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Math Help - how do i prove this summation?

  1. #1
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    how do i prove this summation?

    I need a simple solution on how to prove this so that i can solve a convolution problem:

     \displaystyle \sum_{k=-\infty}^{\infty} x(k)\delta(n-k-m) = x(n-m)
    where m is a positive integer

    thank you and your help is appreciated
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  2. #2
    A Plied Mathematician
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    Assuming your \delta is the Kronecker delta, and not the Dirac delta, the proof is actually quite straight-forward, and follows from the definition of the Kronecker delta, which is as follows:

    \delta(n)=\begin{cases}1,\quad n=0\\ 0,\quad n\not=0\end{cases}.

    So your delta function is 1 when what happens? What implications does that have for your summation?
    Last edited by Ackbeet; January 15th 2011 at 01:58 PM. Reason: Commas.
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  3. #3
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    thank you for replying to my thread

    the given is a convolution of a discrete signal which is shown in the equation below
    x(n)*\delta(n-m) = \displaystyle \sum_{k=-\infty}^{\infty} x(k)\delta(n-k-m) = x(n-m)

    i need to find how it equates to x(n-m) and i quite need the solution for it.
    i am newb in summation and forgot about how to do it
    and yes that is the delta dirac symbol not the Kronecker delta


    your help is appreciated
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  4. #4
    A Plied Mathematician
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    Oops, I think I didn't define the Kronecker delta correctly. I think what I defined actually was the discrete Dirac delta. The Dirac delta, in this case, "picks out" one term in a sum (this is actually how the continuous version of the Dirac delta distribution is defined: with an integral). So, you have the following:

    \delta(n-k-m)=\begin{cases}1,\quad n-k-m=0\\0,\quad n-k-m\not=0\end{cases}=\begin{cases}1,\quad n-m=k\\0,\quad n-m\not=k\end{cases}.

    So the delta in your sum is 1 only when k=n-m, and zero elsewhere, which means that every single term in your sum is zero except for the one when k=n-m. Do you see where this is going?
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