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</math>
 
</math>
 
:Can you try that? -pm
 
:Can you try that? -pm
 +
===Answer 4===
 +
:<math>
 +
\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}
 +
</math>
 
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]
 
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Revision as of 12:11, 26 August 2011

Simplify this summation

$ \sum_{n=-\infty}^\infty n \delta [n]  $

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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} $

Back to ECE438 Fall 2011 Prof. Boutin

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Ryne Rayburn