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LOGO

LAB 6

DFT

iugaza2010.blogspot.commelasmer@gmail.com

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Given: [2,2,3,1,8]

and ( ) ( ) * , k=1,2,3,4,5

Write a Matlab code to get the value of y

n

x

y k x n k

QUIZ

x=[2,2,3,1,8];

y=zeros(1,5);

for k=1:5;

Sum=0;

for n=1:5;

y(k)=y(k)+x(n)*k;

end

y1(k)=sum;

end

y = 16 32 48 64 80

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21

100

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100

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100

2

100100

2 k

N

k

1 2 3

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DTFT DFT

Discrete Fourier Transform (DFT)

( ) ( ) ( ) | ( ) j

j j n

z e n

DTFT X e X z x n e

2

1( ) ( ) ( )

2

j j ninverse DTFT x n X e e d

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0

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(DFT) ( ) ( ) , k=0,1,2,.......,N 1

1(inverse DFT) x(n)= ( ) , n=0,1,2,.......,N 1

nkN jN

n

nkN jN

k

X k x n e

X k eN

2

( ) ( ) |k

N

X k X

Write your own Matlab function to implement

the DFT of equation X = DFTsum(x)

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k=0,1,2,..(DFT) ( ) ( ) , .....,N 1 nkN j

N

n

X k x n e

Part1: DFTsum and IDFTsum

LOGOfunction Y=DFTsum(x)

N=length(x);

Y=zeros(1,N);

for k=0:N-1;

for n=0:N-1;

Y(k+1)=Y(k+1)+x(n+1).*exp(-2*j*n*k*pi/N);

end

end

•Test your routine DFTsum by computing

X (k) for:

(a) x(n) = {1,2,3,1}

(b) x(n) = cos(2πn/10) for N = 10.

(a) x(n) = {1,2,3,1}

(b) x(n) = cos(2πn/10) for N = 10.

(b)

n=0:9

x=cos(2*pi*n/10);

XK=DFTSum(x)

stem(n,abs(XK))

%XK=fft(x)

(a)

n=0:3

x=[1,2,3,1];

XK=DFTSum(x)

stem(n,abs(XK))

%XK=fft(x)

XK={ 7 , -2- 1i , 1 , -2+ 1i}

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1 2 3

2 3

2 32 2 2

2

2

x(n)= (n)+2 (n 1)+3 (n 2)+ (n 3)

X(z)=1+2Z +3Z Z

X( ) | =1+2e 3e e

( ) ( ) | ( ) | 1+2e 3e e , k=0,1,2,3

j

j j j

z e

k k kj j j

kk

N

X k X X

DTFT :

DFT :

x(n)={ ,1 2,3,1}

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32 2

32 2

( ) 1+2e 3e e , k=0,1,2,3

1 2 3 1 7

1 1+2e 3e e 2

2 1

3 2

( ) {7, 2 ,1, 2 }

k kj j

j k

j jj

X k

X(k = 0) =

X(k = ) = j

X(k = ) =

X(k = ) = j

X k j j

Write a matlab function for computing the inverse DFT.

x = IDFTsum(X)

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1(inverse DFT) x(n)= ( ) , n=0,1,2,.......,N 1

nkN jN

k

X k eN

function y=IDFTsum(X)

N=length(X)

y=zeros(1,N);

for n=0:N-1

for k=0:N-1

y(n+1)=y(n+1)+(X(k+1).*exp(2*j*n*k*pi/N)/N);

end

end

XK=[ 7 , -2- 1i , 1 , -2+ 1i]

x=IDFTSum(XK)

%x=ifft(XK)

x = ,21 ,3,1

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(A)

(B) Choose

then Check by Matlab

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Part2: Circular convolution

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Circular symmetries of a sequence

Circular convolution

The circular convolution is very similar to normal

convolution apart from that the signal is shifted using

circular shift.

