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Heavyside convolution

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the convolution
    [tex] (\theta \ast \theta)(x) [/tex]

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

    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

    [tex] (\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 [/tex]
    [tex] = \int_0^{\infty} \theta(x - y) \ dy = y \theta(x - y) [/tex]

    However my result should be
    [tex] = y \theta(y) [/tex]

    Gawdemmit, I can't spot my fault here. What am I missing?
    Cheers.
     
  2. jcsd
  3. Feb 5, 2012 #2

    vela

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    Your integration with respect to y can't end up with a y in the result. How are you getting to your final result?
     
  4. Feb 5, 2012 #3
    You're right, It should be

    [tex] \int_0^y \theta(x - y) dy [/tex]
     
  5. Feb 5, 2012 #4

    vela

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    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.
     
  6. Feb 5, 2012 #5
    [tex]
    \theta \ast \theta (x) = \int_{-\infty}^{\infty}{\theta(x - y) \, \theta(y) \, dy}
    [/tex]
    Look where the argument of each of the functions becomes zero:
    [tex]
    y = 0, \ y = x
    [/tex]
    So, if [itex]x < 0[/itex], the intervals in question are:

    [tex]
    \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}
    [/tex]

    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 [itex]x < 0[/itex].

    Now, suppose [itex]x > 0[/itex]. The intervals in question are:
    [tex]
    \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}
    [/tex]
    Now, the only interval when the product is non-zero is [itex](0, x)[/itex]. Thus, the convolution integral reduces to:
    [tex]
    \theta \ast \theta(x) = \left\lbrace \begin{array}{ll}
    0 &, x < 0 \\

    \int_{0}^{x}{d y} &, x > 0
    \end{array}\right.
    [/tex]

    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.
     
  7. Feb 5, 2012 #6
    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
     
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