Contents
Simplify this summation
$ \sum_{n=-\infty}^\infty n \delta [n] $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
The answer is 0 ?
- Instructor's comment: Yes, it's zero. Can you justify your answer? -pm
Answer 2
The delta function is zero everywhere except when n=0 and since we are multiplying the delta by n the answer would thus be 0.
- Instructor's comment: Yes, that's the idea. Now can you justify your answer "in math" instead of "in words"? -pm
Answer 3
The answer is zero since impulse function is 0 everywhere except n = 0.
- Instructor's comments: Ok, I guess I am going to have to be a bit more specific. I would like to see a way to answer this question as a sequence of small changes to this expression until you get to zero. Something like
- $ \begin{align} \sum_{n=-\infty}^\infty n \delta [n] &= \text{ blah } \\ &= \text{ blih} \\ &= 0 \end{align} $
- Can you try that? -pm
Answer 4
- $ \begin{align} \sum_{n=-\infty}^\infty n \delta [n] &= ...(-2)\delta[-2] + (-1)\delta[-1] + 0\delta[0] + \delta[1] + 2\delta[2] ... \\ &= ...(-2)\cdot 0 + (-1)\cdot 0 + 0\cdot 1 + 0 + 2\cdot 0... \\ &= ...0 + 0 + 0 + 0 + 0... \\ &= 0 \end{align} $
- Instructor's comments: It's awesome to see Purdue alums participate in this collective problem solving! As expected from a graduate of our ECE program, this solution does not contain any mistake. Now a challenge: can anybody do it without using any "dot dot dot"? -pm
Answer 5
Unless you want a deeper proof of the discrete sifting property:
- $ \begin{align} \sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \Rightarrow \sum_{n=-\infty}^\infty f[n] \delta [k-n] \Bigg|_{f[n]=n, \ k=0} = \sum_{n=-\infty}^\infty n \delta [n] = 0 \end{align} $
