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

$\displaystyle \displaystyle \sum_{k=-\infty}^{\infty} x(k)\delta(n-k-m) = x(n-m)$

where m is a positive integer

thank you and your help is appreciated

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- Jan 15th 2011, 11:47 AMTechnicianEngineerhow do i prove this summation?
I need a simple solution on how to prove this so that i can solve a convolution problem:

$\displaystyle \displaystyle \sum_{k=-\infty}^{\infty} x(k)\delta(n-k-m) = x(n-m)$

where m is a positive integer

thank you and your help is appreciated - Jan 15th 2011, 01:57 PMAckbeet
Assuming your $\displaystyle \delta$ 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:

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

So your delta function is 1 when what happens? What implications does that have for your summation? - Jan 15th 2011, 03:07 PMTechnicianEngineer
thank you for replying to my thread

the given is a convolution of a discrete signal which is shown in the equation below

$\displaystyle x(n)*\delta(n-m) = \displaystyle \sum_{k=-\infty}^{\infty} x(k)\delta(n-k-m) = x(n-m)$

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 - Jan 15th 2011, 03:28 PMAckbeet
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:

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

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