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Basic Convolution - if someone wouldn't mind checking

  1. Apr 8, 2007 #1
    I have two convolution problems, that I would like to be sure are right. If someone wouldn't mind the fun job of checking these, that would be great.


    Compute the convolution of [itex] x(t) [/itex] and [itex] h(t) [/itex] where:

    [tex] x(t) = u(t) [/tex]
    [tex] h(t) = e^{-at}u(t), \,\,\, a>0 [/tex]

    [tex] x(t) = u(t) - u(t-T_1) [/tex]
    [tex] h(t) = u(t) - u(t-T_2), \,\,\, T_1 > T_2 [/tex]

    Let [itex] y(t) [/itex] be the convolution of x(t), h(t)


    [tex] t < 0 [/tex]:
    y(t) =0

    [tex] t \geq 0 [/tex]:
    [tex] y(t) = \int_{-\infty}^t e^{-a(t-\tau)}d\tau = \frac{1}{a}[/tex]


    [tex] t < 0 [/tex] or [tex] t-T_2 > 0 [/tex]:
    y(t) =0

    [tex] t \geq 0 [/tex] and [tex] t-T_2 < 0 [/tex]:
    [tex] y(t) = \int_0^t d\tau = t [/tex]

    [tex] t-T_2 > 0 [/tex] and [tex] t<T_1 [/tex]:
    [tex] y(t) = \int_{t-T_2}^t d\tau = T_2 [/tex]

    If I need to show more steps, please let me know. Thanks!
  2. jcsd
  3. Apr 8, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Is u(t) a specific function- say, the unit step function?
  4. Apr 8, 2007 #3
    My apologies. I thought that was the standard notation. Yes, u(t) is the step function.

    EDIT: I see how I should have said that to begin with. It is rather vague in the opening question.
  5. Apr 8, 2007 #4

    D H

    Staff: Mentor

    Please show some additional work, as I get answers different from yours.
  6. Apr 8, 2007 #5
    I don't think your first integral is correct. For [itex]t \geq 0 [/itex],

    [tex] \int_{0}^{t}x(t) h(t-\tau) \ d \tau = \int_{0}^{t} e^{-a(t-\tau)} \ d \tau[/tex].
  7. Apr 8, 2007 #6

    D H

    Staff: Mentor

    Corneo, I did not post that result because the intent of the homework section is the help the student get to the correct answer. If you give the correct answer, how does that help the student?
  8. Apr 8, 2007 #7
    Is what I posted considered too much help? I just wanted to show the OP the setup was wrong. I didn't think I solved the problem already. But I guess I can keep this in mind next time.
  9. Apr 8, 2007 #8
    I think you were fine. I redid the first one and got this:

    [tex] y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau [/tex]

    Setting up the bounds gives:
    [tex] x(\tau) [/tex] --- a step function in the [itex] \tau [/itex] domain.
    [tex] h(t-\tau) [/tex] --- (this is kind of hard to describe), but... a inverted decaying exponential that has a value from [itex] -\infty [/itex] to [itex] t [/itex] in the [itex] \tau [/itex] domain.

    We now look at the overlap regions:

    [tex] t<0 [/tex] , [tex]y(t) = 0 [/tex]

    [tex] t \geq 0 [/tex] , [tex] y(t) = \int_0^t e^{-a(t-\tau)}d\tau = \frac{1}{a}(1- e^{-at}) [/tex]

    Thus, [tex] y(t) = \frac{u(t)}{a} (1 - e^{-at}) [/tex]

    Does this look better?
  10. Apr 8, 2007 #9
    For the second problem.

    x(t) = u(t) - u(t - T1)
    y(t) = u(t) - u(t-T2)


    [tex] x(\tau) [/tex] is a function that is 1 from 0 to T1
    [tex] h(t-\tau) [/tex] is a function that is 1 from t-T2 to t

    Thus the bounds are:

    [tex] t< 0 [/tex]: no overlap, [tex] y(t) = 0 [/tex]

    [tex] t-T_2 > T_1 [/tex]: no overlap, [tex] y(t) = 0 [/tex]

    [tex] t>0 [/tex] AND [tex] t-T_2 < 0 [/tex]: partial overlap (left side)
    [tex] y(t) = \int_0^t d\tau = t [/tex]

    [tex] t-T_2 > 0 [/tex] AND [tex] t< T_1 [/tex] h(t-tau) is fully enclosed within x(tau)
    [tex] y(t) = \int_{t-T_2}^t d\tau = T_2 [/tex]

    [tex] t-T_2 < T_1 [/tex] AND [tex] t> T_1 [/tex]: partial overlap (right side)
    [tex] y(t) = \int_{t-T_2}^{T_1} d\tau = T_1 + T_2 -t [/tex]

    Is this right? Thanks again you guys
  11. Apr 8, 2007 #10
    This one looks right.
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