Lecture 8 Objectives - eee.iub.edu.bd

IUB Dr. Abdur Razzak 1

Digital Signal Processing

Objectives

To learn and understand DFT - the frequency domain

sampling & reconstruction of

discrete-time signal.

FFT – an efficient computation

technique of DFT.

Lecture – 8

Discrete Fourier Transform

& Fast Fourier Transform

ECR 305_L8

nk

N

N

n

WnxnxkX ~~DSFˆ~ 1

0

nk

N

N

k

WkXN

kXnx

~1~

IDFSˆ~1

0


Introduction

In lecture-4 & 5, we have studied transform-domain

representations of discrete signals. The discrete-time Fourier transform (DTFT) provides the frequency-

domain () representation for absolutely summable sequences (see

lecture-5).

The z-transform provides a generalized frequency-domain (z)

representation for arbitrary sequences (see lecture-4).

These transforms have two features in common: First, the transforms are defined for infinite-length sequences.

Second, and the most important, they are functions of continuous

variables ( or z).

From the numerical computation viewpoint (or from MATLAB's

viewpoint), these two features are troublesome because one has

to evaluate infinite sums at uncountably infinite frequencies

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Introduction (contd..)

To use MATLAB, we have to truncate sequences and then

evaluate the expressions at finitely many points, which are the

approximations to the exact calculations. In other words, the

discrete-time Fourier transform and the z-transform are not

numerically computable transforms.

Therefore we need a numerically computable transform, which is

obtained by sampling the discrete-time Fourier transform in the

frequency domain. This is known Discrete Fourier Transform.

From Fourier analysis we know that a periodic sequence can

always be represented by a linear combination of harmonically

related complex exponentials (which is a form of sampling). This

gives us the Discrete Fourier Series (DFS) to finite-duration

sequence which leads to a new transform, called DFT.

ECR 305_L8


Discrete Fourier Series

Let us consider a periodic sequence

where N is the fundamental period of the sequence.

From Fourier analysis, we know that the periodic function can be

synthesized as a linear combinations of complex exponentials

whose frequencies are multiples (or harmonics) of the

fundamental frequency (which in our case is 2/N).

From the frequency-domain periodicity of the discrete-time

Fourier transform, we conclude that there are a finite number of

harmonics; the frequencies are { , k = 0, 1, 2,……, N–1}.

kNnxnx ~~

kN

2

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Discrete Fourier Series (contd..)

Therefore, the discrete Fourier series representation of a periodic

sequence can be expressed as

where are called the discrete

Fourier series coefficients, which are given by

1,...,2,1,0,~~21

0

NkenxkXnk

Nj

N

n

1,...,2,1,0,~1~

21

0

NnekXN

nxkn

Nj

N

k

nx~

1,...,2,1,0,~

NkkX

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IDFS

DFS



Using , previous 2 equations can be expressed as

Analysis or a DFS equation:

Synthesis or an inverse DFS equation:

Nj

N eW

2

nk

N

N

n

WnxnxkX ~~DFSˆ~ 1

0

nk

N

N

k

WkXN

kXnx

~1~

IDFSˆ~1

0

ECR 305_L8



In matrix form, the DFS and IDFS equations can be rewritten as

where the matrix called a DFS matrix given by

NNN xWX NNNNN

NXWXWx

*1- 1

211

11

10

1

1

111

N

N

N

N

N

NN

kn

NnkNN

WW

WWk

n

W

...

...::

...

...

ˆW ,

NW

ECR 305_L8

1-

1

0

Nx

x

x

N:

x

1

1

0

N

X

X

N

X

:X


Example-1

Find DFS representation of .

Solution: Given N = 4. So

Now

Hence

Similarly,

,.....3,2,1,0,3,2,1,0,3,2,1,0....,~ nx

32104

3

0

,,,,~~

kWnxkX nk

n

jjeeW jj 2/sin2/cos2/4/2

4

632103~2~1~0~~0~ 0

3

0

xxxxjnxXn

n

jjjjjnxXn

n

223210~1~ 32

3

0

23210~2~ 6422

3

0

jjjjnxXn

n

jjjjjnxXn

n

223210~3~ 9633

3

0

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Relation to z-transform

Construct a periodic sequence by periodically repeating x(n)

with a period N,

The DFS of is given by,

Comparing with z-transform, ,we have

which means that DFS represents N evenly spaced samples

of the z-transform X(z) around the unit circle.

