"Poorya " <pooryaiust.2004@gmail.com> wrote in message <jvq3ok$5bn$1@newscl01ah.mathworks.com>...
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> I mean you cannot just simply square the vectors and take their norms and conclude Parseval is holding.
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You can if you're using the discrete domain definition of Parseval, see here
http://en.wikipedia.org/wiki/Parseval's_theorem#Applications![]()
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> I mean you cannot just simply square the vectors and take their norms and conclude Parseval is holding.
==============
You can if you're using the discrete domain definition of Parseval, see here
http://en.wikipedia.org/wiki/Parseval's_theorem#Applications