Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!sdd.hp.com!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!att!pacbell.com!tandem!netcom!mcmahan
From: mcmahan@netcom.COM (Dave Mc Mahan)
Newsgroups: sci.electronics
Subject: Re: A question about the Nyquist theorm
Keywords: Nyquist Theorem
Message-ID: <26575@netcom.COM>
Date: 2 Mar 91 22:39:50 GMT
References: <625@ctycal.UUCP> <11515@pasteur.Berkeley.EDU> <1347@cameron.egr.duke.edu>
Organization: Dave McMahan @ NetCom Services
Lines: 29


 In a previous article, rea@egr.duke.edu (Rana E. Ahmed) writes:
>Suppose we sample a pure sine wave of frequency 'f' at the Nyquist rate,
>i.e., at 2f samples/sec (exact), such that we start sampling the sine wave
>at the time of its zero crossing. Thus, if we assume uniform sampling,
>then all subsequent samples will have values equal to zero, i.e., x[m]=0 for 
>all m (Assuming instantaneous sampling, i.e., no Hold Time for samples).
>Intutively, if we pass these samples (each of zero voltage (say)) through an 
>ideal low-pass filter, then we should expect to get zero voltage at the output
>of filter. In other words, reconstructed signal voltage =0 for all t. 
>(see also the formula for x(t) above ).
>How can we recover the pure sine in this sampling strategy? 
>Am I missing something ??  

You have to go back and carefully read the criteria that Nyquist stated.  He
did NOT state that you can sample AT twice the lowest frequency for perfect
representation.  He stated that you have to sample at GREATER THAN twice the
lowest frequency for perfect representation.  You are quite correct in your
analysis above.  However, it doesn't meet Nyquist criteria for the original
waveform, so you will distort the information content of the original
waveform.

   -dave



-- 
Dave McMahan                            mcmahan@netcom.com
					{apple,amdahl,claris}!netcom!mcmahan