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\sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \Rightarrow | \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 | \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} | ||
| + | </math> | ||
| + | :Instructor's comments: You correctly observed that the sifting property applies in this case. Good job! Your answer is correct, but perhaps it would flow better (more logically) if you changed the other of the first two inequalities: | ||
| + | ::<math> | ||
| + | \begin{align} | ||
| + | \sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \text{ (sifting property)}\\ | ||
| + | \Rightarrow\sum_{n=-\infty}^\infty n \delta [n] &= \sum_{n=-\infty}^\infty f[n] \delta [k-n] \Bigg|_{f[n] =n, \ k=0} &= 0 | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 10:15, 27 August 2011
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} $
- Instructor's comments: You correctly observed that the sifting property applies in this case. Good job! Your answer is correct, but perhaps it would flow better (more logically) if you changed the other of the first two inequalities:
- $ \begin{align} \sum_{n=-\infty}^\infty f[n] \delta [k-n] &= f[k] \text{ (sifting property)}\\ \Rightarrow\sum_{n=-\infty}^\infty n \delta [n] &= \sum_{n=-\infty}^\infty f[n] \delta [k-n] \Bigg|_{f[n] =n, \ k=0} &= 0 \end{align} $
