Convultion with Delta Function

  • Thread starter Caspian
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  • #1
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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.

Thanks!
 
Last edited:

Answers and Replies

  • #2
AKG
Science Advisor
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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.
 
  • #3
15
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Thanks, AKG!

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

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