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Circular Folding and Circular Convolution: Concepts in Signal Processing, Slides of Digital Signal Processing

An explanation of circular folding and its equivalence to circular plotting in clockwise fashion. It also discusses the concept of circular convolution in the context of linear filtering and discrete fourier transform (dft). Instructions for solving problems related to circular convolution and finding the idft of circular convolution.

Typology: Slides

2011/2012

Uploaded on 07/20/2012

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Circular folding of a sequence is given as
'( ) ( ,mod )
'( ) (( ))
Circular folding is equivalent to
plotting ( ) in clockwise fashion!
N
xn x n N
xn x n
xn


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Download Circular Folding and Circular Convolution: Concepts in Signal Processing and more Slides Digital Signal Processing in PDF only on Docsity!

Circular folding of a sequence is given as'( )^ (^ ,mod

x^ n^ x^ n^ '( )^ ((^ ))^ N Circular folding is equivalent toplotting^ ( ) in clockwise fashion!

N

^  x n x^ n^  x n

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^ ^

^  4

Let^ ( )^ 1,2,^ 2,

Then^ ((^ ))^ 1,1,

x n^

x^ n   ^ ^ ^  (0) (^1) x   (1)^2 x  (2) (^2) x   (3)^1 x 

'(1)^1 x  '(2) (^2) x  ^ '(0)^1 x^ ^  '(3)^2 x 

( )x n^

((^ ))x n^4 docsity.com

^ We know if corresponding spectra of two signalsare multiplied in frequency domain, the twosignals undergo^ convolution

in time domain. ^ If spectra consist of Discrete Fourier transform(DFT) their multiplication corresponds to newtype of convolution in time domain called

Circular convolution.

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^ FindX(k) DFT of x(n) andY(k) DFT of y(n) ^ Then findZ(k) = X(k)Y(k) ^ Then takeIDFT Z(k)which is equivalent to circular convolution of x(n) andy(n)

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3 21 3 3 0 21 3 1 2 0 2

2 2 1 1

1 3 1

2 0 0 0 21 1 (^

) 3 1

2 0 0 0

The IDFT of^ ( )^1 (^ )^ ( )^1 (^ )^ ( )

kN j mN k kN j^ mN ( )k k^1 ( ) ( ) ( ) 1 ( ) ( ) ( )

k^ k N^ N^

Nj n j^ l^ j^

m N^ N^

N k^ n^

l k N^ N^ N^

j^ m^ n^ l N

X^ k x m X^ k eN x m X^ k X^ k eN x m x^ n e^ n l^ k

x^ l e^ e

N x m x^ n^ x^ l^

e

 N

 

      ^ ^

  ^ ^

 ^ ^ ^

 

  ^   ^ ^ 

^ ^

 ^   ^ 

   ^ 

 (^1)   

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21 (^ (^ ) 0 21 ( ) 0 21 ( ) 0
)^ 0,^ ,2^ ,...
Now^
0 (^ )^
or^0
or^ ((^ ))
kN j m^ n^ l N k kN j m n^ l N k kN j m n^ l^ or 0

N

N^ m N k
n^ l^ N^ Ne otherwiseN m n^ l^ pN where p
e^ otherwiseN^ l^
m^ n^ pN^ l^ m
n
          e o
^ ^    ^ ^    ^ ^ ^

   So we substitute^ ((^1 13 1 20

(^ )^ ( )^ ((^
therwisel m n^ pN^ m^ n^ N N N ))N n l
x^ m^ x^ n^ x^

   ^ ^ ^ ^    m^ n   

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N-point DFT

N-point DFT Let ( )^

( ) Then ((^ ))^

kj p N( ) x n^ X k x^ n^ p^ N

 X k e  ^ 

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^ DFT is computationally very exhaustive! ^ Number of complex Multiplications requiredare^ for^ N -point DFT.^2 N ^ Just Imagine if N = 4096, how manymultiplications are needed?

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^ Assume that we are calculating

N^ point DFT where N is a power of 2.  We divide our N-point signal into two signalsas follows:( )^ (2 )^1 ( )^ (2^ 1),^2

0,1,2,...,^12 f^ n^ x^ n and

N  f n x^ n^ n^ ^ ^ 

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1 0 odd/ 2^1

/ 2^1 ( 2 ) ( 2^

0

0 2 / 2 / 2^1

/ 2^11 / 2 2 / 2 0

0 (^ )^ (^ )^

0,1, 2...,^1
(^ )^ (^ )^
(^ )
(^ )^ (2^ )^
(2^ 1)

But^ T herefore(^ )^ (^ )^

(^ )

N^ k N n kn^

knN N n even^ n N^

Nk m k^

m N^

N m^

m N^ N N^

Nkm k k N N N m^

m X^ k^ x n W^

k^ N X^ k^ x n W^

x n W X^ k^ x^ m W

x^ m^

W

W^ W X k f^ m W^   ^ ^ W^ f^ m W

 ^

 ^

 ^

 ^

^ 
^  ^
^  

 ^  ^  

 ^

 (^ )^ (^ )^ (^ )^1

m kX k F k W F^ k N

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