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[[Category:ECE438Fall2011Boutin]] | [[Category:ECE438Fall2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
| − | = Simplify this summation | + | [[Category:discrete Dirac delta]] |
| − | + | ||
| + | <center><font size= 4> | ||
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Review of summations | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==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} \delta [n-k] = u[n]-u[n-16]</math> | <math> = \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16]</math> | ||
| + | :<span style="color:green">Instructor's comments: Great job! Note that you could display your answer like this: </span> | ||
| + | ::<math> \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} | ||
| + | </math> | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
| − | + | ::<math> \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}</math> | ||
| + | |||
| + | |||
| + | because <math>\delta[n-k] = 0</math> when <math> k < 0 </math> | ||
| + | |||
| + | actualy <math>\delta[n-k] = 0</math> when <math> n!=k </math> | ||
| + | |||
| + | |||
| + | <math> \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} | ||
| + | </math> | ||
| + | |||
| + | ===Answer 4=== | ||
| + | <math> \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}</math> | ||
| + | :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... | ||
| + | :<math>\sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16]</math> | ||
| + | |||
| + | ===Answer 5=== | ||
| + | <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] | ||
| + | = \sum_{k=0}^{15} \delta [n-k] | ||
| + | </math> | ||
| + | :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. | ||
| + | <math> = u[n]-u[n-16] </math> | ||
| + | ===Answer 6=== | ||
| + | |||
| + | <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> | ||
| + | |||
| + | ===Answer 7=== | ||
| + | <math>u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=0}^{15} \delta [n-k] = u[n] - u[n-16] </math> | ||
| + | |||
| + | ===Answer 8=== | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | ===Answer 9=== | ||
| + | <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> | ||
| + | |||
| + | ===Answer 10=== | ||
| + | <math>\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}</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 08:24, 11 November 2013
Practice Question on "Digital Signal Processing"
Topic: Review of summations
Contents
Question
Simplify this summation:
$ u[n] \sum_{k=-7}^{15} \delta [n-k]. $
(Justify your answer.)
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} $
