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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}  u[k]\delta [n-k]</math>
 
===Answer 3===
 
===Answer 3===
 
write it here.
 
write it here.
 
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[[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 05:59, 3 September 2011

Simplify this summation

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

(Justify your answer.)


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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}  u[k]\delta [n-k] $

Answer 3

write it here.


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