If we let : $$ x(t)=\begin{cases} 1&\text{if $0<t<1$}\\ 0&\text{if otherwise} \end{cases} $$ and $$ h(t)=x(t/a)=\begin{cases} 1&\text{if $0<t<a$}\\ 0&\text{if otherwise}\end{cases} $$ where $0<a\leq 1$ I wish to find convolution of $(x*h)(t)$ without graphing. We have that : $$ \max(0,t-a)<\tau<\min(1,t) $$

Is the following assertion correct? $$ \max(0,t-a)=\begin{cases} 0&\text{if $0<t<a$}\\ t-a&\text{if $a<t<1$} \end{cases} \quad\text{and}\quad \min(t,1)=\begin{cases} 1&\text{if $1<t<1+a$}\\ t&\text{if $0<t<1$} \end{cases} $$


1 Answer 1


By definition:

$$x(t)*h(t)=\int_{-\infty}^\infty x(\tau)h(t-\tau) \ \mathrm{d}\tau$$

If any of the functions inside the integral equals 0, then the product does as well. We know that:

$$x(\tau)\neq0 \iff \tau\in(0,1)$$ $$h(t-\tau)\neq0 \iff t-\tau\in(0,a) \iff \tau \in (t-a,t)$$

So the convolution integral can be simplified:

$$x(t)*h(t)=\int_{-\infty}^\infty x(\tau)h(t-\tau) \ \mathrm{d}\tau = \int_{\max(0, t-a)}^{\min(1,t)} x(\tau)h(t-\tau) \ \mathrm{d}\tau $$

Just as you had found out.

Notice that this works just because $0<a\leq 1$. If that was not the case, then some other things should be taken into account.

  • $\begingroup$ @MattL. If $t<0$ then $x(\tau)$ and $h(t-\tau)$ would be zero for any value of $\tau$ in the interval of integration. I don't see why you say that the result of the integral would be $t$. $\endgroup$
    – Tendero
    Commented Mar 13, 2021 at 23:21
  • $\begingroup$ You're absolutely right. Somehow I overlooked that you left the integrand unchanged. I've often seen that type of integral just with $d\tau$, which is not the correct result because the integral isn't equal to zero when its lower bound exceed its upper bound. Sorry for the confusion. $\endgroup$
    – Matt L.
    Commented Mar 14, 2021 at 13:59

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