nx~

nx~

elsewhere

10

,0

,~

Nnnx

nx

n

kN

jN

n

nkN

jN

n

enxenxkX

21

0

21

0

~~

nN

nznxzX

~1

0

kN

j

ezzXkX 2

~

kX~

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Relation to DTFT

The DTFT of x(n) is given by

Comparing DFS & IDFS equations, we have

Let and , then the DFS

which means that the DFS is obtained by evenly sampling the

DTFT with the sampling interval .

njN

n

enxX

1

0

k

Nk

N

njN

n

nkN

jN

n

XenxenxkX

2

2

1

0

21

0

~~~

N

2ˆ1 1

2ˆ

kk

Nk

1 jkjeXeXkX k

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Discrete Fourier transform

The discrete Fourier series provided us a mechanism for numerically

computing the discrete-time Fourier transform. It also alerted us to a potential

problem of aliasing in the time domain.

Mathematics dictates that the sampling of the discrete-time Fourier transform

result in a periodic sequence x(n). But most of the signals in practice are not

periodic. They are likely to be of finite duration. How can we develop a

numerically computable Fourier representation for such signals?

Theoretically, we can take care of this problem by defining a periodic signal

whose primary shape is that of the finite-duration signal and then using the

DFS on this periodic signal.

Practically, we define a new transform called the Discrete Fourier Transform

(DFT), which is the primary period of the DFS. This DFT is the ultimate

numerically computable Fourier transform for arbitrary finite-duration

sequences.

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Discrete Fourier transform (contd..)

Discrete Fourier transform (DFT) of an N-point sequence:

or

Inverse discrete Fourier transform (IDFT) of an N-point

sequence:

elsewhere

10

,0

,~

DFTˆ

NkkX

nxkX

10,1

0

NkWnxkX nk

N

N

n

10,1~

1

0

NnWkXN

nx nk

N

N

k

ECR 305_L8


Properties of DFT

Linearity:

Circular folding: If an N-point sequence is folded, then the

result would not be an N-point sequence, and it would not be

possible to compute DFT. Therefore, we use the modulo-N

operation on the argument (-n) and define folding by

The DFT of a circular folding is given by

nxbnxanbxnax 2121 DFTDFTDFT

11

0

,

,0

Nn

n

nNx

xnx N

11

0

,

,0DFT

Nk

k

kNX

XkXnx NN

ECR 305_L8


Properties of DFT (contd..)

Circular shifting: An N-point DFT of a finite duration sequence

x(n) of length N is equivalent to the N-point DFT of a periodic

sequence xp(n), which is obtained by periodically extending x(n),

Now suppose that we shift xp(n) by k units to the right,

Then the circular shift is

Note: Counter clockwise +ve, clockwise -ve direction.

r

p rNnxnx

ECR 305_L8

r

pp rNknxknxnx

otherwise

10

0

Nnnxnx

p

,

,


Properties of DFT (contd..)

Circular convolution: Consider 2 N-point sequences x1(n) and

x2(n). The respective N-point DFTs are

Then

Then the IDFT of X3(k), called the circular convolution, given by

ECR 305_L8

1210

2

1

1

0

1

NkenxkXnk

NjN

n

,...,,,,

1210

2

2

1

0

2

NkenxkXnk

NjN

n

,...,,,,

kXkXkX 213

1,...,2,1,01

0

213

NmnmxnxmxN

n

N


Example-2

Determine the circular shifting of the sequence by 2 units:

Solution:

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4321 ,,,nx


Example-2 (contd..) ECR 305_L8


Example-3

Determine the circular folding of the sequence:

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135797531 ,,,,,,,,nx


Example-4

Determine the circular convolution of the following two

sequences:

Solution:

ECR 305_L8

12121 ,,,nx

43212 ,,,nx

1,...,2,1,0where

1

0

213

Nm

nmxnxmxN

n

N

16,14,16,143 mx


Example-4 (contd..) ECR 305_L8

1,...,2,1,0where

1

0

213

Nm

nmxnxmxN

n

N

16,14,16,143 mx


DCT & IDCT ECR 305_L8

The DCT is given by

The IDCT is given by

where

N

jn

N

imnmSjCiC

