

w0(0) = h(0)x(0) + h(1)x(3) + h(2)x(2) � Rejectw0(1) = h(0)x(1) + h(1)x(0) + h(2)x(3) � Rejectw0(2) = h(0)x(2) + h(1)x(1) + h(2)x(0) = y[2] � Savew0(3) = h(0)x(3) + h(1)x(2) + h(2)x(1) = y[3] � Save


50

Overlap-Save Method

w1(0) = h(0)x(2) + h(1)x(5) + h(2)x(4) � Rejectw1(1) = h(0)x(3) + h(1)x(2) + h(2)x(5) � Rejectw1(2) = h(0)x(4) + h(1)x(3) + h(2)x(2) = y(4) � Savew1(3) = h(0)x(5) + h(1)x(4) + h(2)x(3) = y(5) � Save

w2(0) = h(0)x(4) + h(1)x(7) + h(2)x(6) � Rejectw2(1) = h(0)x(5) + h(1)x(4) + h(2)x(7) � Rejectw2(2) = h(0)x(6) + h(1)x(5) + h(2)x(4) = y(6) � Savew2(3) = h(0)x(7) + h(1)x(6) + h(2)x(5) = y(7) � Save

� It should be noted that to determine y(0) and y(1), we need to form x−1(n): x−1(0) = 0, x−1(1) = 0, x−1(2) = x(0) , x−1(3) = x(1) and compute w−1(n) = h(n) 4 x−1(n) for 0 ≤ n ≤ 3, reject w−1(0) and w−1(1), and save w−1(2) = y(0), and w−1(3) = y(1)


Overlap-Save Method

� General Case : Let h(n) be a length-L sequence� Let xm(n) denote the m-th section of an infinitely long

sequence x[n] of length L and defined byxm(n) = x(n + m(L − m + 1)), 0 ≤ n ≤ L − 1 with M < L

� Let wm(n) = h(n) xm(n)� Then, we reject the first M − 1 samples of wm(n) and “abut”

the remaining L − M + 1 samples of wm(n) to form yL(n), the linear convolution of h(n) and x(n)

� If ym[n] denotes the saved portion of wm(n), i.e.,

� Then yL(n + m(L − M + 1)) = ym(n), M − 1 ≤ n ≤ l − 1


0, 0 2( )

( ), -1 2m

n My n

w n M n L

≤ ≤ −=

≤ ≤ −

51

Overlap-Save Method

� The approach is called overlap-save method since the input is segmented into overlapping sections and parts of the results of the circular convolutions are saved and abutted to determine the linear convolution result


Overlap-Save Method


52

� Let denote the sequence to be analyzed. Limiting the duration of the sequence to L samples, in the interval

, is equivalent to multiplying by a rectangular window of length L. That is

where


Frequency Analysis of Signals Using the DFT

{ ( )}x n

0 1n L≤ ≤ − { ( )}x n( )w n

ˆ( ) ( ) ( )x n x n w n=

1, 0 1( )

0, otherwise

n Lw n

≤ ≤ −=

� Suppose that the sequence consists of a signal sinusoid, that is

� Then the Fourier transform of the finite-duration sequence can be expresses as

where is the Fourier transform of the window sequence, which is



( )x n

0( ) cosx n nω=

( )x n

0 0

1ˆ ( ) [ ( ) ( )]2

X W Wω ω ω ω ω= − + +

( )W ω

( 1)/2sin( / 2)( )

sin( / 2)j LL

W e ωωω

ω− −=

53

� Magnitude spectrum for L=25 and N=2048, illustrating the occurrence of leakage.



�



0 1 2( ) cos cos cosx n n n nω ω ω= + +

54

Signal Transform

• Motivation: – Represent a vector (e.g. a block of image samples) as

the superposition of some typical vectors (block patterns)

+t1 t2 t3 t4


Transform Coding of a Image


55

1-D 16-Pont DFT Basis Vectors


Disadvantages of DFT in Signal Coding

• Fourier Transform of a real function results incomplex numbers

• May result in artifacts due to discontinuityat the block boundary

Discontinuities (high freq. components)


56

From DFT to DCT

• DFT of any real and symmetric sequence contains only real coefficients corresponding to the cosine terms of the series

• Construct a new symmetric sequence y(n) of length 2N out of x(n) of length N

• Y(n) is symmetrical about n = N - (1/2)

= ≤ ≤ −

= − − ≤ ≤ −

( ) ( ),0 1,

( ) (2 1 ), 2 1.

y n x n n N

y n x N n N n N


From DFT to DCT

• DCT has a higher compression ration than DFT• DCT avoids the generation of spurious spectral

components

No discontinuities


57

From DFT to DCT

− +

=

− −+ +

= =

=

= +

∑

∑ ∑

12 1 ( )2

20

1 11 2 1( ) ( )2 2

2 20

( ) ( ) ,

( ) ( )

N n k

Nn

N Nn k n k

N Nn n N

Y k y n W

y n W y n W

π

π

− −+ +

= =

− −+ − +

= =

−

=

−

= + − −

= +

+=

≤ ≤ − =

∑ ∑

∑ ∑

∑

1 11 2 1( ) ( )2 2

2 20

1 11 1( ) [ 2 ( ) ]2 2

2 20 0

1

0

22

2

( ) ( 2 1 )

( ) ( )

( 2 1)2 ( ) c o s ,

2

0 2 1,

N Nn k n k

N Nn n N

N Nn k N n k

N Nn n

N

n

jN

N

x n W x N n W

x n W x n W

n kx n

N

k N a n d W e


1-D N-Point DCT

π

π

−

=

−

=

+ =

= −

+ = = −

∑

∑

�

�

1

0

1

0

( 2 1)( ) ( ) ( ) c o s ,

2

0 ,1, , 1,

( 2 1)( ) ( ) ( ) c o s ,

2

0 ,1, , 1,

N

n

N

k

n kF k C k f n

N

k N

n kf n C k F k

N

n N

= = = −�1 2

(0) , ( ) , 1,2, , 1C C k k NN N

where

• The constants are often defined differently


58

1-D N-Point DCT

( )cos (2 1)0 /16n π+

( )cos (2 1)1 /16n π+

( )cos (2 1)2 /16n π+

( )cos (2 1)3 /16n π+

( )cos (2 1)4 /16n π+

( )cos (2 1)5 /16n π+

( )cos (2 1)6 /16n π+

( )cos (2 1)7 /16n π+


Example of 1-D DCT

0.00 200.00 400.00 600.00

0.00

40.00

80.00

120.00

160.00

200.00Row 256 of Lena

0.00 200.00 400.00 600.00

0.00

500.00

1000.00

1500.00

2000.00

2500.00

absolute DCT values of Lena row 256


59

Illustration of Image Coding Using 2-D DCT


DCT-Based Image Coding with Different Quantization Levels

• DCT coding with increasingly coarse quantization, block size 8x8

quantizer step-size for AC coefficient: 25




60

Image Coding with Different Numbers of DCT Coefficients

original

with 8/64 coefficients




DCT-Based Coding: JPEG

8x8 blocks

Entropyencoder

Compressedimage data

QDPCM

Zigzagscan

Quantizationtable

Tablespecification

Entropydecoder

Compressedimage data

IQDPCM

Zigzagscan

Quantizationtable

Tablespecification

Sourceimage data

Reconstructedimage data

DCTAC

DC

8x8 blocks

IDCT

DC

AC


Documents

DSP-chap7 2010cwlin/courses/dsp/notes/DSP... · 2010. 6. 11. · 2 Before we introduce the DFT, we consider the sampling of the Fourier transform of an aperiodic discrete-time sequence