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Help with convolution bounds

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Mainly concerned with part (a). Here's the answer:

    I understand where the answers inside the bracket came from, but I don't understand how they got their bounds (-infinity to 3, 3 to 5, and 5 to infinity)

    2. Relevant equations
    x is the impulse function here so y(t) = ∫h(τ)x(t-τ)dτ [-∞,∞]

    3. The attempt at a solution
    Since x was chosen to be the impulse function, I start by reflecting it across the y-axis and then shifting it by t to get the graph x(t-τ):
    (minor mistake in my graph here: that τ-5 should be t-5 and the τ-3 should be t-3)

    I then start "sliding" this graph into h(τ):

    And this is where my confusion starts. If I perform the integration it'll be:

    ∫e^(-3τ)dτ [0, t-3]

    and I get the answer they have for when 3 < t ≤ 5

    So basically my question is why is it 3 < t ≤ 5 here?
  2. jcsd
  3. Feb 16, 2015 #2


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    I don't understand your method of 'sliding'.
    I wrote out the integrand as ##(u(t-3-\tau)-u(t-5-\tau))u(\tau)e^{-3\tau}.d\tau##. It is clear that this vanishes outside ##t-3>\tau>t-5##. Write the resulting integral.
  4. Feb 16, 2015 #3
    I'm using the "flip and shift method" like here: http://www-rohan.sdsu.edu/~jiracek/DAGSAW/4.2.html

    I think I may understand the bounds now though.
    In the picture I have illustrated where the integrand is from 0 to t-3, it's only valid when t-5 < 0 and t - 3> 0. In other words, 3 < t ≤ 5.

    When I slide it in some more so that t-5 is now included under the graph h(τ), the integrand would be from t-5 to t-3, and it's then valid from t = 5 to infinity since h(τ) goes on forever
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