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<math>r_{yy}(n)=h(-n)*h(n)*r_{xx}(n)</math> - Autocorrelation
 
<math>r_{yy}(n)=h(-n)*h(n)*r_{xx}(n)</math> - Autocorrelation
  
* means convolution.
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<math>*</math> means convolution.
  
  
 
If I am wrong, please reply ASAP.--[[User:Kim415|Kim415]] 09:59, 13 April 2009 (UTC)
 
If I am wrong, please reply ASAP.--[[User:Kim415|Kim415]] 09:59, 13 April 2009 (UTC)
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I think that "Cross Correlation" is <math> r_{xy}(n)=h(-n)*r_{xx}(n)</math>
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not,<math> r_{xy}(n)=h(n)*r_{xx}(n)</math>.
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Because,<math> r_{xy}(n)=x(n)*y(-n)=x(n)*(x(-n)*h(-n))=r_{xx}(n)*h(-n)</math>
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You can verify that in Proakis's book (page #127)
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--[[User:Kim682|Kim682]] 16:24, 23 April 2009

Latest revision as of 10:25, 23 April 2009

You know how to do the autocorrelation and cross-correlation.


$ r_{xy}(n)=h(n)*r_{xx}(n) $ - Cross Correlation

$ r_{yy}(n)=h(-n)*h(n)*r_{xx}(n) $ - Autocorrelation

$ * $ means convolution.


If I am wrong, please reply ASAP.--Kim415 09:59, 13 April 2009 (UTC)


I think that "Cross Correlation" is $ r_{xy}(n)=h(-n)*r_{xx}(n) $

not,$ r_{xy}(n)=h(n)*r_{xx}(n) $.

Because,$ r_{xy}(n)=x(n)*y(-n)=x(n)*(x(-n)*h(-n))=r_{xx}(n)*h(-n) $

You can verify that in Proakis's book (page #127)

--Kim682 16:24, 23 April 2009

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