Department of Computer Eng. Sharif University of Technology Discrete-time signal processing Chapter...

Department of Computer Eng.

Sharif University of Technology

Discrete-time signal processing

Chapter 8:Chapter 8:THE DESCRETE FOURIER THE DESCRETE FOURIER

TRANSFORMTRANSFORM

Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer and Buck, ©1999-2000 Prentice Hall Inc.

8.0 Introduction

• For finite-duration sequences, it is possible to develop an alternative Fourier representation referred to as the discrete Fourier transform(DFT).

• It is a sequence rather than a continuous variable.

• It corresponds to samples, equally spaced in frequency, of the Fourier transform of the signal.

• The importance of DFT is because efficient algorithms exists for the computation of the DFT (FFT).

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Chapter 8: The Discrete Fourier Transform

8.1 Discrete Fourier Series

• Given a periodic sequence with period N so that

• The Fourier series representation of continuous-time periodic signals require infinite complex exponentials.

• But for discrete-time periodic signals we have

• Due to the periodicity of the complex exponential we only need N exponentials for discrete time Fourier series

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[ ] [ ]x n x n rN

2 / 2 / 2 2 /j N k mN n j N kn j mn j N kne e e e

1

2 / 2 /

0

1 1[ ]

Nj N kn j N kn

k N k

x n X k e X k eN N

[ ]x n


Discrete Fourier Series Pair

• A periodic sequence in terms of Fourier series coefficients

• The Fourier series coefficients can be obtained via

• For convenience we sometimes use

• Analysis equation

• Synthesis equation

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1

2 /

0

[ ]N

j N kn

n

X k x n e

1

2 /

0

1[ ]

Nj N kn

k

x n X k eN

2 /j NNW e

1

0

1[ ]

Nkn

Nk

x n X k WN

1

0

[ ]N

knN

n

X k x n W


Example 1 1

• DFS of a periodic impulse train

• Since the period of the signal is N

• We can represent the signal with the DFS coefficients as

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1[ ]

0r

n rNx n n rN

else

1 1

2 / 2 / 2 / 0

0 0

[ ] [ ] 1N N

j N kn j N kn j N k

n n

X k x n e n e e

1

2 /

0

1[ ]

Nj N kn

r k

x n n rN eN


Example 2: Duality in the DFT

• Here let the Fourier series coefficients be the periodic impulse train Y[k] , and given by this equation:

• Substituting Y[k] in to DFS equation gives

• Comparing this result with the results for example1 we see that Y[k]=Nx[k] and y[n]=X[n].

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[ ] [ ]r

Y k N k rN

10

0

1[ ] [ ] 1

Nkn

N Nk

y n N k W WN

Example 3

• DFS of an periodic rectangular pulse train

• The DFS coefficients

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2 /10 54

2 /10 4 /10

2 /100

sin / 21

sin /101

j kj kn j k

j kn

keX k e e

ke


8.2 Properties of DFS

• Linearity (all signals have the same period)

• Shift of a Sequence

• Duality

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1 1

2 2

1 2 1 2

DFS

DFS

DFS

x n X k

x n X k

ax n bx n aX k bX k

2 /

2 /

DFS

DFS j km N

DFSj nm N

x n X k

x n m e X k

e x n X k m

DFS

DFS

x n X k

X n Nx k


0 1m N

Periodic Convolution

• Take two periodic sequences

• Let’s form the product

• The periodic sequence with given DFS can be written as

• Periodic convolution is commutative

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1 1

2 2

DFS

DFS

x n X k

x n X k

3 1 2X k X k X k

1

3 1 20

N

m

x n x m x n m

1

3 2 10

N

m

x n x m x n m


Periodic Convolution Cont’d

• Substitute periodic convolution into the DFS equation

• Interchange summations

• The inner sum is the DFS of shifted sequence

• Substituting

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1

3 1 20

N

m

x n x m x n m

1 1

3 1 20 0

[ ] [ ]N N

knN

n m

X k x m x n m W

1 1

3 1 20 0

[ ] [ ]N N

knN

m n

X k x m x n mW

1

2 20

[ ]N

kn kmN N

n

x n mW W X k

1 1 1

3 1 2 1 2 1 20 0 0

[ ] [ ] [ ]N N N

kn kmN N

m n m

X k x m x n mW x m W X k X k X k


Graphical Periodic Convolution

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Symmetry Properties

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Symmetry Properties Cont’d

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8.3 The Fourier Transform of Periodic Signals

