# Finding Interval of Integration

If we let : $$x(t)=\begin{cases} 1&\text{if 0 and $$h(t)=x(t/a)=\begin{cases} 1&\text{if 0 where $$0 I wish to find convolution of $$(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

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. If that was not the case, then some other things should be taken into account.

• @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 $t$. – Tendero Mar 13 at 23:21
• 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. – Matt L. Mar 14 at 13:59