Practice summation finite inpulse train - Rhea

Question

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}  \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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