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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) | ||
| + | |||
| + | --[[User:Kim682|Kim682]] | ||
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
