0
$\begingroup$

I know that if I sampling with impulse train so I get in the frequency plane X(f)*h(f) (when x(f) is my signal, * means convolution and h(f) is fourier transform of impulse train).

what the difference between this way to x(t=n*Ts) sampling way?

If I can get explenation in the "frequency plane" it will be great.

$\endgroup$
1
$\begingroup$

I cannot recall a meaningful difference between the ideal impulse train sampling and the uniform sampling relation indicated by the expression $t_n = n T_s$.

Given a continuous-time bandlimited signal $x_c(t)$, when you sample this with an ideal impulse train $\delta_{T_s}(t) = \sum_k \delta(t - k T_s)$, then the relation between the obtained discrete-time sequence $x[n]$ and the continuous-time signal will be given by :

$$ x[n] = x_c(t_n) = x_c(n T_s)$$

Hence, the latter expression is a consequence of the former operation.

$\endgroup$
  • $\begingroup$ I think the only difference is that using deltas allows one to write the interpolator as a low-pass filter: if the filter impulse response is $\text{sinc}(t/T)$, and the input is $\sum x(nT) \delta(t-nT)$, then the Shannon interpolation formula follows immediately. $\endgroup$ – MBaz Nov 21 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.