1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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:

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

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.