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ilolu 1

GETTING THE RESULT OF LINEAR CONVOLUTION USING DFT

Let []and []be -point and -point sequences and []be their linearconvolution.

Let[]and []be -point DFTs of []and [].The IDFT of [] [][]is the same as the linear convolution of []and[]

if

+ 1.

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ilolu 2

Ex:Let []and []be 5-point and 9-point sequences, respectively. At leasthow many point DFT has to be used to get the result of their linear convolution?

Answer: 5+9-1=13 point DFT.

However, if, for example, 10-point DFT used then

The result has 10 samples in total

The first 13-10 = 3 samples out of these 10 samples are not equal to the

those of the linear convolution

The remaining 10-3 samples are equal to those of the linear convolution.

Therefore only 7 samples of the result of linear convolution (13 samples)

will be obtained.

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ilolu 3

MATLAB Example

x = [1 2 3]; y = [-1 2 1 2]; z = conv(x,y); z = -1 0 2 10 7 6

X = fft(x,6)

X = 6 0.5-4.3i -1.5 + 0.9i 2 -1.5-0.9i 0.5+4.3i

Y = fft(y,6)

Y = 4 -2.5 - 2.6i -0.5 - 0.9i -4 -0.5 +0. 9i -2.5 + 2.6i

Z=X.*Y

Z =24 -12.5 +9.5i 1.5 + 0. 9i -8 1.5 -0. 9i -12.5 - 9.5i

z_=ifft(Z) z = -1 0 2 10 7 6

X = fft(x,5)

X =6 -0.8 -3.7i 0.3+1.7i 0.3-1.7i -0.8+3. 7i

Y = fft(y,5)

Y =4 -2.8 - 1.3i -1.7 - 2.1i -1.7 + 2.1i -2.8 + 1.3i

Z = X.*Y

Z =24 -2.5 + 11.4i 3.0 - 3.5i 3.0 + 3.5i -2.5 - 11.4i

z_= ifft(Z) z_ = 5 0 2 10 7

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ilolu 4

7) Sampling the DTFT

XN

jFTD

eXT

nx

length

Let

1,...,1,0 2

NkeXkXN

k

j

where is arbitrary, i.e. not necessarily !

[] ? ? ?(

-point IDFT of

[])

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ilolu 5

[] []= 0 , 1 , , 10 ..

[]= []

= =

[] = []=

[] [ ]=

[ ]

=

Therefore

..0

1,...,1,0

wo

NnmNnxnx m

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ilolu 6

Ex: 6, 4

56 3 3 3

2

2 2 2

1 sin 3

1sin

2

j j j jj

j

jj j j

e e e eX e e

ee e e

For N = 4

0 6 1 1 2 0 3 1X X j X X j

and

40

: 0 0 2 2 1 1 0 0n

x n

x n

5

m

mnx 4

5

5

nx

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ilolu 7

Ex: Demonstration of the proof. 8 , 5

n

nj

n

njj

enx

enxeX

7

0

4

5

3

5

2

5

5

7

5

6

5

5

5

4

5

3

5

2

55

5

5

2

4

3

72

61

50

76543210

2

5

1

5

0

5

k

k

k

k

W

k

W

k

W

kkkkk

n

kn

k

j

Wx

Wx

Wxx

Wxx

xx

WxWxWxWxWxWxWxx

Wnx

eXkX

kkk

This is the 5-point DFT of

44

33

722

611

500

xx

xx

xxx

xxx

xxx

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ilolu 8

8) Multiplication in Time Domain

DFS

Let 1x n and 2x n be periodic with common period N.

1

DFS

3 1 2 3 1 2

0

1 N

l

x n x n x n X k X l X k lN

DFT

Let 1x n and 2x n be two finite length sequences and 1X k and 2X k their N-

point DFTs, 1 2max ,N N N .

1

-point DFT

3 1 2 3 1 2

0

10,1,..., 1

N

N

Nl

x n x n x n X k X l X k l k NN

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ilolu 9

IMPLEMENTING LTI SYSTEMS USING DFT

The output of an LTI system can be computed using DFT:

Multiply the DFTs of input and impulse response and compute the IDFT

There are efficient DFT computation techniques.

We consider FIR systems with long inputs.

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ilolu 10

BLOCK PROCESSING

Input is decomposed into short segments and the output is computed by

properly combining the responses to the short segments.

Otherwise, it is not practical to consider DFTs of very long sequences.

Furthermore the output of the system will be delayed excessively since all the

input has to be collected first..

h[n]: length P.

The input is decomposed into blocks of length L.

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Two Methods

Overlap-Add

DFT length: L+P-1 point

Overlap-Save

DFT length: Lpoint

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OVERLAP-ADD METHOD

Lets start with a linear convolution example

,...0,0,3,2,1,0,0...,:0n

nh 3-point sequence (P= 3)

,...0,0,1,1,2,2,1,1,0,0...,:0

n

nx

Linear convolution: 6+3-1 = 8 points

.. .0,0,3,1,7,1,1,3,1,1,0,0...,:0

n

ny

0n

31711311

321

321

642

642

321

321

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ilolu 13

By Overlap-add:

,...0,0,3,2,1,0,0...,:0n

nh 3-point sequence (P= 3)

,...0,0,1,1,2,2,1,1,0,0...,:0

n

nx (In general, infinite length, starts at n = 0)

Decompose the input, x[n], into L-point segments.

Lets choose L= 3; could be something else...

