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!

Convolution integral problem

  1. Apr 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi all, I hope you all can help me
    so I'm studying for my signals course and I encounter this example in the book, and the answer is there but the solution isn't..... The convolution integral exists for 3 intervals and I could evaluate the first two just fine..... however I can't find the limits of integration of the third.
    The third one corresponds to the fourth in the picture which is 2T < t < 3T
    The question and answer is shown in the picture attached

    Thank you!!

    2. Relevant equations
    No relevant equations

    3. The attempt at a solution
    I believe that the lower limit should be -2T + t, and the upper limit should be to T.
     

    Attached Files:

  2. jcsd
  3. Apr 20, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Limit for what ? This will become clear when you post your relevant equation: the definition of a convolution. Per that definition, the integration limits for ##\tau## are ##-\infty## and ##+\infty##. Of course for most ##\tau## the contribution is zero. Except for ##\tau\in [0,T]##
     
  4. Apr 20, 2017 #3
    Yes actually the relevant equation would be
    x(t) * h(t) = (from -∞ to +∞) ∫x(k)h(t - k)dk where K was used in this case instead of tau to avoid confusion.......
    However as shown in the file attached there is more than one interval of integration.......
    I need the limits of the fourth one where 2T < t < 3T
     
  5. Apr 21, 2017 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You mean: 2T < t - k < 3T

    Remember: k is your integration variable, NOT t ! And its limits are ##-\infty## and ##+\infty## .
     
  6. Apr 22, 2017 #5
    true, it's limits are from -∞ to ∞
    however, what is the range on which evaluated integral is nonzero....... when 2T < t (is constant) < 3T
    if you check the picture I attached there are many intervals on which the integral is evaluated
    on the fourth interval 2T < t < 3T, when is the range on which the integral is nonzero?
     
  7. Apr 22, 2017 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The point is that k is the integration variable and t - k is the argument of the function h(). You want to intgrate the section where the argument is in ##[2T, 3T]##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted