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
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
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?
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
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?