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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 $
$ \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 $
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} $
