| Line 16: | Line 16: | ||
::<math>\sum_{k=-7}^{15} u[n]\delta [n-k].</math> | ::<math>\sum_{k=-7}^{15} u[n]\delta [n-k].</math> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | + | We know that <math>x[n]\delta[n-k]=x[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] | ||
===Answer 3=== | ===Answer 3=== | ||
write it here. | write it here. | ||
---- | ---- | ||
[[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 04:59, 3 September 2011
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} u[k]\delta [n-k] ===Answer 3=== write it here. ---- [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] $
