Nyquist Condition question.

Started by A.E lover August 9, 2007
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

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