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Heaviside function and Unit impulse (delta) help

  1. Apr 8, 2009 #1
    Can someone give me quick refresher on what happens when you multiply the heaviside function with the unit impulse?

    Typically, the unit step function multiplied by anything simply delays it by the offset in the unit step function. The unit impulse function makes the value defined at only one point. At least I believe this is so... for example:
    (n+1)U(n)*d(n-2) = (n-1)u(n-2) right?
  2. jcsd
  3. Apr 11, 2009 #2
    This may be somewhat late, but your final equation does not technically make sense, though I believe you are on the correct track. The Dirac delta function is not a function at all, but rather a distribution, and thus only makes sense as an operator on a functional space. Consequently, you can only evaluate it when you have it as part of an inner-product.

    Furthermore, it does not simply make the substitution [itex] n \to n-2 [/itex]. It evaluates the function at the root of the argument, so that
    [tex] \begin{align*} \int_{-\infty}^\infty (x+1)U(x)\delta(x-2) dx &= \left. (x+1)U(x) \right|_{x=2}\\
    &= 3 \end{align*}[/tex]
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