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**To**:**Loopers-Delight@loopers-delight.com****From**:**a k butler <akbutler@tiscali.co.uk>****Subject**:**Re: Loopers-Delight-d Digest V05 #834****Date**:**Wed, 21 Dec 2005 22:45:11 +0000**

At 22:24 21/12/05, you wrote: > > sample a 22kHz signal at 44.1kHz > > > the result is identical to sampling a 22.05kHz tone > > which is amplitude modulated at 500Hz > > > agree? (if no, then try working it out on paper) just draw the waveform, and put the sample points on ...easy then you'll see it sorry for being unclear, I didn't mean to do the arithmetic > >Unless my math is wrong, I don't agree. A 22khz tone sampled at >44.1khz produces a signal containing the fundamental (22k) plus >additional components at 22.1k, 66.1k, 66.2k, 110.2k, 110.3k, >154.3k, ad infinitum. Do you agree? no, for one thing this ignores aliasing. secondly, only 22.1k is within the limits imposed by the sampling frequency. thirdly, ;-) it's getting late here and I'm a bit tired to crunch the numbers, I will do later, but once you see the samples plotted the maths won't matter >I may be misinterpreting "the result is identical to sampling a 22.05kHz >tone >which is amplitude modulated at 500Hz". I'm assuming you're first >modulating a 22.05k tone at 500hz, producing the original 22.05k >tone plus sidebands at 22.55k and 21.55k. Sample that signal and >you don't get the same result as the 22k tone sampled at 44.1k. > >Which is wrong? My math, my interpretation, or both? > > >The ideal Nyquist sampling theorem is based on the ideal Fourier >transform - ideal Fourier transforms are lossless aren't >they? Granted it all falls apart when you try to put it into practice. um, I already dealt with that. Ideal fourier presumes a periodic signal of known frequency. So it falls apart in theory, even before the practice andy

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