I am having trouble evaluating the convolution of two signals using the convolution integral.I want to find the convolution of two signals x and h where,
$$
x(t) = \begin{cases}
e^{-at} & \text{$t > 0$} \\
0 & \text{$t < 0$ } \\
\end{cases}
$$
$$
h(t) = \begin{cases}
e^{-bt} & \text{$t > 0$} \\
0 & \text{$t < 0$ } \\
\end{cases}
$$
using the convolution integral
$$
y(t) = x(t)*h(t) = \int_{-\infty}^{\infty} h(\tau)x(t – \tau) d\tau
$$
Which will mean that:
$$
h(\tau) = \begin{cases}
e^{-b\tau} & \text{$\tau > 0$} \\
0 & \text{$\tau < 0$ } \\
\end{cases}
$$
$$
x(t – \tau) = \begin{cases}
e^{-at}e^{a \tau} & \text{$t > \tau$} \\
0 & \text{$t < \tau$ } \\
\end{cases}
$$
But how do I proceed from here? I don't know how to handle the $x(t – \tau)$ function which is non-zero only when $t > \tau$.
Best Answer
Write each of the signals as
$$e^{-k \tau} \theta(\tau)$$
where $k$ is either of $a$ or $b$, and $\theta(\tau)$ is the Heaviside step function, zero when $\tau < 0$ and $1$ when $\tau > 0$. The convolution integral may then be written as
$$\int_{-\infty}^{\infty} d\tau \, e^{-a \tau} \theta(\tau) \, e^{-b (t-\tau)} \theta(t-\tau)$$
Now, the product of the two Heavisides in the integral is zero outside the interval $[0,t]$. Therefore, we may write the convolution integral as
$$\int_0^t d\tau \, e^{-a \tau} \, e^{-b (t-\tau)} = e^{-b t} \int_0^t d\tau \, e^{-(a-b) \tau} $$
which is
$$\frac{1}{a-b} e^{-b t} \left (1-e^{-(a-b) t} \right ) = \frac{e^{-b t}-e^{-a t}}{a-b}$$