# Evaluating the convolution using the convolution integral

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 $t > \tau$.

• The integral effectively is $\int_0^t h(\tau)x(t-\tau)\,d\tau$. That can be evaluated elementarily. – Daniel Fischer Oct 11 '13 at 12:04
• @KillaKem Although the post is a bit old, I added a method involving Fourier transform. Cheers! – Brightsun Jan 15 '16 at 22:36

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

$$\frac{1}{a-b} e^{-b t} \left (1-e^{-(a-b) t} \right ) = \frac{e^{-b t}-e^{-a t}}{a-b}$$

Assuming that $a,b>0$, then, $$\widehat f_a(\omega) = \int_0^{+\infty}e^{-(a+i\omega)t}dt=\frac{1}{a+i\omega}.$$ So \begin{align} \widehat{f_a\ast f_b}=&\frac{1}{(a+i\omega)(b+i\omega)}\\ f_a\ast f_b =& \frac{1}{2\pi}\int_{\mathbb R}\frac{e^{i\omega t}}{(a+i\omega)(b+i\omega)}d\omega. \end{align} If $t>0$, we can compute this using a half-circular contour $C_+$ in the upper-half plane: such an integral is equal to $i2\pi$ times the sum of the residues at $a$ and $b$, that lie in the upper-half plane, where the integrand has simple ploes for $a\neq b$. This gives \begin{align} -\frac{1}{2\pi}\oint_{C_+}\frac{e^{i\zeta t}}{(\zeta-ia)(\zeta-ib)}d\zeta= -\frac{i2\pi}{2\pi}\left( \frac{e^{-at}}{ia-ib}+\frac{e^{-bt}}{ib-ia}\right)=\frac{e^{-bt}-e^{-at}}{a-b}; \end{align} as the radius $R$ of the half-circle tends to infinity only the real line integral contributes: $$(f_a\ast f_b)(t)=\frac{e^{-bt}-e^{-at}}{a-b}.$$ In case $a=b$, the residue of the double pole yields precisely $(f_a\ast f_b)(t)=te^{-at},$ which is just the limit of the above expression as $b\to a$.

If $t\le0$, we can use a similar contour but in the lower-half plane, getting zero since there are no singularities there.

Finally $$(f_a\ast f_b)(t) = \frac{e^{-bt}-e^{-at}}{a-b}\theta(t).$$

Use LAPLACE Transform property

$$e^{-at} u(t) \rightleftharpoons \frac{1}{s+a}$$

$$e^{-bt} u(t) \rightleftharpoons \frac{1}{s+b}$$

$$X(s) = \frac{1}{(s+a)(s+b)}$$

Now partial fraction

$$X(s) = \frac{1}{(s+a)(s+b)} = \frac{A}{s+a} + \frac{B}{s+b}$$

$$A = \frac{1}{b-a}$$

$$B =\frac{1}{a-b}$$

$$x(t) = \frac{1}{b-a} [e^{-at} - e^{-bt}] u(t)$$