Unit impulse signal. Evaluating an integral

Discussion in 'Math' started by Jess_88, Aug 14, 2011.

  1. Jess_88

    Jess_88 Thread Starter Member

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    Hey guys.

    I have the following example in my notes but there is no working.
    Could someone please explain how to reach to solution?
    Is he acutely using integration or is he just looking at the "sift property" and noting values of t0, t1, t2 to come to a conclusion.

    Here is the question
    [​IMG]

    The Solution
    [​IMG]

    Sifting property noted earlier in my notes
    [​IMG]

    thanks a lot guys :)
  2. someonesdad

    someonesdad Senior Member

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    It's shifting property, not sifting property. If it was sifting, you'd use it in the kitchen with flour. :p

    The solution is staring you in the face. One way to think of the delta function is that it is a continuous analog of the Kronecker delta. It is often used to evaluate an expression at a particular point. Thus, in the example, the function x is evaluated at t = 4.

    Here's the analogy. For a discrete sum using the Kronecker delta (and I'm using the Einstein summation convention),

    x_i = \delta_{ij} x_j.

    The continuous analog is

    \int x(t) \delta(t - i) dt = x(i)

    Assume the integration is over the real line.
  3. guitarguy12387

    guitarguy12387 Active Member

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  4. someonesdad

    someonesdad Senior Member

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    Absolutely not -- truth is truth. I just learned about the Dirac delta function in the 60's and never heard of the sifting property -- looks like they invented it after I studied it. Gave me a good laugh and I appreciate the education!

    Oh, another explanation is that I missed that day in class. :p
  5. guitarguy12387

    guitarguy12387 Active Member

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    Hehehe

    Was that when frequency was spec'd in CPS instead of HZ? :D
  6. Jess_88

    Jess_88 Thread Starter Member

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    good stuff.
    Yeah I was wondering what was going on there. My lecture notes said "Sifting" like 20 times haha.
    cheers guys
    :)

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