# 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? Commented Feb 9, 2022 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 $h(t)$? Commented Feb 9, 2022 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. Commented Feb 9, 2022 at 16:31
• Then what should be the case if the above statement isn't true? Is it not true only under certain circumstances? Commented Feb 10, 2022 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. Commented Feb 13, 2022 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.