NjiT

N

m

N

n 2

12cos

2

12cos,

2,

1

0

1

0

otherwise

0

12

1

k

kC

N

jn

N

imjiTjCiC

NnmS

N

j

N

i 2

12cos

2

12cos,

2,

1

0

1

0


Fast Fourier Transform

Although DFT is a computable transform, the numerical

computation of the DFT is time consuming and its straightforward

implementation is very inefficient especially for large sequences.

Therefore, several algorithms have been developed to efficiently

compute the DFT. These are collectively called Fast Fourier

Transform (FFT) algorithms.

In 1965, Cooley and Tukey showed a procedure to substantially

reduce the amount of computations involved in the DFT, which

led to the explosion of the application of the DFT including in the

digital signal processing area and to the development of other

efficient algorithms. These are collectively known as Fast Fourier

Transform (FFT).

ECR 305_L8


Fast Fourier Transform (contd..)

10,1

0

NkWnxkX nk

N

N

n

Consider an N-point sequence x(n). Its N-point DFT is given by

where . To obtain one sample of X(k), we need N complex

multiplications and (N–1) complex addition.

Hence to obtain a complete set of DFT coefficients, we need N2

complex multiplications and N(N-1) N2 complex additions. Also one

has to store N2 complex coefficients or generate internally at an

extra cost.

Therefore, the number of DFT computation for an N-point sequence

depends quadratically on N, i.e. N4 . Generally processing time for one

addition is much less than for one multiplication.

Njnk

N eW /2

nk

NW

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Fast Fourier Transform (contd..)

Goal of an efficient computation: In an efficiently designed

algorithm the number of computation should be constant per data

sample, and therefore the total number of computation should be

linear with respect to N.

The quadratic dependence of N can be reduced by realizing that

most of the computations (which are done again and again) can

be eliminated using the periodicity property

and the symmetry property

of the factor .

nk

N

Nnk

N WW 2/

Nkn

N

Nnk

N

nk

N WWW

nk

NW

ECR 305_L8


Example-5

Compute a 4-point DFT of x(n) = {1, 1, 1, 1} and develop an

efficient algorithm for its computation.

The above computation can be done in the matrix form

which requires 16 complex multiplications.

Efficient approach: Using periodicity

jeWkWnxkX jnk

n

4/2

44

3

0

;30,

3

2

1

0

3

2

1

0

9

4

6

4

3

4

0

4

6

4

4

4

2

4

0

4

3

4

2

4

1

4

0

4

0

4

0

4

0

4

0

4

x

x

x

x

WWWW

WWWW

WWWW

WWWW

X

X

X

X

jWWW

jWWWW

3

4

6

4

2

4

9

4

1

4

4

4

0

4

;1

;1

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Example-5 (contd..)

Putting these values into previous equation, we have

Using symmetry we obtain

3

2

1

0

11

1111

11

1111

3

2

1

0

x

x

x

x

jj

jj

X

X

X

X

21

21

21

21

312032103

312032102

312032101

312032100

hh

gg

hh

gg

xxjxxjxxjxxX

xxxxxxxxX

xxjxxjxxjxxX

xxxxxxxxX

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4-point DFT (contd..)

Hence an efficient algorithm is

These requires only 2 complex multiplications, which is a

considerably smaller number. A signal flow graph structure of

this algorithm is shown below:

212

211

212

211

331

220

131

020

21

jhhXxxh

ggXxxh

jhhXxxg

ggXxxg

stepstep

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Assignment-7 (due on next class)

Problems : 7.8, 7.12, 7.14, 7.17, 7.24, 7.25

ECR305_L8


MATLAB implementation

MATLAB

Examples

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MATLAB implementation of DFS

The DFS is a numerically computable representation for which

we need two nested for … end loop (one for summation and

another for DFS coefficients), which is not efficient in MATLAB.

