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Like the title. We already know $x[n]$ and $h[n]$ are the discrete time signal of $x(t)$ and $h(t)$ respectively, which are continuous time signals. And also $x(t)*h(t)=y(t)$, $x[n]*h[n]=y[n]$. Prove that $y[n]$ is the discrete time signal of $y(t)$.

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    $\begingroup$ What have you tried so far and where are you stuck? Have you considered working out the continuous convolution, sampling it and comparing it to the discrete case? $\endgroup$ Feb 9 at 1:43
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    $\begingroup$ Let $x(t) = \delta(t - T_s / 2)$, where $T_s$ is the sampling interval. Let $h(t) = u(t) a e^{-a t}$. Then $y(t) = u(t - T_s / 2) a^{a T_s/2} e^{-a t}$. Now observe that $x[n] = 0 \forall n$, and that if you convolve it with anything the result will be zero. Thus, $y[n] = 0$ and -- oh crap. Did the prof put any constraints on $x(t)$ and $h(t)$? $\endgroup$
    – TimWescott
    Feb 9 at 5:33
  • $\begingroup$ What is your reason for believing that the result that you are asked to prove is true? It might be easier to come up with a counterexample to the alleged result. $\endgroup$ Feb 9 at 16:31
  • $\begingroup$ Then what should be the case if the above statement isn't true? Is it not true only under certain circumstances? $\endgroup$
    – jsnjztr
    Feb 10 at 22:31
  • $\begingroup$ @jsnjztr didn't you read my comment? I disprove the hypothesis that $y[n]$ is the discrete equivalent of $y(t)$ in every case, by showing you a case where that is not true. $\endgroup$
    – TimWescott
    Feb 13 at 2:45

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Say continuous signals $x(t)$, $h(t)$, $y(t)$, sampled continuous signals $x_s(t)$, $h_s(t)$ satisfy

$$ y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau \tag{1} $$

$$ \begin{aligned} x_s(t) = \begin{cases} x(t) & t=nT\\ 0 & \text{otherwise} \end{cases} \end{aligned}\tag{2} $$

$$ \begin{aligned} h_s(t) = \begin{cases} h(t) & t=nT\\ 0 & \text{otherwise} \end{cases} \end{aligned}\tag{3} $$

where $T$ is the sampling period. The continuous-time convolution of $x_s(t)$ and $h_s(t)$ is given by $$ \hat{y}(t) = \int_{-\infty}^{\infty} x_s(\tau)h_s(t-\tau)d\tau \tag{4} $$

Now we have discrete signals $x[n]$, $h[n]$ are the discrete version of sampled continuous signals $x_s(t)$ and $h_s(t)$. $y[n]$ is the discrete convolution of $x[n]$ and $h[n]$ $$ y[n] = \sum_{m=-\infty}^{\infty} x[m] h[n-m] \tag{5} $$

If you are talking about continuous-time convolution between two sampled continuous signals, which is $\hat{y}(t)$, equals to the discrete-time convolution between two discrete signals, which is sampled $y[n]$, that's true because $x_s(t)$ and $h_s(t)$ are zero unless $t=nT$ and the integral in Eq. (4) is reduced to summation.

But $y(t)$ is a full integral whose discrete signal is not equal to $\hat{y}(t)$ nor $y[n]$. These two pictures are going to help you understand.

Eq. (1): $$y(t_0) = \int_{\tau=0}^5 x(\tau)h(t_0-\tau) d\tau$$

enter image description here

Eq. (4) and Eq. (5): $$y[n_0] = \sum_{m=0}^5 x[m]h[n_0-m]$$

enter image description here

I think this question is related.

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