Solved: Heavyside Convolution: Calculate (\theta \ast \theta)(x)

[Sigurdsson]

Homework Statement

Calculate the convolution
(\theta \ast \theta)(x)

Homework 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

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.

 

[vela]

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

 

[Sigurdsson]

You're right, It should be

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

 

[vela]

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.

 

[Dickfore]

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

<br /> \begin{array}{c|c|c|c|c|c|c|c|}<br /> &amp; -\infty &amp; &amp; x &amp; &amp; 0 &amp; &amp; \infty \\<br /> \hline<br /> \theta(y) &amp; &amp; - &amp; - &amp; - &amp; 0 &amp; + &amp; \\<br /> \hline<br /> \theta(x - y) &amp; &amp; + &amp; 0 &amp; - &amp; - &amp; - &amp;<br /> \end{array}<br />

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 &lt; 0.

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

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.

 

[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