# Convolution sine function and unit step

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?

$$y =\int\limits_{0}^{t}e^{-t}sin(2\tau)e^{\tau}d\tau$$

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?

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?

Yes. $u(t)$ is zero for $t<0$.

Or is the answer simply: $$y =\int_{0}^{t}e^{-t}sin(2\tau)e^{\tau}d\tau$$

Yes. it can be written as$$y =\int_{0}^{t}sin(2\tau)e^{-(t-\tau)}d\tau$$

Note that there is an important thing to consider after this step: The convolution can be defined in two ways: $$f*g=\int_{-\infty}^{\infty}{f(\tau)g(t-\tau)d\tau}=\int_{-\infty}^{\infty}{g(\tau)f(t-\tau)d\tau}$$ While the final result is the same, one of these two are easier to calculate. Do you know which one?