Nyquist Frequency CriteriaDate: 08/04/2003 at 21:25:13 From: Paul Subject: Nyquist Frequency Criteria I was asked by my professor, "Why do we need two points or samples per cycle to define a frequency component in the original data?" I said because the Nyquist Frequency Criteria says so. He said yes, but, why? I have tried to figure out why the number is 2 but I have had no luck. Any explanation would be great. Date: 08/04/2003 at 22:30:27 From: Doctor Douglas Subject: Re: Nyquist Frequency Criteria Hi Paul. Thanks for writing to the Math Forum. Suppose you have a 1 kHz sine wave y = A sin(wt). And you sample this wave at EXACTLY 1 kHz (f = 1 kHz = w/(2*pi)). All of the samples will have the same y-value; if you sample at the maximum, all of the samples will be y[i] = A. And if you sample at a zero-crossing, all of the samples will be y[i] = 0. And so on. Hence there is nothing to distinguish the sine wave y = A sin(wt) from a constant function y = C. With only one sample per cycle, one cannot distinguish the frequency f = 2*pi*w from the DC (f=0) component. This is why you need (more than) 2 points per cycle. Two points are barely enough, but it's possible with exactly 2 points per cycle to sample the two zero crossings in each cycle - which, again, is indistinguishable from zero. But if you have more than 2 points per cycle, then this cannot happen. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 08/05/2003 at 00:40:00 From: Paul Subject: Thank you (Nyquist Frequency Criteria) Doctor Douglas, Thanks for your help. What was once clear as mud, is now clear as water. Paul |
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