Dsp Using Matlab® - 7

Lecture 7: The Discrete Fourier Transform

Mr. Iskandar Yahya

Prof. Dr. Salina A. Samad

IntroductionDiscrete-time Fourier Transform

provides the frequency-domain (w) representation for absolutely summable sequences Hence, useful signals in practice for which the discrete-time

Fourier transform does not exist [ex) u(n), nu(n)] The transient response of a system due to initial conditions

or due to changing inputs cannot be computed using the discrete-time Fourier transform approach

z-Transformprovides a generalized frequency-domain (z)

representation for arbitrary sequences Extension of the discrete-time Fourier transform

IntroductionTwo features of these transforms

The transforms are defined for infinite-length sequencesThey are functions of continuous variables (w 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

-> To use MATLAB, we have to truncate sequences and then evaluate the expressions at finitely many points

-> The evaluations were obviously approximations to the exact calculations

-> In other words, the discrete-time Fourier transform and the z-transform are not numerically computable transforms

IntroductionNumerical computable transform

DFS (Discrete Fourier Series)DFT (Discrete Fourier Transform)FFT (Fast discrete Fourier Transform)

Because the numerical computation of the DFT for long sequences is prohibitively time consuming, several algorithms have been developed to efficiently compute the DFT

Discrete Fourier SeriesPeriodic sequence

Discrete Fourier series representation of periodic sequence

synthesis or

an inverse DFS equation

The discrete Fourier series coefficients analysis or an DFS

sequencetheofperiodlfundamentaN

knkNnxnx

:

,),(~)(~

,...,n,W)k(X~

N

...,,n,e)k(X~

N)]k(X

~[IDFS)n(x~

N

k

nkN

N

k

knN

j

101

101

1

0

1

0

2

,...,k,W)n(x~

...,,k,e)n(x~)]n(x~[DFS)k(X~

N

n

nkN

N

n

nkN

j

10

10

1

0

1

0

2

Nj

eWwhere

2

Discrete Fourier SeriesExample 1: Find DFS representation of the

priodic sequence given below:

Solution:

,...},,,,,,,,,,,{...,)n(x~ 321032103210

,...)(x~)(x~)(x~)(x~)j)(n(x~W)n(x~)(X~

j)j)(n(x~W)n(x~)(X~

)j)(n(x~W)n(x~)(X~

j)(x~j)(x~)(x~j)(x~)j)(n(x~W)n(x~)(X~

)(x~)(x~)(x~)(x~W)n(x~)(X~

jsinjcoseW

Nissequenceabovetheofperiodalfundanment

,...,k,W)n(x~)k(X~

nn

n

n

nn

n

n

nn

n

n

nn

n

n

n

n

j

N

n

nkN

632104

223

22

2232101

632100

22

4

10

3

043

0

44

3

033

0

34

3

023

0

24

3

0

3

0

14

3

0

04

4

2

4

1

0

Discrete Fourier SeriesSolution in MATLAB:

We can use for…end loop to implement X(k) summation and compute all DFS coefficients. But this nested two for…end loop is inefficient in MATLAB.

An efficient way is to use a matrix-vector multiplication for each of the relations in the manual solution:

n

k

WW

WW

}W{W

X~

WN

x~x~WX~

)N()N(N

)N(N

)N(NN

Nn,kkn

NN

*NN

1111

1111

10

1

1

111

1

Discrete Fourier SeriesMatlab implementation:

Create a function called dfs:

function [Xk] = dfs(xn,N)

% Computes Discrete Fourier Series Coefficients

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

% [Xk] = dfs(xn,N)

% Xk = DFS coeff. array over 0 <= k <= N-1

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

% N = Fundamental period of xn

%

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

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

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

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

WNnk = WN .^ nk; % DFS matrix

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

Discrete Fourier Series Implementation for example 1 in the main window:

Check with the manual approach done earlier, should be the same.

xn = [0,1,2,3]; N = 4;

Xk = dfs(xn,N)

