Consider the following : $$ x(t)=u(t)-2u(t-2)+u(t-5)\qquad\text{and}\qquad h(t)=e^{2t}u(1-t) $$ I wish to compute $(x*h)(t)$ without graphing it.
So as a first step, I would rewrite $x(t)$ and $h(t)$ as : $$ x(t)= \begin{cases} 0&\text{if $t\geq 5$}\\ -1&\text{if $2\leq t<5$}\\ 1&\text{if $0\leq t <2$}\\ 0&\text{if otherwise} \end{cases} \qquad\text{and}\qquad h(t)=\begin{cases} e^{2t}&\text{if $t\leq1$}\\ 0&\text{if otherwise} \end{cases} $$ Now, I must $x(t)\mapsto x(\tau)$ and $h(t)\mapsto h(t-\tau)$. So we get : $$ x(\tau)= \begin{cases} 0&\text{if $\tau\geq 5$}\\ -1&\text{if $2\leq \tau<5$}\\ 1&\text{if $0\leq \tau <2$}\\ 0&\text{if otherwise} \end{cases} \qquad\text{and}\qquad h(t)=\begin{cases} e^{2(t-\tau)}&\text{if $\tau\geq t-1$}\\ 0&\text{if otherwise} \end{cases} $$ Now from what I know I must restrict the domain of : $$ \tau\geq 5 $$ $$ 2\leq \tau<5 $$ $$ 0\leq \tau <2 $$ $$ \tau>t-1 $$ Therefore, I have $\max(0,2,5,t-1)<\tau<\min(...)$ I don't know how to find the upper bound which is minimum of upper bounds of the inequalities listed above. I would hope someone can assist me in computing the convolution.
Use linearity. (You may have seen linearity as two properties: distributivity of convolution over addition and associativity with scalar multiplication. Notice that the latter makes convolution distribute over addition and subtraction.) Treat $x(t)$ as a linear combination of $u(t)$, $u(t-2)$, and $u(t-5)$. Compute the convolution of $h$ with each of these. Then your answer is the fist convolution, minus twice the second one, plus the third one.