How does one solve convolution $f(t)*g(t)$ where $f(t) = u(t) - u(t-2)$ and $g(t) = e^{-2t}u(t)$ where $u(t)$ is heaviside (unit) step function?
I tried using Fourier transform of both functions to simplify calculations, but it does not really seem to do any simplification. I do know that convolution is distributive, associative and so on.
$f$ is a square, i.e. the constant $1$ for $0<t<2$ and $0$ elsewhere. $g$ is a negative exponential starting at $0$.
Now slide $f$ from the left (where $f\times g=0$), until its right edge (say at $v$) arrives at $0$. Continuing from there, the nonzero product extends from $0$ to the right edge, i.e. from $0$ to $v$, until the left edge arrives at $0$. And then the nonzero product runs from $v-2$ to $v$.
$$\begin{align}v<0&\to 0 \\ 0<v<2&\to\int_0^v e^{-2t}dt \\ 2<v&\to\int_{v-2}^ve^{-2t}dt \end{align}$$