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[[Category:ECE438Fall2011Boutin]]
 
[[Category:ECE438Fall2011Boutin]]
 
[[Category:problem solving]]
 
[[Category:problem solving]]
= Simplify this summation=
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[[Category:discrete Dirac delta]]
<math>u[n] \sum_{k=-7}^{15}  \delta [n-k]. </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>u[n] \sum_{k=-7}^{15}  \delta [n-k]. </math>
 +
 
 
(Justify your answer.)
 
(Justify your answer.)
 
----
 
----
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So then: <math>u[n] \sum_{k=-7}^{15}  \delta [n-k] =  \sum_{k=-7}^{15}  u[n]\delta [n-k] = \sum_{k=-7}^{15}  u[k]\delta [n-k]</math>
 
So then: <math>u[n] \sum_{k=-7}^{15}  \delta [n-k] =  \sum_{k=-7}^{15}  u[n]\delta [n-k] = \sum_{k=-7}^{15}  u[k]\delta [n-k]</math>
         <math> = \sum_{k=0}^{15}  u[k]\delta [n-k]
+
         <math> = \sum_{k=0}^{15}  \delta [n-k] = u[n]-u[n-16]</math>
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:<span style="color:green">Instructor's comments: Great job! Note that you could display your answer like this: </span>
 +
::<math> \begin{align}
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u[n] \sum_{k=-7}^{15}  \delta [n-k] & =  \sum_{k=-7}^{15}  u[n]\delta [n-k] \\
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&= \sum_{k=-7}^{15}  u[k]\delta [n-k]\\
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& = \sum_{k=0}^{15}  \delta [n-k] \\
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& = u[n]-u[n-16]
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\end{align}
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</math>
 +
 
 
===Answer 3===
 
===Answer 3===
write it here.
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::<math> \begin{align}
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u[n] \sum_{k=-7}^{15}  \delta [n-k] & =  \sum_{k=-7}^{15}  u[n]\delta [n-k] \\
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&= \sum_{k=-7}^{15}  u[k]\delta [n-k]\\\end{align}</math>
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 +
 
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because <math>\delta[n-k] = 0</math> when <math> k < 0 </math>
 +
 
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actualy <math>\delta[n-k] = 0</math> when <math> n!=k </math>
 +
 
 +
 
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<math> \begin{align} \sum_{k=-7}^{15}  u[k]\delta [n-k] & = \sum_{k=0}^{15}  \delta [n-k] \\
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& = u[n]-u[n-16]
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\end{align}
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</math>
 +
 
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===Answer 4===
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<math> \begin{align}
 +
u[n] \sum_{k=-7}^{15}  \delta [n-k] & =  \sum_{k=-7}^{15}  u[n]\delta [n-k] \\
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&= \sum_{k=-7}^{15}  u[k]\delta [n-k]\\\end{align}</math>
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:but <math>\delta[n-k] = 0</math> when <math> k < 0 </math>
 +
 
 +
same as Answer3, <math>\delta[n-k] = 0</math> when <math> n!=k </math>
 +
 
 +
so that gives us...
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:<math>\sum_{k=0}^{15}  \delta [n-k] = u[n]-u[n-16]</math>
 +
 
 +
===Answer 5===
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<math> u[n] \sum_{k=-7}^{15}  \delta [n-k] = \sum_{k=-7}^{15}  u[n]\delta [n-k]
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      = \sum_{k=-7}^{15}  u[k]\delta [n-k]
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      = \sum_{k=0}^{15}  \delta [n-k]
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</math>
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:The range of summation now changes to 0 to 15 because of the unit function u[k].
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:This is same as a unit impulse from 0 to 15.
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<math> = u[n]-u[n-16] </math>
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===Answer 6===
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<math>u[n] \sum_{k=-7}^{15}  \delta [n-k]= \sum_{k=-7}^{15}u[n]\delta[n-k]=\sum_{k=0}^{15}\delta[n-k]=u[n]-u[n-16]</math>
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 +
===Answer 7===
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<math>u[n] \sum_{k=-7}^{15}  \delta [n-k] = \sum_{k=0}^{15}  \delta [n-k] = u[n] - u[n-16] </math>
 +
 
