# Sifting Property of the Impulse Function

1. Nov 21, 2011

### El Moriana

1. The problem
I have a problem grasping what the point of the sifting property of the Dirac function is.
It isolates the value at a point in a function, right?
Doesn't just substituting that point into the function do exactly the same thing?

2. Relevant equations

Sifting poperty:
if $f(t)$ is continuous at $t=a$ then

$\int_{-∞}^∞f(t)δ(t-a)dt= f(a)$

from: Glyn James, Advanced Modern Engineering Mathematics (3rd Ed), Section 2.5, p.155

2. Nov 21, 2011

### lanedance

so based on the properties of the delta function you know
$$\int \delta (x) dx = 1$$

A handwaving explanation is that if f is continuous and if you zoom in on a small enough region $(x-\epsilon, x+\epsilon)$, then f(x) will be close to constant on this region.

The delta function zero everywhere except at x=a and the integral evaluates to exactly the value of the function at the point x=a
$$\int f(x)\delta (x-a) dx = f(a)$$

It is the same as substituting into the function for f(a), and this is exactly what the inequality tells you. It is a useful propperty of the delta function for various mathematical manipulations

3. Nov 21, 2011

### El Moriana

First of all, thanks for the reply.

I understand how the shifting property works and how it equates to the function at point x=a.

What I don't see are the "various mathematical manipulations", its usefulness.
Under what circumstances does it make life simpler to write $\int f(t)δ(t-a)dt$ than simply $f(a)$?

4. Nov 22, 2011

### lanedance

Hows your probability? one that springs to mind is as follows...

say you have two independent random variables X & Y, with joint pdf $p_{X,Y}(x,y) = p_{X}(x)p_{Y}(y)$

And say you want to find the probability distirbution for $Z= X+Y [\itex] Well p(z) will only have contibutions when z= x+y, or z-x-y=0, so you can write $$p_{Z}(z) = \int \int p_{X}(x)p_{Y}(y) \delta(z-x-y)dxdy$$ $$p_{Z}(z) = \int \int p_{X}(x)p_{Y}(z-x) dx$$ which shows the distribution of the sum of two RVs is given by the convolution of their distributions 5. Nov 22, 2011 ### lanedance In fact its almost the opposite, in that if you end up with an integral with a delta function in it, you can use the above fact to simplify the expression 6. Nov 23, 2011 ### El Moriana I'm starting to get it. Though, looking at it, would your example not result in: [itex] p_{Z}(z) = \int p_{X}(x)p_{Y}(x-z) dx$

due to the sign of the $y$ and the one integral being cancelled?

I take it you would also be able to use this to eliminate the integral w.r.t $x$ if it turns out to be more convenient?

$p_{Z}(z) = \int p_{Y}(y)p_{X}(y-z) dy$

7. Nov 23, 2011

### lanedance

Not quite you can only use it in this case because you only want contributions from where z=x+y. When integrating over y, this occurs when y=z-x in the delta function.