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Delta Function

  1. Jul 18, 2006 #1
    This isn't really homework, I'm just curious. So I'm dealing with the delta dirac function, and I was just wondering what would happen if we had two functions.

    So the sampling property,

    [tex] \int_{-\infty}^{\infty} f(t)\delta(t-a)\,\,dt = f(a) [/tex]

    Now what if we had:
    [tex] \int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt =?[/tex]

    What would happen?
    My guess would be the following:

    If [itex] a = b [/itex]:
    [tex] \int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt =\int_{-\infty}^{\infty} f(t)\delta(a)\,\,dt =\int_{-\infty}^{\infty} f(t)\delta(b)\,\,dt [/tex]

    If [itex] a \neq b [/itex]:
    [tex] \int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt = 0 [/tex]

    Or is my answer or question just nonsense?

  2. jcsd
  3. Jul 18, 2006 #2


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    [tex] \int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt = \int_{-\infty}^{\infty} F(t)\delta(t-a)\,\,dt = F(a) = 0 [/tex]

    where I have put [tex]F(t) = \delta(t-b)f(t)[/tex] so that [tex]F(a) = \delta(a-b)f(a)=0[/tex] if [tex]a\neq b[/tex]
  4. Jul 18, 2006 #3
    Thanks benorin!

    That was (for lack of a better word) nifty how you just used the property that f(t)\delta(t-a)=f(a) by defining a function that encapsulated what was necessary to use the property :smile: Really cool.

    Well, I'm actually glad I asked the question then.

    Thanks again.
  5. Jul 19, 2006 #4


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    If you want to combine delta functions, of somewhat more importance is the fact that
    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\delta(x-a)\delta(y-b)dxdy= f(a,b)[/tex]
  6. Jul 19, 2006 #5

    So I'm guessing that works as follows:
    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\delta(x-a)\delta(y-b)dxdy= \int_{\infty}^{\infty}dy \, \delta(y-b)\, [ f(a,y) ] = f(a,b) [/tex]

    Where, [itex] \delta(y-b) [/itex] is held constant while integrating through [itex] dx [/itex].
    Last edited: Jul 19, 2006
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