How to obtain the convolution directly (not graphical) of the two functions $e^{-t}u(t)$ and $e^{-2t}u(t)$?

Question

I'm having trouble solving this convolution integral graphically. I don't understand where I stop sliding my function $h(t-\lambda)$ since $x(t)$ doesn't have a boundary as lambda approaches infinity so I don't know how many integrals I have to take. The two functions I have to take the convolution of are:

$$x(t) = e^{-t}u(t)$$ and $$h(t) = e^{-2t}u(t)$$

So my question is: Is there a way to take the convolution directly using the definition of the convolution? I don't know how to solve that particular integral can someone help me with it. My professor didn't really go indepth in solving the convolution directly instead he went for the graphical method. He explained that usually it's really difficult to solve the convolution directly and that the graphical method works most of the time.

Answer 1

Yes, you can find it directly as:

$$\displaystyle \int_{-\infty}^{\infty} e^{- \tau}~u(\tau)~e^{-2(t-\tau)}~u(t-\tau)~d\tau = e^{-2t}~\int_0^t e^{\tau}~d\tau = e^{-2t}(e^t-1)$$

A plot shows:

When we have functions $f(t)u(t)$ and $g(t)u(t)$ with the Heaviside Unit Step Function, we can just write:

$$\displaystyle (f*g)(t) = \int_0^t f(\tau)~g(t-\tau) ~ d\tau$$

Having said all of that, I think it is very important to understand what is going on graphically. I recommend spending time with the examples, particulary $3.4.1$, Example $1$ as they solve the general example to yours and do it both ways. It is critical to understand the graphical method as it can keep you away from unrecognizable integrals.

This is also a useful Convolution Table. Especially review "Convolution using graphical method (1)".