# how to prove that the convolution between two discrete signals is the discrete signal of convolution between two continuous signals

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 $$y(t)$$.

• 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? Feb 9 at 1:43
• 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 $h(t)$? Feb 9 at 5:33
• 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. Feb 9 at 16:31
• Then what should be the case if the above statement isn't true? Is it not true only under certain circumstances? Feb 10 at 22:31
• @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. Feb 13 at 2:45

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

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

I think this question is related.