• Periodic sequences are not absolute or square summable– Hence they don’t have a Fourier Transform

• We can represent them as sums of complex exponentials: DFS• We can combine DFS and Fourier transform• Fourier transform of periodic sequences

– Periodic impulse train with values proportional to DFS coefficients

– This is periodic with 2 since DFS is periodic

• The inverse transform can be written as

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2 2j

k

kX e X k

N N

2 2

0 0

212

00

1 1 2 2

2 2

1 2 1

j j n j n

k

kN j nj n N

k k

kX e e d X k e d

N N

kX k e d X k e

N N N


Example

• Consider the periodic impulse train

• The DFS was calculated previously to be

• Therefore the Fourier transform is

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[ ]r

p n n rN

1 for all kP k

2 2j

k

kP e

N N


Relation between Finite-length and Periodic Signals

• Consider finite length signal x[n] spanning from 0 to N-1• Convolve with periodic impulse train

• The Fourier transform of the periodic sequence is

• This implies that

• DFS coefficients of a periodic signal can be thought as equally spaced samples of the Fourier transform of one period

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[ ] [ ] [ ] [ ]r r

x n x n p n x n n rN x n rN

2

2 2

2 2

j j j j

k

kjj N

k

kX e X e P e X e

N N

kX e X e

N N

2

2

kj jN

k

N

X k X e X e

Example

• Consider the following sequence

• The Fourier transform

• The DFS coefficients

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2 /j NNW e

2 sin 5 / 2

sin / 2j jX e e

4 /10 sin / 2

sin /10j k k

X k ek


8.4 Sampling the Fourier Transform

• Consider an aperiodic sequence with a Fourier transform

• Assume that a sequence is obtained by sampling the DTFT

• Since the DTFT is periodic resulting sequence is also periodic• We can also write it in terms of the z-transform

• The sampling points are shown in figure• could be the DFS of a sequence• Write the corresponding sequence

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2 /

2 /

j N kj

N kX k X e X e

[ ] DTFT jx n X e

2 /

2 /N k

j N k

z eX k X z X e

X k

1

2 /

0

1[ ]

Nj N kn

k

x n X k eN


Sampling the Fourier Transform Cont’d

• The only assumption made on the sequence is that DTFT exist

• Combine equation to get

• Term in the parenthesis is

• So we get

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j j m

m

X e x m e

1

2 /

0

1[ ]

Nj N kn

k

x n X k eN

2 /j N kX k X e

12 / 2 /

0

12 /

0

1[ ]

1

Nj N km j N kn

k m

Nj N k n m

m k m

x n x m e eN

x m e x m p n mN

1

2 /

0

1 Nj N k n m

k r

p n m e n m rNN

[ ]r r

x n x n n rN x n rN



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• 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

• 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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0 1

0

x n n Nx n

else


8.5 The Discrete Fourier Transform

• Consider a finite length sequence x[n] of length N

• For given length-N sequence associate a periodic sequence

• 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

• To maintain duality between time and frequency– We choose one period of as the Fourier transform of x[n]

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0 outside of 0 1x n n N

r

x n x n rN

mod NN

X k X k X k

X k

0 1

0

X k k NX k

else

mod NN

x n x n x n


The Discrete Fourier Transform Cont’d

• The DFS pair

• The equations involve only on period so we can write

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1

2 /

0

1[ ]

Nj N kn

k

x n X k eN

1

2 /

0

[ ]N

j N kn

n

X k x n e

12 /

0

[ ] 0 1

0

Nj N kn

n

x n e k NX k

else

1

2 /

0

10 1

[ ]

0

Nj N kn

k

X k e n Nx n N

else


The Discrete Fourier Transform Cont’d

• The DFT pair can also be written as

• The Discrete Fourier Transform

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[ ]DFTX k x n

12 /

0

[ ] 0 1

0

Nj N kn

n

x n e k NX k

else

1

2 /

0

10 1

[ ]

0

Nj N kn

k

X k e n Nx n N

else

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]

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42 /5

0

2

2 /5

1

15 0, 5, 10,...

