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Sifting Property of the Impulse Function

  1. Nov 21, 2011 #1
    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 [itex]f(t)[/itex] is continuous at [itex]t=a[/itex] then

    [itex]\int_{-∞}^∞f(t)δ(t-a)dt= f(a)[/itex]

    from: Glyn James, Advanced Modern Engineering Mathematics (3rd Ed), Section 2.5, p.155
     
  2. jcsd
  3. Nov 21, 2011 #2

    lanedance

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    so based on the properties of the delta function you know
    [tex]\int \delta (x) dx = 1 [/tex]

    A handwaving explanation is that if f is continuous and if you zoom in on a small enough region [itex] (x-\epsilon, x+\epsilon) [/itex], 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
    [tex]\int f(x)\delta (x-a) dx = f(a) [/tex]

    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
     
  4. Nov 21, 2011 #3
    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 [itex]\int f(t)δ(t-a)dt[/itex] than simply [itex]f(a)[/itex]?
     
  5. Nov 22, 2011 #4

    lanedance

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

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

    And say you want to find the probability distirbution for [itex] Z= X+Y [\itex]

    Well p(z) will only have contibutions when z= x+y, or z-x-y=0, so you can write
    [tex] p_{Z}(z) = \int \int p_{X}(x)p_{Y}(y) \delta(z-x-y)dxdy [/tex]
    [tex] p_{Z}(z) = \int \int p_{X}(x)p_{Y}(z-x) dx [/tex]

    which shows the distribution of the sum of two RVs is given by the convolution of their distributions
     
  6. Nov 22, 2011 #5

    lanedance

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    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
     
  7. Nov 23, 2011 #6
    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 [/itex]

    due to the sign of the [itex]y[/itex] and the one integral being cancelled?

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

    [itex] p_{Z}(z) = \int p_{Y}(y)p_{X}(y-z) dy [/itex]
     
  8. Nov 23, 2011 #7

    lanedance

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    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.
     
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