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Homework Help: I can't understand why δ[2n] = δ[n] and δ(2t) = 1/2 * δ(t)

  1. Jul 8, 2011 #1
    where δ is the dirac delta function, I'm using the Greek alphabet on my keyboard

    how can I prove this?

    for the continuous time, we have that

    [tex] \int_{-\infty}^{+\infty}\delta (t) dt = 1 [/tex]

    so by having


    [tex] \int_{-\infty}^{+\infty}\delta (2t) dt = 1 [/tex]

    I use [tex] 2t = u <=> 2dt = du <=> dt = du/2 [/tex]


    [tex] \int_{-\infty}^{+\infty}\delta (2t) dt = 1 => \frac{1}{2} \int_{-\infty}^{+\infty}\delta (u) du = 1 [/tex]

    but since u is just a variable, we have that

    [tex] \int_{-\infty}^{+\infty}\delta (u) du = 2[/tex]

    now about δ[2n] = δ[n]

    in discrete time the summation over the integral of [tex] (-\infty, +\infty) [/tex]

    is 1.

    now, what does δ[2n] mean?

    3. The attempt at a solution

    what I understand is that, when we have normal functions in discrete time, and for example we know that from -2 to 2 the values are not zero we say that




    and the step now will be not 1, but 0.5, but in discrete time, we can see the values only when n is an integer, hence we will lose some of them

    now for dirac delta we know that for n = 0 it is 1. hence we have


    hence again, for n = 0, it will be not 0...

    but I'm not really sure if this logic is correct

    I mean, why can't I use the same thing for the continuous time and say that δ(2t) = δ(t)?
    Last edited: Jul 8, 2011
  2. jcsd
  3. Jul 8, 2011 #2


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    It is not true that [itex]\displaystyle \int_{-\infty}^{+\infty}\delta (2t) dt = 1 \,.[/itex]

    But it is true that [itex]\displaystyle \int_{-\infty}^{+\infty}\delta (t) dt =\int_{-\infty}^{+\infty}\delta (u) du = 1 \,.[/itex] The variables u & t are what we call 'Dummy' variables.

    Therefore, [itex]\displaystyle \int_{-\infty}^{+\infty}\delta (2t) dt =\frac{1}{2}\int_{-\infty}^{+\infty}\delta (u) du = \frac{1}{2} \,.[/itex]
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