Hi all,
(1) I know the famous Nyquist Condition, f(t) with bandwidth B is
sampled without aliasing if Fs>2B.
Today I read another thing called Nyquist Condition which says:
for a continuous time signal x(t), take x(t) convolute with itself and
then sample the obtained signal at Fs,
if the sampled signal =delta[n], then Fs satisfies the Nyquist
condition
i.e: g(t)= x(t) (*) x(t).
g[n]=g(n/Fs)=delta(n).
Is this condition equal to the famous Nyquist condition in (1)?
Thanks

Posted by mnentwig●August 10, 20072007-08-10

Hello,
I won't say "yes" or "no". But some comment nonetheless, possibly I'm not
telling anything new here:
The 2nd statement relates in my interpretation to the question of
intersymbol interference, and convolving the pulse with itself may form a
matched filter. The first statement is aliasing.
Now if your system frequency response is symmetrical around the symbol
rate, as for example a raised-cosine filter or its special case the ideal
lowpass, then the pulse appears only at one sample as you stated.
I might come up with an example where the two are not the same (i.e. no
aliasing but ISI), depending on how I read the question. But as said, one
typical case is alias free, namely sampling a signal at symbol rate. You
get delta pulses precisely on the sampling instants, otherwise it's all
zero. That's actually a trivial statement :)
Cheers
Markus

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