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Scaling Property of the Dirac Delta Function

by thegreenlaser
Tags: delta, dirac, function, property, scaling
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thegreenlaser
#1
May23-11, 10:17 PM
P: 465
1. The problem statement, all variables and given/known data
Prove that
[tex]\displaystyle \int_{-\infty}^{\infty} \delta (at - t_0) \ dt = \frac{1}{ | a |} \int_{-\infty}^{\infty} \delta (t - \frac{t_0}{a}) \ dt[/tex]
For some constant a.

3. The attempt at a solution
Edit: Looking at this again, I really don't understand where this is coming from. Everywhere I've read has just said to do a change of variable with u = at, but performing this change of variable, I get
[tex] \displaystyle \frac{1}{a} \int_{-\infty}^{\infty} \delta (u - t_0) \ du [/tex]
I don't really understand where the absolute value or the [tex]t_0 / a[/tex] come from.
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Dick
#2
May23-11, 10:24 PM
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The order of the limits -infinity to +infinity depend on the sign of a. Think about that. That's where the absolute value comes from.
thegreenlaser
#3
May23-11, 10:46 PM
P: 465
Oh, right, I forgot to change my integration limits when I did the change of variable. Thanks. What about the [tex]\frac{t_0}{a}[/tex] term?

nrqed
#4
May23-11, 11:46 PM
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Scaling Property of the Dirac Delta Function

Quote Quote by thegreenlaser View Post
Oh, right, I forgot to change my integration limits when I did the change of variable. Thanks. What about the [tex]\frac{t_0}{a}[/tex] term?



SInce you defined u = a t, you have to define as well a [tex] u_o = a t_o [/tex] so that your integral over u will contain a delta [tex] \delta(u- u_o/a) [/tex]. Then you rename [tex] u \rightarrow t [/tex] and you get their final expression.
thegreenlaser
#5
May24-11, 12:43 AM
P: 465
^Well, I guess that makes sense because the delta function should be non-zero when [tex]at-t_0 = 0[/tex] i.e. when [tex]t = t_0 / a[/tex] However, I don't quite understand from a general mathematical standpoint why the t_0 constant needs to be replaced when t is replaced. It doesn't seem to me that this is required generally. Like, for example if I have
[tex]f(x) = x^2[/tex]
and I want to evaluate
[tex]\int f(2x - x_0) \ dx[/tex]
I could do a change of variables, like
[tex]u = 2x[/tex]
but it makes no difference in the evaluation of the integral whether or not I make the substitution
[tex]u_0 = 2x_0[/tex]

(I did the integration on my whiteboard, but I'm not going to bother posting it, cause I'm just hoping this will help someone see where my thinking is wrong)
I understand that this is no-where near a 'dis-proof' of the fact that the substitution is required generally. I do believe you, I'm just curious why this is required.
L-x
#6
May24-11, 05:40 AM
P: 61
I *think* it's because you're redefining u->t. Normally when using a substitution you just either substitute back in at the end or change the limits over which you are integrating. If you redefine u so it is a times smaller than it was before, you also need to make sure all the constants are scaled by the same amount (you don't notice this scaling for the limits of the integral as infinity/a is still infinity.
Dick
#7
May24-11, 07:17 AM
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Quote Quote by thegreenlaser View Post
^Well, I guess that makes sense because the delta function should be non-zero when [tex]at-t_0 = 0[/tex] i.e. when [tex]t = t_0 / a[/tex] However, I don't quite understand from a general mathematical standpoint why the t_0 constant needs to be replaced when t is replaced. It doesn't seem to me that this is required generally. Like, for example if I have
[tex]f(x) = x^2[/tex]
and I want to evaluate
[tex]\int f(2x - x_0) \ dx[/tex]
I could do a change of variables, like
[tex]u = 2x[/tex]
but it makes no difference in the evaluation of the integral whether or not I make the substitution
[tex]u_0 = 2x_0[/tex]

(I did the integration on my whiteboard, but I'm not going to bother posting it, cause I'm just hoping this will help someone see where my thinking is wrong)
I understand that this is no-where near a 'dis-proof' of the fact that the substitution is required generally. I do believe you, I'm just curious why this is required.
It makes no difference in the evaluation of the integral of delta(t-t0/a). That integral is 1 regardless of what the additive constant is. The constant makes a difference when you want to integrate g(t)*delta(t-t0/a) for some function g(t). Then you get g(t0/a), because as you've said the delta function is concentrated at t=t0/a.
thegreenlaser
#8
May24-11, 03:59 PM
P: 465
Oh, I see. So, just to make sure I understand this, is this another way to look at it?

Given
[tex]\int_{-\infty}^{\infty} g(t) \delta(at - t_0) \ dt[/tex]
let
[tex]u = at[/tex]
Subbing this in gives
[tex]\frac{1}{|a|}\int_{-\infty}^{\infty} g\left(\frac{u}{a}\right) \delta(u - t_0) \ du[/tex]
which is then equal to the function g(u/a) evaluated at u=t0, where the argument of the delta function is zero. So then,
[tex]\frac{1}{|a|}\int_{-\infty}^{\infty} g\left(\frac{u}{a}\right) \delta(u - t_0) \ du
= \left. \frac{1}{|a|} g\left(\frac{u}{a}\right) \right|_{u=t_0}
= \frac{1}{|a|} g\left(\frac{t_0}{a}\right)[/tex]
which is the desired result. Whereas with the other way, you would leave g(t) be and then change the t0 variable, so that the result is (1/|a|)*g(t) evaluated at t=t0/a, which gives the same thing.

Sorry to keep going with this, I just really want to make sure I understand this. I'm having trouble conceptually with the delta function.
Dick
#9
May24-11, 05:41 PM
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I think you've got it.


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