(New page: Category:ECE438Fall2011Boutin Category:problem solving = Simplify this summation= <math>\sum_{n=-\infty}^\infty n \delta [n] </math> ---- ==Share your answers below== You will rec...)
 
 
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[[Category:ECE438]]
 
[[Category:ECE438Fall2011Boutin]]
 
[[Category:ECE438Fall2011Boutin]]
 
[[Category:problem solving]]
 
[[Category:problem solving]]
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= Simplify this summation=
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[[Category:discrete Dirac delta]]
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<math>\sum_{n=-\infty}^\infty n \delta [n] </math>
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 +
<center><font size= 4>
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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</font size>
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 +
Topic: Review of summations
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 +
</center>
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----
 +
==Question==
 +
Simplify this summation
 +
 
 +
<math>\sum_{n=-\infty}^\infty n \delta [n] </math>
 
----
 
----
 
==Share your answers below==
 
==Share your answers below==
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----
 
----
 
===Answer 1===
 
===Answer 1===
−
Write it here.
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The answer is 0 ?
 +
:<span style="color:green">Instructor's comment: Yes, it's zero. Can you justify your answer? -pm</span>
 
===Answer 2===
 
===Answer 2===
−
Write it here.
+
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.  
 +
:<span style="color:green">Instructor's comment: Yes, that's the idea. Now can you justify your answer "in math" instead of "in words"? -pm
 
===Answer 3===
 
===Answer 3===
−
write it here.
+
The answer is zero since impulse function is 0 everywhere except n = 0.
 +
:<span style="color:green">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 </span>
 +
:<math>
 +
\begin{align}
 +
\sum_{n=-\infty}^\infty n \delta [n] &= \text{ blah } \\
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&= \text{ blih} \\
 +
&= 0
 +
\end{align}
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</math>
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:<span style="color:green">Can you try that? -pm </span>
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===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] ... \\
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&= ...(-2)\cdot 0 + (-1)\cdot 0 + 0\cdot 1 + 0 + 2\cdot 0... \\
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&= ...0 + 0 + 0 + 0 + 0... \\
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&= 0
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\end{align}
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</math>
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:<span style="color:green">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 </span>
 +
===Answer 5===
 +
Unless you want a deeper proof of the discrete sifting property:
 +
:<math>
 +
\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}
 +
</math>
 +
:<span style="color:green">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: </span>
 +
::<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}
 +
</math>
 +
----
 +
<span style="color:green">Instructor's comment: Can anybody write a solution based on this little "trick" I have used several times in class already? -pm </span>
 +
----
 +
===Answer 6===
 +
We know that <math>x[n]\delta [n-k] = x[k]\delta [n-k]</math>
 +
 
 +
Then: using <math>x[n]=n</math> and <math> k=0 </math> we get:
 +
<math>\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0</math>
 +
:<span style="color:green">Instructor's comments: I think this solution is nice and sweet. It does use the "trick" I showed in class. To get full credit, it would be sufficient to write:</span>
 +
::<math>\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0</math>
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:<span style="color:green">-pm </span>
 +
==Answer 7==
 +
<math>\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0</math>
 +
because everywhere else besides when n=0,
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<math> \delta[n] = 0</math>
 +
 
 +
===Answer 8===
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<math>\delta[n] = 1 </math> only when n = 0.
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:Since <math>\delta[n] = 0 </math> when n is not equal to 0,
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:<math>
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\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0
 +
</math>
 +
 
 +
===Answer 9===
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When <math>n\neq0 </math>, <math>\delta[n] = 0 </math>
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:Also when n = 0, <math>\delta[n] = 1 </math> so...
 +
:<math>
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\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0
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</math>
 +
 
 +
===Answer 10===
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:<math>\delta[n] = 0 </math> everywhere except when n = 0. Therefore, at n = 0, the value is 0.
 +
:<math>
 +
\sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0
 +
</math>
 +
 
