how do i prove this summation?

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

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

thank you and your help is appreciated
• Jan 15th 2011, 01:57 PM
Ackbeet
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 PM
TechnicianEngineer

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

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