I am currently studying for a test and I have this example:

Given the impulse response of a system: $$ h(t) = \left \{ \begin{matrix} 1,& 0 \le t \le1\\ 0, & \mbox{elsewhere} \end{matrix} \right . $$

find the output signal of the system for the input signal: $$x(t) = u(t)u(2-t)2t$$

Here is my progress so far:

But after I put the limits of 1s and 2s, I realized that I had already dealt with the case where part of $h(t-\tau)$ is inside $x(\tau)$ so I got confused. How should I continue?


1 Answer 1


So we have

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

We go with the first form. That means we have to time flip $h(t)$, slide it over $x(t)$ and integrate. Since $h(t)$ has only support on $[0,1]$ we can write this as

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

Furthermore since $h(t) = 1$ inside $[0,1]$ that simplifies to $$y(t) = \int_{t-1}^{t}x(\tau)d\tau $$

Since $x(t)$ has finite support on $[0,2]$ we can split this into three sections.

  1. $[0,1]$: partial overlap on the left
  2. $[1,2]$: full overlap
  3. $[2,3]$: partial overlap on the right

and adjust the bounds of the integral accordingly.

$$y_{[0,1]} = \int_{0}^{t}x(\tau)h(\tau-t)d\tau = \tau^2 \biggr|_{0}^t = t^2 $$

$$y_{[1,2]} = \int_{t-1}^{t}x(\tau)h(\tau-t)d\tau = \tau^2 \biggr|_{t-1}^t = 2t-1 $$

$$y_{[2,3]} = \int_{t-1}^{2}x(\tau)h(\tau-t)d\tau = \tau^2 \biggr|_{t-1}^2 = 3+2t-t^2 $$

And putting it all together:

$$ y(t) = \begin{cases} t^2 & 0 \leq t \leq 1 \\ 2t-1 & 1 \leq t \leq 2 \\ 3+2t-t^2 & 2 \leq t \leq 3 \\ 0 & \text{elsewhere} \end{cases} $$

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