Difference between revisions of "Practice summation finite inpulse train" - Rhea

Latest revision as of 08:24, 11 November 2013

Practice Question on "Digital Signal Processing"

Topic: Review of summations

Question

Simplify this summation:

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

(Justify your answer.)

Share your answers below

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 $$ k < 0 $$

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

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

Back to ECE438 Fall 2011 Prof. Boutin

@@ Line 2: / Line 2: @@
 [[Category:ECE438Fall2011Boutin]]
 [[Category:problem solving]]
-= Simplify this summation=
+[[Category:discrete Dirac delta]]
- <math>u[n] \sum_{k=-7}^{15}  \delta [n-k]. </math>
+<center><font size= 4>
+'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
+</font size>
+Topic: Review of summations
+</center>
+----
+==Question==
+ Simplify this summation:
+<math>u[n] \sum_{k=-7}^{15}  \delta [n-k]. </math>
 (Justify your answer.)
 ----
@@ Line 30: / Line 43: @@
 ===Answer 3===
-write it here.
+::<math> \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}</math>
+because <math>\delta[n-k] = 0</math> when <math> k < 0 </math>
+actualy <math>\delta[n-k] = 0</math> when <math> n!=k </math>
+<math> \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}
+</math>
+===Answer 4===
+<math> \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}</math>
+:but <math>\delta[n-k] = 0</math> when <math> k < 0 </math>
+same as Answer3, <math>\delta[n-k] = 0</math> when <math> n!=k </math>
+so that gives us...
+:<math>\sum_{k=0}^{15}  \delta [n-k] = u[n]-u[n-16]</math>
+===Answer 5===
+<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]
+       = \sum_{k=0}^{15}  \delta [n-k]
+</math>
+: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.
+ <math> = u[n]-u[n-16] </math>
+===Answer 6===
+ <math>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]</math>
+===Answer 7===
+<math>u[n] \sum_{k=-7}^{15}  \delta [n-k] = \sum_{k=0}^{15}  \delta [n-k] = u[n] - u[n-16] </math>
+===Answer 8===
+<math>
+\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}
+</math>
+===Answer 9===
+<math>u[n] \sum_{k=-7}^{15}  \delta [n-k] = \sum_{k=-7}^{15}  u[n]\delta [n-k] </math>
+<math> = \sum_{k=0}^{15} \delta [n-k] = u[n]-u[n-16] </math>
+===Answer 10===
+<math>\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}</math>
 ----
 [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]