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FT OF PERIODIC SIGNALS!

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purnapragna

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hi i want to know how to find the convolution of periodic signals using FT of them.

thnx

purna!
 

flatulent

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Convolution in one domain is identical to multiplication in the other domain. So multiply in the frequency domain and then convert the results to the time domain.
 

GroundCtrl

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There are some difference between FT in discrete and continious time domains.

1. In continious time domain no convolution between two periodocal signals exists because of the convolution integral grows infinetily.

2. More simly case the convolution between one periodic signal and one non-periodic signal (like a filter impulse reaction). In continous time domain no FT (in classical functions) exists for periodic signals. But for this case exists FOURIER SEQUENCE. This sequence can be converted to FT by using delta-functions. You have to multiply all coefficietns of Fourier sequence of periodic signal by sequence of delta-functions. The result of this operation is FT for periodic signal in terms of general functions (more common case than classical functions). For complete convolytion you have to multiply this FT of periodic signal to FT of non-periodic signal (this FT exists in all practical cases) and convert the result from frequency domain to time domain.
 

purnapragna

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i mean for example how do you convolve \[\sin(\omega_{0}t) and \]\[\cos(\omega_{0}t) \] in frequncy domain. We already know that convolution in time domain is multiplication in frequency domain. But for the above mentioned signals multiplication is not possible as their FT have impulses amd multiplication of impulses is not defined.

can anybody clarify me?

thnx

purna!
 
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