given the following signals:
$$ x(t) = \begin{cases} A, & 0 < t < T/2 \\ -A, & T/2 < t < T \\ 0, & \text{elsewhere} \end{cases} $$
and
$$h(t) = e^{-t}, 0 < t < \infty$$
Then finding the convolution of the two:
$$ y(t) = x(t)*h(t)$$ $$ y(t) = \int_{-\infty}^\infty x(\tau)h(t-\tau)\,d\tau$$
case 1:
$$t < 0 $$ $$y(t) = 0 $$
case 2:
$$t > 0, t < T/2 $$ $$ y(t) = \int_0^t Ae^{-t} e^\tau \, d\tau$$ $$ y(t) = Ae^{-t}\int_0^{t}e^\tau \, d\tau$$ $$ y(t) = Ae^{-t}[e^{t} - 1] $$
case 3 (this one I'm not to sure of):
$$t > T/2, t < T $$ $$ y(t) = \int_{T/2}^T -Ae^{-t} e^\tau \, d\tau$$ $$ y(t) = -Ae^{-t}\int_{T/2}^T e^\tau \, d\tau$$ $$ y(t) = -Ae^{-t}[e^T - e^{T/2} ] $$
case 4:
$$t > T$$ $$ y(t) = 0$$
I'm going to upload pictures as soon as I can. Does the following look correct so far? Have to fix it up a bit. Forgot about the boundary for h(t).
First note that $$0<t-\tau<\infty \iff \tau < t <\infty. $$ So for $0<t<T/2$ we $0<\tau<t$ and hence $$y(t) = \int_0^t A e^{-(t-\tau)}\,\mathsf d\tau = A\left(1-e^{-t}\right). $$ Similarly, for $T/2<t<T$ we have $T/2<\tau<t$, and hence $$y(t) = \int_{T/2}^t -A e^{-(t-\tau)}\,\mathsf d\tau = -A\left(1-e^{-\left(t-T/2\right)}\right). $$