🚀 go-pugleaf

In article <11515@pasteur.Berkeley.EDU> jbuck@galileo.berkeley.edu (Joe Buck) writes:
>In article <625@ctycal.UUCP>, ingoldsb@ctycal.UUCP (Terry Ingoldsby) writes:
>|> Pursuing the discussion of the Nyquist theorem, I have a question
>|> about practical sampling applications.  If you have a sine wave at
>|> frequency f, which you sample at just over 2f samples per second then
>|> the Nyquist theorem is satisfied.  I know that by performing a Fourier
>|> transform it is possible to recover all of the signal, i.e. deduce that
>|> the original wave was at frequency f.
>
>
>If you have a signal with no frequency components higher than f = 1/2T,
>where T is the spacing between samples, then the original waveform x(t)
>may be found exactly at any point by computing the sum
>
>x(t) = sum from m=-infinity to infinity x[m] * sinc (pi (t - m*T) / T)
>
>where sinc(x) is just sin(x)/x (note: sinc(0) is 1).
>
>Joe Buck

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 ??  

Comments are appreciated.

Rana Ahmed
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