Xk =

6.0000 -2.0000 + 2.0000i -2.0000 - 0.0000i -2.0000 - 2.0000i

Discrete Fourier SeriesTo implement the synthesis equation, we can write a

function called “idfs”:

function [xn] = idfs(Xk,N)

% Computes Inverse Discrete Fourier Series

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

% [xn] = idfs(Xk,N)

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

% Xk = DFS coeff. array over 0 <= k <= N-1

% N = Fundamental period of Xk

%





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

xn = (Xk * WNnk)/N; % row vector for IDFS values

Discrete Fourier SeriesExample 2: A periodic “square wave” sequence is given by:

N : fundamental period, L/N : duty cycle

A) determine the expression for magnitude:

,...,m;N)m(nLmN

LmNnmN)n(x~ 10

11

1

0

1

)kN

sin(

)LkN

sin(e

ee

ee

e

e

e

e

otherwise

,...N,k

e

e

L

)e(ee)n(x~)n(X~

k)L(Nj

kNjk

Nj

LkNjLk

Nj

kNj

LkNj

kN

j

LkN

j

kN

j

LkN

j

L

nn

kN

jL

n

nkN

jN

n

nkN

j

1

2

2

2

2

1

0

21

0

21

0

2

1

1

0

1

1

otherwise

,...N,k

)N/ksin(

)N/kLsin(L

)n(X~ 0

Discrete Fourier SeriesB) Plot magnitude:

% Part (b)1

L = 5; N = 20;

xn = [ones(1,L), zeros(1,N-L)];

Xk = dfs(xn,N);

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

k = [-N/2:N/2];

subplot(2,2,1); stem(k,magXk); axis([-N/2,N/2,-0.5,5.5])

xlabel('k'); ylabel('Xtilde(k)')

title('DFS of SQ. wave: L=5, N=20')

% Part (b)2

L = 5; N = 40;


Xk = dfs(xn,N);


k = [-N/2:N/2];




% Part (b)3

L = 5; N = 60;


Xk = dfs(xn,N);


k = [-N/2:N/2];




% Part (b)4

L = 7; N = 60;


Xk = dfs(xn,N);


k = [-N/2:N/2];




Discrete Fourier SeriesB) Plot magnitude:

-10 -5 0 5 10

0

1

2

3

4

5

k

Xtild

e(k)

DFS of SQ. wave: L=5, N=20

-20 -10 0 10 20

0

1

2

3

4

5

k

Xtild

e(k)


-20 0 20

0

1

2

3

4

5

k

Xtild

e(k)


-20 0 20

0

2

4

6

k

Xtild

e(k)


Discrete Fourier SeriesRelations to the z-transform

x(n) is a finite-duration sequence of duration N

We construct a periodic sequence by periodically repeating x(n) with period N

1

0

10

0N

nnz)n(x)z(X

elsewhere

Nnnonzero)n(x

)n(x~

nN

n

kN

jN

n

nkN

j]e)[n(xe)n(x~)z(X

)n(x~ofDFS

elsewhere

Nn)n(x~)n(x

1

0

21

0

2

10

0

kN

jez

)z(X)k(X~

2

Discrete Fourier SeriesRelations to the DTFT

DFS is obtained by evenly sampling the DTFT at w1=2/N intervals

Example 3: Let: X(n) = {0,1,2,3} A) Compute DTFT:

B) Sampling at kw1=2/4 k, k=0,1,2,3

kN

w

jw )e(X)k(X~

2

wjwjjwjwnjw eeee)n(x)e(X 32 32

)(X~

jeee)e(X

)(X~

eee)e(X

)(X~

jeee)e(X

eee)e(X

/j/j/j/j

/j/j/j/j

/j/j/j/j

jjjj

32232

2232

12232

632132

4634624646

4434424444

4234224242

030200

Discrete Fourier SeriesDFT

The sampling of the DTFT result in a periodic sequence But most of all 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 x(n) : finite-duration sequence that has N samples over

0<=n<=N-1 as an N-point sequence : periodic signal of period N, created using the N-point

sequence x(n))n(x~

)Nmodn(x)n(x~)rNn(x)n(x~

r

Discrete Fourier SeriesWe use the following convenient notation to denote the

modulo-N operation:

