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Convultion with Delta Function

  1. Feb 25, 2007 #1
    I can remember from Differential Equations that any function convolved with a delta function results in a copy of the function located at the impulese.

    That is, [tex]x(t) * \delta(t-5) = x(t-5)[/tex]

    However, I can't remember why. This is really irritating me since I need to use this concept for my courses, yet I can't remember why this is true. This makes sense... but I get stuck when trying to evaluate the following integral:

    [tex]\int_0^t \delta(t - \tau) d \tau[/tex]

    Any help would be appreciated.

    Last edited: Feb 25, 2007
  2. jcsd
  3. Feb 25, 2007 #2


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    How is the convolution defined? Also, remember [itex]\delta (t-5) = (\delta \circ \epsilon )(t)[/itex] where [itex]\epsilon : t \mapsto t-5[/itex]. I haven't worked with convolutions, but looking it up the definition on Mathworld, the equation above seems obvious:

    [tex]x * (\delta \circ \epsilon ) = \int _{-\infty} ^{\infty} x(\tau )(\delta \circ \epsilon )(t - \tau )d\tau [/tex]

    [tex]= \int _{-\infty} ^{\infty} x(\tau )\delta ((t - \tau) - 5)d\tau [/tex]

    [tex]= \int _{-\infty} ^{\infty} x(\tau )\delta ((t - 5) - \tau)d\tau [/tex]

    [tex]= x(t - 5)[/tex]

    where the last line follows by the definition of the delta function.
  4. Feb 25, 2007 #3
    Thanks, AKG!

    I don't know why that waws stumping me -- I really appreciate your help.
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