1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help with discrete convolution sum

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Find ##x[n] \ast h[n]## when ##x[n] = 3 u[2-n]## and ##h[n] = 4\left( \frac{1}{2} \right)^{n+2}u[n+4]## where ##u[n-k]## is the unit step function.

    2. Relevant equations
    None really

    3. The attempt at a solution

    So I know this is probably simple but I am confused.

    So the convolution is written as,

    [tex]\sum_{k=-\infty}^{\infty} 4\left( \frac{1}{2} \right)^{k+2}u[k+4](3)u[2-n+k][/tex]

    This simplifies to,

    [tex]3\sum_{k=-\infty}^{\infty} \left( \frac{1}{2} \right)^{k}u[k+4]u[2-n+k][/tex]

    The first step function of ##u[k+4]## means that the terms in the sum corresponding to ## k < -4 ## are zero, therefore this becomes the bottom limit.

    [tex]3\sum_{k=-4}^{\infty} \left( \frac{1}{2} \right)^{k}u[2-n+k][/tex]

    This form is confusing to me. I don't see an obvious limit on k except for that its LOWER limit is ##n-2##. What I tried to do is do some change of variables, such as ##k'=2-n+k##, this changes the lower limit to -2-n and the upper limit to infinity. So I have...

    [tex]3\sum_{k'=-2-n}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}u[k'][/tex]

    So this means that ##k'## must be larger than zero, which means ## n < -2##. So in this regime, the step function is 1 and I get..

    [tex]3\sum_{k'=0}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}=3\left( \frac{1}{2} \right)^{n-1}[/tex]

    Am I doing this correctly? I am just not sure. Thank you in advance for anyone that takes the time to read this!
     
  2. jcsd
  3. Jan 23, 2016 #2

    fresh_42

    Staff: Mentor

    I don't see how the substitution is any improvement. Since both step functions have to be simultaneously ##1## I get:
    [tex]3\sum_{k=max\{-4, n-2\}}^{\infty} \left( \frac{1}{2} \right)^{k}[/tex]
    which leaves you two cases. Is ##n## a natural number or another integer?
     
  4. Jan 23, 2016 #3
    ##n## is an integer. And I believe must be smaller than -2 from what I showed here. So that would mean the max value in the lower limit should be -4? Then I could just expand out until k=0 and use a geometric sum.
     
  5. Jan 23, 2016 #4

    fresh_42

    Staff: Mentor

    Yes. Except that I'm not certain about your assumption on ##n##. But that will only give you a factor depending on ##n##. If you can calculate it for ##k = -4## you should as well be able to calculate it for ##k=n-2## regardless what ##n## is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need help with discrete convolution sum
  1. Convolution help. (Replies: 4)

Loading...