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Dirac delta function

  1. Feb 1, 2006 #1
    Just have a question about the dirac delta function. I understand how you would write it if you want to shift it but how would you scale it assuming we are using discrete time. Would you write 2*diracdelta[n] or diracdelta[2n]. Also, would that increase it or reduce it by 2 meaning that instead of it being 1 at n=0, it would be 2 instead or would it be 1/2. Does it work the same way as scaling other functions in other words? Thank you!
     
  2. jcsd
  3. Feb 1, 2006 #2
    Remember the defining property of the Dirac delta:

    [tex] \int_{-\infty}^{\infty} \delta(x)dx = 1[/tex]

    Thus

    [tex] \int_{-\infty}^{\infty} \delta(ax)dx = \frac{1}{a} \int_{-\infty}^{\infty} \delta(y)dy = \frac{1}{a}[/tex]

    So we can think of multipling the argument by a as being the same thing as dividing the function by a:

    [tex]\delta(ax) = \frac{1}{a}\delta(x)[/tex]
     
    Last edited: Feb 1, 2006
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