,...0,0,1,1,2,2,1,1,0,0...,:21

xx

nx

Compute the 5-point(L+P-1) DFTs ofx1and x2 X1and X2

Compute the 5-point DFT of h H

Compute

y1= IDFT{H X1}

,...0,0,6,1,3,1,1,0,0...,:1

y

and

y2= IDFT{H X2}

,...0,0,3,1,7,5,2,0,0,0,0...,:2

y

Compute y[n] by overlap-add, i.e,

0n

31711311

31752

61311

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ilolu 14

Overlap Add Formally

Let the impulse response of the system be of length Psamples.

Decompose the input x n into nonoverlaping, consecutive blocks of length L.

thr block is

0

0 . .r

x n rL n Lx n

o w

so that

0

r

r

x n x n rL

Compute the (L+P-1)-point DFT of h n , H k

For the thr block, the output is found by (L+P-1)-point IDFT of r rY k H k X k ,

ry n

0 1 2, , , ...y n y n y n

The overall output is formed as 0 1 20

2 ...rr

y n y n rL y n y n L y n L

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Overlap Add Pictorially

samples

L

1 samples

L P

input segments

output segments

0n

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ilolu 16

OVERLAP-SAVE METHOD

Ex:

,...0,0,3,2,1,0,0...,:0n

nh 3-point sequence (P= 3)

,...0,0,1,1,2,2,1,1,0,0...,:0

n

nx (In general, infinite length, starts at n = 0)

Decompose the input, [], into -point segments ( ).Choose 4, could be something else...In overlap-save method, we use -point DFTsTherefore, the first + 1 1samples for each output segmentwill be incorrect. In this example, the first 2 samples...

Because of the two incorrect samples at each output block, the first segment is

choosen to start at 1 2

,...0,0,1,1,2,2,1,1,0,0,0,0...,:

1

x

nx

then, the second segment is chosen as

,...0,0,1,1,2,2,1,1,0,0,0,0...,: nx

which overlaps wih the previous segmentby two samples

1x

2x

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ilolu 17

Compute the 4-pointDFTs of , , , , , . . . , , , , , . . .Compute the 4-point DFT of Compute ,...0,0,1,1,3,1,0,0...,:1 y ,...0,0,1,3,5,3,0,0...,:2 y ,...0,0,7,1,5,1,0,0...,:3 y

,...0,0,3,1,1,1,0,0...,:4

y ,...0,0,0,0,0,0,0,0...,:5y Compute []by saving the correct samples,

...0,0,3,1,7,1,1,3,1,1,0,0...,:0

n

ny

0n

31711311

3111

7151

1353

1131

saved

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ilolu 18

OVERLAP SAVE FORMALLY

Let < + 1 The first + 1 samples of -point IDFT are incorrect.In particular, if , 1samples are incorrect.So, choose the first input segment, nx

0, as

samples

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

L

PLxxxxPxPx

i.e, 1010 LnPnxnx

Compute its L-point DFT kX0

Compute the L-point DFT of h kH

Compute ny0 L-point IDFT of kHkX0 .

Discard the first (P-1) samples to get the first ouput segment

0

0

1 1

0 . .

y n P n Ly n

o w

then the first (L-(P-1)) samples of the output are

0 1 0,1,..., 1 1y n y n P n L P

Then take the second input segment as

samples

122...,,11

L

PLxPPLx

i.e. 122...,,22 PLxPLx

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ilolu 19

and continue the process...

Overlap Save Pictorially

samples

L

input segments

output segments

0n

1 zerosP

input segments

1 samplesP

-1 samples discardedP

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ilolu 20

LINEAR CONVOLUTION AND CIRCULAR CONVOLUTION

We know the following

[] ()[] ()

(

) (

)(

)

[] ()} [] []

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ilolu 21

Now, the question is [] ?

[] [][] []

where [] ()()=

0 , 1 , , 1

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ilolu 22

We know that IDFT yields

[] { [ ]

= 0 , 1 , , 10 ..

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Therefore -point circular convolution of []and []can also be computedvia linear convolution:

Compute

[] [] []Then, you can find [] []as

[] [] {

[ ]= 0 , 1 , , 10 ..

or as

[] [] [][]

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ilolu 24

Ex:

[] 0 0 1= 2 0 0

[] 0 0 2= 1 1 0 0 a) Let[]and []be 5-point DFTs of[]and[], respectively.Find the sequence[] [] [].b) Let[]and []be 3-point DFTs of[]and[], respectively. Find thesequence [] [] []c) Let

[] (

)|= 0 , 1

[] ()|= 0 , 1Find the sequence []obtained by applying 2-point inverse DFT operation to[] [].

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a) [] [] []

[] ( )

= 0,1,2,3,4

[0] () + [1] ( 1) 0,1,2,3,4

Therefore

[] 0 0 2= 3 3 2 0 0

Or, since the linear convolution of []and[]yields (3+2-1) 4-pointsequence and the DFTs are 5-point, 5-point circular convolution and linear

convolution yields the same result.

[] [] [] [] [][]

02332

02240

00112

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ilolu 26

b) [] [] []

[] ( )

= 0,1 ,2

[0] () + [1] ( 1) 0 ,1 ,2

Therefore

[] 0 0 3= 3 0 0 or

[] [] []

{

[ 3]= 0,1 ,2

0 ..

330

2332

2332

2332

210

nnn

330

242

112

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c)

[

] {

12 [][]

= 0 , 1

0 . .

{

[ 2]= 0 , 1

0 . .

where [] [] [](linear conv.)

Therefore

[] 0 0 1= 3 0 0

31

2332

2332

2332

10

nn