An efficient implementation in MATLAB would be to use a

matrix-vector multiplication same as we used for DTFT.

Let and denote the column vectors corresponding to the

primary periods of sequences and , respectively.x~ X~

nx~ kX~

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MATLAB implementation of DFS (contd..)

Then DFS and IDFS equations can be rewritten as

where the matrix called a DFS matrix given by

xWX ~~N

XWx* ~1~N

N

211

11

1,0

...1

...::

...1

1...11

ˆ

N

N

N

N

N

NNkn

NnkNN

WW

WWk

n

WW

NW

ECR 305_L8


dfs function

function [Xk] = dfs(xn,N)

% -------------------------------------------------------------------------

% Compute discrete Fourier series coefficients

% Xk : DFS coefficient array over 0 <= k <= N-1

% xn : one period of periodic signal over 0 <= n <= N-1

% N : fundamental period of xn

% -------------------------------------------------------------------------

n = 0:N-1; % row vector for n

k = 0:N-1; % row vector for k

WN = exp(-j*2*pi/N); % Wn factor

nk = n'*k; % N by N matrix of nk values

WNnk = WN.^nk; % DFS matrix

Xk = xn*WNnk; % row vector for DFS coefficient

% ------------------------------------------------------------------N

j

N eW

2

xWX ~~N

kn

NnkNN W 1,0ˆ W

ECR 305_L8


idfs function

Nj

N eW

2

kn

NnkNN W 1,0ˆ W

XWx* ~1~N

N

function [xn] = idfs(Xk,N)

%=============================================

% Compute inverse discrete Fourier series coefficients

% Xk : DFS coefficient array over 0 <= k <= N-1

% one period of periodic signal over 0 <= n <= N-1

% N : fundamental period of xn

% -------------------------------------------------------------------------





WNnk = WN.^(-nk); % DFS matrix

xn = (1/n).*(Xk*WNnk); % row vector for DFS coefficient

% -------------------------------------------------------------------------

ECR 305_L8


MATLAB example 5.2b

A periodic square wave sequence is given by

where N is the fundamental period and L/N is the duty cycle.

Plot the magnitude for L = 7 and N = 60.

11

1

0

,1~

NmnLmN

LmNnmNnx

kX~

ECR 305_L8


MATLAB example 5.2b (contd..)

% File name: ex5p2.m

% Definition

L = 7; N = 60; k = -N/2:N/2; xn = [ones(1,L), zeros(1,N-L)];

% Calculation

Xk = dfs(xn,N); magXk = abs([Xk(N/2+1:N) Xk(1:N/2+1)]);

% Plotting

Subplot(2,1,1); stem(k, magXk);

title('DFS of square wave'); xlabel('k'); ylabel('Xtilde(k)');

ECR 305_L8


MATLAB implementation of DFT

Then DFT and IDFT equations can be rewritten as

where the matrix called a DFT matrix given by

xWX N

XWx*

NN

1

211

11

1,0

...1

...::

...1

1...11

ˆ

N

N

N

N

N

NNkn

NnkNN

WW

WWk

n

WW

NW

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dft function

function [Xk] = dft(xn,N)

% -------------------------------------------------------------------------

% Compute discrete Fourier transform

% Xk : DFT coefficient array over 0 <= k <= N-1

% xn : N-point finite duration sequence

% N : Length of DFT

% -------------------------------------------------------------------------





WNnk = WN.^nk; % DFT matrix

Xk = xn*WNnk; % row vector for DFT coefficient

% ------------------------------------------------------------------N

j

N eW

2

xWX N

kn

NnkNN W 1,0ˆ W

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idft function

Nj

N eW

2

kn

NnkNN W 1,0ˆ W

XWx*

NN

1

function [xn] = idft(Xk,N)

%=============================================

% Compute inverse discrete Fourier transform

% Xk : DFT coefficient array over 0 <= k <= N-1

% xn: N-point sequence over 0 <= n <= N-1

% N : Length of DFT

% -------------------------------------------------------------------------





WNnk = WN.^(-nk); % DTS matrix

xn = (1/N).*(Xk*WNnk); % row vector for DFT coefficient

% -------------------------------------------------------------------------

