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Nyquist Frequency Criteria

Date: 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 

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 

Associated Topics:
College Physics

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