🚀 go-pugleaf

In article <1347@cameron.egr.duke.edu>, 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 ??  

Yes.  The theorem has a "<" and you're assuming "<=".  Sampling produces aliasing.
If the Nyquist frequency is f, then a sine wave at frequency q looks exactly
like a sine wave at frequency q+mf, where m is any integer.  So if you want
to recover your sine wave at frequency f, sampling at 2f isn't enough: you
need 2f + delta (for an arbitrarily small delta).

--
Joe Buck
jbuck@galileo.berkeley.edu	 {uunet,ucbvax}!galileo.berkeley.edu!jbuck