Topic: Review of summations

## Question

Simplify this summation:


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

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

We know that $x[n]\delta[n-k]=x[k]\delta[n-k]$

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

        $= \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16]$

\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}

\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}

\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...

$\sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16]$

$u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=-7}^{15} u[n]\delta [n-k] = \sum_{k=-7}^{15} u[k]\delta [n-k] = \sum_{k=0}^{15} \delta [n-k]$

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.
$= u[n]-u[n-16]$


$u[n] \sum_{k=-7}^{15} \delta [n-k]= \sum_{k=-7}^{15}u[n]\delta[n-k]=\sum_{k=0}^{15}\delta[n-k]=u[n]-u[n-16]$


$u[n] \sum_{k=-7}^{15} \delta [n-k] = \sum_{k=0}^{15} \delta [n-k] = u[n] - u[n-16]$

\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}

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

$= \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16]$

\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}