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In Reply to: Re: No thats not an issue with Synchronous Rate Conversion posted by Todd Krieger on May 16, 2006 at 01:27:01:
Consider.If I do f&y (& meaning convolution), f a filter, y the signal in the time domain...
If I then calculate F (fourier transform of the filter) and Y (fourier transform of the signal), then F*Y is the exact dual of f&y.
(* being point by point multiply, complex)If I then do IFFT(F*Y), I get exactly f&y. You need some guard band to prevent wraparound, but that's easy to calculate.
Perhaps you're mistaking this for something else?
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Follow Ups
- You are aware of "fast convolution" aren't you? - real_jj 12:54:26 05/25/06 (0)
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