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MATLAB example 5.6

Let x(n) be an N-point sequence:

(a) Compute the DTFT and plot its magnitude and phase.

(b) Compute the 4-point DFT of x(n).

otherwise

30

,0

,1

n

nx

% Definition

n1=0; n2=3; n=n1:n2; M = 200; k = 0:M;

w=(4*pi/M)*k; N = 4; m = 0:N-1;

x = [1,1,1,1]; xn = [1,1,1,1];

% DTFT

X = dtft(x,n,k); magX=abs(X);

angX=angle(X)*180/pi;

% DFT

Xk = dft(xn,N); magXk = abs(Xk);

PhaXk = angle(Xk)*180/pi;

% Plotting

Subplot(2,1,1); stem(m,magXk);

title('Magnitude','fontsize',15);

xlabel('k','fontsize',15);

ylabel('X(k)','fontsize',15);

grid; hold on

plot(w/pi,magX,'--'); axis([0 4 0 5]);

Subplot(2,1,2); stem(m,PhaXk);

title('Angle','fontsize',15);


ylabel('Degrees','fontsize',15);

grid; hold on

plot(w/pi,angX,'--'); axis([0 4 -200 200]);

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MATLAB example 5.6 (contd..) ECR 305_L8


MATLAB example 5.7

How can we obtain further samples of X(ej)?

Ans:

% Definition

n1=0; n2=7; n=n1:n2; M = 200; k = 0:M;

w=(8*pi/M)*k; N = 8; m = 0:N-1;

x = xn = [ones(1,4),zeros(1,4)];

% DTFT

X = dtft(x,n,k); magX=abs(X);

angX=angle(X)*180/pi;

% DFT

Xk = dft(xn,N); magXk = abs(Xk);


% Plotting

Subplot(2,1,1); stem(m,magXk);




grid; hold on

plot(w/pi,magX,'--'); axis([0 4 0 5]);

Subplot(2,1,2); stem(m,PhaXk);




grid; hold on

plot(w/pi,angX,'--'); axis([0 4 -200 200]);

0,0,0,0,1,1,1,1nx <= By padding zeros

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MATLAB example 5.7 (contd..) ECR 305_L8


MATLAB example 5.7 (contd..)

How can we obtain further samples of X(ej)?

Ans:

0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1nx

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MATLAB problem 5.12

(a) Design a MATLAB function to implement an N-point circular

folding operation x2(n) = x1((-n))N.

(b) determine the circular folding of the sequence:

x1(n) = {1,3,5,7,9,-7,-5,-3,-1}

function x2 = circfold(x1,N)

n = 0:N-1;

x2 = x1(mod(-n,N)+1);

% Function

x1 = [1,3,5,7,9,-7,-5,-3,-1];

N = length(x1);

% Circular fold of x1

x2 = circfold(x1,N);

% Plotting

subplot(2,1,1)

stem(0:N-1,x1); title ('x_1(n)','fontsize',15);

xlabel('n','fontsize',15); ylabel('x(n)','fontsize',15);

subplot(2,1,2)

stem(0:N-1,x2); title ('x_2(n)','fontsize',15);

xlabel('n','fontsize',15);

ylabel('x((-n))_N','fontsize',15);

ECR 305_L8


MATLAB problem 5.12 (contd..) ECR 305_L8


MATLAB example

Let x(n) = 10(0.9)n, 0 n 10. Plot x((-n))11.

% File name: 5p12b.m

% Function x

n = 0:10;

x = 10*(0.9).^n;

% Circular fold x

y = circfold(x,11);

% DFT of y

Y = dft(y,11);

% Plotting

subplot(2,1,1)

stem(n,real(Y))

title ('Real{DFT[(x((-n))_1_1]}','fontsize',15);


subplot(2,1,2)

stem(n,imag(Y))

title ('Imaginary{DFT[(x((-n))_1_1]}','fontsize',15);


ECR 305_L8


MATLAB example (contd..) ECR 305_L8


MATLAB example 5.6-fft

Let x(n) be an N-point sequence:

(a) Compute the DTFT and plot its magnitude and phase.

(b) Compute the 4-point DFT of x(n).

otherwise

30

,0

,1

n

nx

% File name: ex5p6fft.m

% Definition

N = 4; m = 0:N-1;

% Function

xn = [1,1,1,1];

% Calculation

Xk = fft(xn,N);

magXk = abs(Xk);


% Plotting

Subplot(2,1,1);

stem(m,magXk');




Subplot(2,1,2);

stem(m,PhaXk');




ECR 305_L8


MATLAB example 5.6-fft (contd..) ECR 305_L8


References

1. John G. Proakis, Digital Signal Processing, Pearson, 4th

Edition, Seventh Impression, 2011. (pp. 449–562)

2. Vinay K. Ingle, and John G. Proakis, Digital Signal

Processing using MATLAB, Thomson Learning

Bookware Companion Series, 2007. (pp. 118–185)

ECR 305_L8


Next class..

Digital Filter

Structure

ECR 305_L8

Lecture 8 Objectives - eee.iub.edu.bd

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