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

# Nyquist Condition question.

Started by ●August 9, 2007

Reply by ●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