Nlengthofwindowgulartanrecacalledis)n(R

else

Nn)n(x~)n(R)n(x~)n(xwhere

operationWindow)n(R)n(x~)n(x

extentionperiodic))n((x)n(x~)Nmodn(x))n((x

N

N

N

N

N

10

0

Discrete Fourier SeriesDiscrete Fourier Transform of an N-point sequence

Inverse Discrete Fourier Transform of an N-point sequence

xWX

Nk,W)n(x)k(X

or

)k(R)k(X~

else

Nn)k(X~

)]n(x[DFT)k(X

N

N

n

nkN

N

10

10

0

1

0

XWN

x

Nn,W)k(XN

)k(X

or

)n(R)n(x~)]k(X[IDFT)n(x

*N

N

k

knN

N

1

101 1

0

Discrete Fourier Series MATLAB implementation

dfs and idfs MATLAB functions can be renamed as the dft and idft functions to implement the discrete Fourier transform computations

function [Xk] = dft(xn,N)

% Computes Discrete Fourier Transform

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

% [Xk] = dft(xn,N)

% Xk = DFT coeff. 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 codfficients

function [xn] = idft(Xk,N)

% Computes Inverse Discrete Transform

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

% [xn] = idft(Xk,N)

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

% Xk = DFT coeff. array over 0 <= k <= N-1

% N = Length of DFT

%





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

xn = (Xk * WNnk)/N; % row vector for IDFT values

Discrete Fourier Series Example 4: Let x(n) be a 4-point sequence:

A) Then, the DTFT:

B) 4-point DFT of x(n)

this are similar to those in example 1

otherwise

n)n(x

30

0

1

02

2

02

2

2

32

3

2

2

2

2

1

11 23

4323

0

)/wsin(

)wsin(when

)/wsin(

)wsin(when

w

w

)e(X

)/wsin(

)wsin()e(X

e)/wsin(

)wsin(

e

eeeee)n(x)e(X

jw

jw

/wjjw

wjwjwjjwjwnjw

j)/sin(j)/cos(eW

k,W)n(xW)n(x)k(X

/j

n

nkN

n

nkN

4242

30

424

3

0 41

04

% Example 4: Simple DFT Example

x = [1,1,1,1]; subplot(1,1,1);

% a) DTFT

w = [0:1:500]*2*pi/500; [H] = freqz(x,1,w);

magH = abs(H); phaH = angle(H); phaH(126)=-47.5841*pi/180;

subplot(2,2,1); plot(w/pi,magH); grid

xlabel('frequency in pi units'); ylabel('|X|'); title('Magnitude of the DTFT')

subplot(2,2,2); plot(w/pi,phaH/pi*180); grid

xlabel('frequency in pi units'); ylabel('Degrees'); title('Angle of the DTFT')

%print -deps2 me0506a.eps

%

% b) 4-point DFT

N = 4; w1 = 2*pi/N; k = 0:N-1; X = dft(x,N);

magX = abs(X), phaX = angle(X)*180/pi

subplot(2,2,3);plot(w*N/(2*pi),magH,'--'); grid

axis([0,4,0,4]); hold on

stem(k,magX); xlabel('k'); ylabel('|X(k)|'); title('Magnitude of the DFT: N=4')

hold off

subplot(2,2,4);plot(w*N/(2*pi),phaH*180/pi,'--'); grid

axis([0,4,-200,200]); hold on

stem(k,phaX); xlabel('k'); ylabel('Degrees'); title('Angle of the DFT: N=4')

%print -deps2 me0506b.eps

0 0.5 1 1.5 20

1

2

3

4

frequency in pi units

|X|

Magnitude of the DTFT

0 0.5 1 1.5 2-200

-100

0

100

200


Degr

ees

Angle of the DTFT

0 1 2 3 40

1

2

3

4

k

|X(k

)|

Magnitude of the DFT: N=4

0 1 2 3 4-200

-100

0

100

200

k

Degr

ees

Angle of the DFT: N=4

Discrete Fourier Series Matlab implementation:

