I'm a little new to signal processing and I'm trying to wrap my head around convolutions.

I know the definition of convolution for a continuous signal is

$$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty}{x(\tau)h(t-\tau) \, \mathrm{d}\tau}$$

Let's say for example that




you end up getting that


which is divergent. I watched a few YouTube videos about it and they always explain how to do convolutions graphically with two rectangles, does anybody mind explaining how to do it with two exponential functions, or point me to somewhere that does?

  • 4
    $\begingroup$ You shouldn't leave out the step function when evaluating the integral. This will give you finite upper and lower integration bounds. $\endgroup$ – Matt L. Oct 16 '16 at 20:24

Just in case if you look for a complete solution based on integration:

$$y(t)=\int_{-\infty}^{+\infty}e^{-4\tau}u(\tau)\times 6e^{-(t-\tau)}u(t-\tau)d\tau$$ Consider these facts:

Both step functions need to be equal to $1$ to have both terms (otherwise the corresponding term, and consequently $y(t)$, become zero). This implies:

  1. $\tau>0$
  2. $t-\tau>0\Rightarrow \tau<t$
  3. From the two above, $t>0$. Otherwise the integral is zero.

We need $1$ and $2$ for the boundaries of integral. They say $\tau \in (0,t)$. Also we need $3$ to know where the output exists. It simply says the integral should be multiplied by $u(t)$. Now that we applied all the given information from the two step functions, we remove them and the integral becomes: $$\begin{align} y(t)&=\left(\int_{0}^{t}e^{-4\tau}\times 6e^{-(t-\tau)}d\tau\right)u(t)\\[10pt] &=\left(6e^{-t}\int_{0}^{t}e^{-3\tau}d\tau\right)u(t)\\[10pt] &=\left(6e^{-t}\left(\frac{e^{-3\tau}}{-3}\biggr\vert^t_0\right)\right)u(t)\\[10pt] &=\left(-2e^{-t}\left(e^{-3t}-1\right)\right)u(t)\\[10pt] &=2(e^{-t}-e^{-4t})u(t) \end{align}$$


You forgot the Heaviside step functions. The convolution integral gives

$$y(t)=\int_0^t 6 e^{-3\tau-t} \, \mathrm{d}\tau = \cdots = \begin{cases} 2 e^{-t} - 2 e^{-4t} & \text{if } t \geq 0\\ 0 & \text{if } t < 0\end{cases}$$

We could also take the Laplace transforms of $h$ and $x$. Multiplying $H (s)$ and $X (s)$ and doing partial fraction expansion,

$$\left(\frac{6}{s+1}\right) \left(\frac{1}{s+4}\right) = \cdots = 2 \left(\frac{1}{s+1} - \frac{1}{s+4}\right)$$

Taking the inverse Laplace transforms, we again obtain $(2 e^{-t} - 2 e^{-4t}) \, u (t)$.


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