# Homework Help: Heavyside convolution

1. Feb 4, 2012

### Sigurdsson

1. The problem statement, all variables and given/known data
Calculate the convolution
$$(\theta \ast \theta)(x)$$

2. Relevant equations
Convolution is defined as:
$$(f \ast g)(x) \equiv \int_{-\infty}^{\infty} f(x - y) g(y) \ dy = \int_{-\infty}^{\infty} f(y) g(x-y) \ dy$$

3. The attempt at a solution
I know this is probably easy for many but I'm really baffled with the outcome. Here we go

$$(\theta \ast \theta)(x) = \int_{-\infty}^{\infty} \theta(x - y) \theta(y) \ dy = \int_{-\infty}^0 \theta(x - y) \underbrace{\theta(y)}_0 \ dy + \int_0^{\infty} \theta(x - y) \underbrace{\theta(y)}_1 \ dy$$
$$= \int_0^{\infty} \theta(x - y) \ dy = y \theta(x - y)$$

However my result should be
$$= y \theta(y)$$

Gawdemmit, I can't spot my fault here. What am I missing?
Cheers.

2. Feb 5, 2012

### vela

Staff Emeritus
Your integration with respect to y can't end up with a y in the result. How are you getting to your final result?

3. Feb 5, 2012

### Sigurdsson

You're right, It should be

$$\int_0^y \theta(x - y) dy$$

4. Feb 5, 2012

### vela

Staff Emeritus
The upper limit should be x, not y, because the step function equals 1 only when x-y > 0. Remember y is a dummy variable in the convolution integral. You should end up with a function of x.

5. Feb 5, 2012

### Dickfore

$$\theta \ast \theta (x) = \int_{-\infty}^{\infty}{\theta(x - y) \, \theta(y) \, dy}$$
Look where the argument of each of the functions becomes zero:
$$y = 0, \ y = x$$
So, if $x < 0$, the intervals in question are:

$$\begin{array}{c|c|c|c|c|c|c|c|} & -\infty & & x & & 0 & & \infty \\ \hline \theta(y) & & - & - & - & 0 & + & \\ \hline \theta(x - y) & & + & 0 & - & - & - & \end{array}$$

Since the Heaviside step function is zero whenever its argument is negative, we must have to "+" in both rows for the product to be non-zero. As you can see, this is never the case for $x < 0$.

Now, suppose $x > 0$. The intervals in question are:
$$\begin{array}{c|c|c|c|c|c|c|c|} & -\infty & & 0 & & x & & \infty \\ \hline \theta(y) & & - & 0 & + & + & + & \\ \hline \theta(x - y) & & + & + & + & 0 & - & \end{array}$$
Now, the only interval when the product is non-zero is $(0, x)$. Thus, the convolution integral reduces to:
$$\theta \ast \theta(x) = \left\lbrace \begin{array}{ll} 0 &, x < 0 \\ \int_{0}^{x}{d y} &, x > 0 \end{array}\right.$$

I think it is pretty straightforward to evaluate the remaining steps, and use the definition of Heaviside's step function to encompass both cases under a single expression.

6. Feb 5, 2012

### Sigurdsson

Thanks guys I got it now.

The way you laid it out Dickfore makes it really clear but it unnerves me to solve an integral like that using just simple logic. I usually go straight for some rigorous integration rules and tricks instead of just visualizing the problem.

Thanks again