 +
 
 +
===Answer 11===
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when n=0, <math>n\delta[n]=0</math>
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when <math>n\neq0 </math>, <math>\delta[n]=0, n\delta[n]=0</math>
 +
therefore <math>\sum_{n=-\infty}^\infty n \delta [n] </math> for all n
 +
 
 +
===Answer 12===
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<math>\sum_{n=-\infty}^\infty n \delta [n] = n \cdot 1  \Bigg|_{n =0} = 0</math>
 +
 
 +
===Answer 13===
 +
Since the range is from <math>-\infty</math> to <math>\infty</math>,the range is totally symmetrical.When we get an <math> n\delta[n]</math>,we will get an <math> -n\delta[-n]</math> with the same amplitude but reverse direction, to which,the sum will be zero. And when n=0,the value of <math> n\delta[n]</math> is zero. Thus the summation in the question is zero.
 +
:<span style="color:green">TA's comments: Are you arguing that <math>n\delta [n]</math> is an odd function of n so that the summation over the range of <math>(-\infty,\ \infty)</math> equals zero? -pm</span>
 +
 
 +
===Answer 14===
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<math>\sum_{n=-\infty}^\infty n \delta [n] = 0 \delta [0] = 0 </math>
 +
 
 +
===Answer 15===
 +
Evaluate: 
 +
:<math>\sum_{n=-\infty}^\infty n \delta [n] </math>
 +
Given: 
 +
:<math>\delta [n] = 0 \; \forall n \neq 0 </math>
 +
Therefore:
 +
:<math>\sum_{n=-\infty}^\infty n \delta [n] = 0\delta [0] = 0 </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]]

Latest revision as of 08:25, 11 November 2013


Practice Question on "Digital Signal Processing"

Topic: Review of summations

Question

Simplify this summation

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

Share your answers below

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} $

Instructor's comment: Can anybody write a solution based on this little "trick" I have used several times in class already? -pm

Answer 6

We know that $ x[n]\delta [n-k] = x[k]\delta [n-k] $

Then: using $ x[n]=n $ and $ k=0 $ we get: $ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $

Instructor's comments: I think this solution is nice and sweet. It does use the "trick" I showed in class. To get full credit, it would be sufficient to write:
$ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $
-pm

Answer 7

$ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $ because everywhere else besides when n=0, $ \delta[n] = 0 $

Answer 8

$ \delta[n] = 1 $ only when n = 0.

Since $ \delta[n] = 0 $ when n is not equal to 0,
$ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $

Answer 9

When $ n\neq0 $, $ \delta[n] = 0 $

Also when n = 0, $ \delta[n] = 1 $ so...
$ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $

Answer 10

$ \delta[n] = 0 $ everywhere except when n = 0. Therefore, at n = 0, the value is 0.
$ \sum_{n=-\infty}^\infty n \delta [n] = \sum_{n=-\infty}^\infty 0 \delta [n] = 0 $


Answer 11

when n=0, $ n\delta[n]=0 $ when $ n\neq0 $, $ \delta[n]=0, n\delta[n]=0 $ therefore $ \sum_{n=-\infty}^\infty n \delta [n] $ for all n

Answer 12

$ \sum_{n=-\infty}^\infty n \delta [n] = n \cdot 1 \Bigg|_{n =0} = 0 $

Answer 13

Since the range is from $ -\infty $ to $ \infty $,the range is totally symmetrical.When we get an $ n\delta[n] $,we will get an $ -n\delta[-n] $ with the same amplitude but reverse direction, to which,the sum will be zero. And when n=0,the value of $ n\delta[n] $ is zero. Thus the summation in the question is zero.

TA's comments: Are you arguing that $ n\delta [n] $ is an odd function of n so that the summation over the range of $ (-\infty,\ \infty) $ equals zero? -pm

Answer 14

$ \sum_{n=-\infty}^\infty n \delta [n] = 0 \delta [0] = 0 $

Answer 15

Evaluate:

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

Given:

$ \delta [n] = 0 \; \forall n \neq 0 $

Therefore:

$ \sum_{n=-\infty}^\infty n \delta [n] = 0\delta [0] = 0 $

Back to ECE438 Fall 2011 Prof. Boutin

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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