Discrete Fourier Series Example 5: DTFT other sample X(ejw) ? For N = 8;

4821

70

00001111

48

7

0 81

08

//w:resolutionFrequency

eW;k,W)n(xW)n(x)k(X

},,,,,,,{)n(x

/jn

nkN

n

nkN

% Example 5: Zero-padding Example

subplot(1,1,1); x = [1,1,1,1];

% a) DTFT

w = [0:1:500]*2*pi/500; [H] = freqz(x,1,w);

magH = abs(H); phaH = angle(H); phaH(126)=-47.5841*pi/180;

% b) 8-point DFT

N = 8; w1 = 2*pi/N; k = 0:N-1;

x = [x, zeros(1,4)]; X = dft(x,N);


subplot(2,1,1);plot(w*N/(2*pi),magH,'--');

axis([-0.1,8.1,-1,5]); hold on

stem(k,magX); xlabel('k'); ylabel('|X(k)|'); title('Magnitude of the DFT: N=8')

hold off

subplot(2,1,2);plot(w*N/(2*pi),phaH*180/pi,'--');

axis([-0.1,8.1,-200,200]); hold on

stem(k,phaX); xlabel('k'); ylabel('Degrees'); title('Angle of the DFT: N=8');

0 1 2 3 4 5 6 7 8-1

0

1

2

3

4

5

k

|X(k

)|


0 1 2 3 4 5 6 7 8-200

-100

0

100

200

k

Degr

ees


Discrete Fourier Series For N = 16;

16216

15

0 151

016 150

0000000000001111

/jn

nkN

n

nkN eW;k,W)n(xW)n(x)k(X

},,,,,,,,,,,,,,,{)n(x

% c) 16-point DFT

subplot(1,1,1)

N = 16; w1 = 2*pi/N; k = 0:N-1;

x = [x, zeros(1,8)];

X = dft(x,N);


subplot(2,1,1);plot(w*N/(2*pi),magH,'--');

axis([-0.1,16.1,-1,5]); hold on

stem(k,magX);

xlabel('k');

ylabel('|X(k)|'); title('Magnitude of the DFT: N=16')

hold off

subplot(2,1,2);plot(w*N/(2*pi),phaH*180/pi,'--');

axis([-0.1,16.1,-200,200]); hold on

stem(k,phaX);

xlabel('k');

ylabel('Degrees'); title('Angle of the DFT: N=16')

%print -deps2 me0507b.eps

0 2 4 6 8 10 12 14 16-1

0

1

2

3

4

5

k

|X(k

)|


0 2 4 6 8 10 12 14 16-200

-100

0

100

200

k

Degr

ees


Discrete Fourier Series Example 6: To illustrate the difference between the high-density

spectrum and the high-resolution spectrum:

A) Determine the

)n.cos()n.cos()n(x 520480

100 n),n(xofDFT

% Chapter 05: Example 6: HiRes Spectrum

%

% a.

% Spectrum based on the first 10 samples of x(n)

n1=[0:1:9];y1=x(1:1:10);

subplot(2,1,1);stem(n1,y1);title('signal x(n), 0 <= n <= 9');xlabel('n')

axis([0,10,-2.5,2.5])

Y1=fft(y1);magY1=abs(Y1(1:1:6));

k1=0:1:5;w1=2*pi/10*k1;

subplot(2,1,2);stem(w1/pi,magY1);title('Samples of DTFT Magnitude');

xlabel('frequency in pi units')

axis([0,1,0,10])

disp('Press RETURN to continue');pause;

%print -deps2 me0508a.eps; subplot

%

0 1 2 3 4 5 6 7 8 9 10

-2

-1

0

1

2

signal x(n), 0 <= n <= 9

n

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10Samples of DTFT Magnitude


Discrete Fourier Series% High density spectrum (100 samples) based on the first 10 samples of x(n)

n3=[0:1:99];y3=[x(1:1:10) zeros(1,90)];

subplot(2,1,1);stem(n3,y3);title('signal x(n), 0 <= n <= 9 + 90 zeros');xlabel('n')

axis([0,100,-2.5,2.5])

Y3=fft(y3);magY3=abs(Y3(1:1:51));

k3=0:1:50;w3=2*pi/100*k3;

subplot(2,1,2);plot(w3/pi,magY3);title('DTFT Magnitude');xlabel('frequency in pi units')

axis([0,1,0,10])


%print -deps2 me0508b.eps; subplot

% b.