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===Answer 8===
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<math>
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\begin{align}
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u[n] \sum_{k=-7}^{15} \delta [n-k] &=  \sum_{k=-7}^{15}  u[n] \delta [n-k] \\
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&= \sum_{k=0}^{15}  \delta [n-k] \\
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&=u[n]-u[n-16]
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\end{align}
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</math>
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===Answer 9===
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<math>u[n] \sum_{k=-7}^{15}  \delta [n-k] = \sum_{k=-7}^{15}  u[n]\delta [n-k] </math>
 +
 
 +
<math> = \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16] </math>
 +
 
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===Answer 10===
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<math>\begin{align}
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= \sum_{k=-7}^{15}  u[n]\delta [n-k] \\
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= \sum_{k=0}^{15}  \delta [n-k] \\
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= u[n]-u[n-16]
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\end{align}</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 09:24, 11 November 2013


Practice Question on "Digital Signal Processing"

Topic: Review of summations


Question

Simplify this summation:

$ u[n] \sum_{k=-7}^{15} \delta [n-k]. $

(Justify your answer.)


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

First off u[n] is nonzero for any value of n >= 0. The delta function is nonzero only for when n-k=0 or n=k. Since n must be >=0 then the values of k must conform to 0=<k<=15. This makes the function behave like u[n]-u[n-15]. I am not sure if this is completely correct.

Instructor's comments. Pretty good! You've got all the elements of the correct justification! Now can you write a justification "in maths" instead of "in words"? -pm
TA's comments. Using distributive property. the equation can be rewritten as
$ \sum_{k=-7}^{15} u[n]\delta [n-k]. $

Answer 2

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

So then: $ u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=-7}^{15} u[n]\delta [n-k] = \sum_{k=-7}^{15} u[k]\delta [n-k] $

        $  = \sum_{k=0}^{15}  \delta [n-k] = u[n]-u[n-16] $
Instructor's comments: Great job! Note that you could display your answer like this:
$ \begin{align} u[n] \sum_{k=-7}^{15} \delta [n-k] & = \sum_{k=-7}^{15} u[n]\delta [n-k] \\ &= \sum_{k=-7}^{15} u[k]\delta [n-k]\\ & = \sum_{k=0}^{15} \delta [n-k] \\ & = u[n]-u[n-16] \end{align} $

Answer 3

$ \begin{align} u[n] \sum_{k=-7}^{15} \delta [n-k] & = \sum_{k=-7}^{15} u[n]\delta [n-k] \\ &= \sum_{k=-7}^{15} u[k]\delta [n-k]\\\end{align} $


because $ \delta[n-k] = 0 $ when $ k < 0 $

actualy $ \delta[n-k] = 0 $ when $ n!=k $


$ \begin{align} \sum_{k=-7}^{15} u[k]\delta [n-k] & = \sum_{k=0}^{15} \delta [n-k] \\ & = u[n]-u[n-16] \end{align} $

Answer 4

$ \begin{align} u[n] \sum_{k=-7}^{15} \delta [n-k] & = \sum_{k=-7}^{15} u[n]\delta [n-k] \\ &= \sum_{k=-7}^{15} u[k]\delta [n-k]\\\end{align} $

but $ \delta[n-k] = 0 $ when $ k < 0 $

same as Answer3, $ \delta[n-k] = 0 $ when $ n!=k $

so that gives us...

$ \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16] $

Answer 5

$ u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=-7}^{15} u[n]\delta [n-k] = \sum_{k=-7}^{15} u[k]\delta [n-k] = \sum_{k=0}^{15} \delta [n-k] $

The range of summation now changes to 0 to 15 because of the unit function u[k].
This is same as a unit impulse from 0 to 15.
$  = u[n]-u[n-16]  $

Answer 6

$ u[n] \sum_{k=-7}^{15}  \delta [n-k]= \sum_{k=-7}^{15}u[n]\delta[n-k]=\sum_{k=0}^{15}\delta[n-k]=u[n]-u[n-16] $

Answer 7

$ u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=0}^{15} \delta [n-k] = u[n] - u[n-16] $

Answer 8

$ \begin{align} u[n] \sum_{k=-7}^{15} \delta [n-k] &= \sum_{k=-7}^{15} u[n] \delta [n-k] \\ &= \sum_{k=0}^{15} \delta [n-k] \\ &=u[n]-u[n-16] \end{align} $

Answer 9

$ u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=-7}^{15} u[n]\delta [n-k] $

$ = \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16] $

Answer 10

$ \begin{align} = \sum_{k=-7}^{15} u[n]\delta [n-k] \\ = \sum_{k=0}^{15} \delta [n-k] \\ = u[n]-u[n-16] \end{align} $


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