# Convolution process confusion

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?

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:

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