Given:
$$h(t) = e^{-t}u(t)$$
$$x(t) = sin(2t)u(t)$$
where $u(t)$ is unit step
and solving for the convolution:
$$y(t) = x(t) * h(t) $$
I'm having troubles with two parts of this problem.
The first part that is confusing me is the unit step input for $x(t)$. Do I ignore the negative half cycles for $sin(2t)$, since the unit step isn't defined there and the product of the two will just be zero?
And if I'm not supposed to ignore it, when t>0 for h(t), do I have to perform two different cases, when sin is positive and when sin is negative?
Or is the answer simply:
$$ y =\int\limits_{0}^{t}e^{-t}sin(2\tau)e^{\tau}d\tau$$
Best Answer
If $h(t)=e^{-t}u(t)$ and $x(t)=\sin(2t)u(t)$, then the convolution $(h*x)(t)$ is given by
$$\begin{align} (h*x)(t)&=\int_{-\infty}^\infty h(t-\tau)x(\tau)\,d\tau\\\\ &=\int_{-\infty}^\infty e^{-(t-\tau)}u(t-\tau)\,\sin(2\tau)u(\tau)\,d\tau\\\\ &=\int_{0}^t e^{-(t-\tau)} \sin(2\tau)\,d\tau\\\\ \end{align}$$
Can you finish now?