where δ is the dirac delta function, I'm using the Greek alphabet on my keyboard(adsbygoogle = window.adsbygoogle || []).push({});

how can I prove this?

for the continuous time, we have that

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

so by having

δ(2t)

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

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

hence

[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

-2<=2n<=2

hence

-1<=n<=1

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

0<=2n<=0

0<=n<=0

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)?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**