0

j k n

n

j k

j k

X k e

e

ek

else


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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8.6 Properties of DFT

• Linearity

• Duality

• Circular Shift of a Sequence

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1 1

2 2

1 2 1 2

DFT

DFT

DFT

x n X k

x n X k

ax n bx n aX k bX k

2 / 0 n N-1

DFT

j k N mDFT

N

x n X k

x n m X k e

DFT

DFT

N

x n X k

X n Nx k


Example: Duality

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Symmetry Properties

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Circular Convolution

• Circular convolution of two finite length sequences

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1

3 1 20

N

Nm

x n x m x n m

1

3 2 10

N

Nm

x n x m x n m


Example

• Circular convolution of two rectangular pulses L=N=6

• DFT of each sequence

• Multiplication of DFTs

• And the inverse DFT

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1 2

1 0 1

0

n Lx n x n

else

21

1 20

0

0

N j knN

n

N kX k X k e

else

2

3 1 2

0

0

N kX k X k X k

else

3

0 1

0

N n Nx n

else


Example

• We can augment zeros to each sequence L=2N=12

• The DFT of each sequence

• Multiplication of DFTs

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DSP

2

1 2 2

1

1

LkjN

kjN

eX k X k

e

22

3 2

1

1

LkjN

kjN

eX k

e


8.7 Linier convolution using the DFT

• Efficient algorithms are available for computing the DFT of finite-duration sequence, therefore it is computationally efficient to implement a convolution of two sequences by the following procedure:

– Compute the N point discrete Fourier transforms X1[k] and X2[k] for the two sequences given.

– Compute the product X3[k]=X1[k]X2[k].– Compute the inverse DFT of X3[k].

• The multiplication of discrete Fourier transforms corresponds to a circular convolution. To obtain a linear convolution, we must ensure that circular convolution has the effect of linear convolution.

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Linear convolution of two finite-length sequences

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3 1 2m

x k x m x n m

Circular convolution as linear convolution

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With aliasing

Without aliasing

DSP

DSP

DSP

Implementing LTI systems using the DFT

• Let us consider an L point input sequence x[n] and a p point impulse response h[n].

• The linear convolution has finite-duration with length L+P-1. • consequently for linear convolution and circular convolution

to be identical, the circular convolution must have the length of at least L+p-1 points.

• i.e. both x[n] and h[n] must be augmented with sequence amplitude with zero amplitude.

• This process is often referred to as zero-padding.

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Implementing LTI systems using the DFT cont’d

• In many applications the input signal is an indefinite duration.

– We have to store all the input data.– No filtered samples calculated until all the input samples have been

collected.– Implementing FFT for such large number of points is impractical.

• The solution is to use block convolution.

• In this method signal is segmented into sections and each section can then be convolved with the finite length impulse response and the filtered sections fitted together in an appropriate way.

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Implementing LTI systems using the DFT cont’d

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Overlap-add method

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0

[ ], 0 1[ ]

0,

rr

r

x n x n rL

x n rL n Lx n

otherwise

0

[ ] [ ]

[ ] [ ]* [ ]

rr

r r

y n x n h n y n rL

y n x n h n

Overlap-add method

• xr[n] has L nonzero points and h[n] is of length P, each of yr[n] has length L+P-1.

• A method to implement linear convolution is the nonzero points in the filtered sections will overlap by P-1 points and these overlap samples must be added to compute linear convolution.

• This method is called overlap-add method.

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Overlap-Save Method DSP

Overlap-save method

• It corresponds to implementing an L-point circular convolution of a P- point impulse response h[n] with an L-point segment xr[n] and identifying the part of singular convolution that corresponds to linear convolution.

• We showed that if an L-point sequence is circularly convolved with a P-point sequence (P<L) then the first P-1 point of result are incorrect.

• This method is called overlap-save method.

45

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Department of Computer Eng. Sharif University of Technology Discrete-time signal processing Chapter...

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