% High resolution spectrum based on 100 samples of the signal x(n)

subplot(1,1,1)

n=[0:1:99];

x=cos(0.48*pi*n)+cos(0.52*pi*n);

subplot(2,1,1);stem(n,x);title('signal x(n), 0 <= n <= 99');xlabel('n')

axis([0,100,-2.5,2.5])

X=fft(x);magX=abs(X(1:1:51));

k=0:1:50;w=2*pi/100*k;

subplot(2,1,2);plot(w/pi,magX);title('DTFT Magnitude');xlabel('frequency in pi units')

axis([0,1,0,60])


%print -deps2 me0508c.eps; subplot

Discrete Fourier Series

0 10 20 30 40 50 60 70 80 90 100

-2

-1

0

1

2

signal x(n), 0 <= n <= 9 + 90 zeros

n

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10DTFT Magnitude


0 10 20 30 40 50 60 70 80 90 100

-2

-1

0

1

2

signal x(n), 0 <= n <= 99

n

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

20

40

60DTFT Magnitude


Fast Fourier Transform (FFT)DFT

DFT is the only transform that is discrete in both the time and the frequency domains,

and is defined for finite-duration sequences Although it is a computable transform, the straightforward

implementation is very inefficient, especially when the sequence length N is large

It also led to the development of other efficient algorithms All these algorithms are collectively known as fast Fourier

Transform (FFT) algorithms N-point DFT

For large N, o(N2) is unacceptable in practice

)N(oC

eWwhere

1Nk0,W)n(x)k(X

2N

N

2j

1N

0n

nkN

Fast Fourier Transform (FFT)Goal of an efficient computation

One algorithm that considers only the periodicity of is the Goertzel algorithm. This algorithm still requires multiplications, but it has certain advantages. This algorithm is described in Chapter 10

DIT-FFT (decimation-in-time) and DIF-FFT (decimation-in-frequency) Symmetry and periodicity properties in reducing the number

of computations)NlogN(oCN

Fast Fourier Transform (FFT)Example: Let us discuss the computations of a 4-point DFT

and develop an efficient algorithm for its computation:

Can be done in matrix form:

which requires 16 complex multiplicationsEfficient Approach : Using periodicity

And substituting in the above matrix form, we get

jeW;3k0,W)n(x)k(X 4/2j4

3

0n

nkN

)3(x

)2(x

)1(x

)0(x

WWWW

WWWW

WWWW

WWWW

)3(X

)2(X

)1(X

)0(X

94

64

34

04

64

44

24

04

34

24

14

04

04

04

04

04

jW;1WW

jWW;1WW3

46

42

4

94

14

44

04

)3(x

)2(x

)1(x

)0(x

j1j1

1111

j1j1

1111

)3(X

)2(X

)1(X

)0(X

Fast Fourier Transform (FFT)Using symmetry, we obtain

Which requires only 2 complex multiplications, which is a considerably smaller number, even for this simple example.

21

21

21

21

2

1

2

1

21

21

21

21

jhh)3(X

gg)2(X

jhh)1(X

gg)0(X

)3(x)1(xh

)2(x)0(xh

)3(x)1(xg

)2(x)0(xg

2step1step

hh

)]3(x)1(x[j)]2(x)0(x[)3(jx)2(x)1(jx)0(x)0(X

gg

)]3(x)1(x[)]2(x)0(x[)3(x)2(x)1(x)0(x)0(X

hh

)]3(x)1(x[j)]2(x)0(x[)3(jx)2(x)1(jx)0(x)0(X

gg

)]3(x)1(x[)]2(x)0(x[)3(x)2(x)1(x)0(x)0(X

Dsp Using Matlab® - 7

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