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

[Jncik]

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

$$\int_{-\infty}^{+\infty}\delta (t) dt = 1$$

so by having

δ(2t)$$\int_{-\infty}^{+\infty}\delta (2t) dt = 1$$

I use $$2t = u <=> 2dt = du <=> dt = du/2$$

hence $$\int_{-\infty}^{+\infty}\delta (2t) dt = 1 => \frac{1}{2} \int_{-\infty}^{+\infty}\delta (u) du = 1$$

but since u is just a variable, we have that

$$\int_{-\infty}^{+\infty}\delta (u) du = 2$$

now about δ[2n] = δ[n]

in discrete time the summation over the integral of $$(-\infty, +\infty)$$

is 1.

now, what does δ[2n] mean?

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)?

 

[SammyS]

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

$$\int_{-\infty}^{+\infty}\delta (t) dt = 1$$

so by having

δ(2t)


$$\int_{-\infty}^{+\infty}\delta (2t) dt = 1$$

I use $$2t = u <=> 2dt = du <=> dt = du/2$$

hence


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

but since u is just a variable, we have that

$$\int_{-\infty}^{+\infty}\delta (u) du = 2$$
...

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

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

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