3 1 2

1

3 1 2

0

3 1 2

( ) ( ) ( )

( ) ( ) (( ))

( ) ( ) ( )

N

N

n

x m x n x n

x m x n x m n

X k X k X k

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1 2

Determine the circular convolution

x (n)={ ,2,3,1} , 1 4 x (n)={ ,3,2,2}

3 1 2

3 1 2

( ) ( ) ( )

( ) ( ) ( )

x m x n x n

X k X k X k

fft(x1) fft(x2)

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3 1 2 3 1 2

1

2

3 1 2

3 1 2

3 1 2

3 1 2

( ) ( ) ( ) ( ) ( ) ( )

( ) { 7 , 2 , 1 , 2 }

( ) { 11 , 2 , 1 , 2 }

(0) (0) (0) (7)(11) 77

(1) (1) (1) ( 2 )(2 ) 5

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

(3) (3) (3) (

x m x n x n X k X k X k

X k j j

X k j j

X X X

X X X j j

X X X

X X X

3

3

( ) { 77 , 5 , 1 ,

( ) {17,19,

2 )(

5 }

22,19}

2 ) 5

X

x m

j

k

j

ifft(X)

x1=[1 2 3 1] ; % sequence x1(n)

x2=[4 3 2 2] ; % sequence x2(n)

X1=fft(x1) ; % DFT of x1(n)

X2=fft(x2) ; % DFT of x2(n)

X=X1.*X2 ; % DFT of x(n)

x3=ifft(X) ; % IDFT of X(k)

3

1

3

2

3

1

2

[7 , 2 , 1 , 2 ] ( )

[ 11 , 2

( ) [17,19,22,19

[

, 1

77 ,

,

5 , 1 , 5 ]

2 ] (

(

)

]

)

X j j X k

X j j X

X X k

x

k

m

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x1=[1 2 3 1] ;x2=[4 3 2 2] ;cconv(x1,x2,4)

Or

ans = 17 19 22 19

Note: try to find the result of:

cconv(x1,x2) = ??????

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Let: h(n)={ ,2,3} , x(n)={ ,2,2,1} , f ind

(a) h(n)*x(n) using conv

(b) h(n) x(n) using fft and ifft

(c) h(n) x(n) using DFTsum and IDFTsum

(d) h(n) x(n) using c n

1

o

1

c v

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3 1 2

3 1 2

( ) ( ) ( )

( ) ( ) ( )

x m x n x n

X k X k X k

3 1 2

3 1 2

( ) ( )* ( )

( ) ( ) ( )

x n x n x n

X z X z X z

Part3:

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x1=[1 2 3 1] ;x2=[4 3 2 2] ;conv(x1,x2)

ans = 4 11 20 19 13 8 2

x1=[1 2 3 1] ;x2=[4 3 2 2] ;cconv(x1,x2,4)

ans = 17 19 22 19

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x1=[1 2 3 1 0 0 0] ;x2=[4 3 2 2 0 0 0] ;cconv(x1,x2,7)

ans = 4 11 20 19 13 8 2

Zero padding

Circular convolution gives the same results as linear convolution

L=4M=4L+M-1=7Length of x1 and x2=7

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DFT repeats itself every N points (Period = N) but we usually display it for n = 0 ,…, N-1

Part4: Periodicity of DFT

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Our plot of the DFT has two disadvantages

•Plotting of the DFT

(1) The DFT values are plotted against k rather than

the frequency w.

(2) DFT goes from 0 to 2π rather than from -π to π

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n=0:127

x=cos(2*pi*n/10);

y=fft(x)

stem(n,abs(y))

N=128;

k=0:127;

w=2*pi*k/N;

stem(w,abs(y))

w(w>=pi)=w(w>=pi)-2*pi

w=sort(w)

y2=fftshift(y)

Stem(w,abs(y2))

:Example

n=0:127

x=cos(2*pi*n/10);

y=fft(x)

stem(n,abs(y))

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0 127k

N=128;

k=0:127;

w=2*pi*k/N;

stem(w,abs(y))

0 1 2 3 4 5 6 70

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0 2 Part1

w(w>=pi)=w(w>=pi)-2*pi

w=sort(w)

y2=fftshift(y)

Stem(w,abs(y2))

Part2

-4 -3 -2 -1 0 1 2 3 40

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>>y=[1 2 3 4 5 6]

>>fftshift(y)

4 5 6 1 2 3

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0 1 2 3 4 5 6 70

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n=[1,2,3,4,5,6];

y=[30,20,10,5,15,20];

stem(n,y)

n2=[1,2,3,-2,-1,0] % n(n>3)-6

n3=[-2,-1,0 ,1 ,2 ,3] % sort(n2)

y2=[5 ,15,20,30,20,10]; % fftshift(y)

stem(n3,y2)

NOTE

>> x=[3,2,1];

>> sort(x)

ans =1 2 3

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0 1 2 3 4 5 6 70

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-3 -2 -1 0 1 2 3 40

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( )y n

2 3( )y n

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LOGO

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