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Convolution Dirac impulse and periodic signal

  1. Nov 2, 2014 #1
    Hi ☺️ i have to do a convolution with a periodic signal and a dirac impulse:
    x(t)=sen(πt)(u(t)−u(t−2))
    h(t)=u(t−1)−u(t−3)
    The first is a periodic graph that intersect axis x in points 0 , 1 and 2 (ecc)
    The se ing is a rectangle ( Dirac impulse ) that intersect AxiS x in points 1 and 3.
    For t−1<0 and t−3>0 convolution doesn't exist
    For me is ∫to 0 from 1 ((senπ(t−τ)dτ) − ∫to 1 from t−1 (senπ(t−τ)dτ))) but my book write ∫to t-1 from 0 (senπtdτ)
    Could someone tell me why he cancel -τ and how to do this ex without Fourier ? Thank you very much and Sorry for my terrible english but i'm italian
     
  2. jcsd
  3. Nov 3, 2014 #2

    vela

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    I wouldn't say the convolution doesn't exist. I think you mean x*h(t) = 0. If so, it's not true that it's equal to 0 for t-3>0.

    I suspect the book simply interchanged the roles of x and h compared to what you did. The convolution is given by
    $$(x*h)(t) = \int_{-\infty}^\infty x(\tau)h(t-\tau)\,d\tau,$$ and since x(t)=0 for t<0 and t>2, you can write
    $$(x*h)(t) = \int_{0}^2 (\sin\pi\tau)h(t-\tau)\,d\tau.$$ The variable that's in the argument of sine should be ##\tau##, not ##t## as you said the book has.
     
  4. Nov 3, 2014 #3

    rude man

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    What is "sen(πt)" ? Is "sen" Italian for "sin"? I will assume so.
    So we have
    f(t) = sin(πt){u(t) - u(t-2)}
    g(t) = u(t-1) -u(t-3)
    You are probably best off with
    f*g = integral from T = -∞ to +∞ of f(t-T)g(T)dT.

    The trick here is to determine how the various u(t) combine. You will need the fact that u(t-T) = u{-(T-t)}.
    If you graph the two u terms (one in f, the other in g) then you will see that the limits of integration change from (-∞, +∞) to (0, t-1).

    The final answer should include a coefficient {u(t-1) - u(t-3)} which has the effect of making the convolution = 0 for t < 1 and > 3.
     
  5. Nov 3, 2014 #4

    rude man

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    ???
    The convolution g*t is a function of t. After you've performed the integration you need a "t" in the sine function I think.
    You also used 2 instead of 3 for the upper limit on t.
     
  6. Nov 3, 2014 #5

    vela

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    That's wrong.

    The sine I was referring to is inside the integral, so no, it's not supposed to be a function of t. If it were, it would be a constant and you could pull it out of the integral.

    Because x(t) is non-zero only between 0 and 2. And it's the upper limit on ##\tau##, not t.
     
  7. Nov 3, 2014 #6

    rude man

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    Oh? How?

    After integrating to constant limits it is a constant except for modulation by u(t) functions. I never suggested it could be "pulled out of the integral".

    The upper limit on T is t-1.

    The OP will have to decide I guess which of us is right. For the record, my answer agrees with what is in his/her textbook.
     
    Last edited: Nov 3, 2014
  8. Nov 3, 2014 #7

    vela

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    Here's a plot of ##x(\tau)## and ##h(t-\tau)## for ##t=4##. Still think the convolution is 0 when t>3?
     

    Attached Files:

  9. Nov 3, 2014 #8

    rude man

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    It is if the problem states it is, which it does: "For t−1<0 and t−3>0 convolution doesn't exist."
     
  10. Nov 3, 2014 #9

    vela

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    I don't know if you're trying to be deliberately obtuse. The OP wrote that the book said the integral was
    $$\int_0^{t-1} \sin \pi t\,d\tau.$$ ##\sin\pi t## is independent of the variable of integration ##\tau##, so that integral is equal to
    $$\int_0^{t-1} \sin \pi t\,d\tau = \sin\pi t \int_0^{t-1}d\tau = (t-1)\sin\pi t.$$ Are you claiming that's correct?
     
  11. Nov 3, 2014 #10

    rude man

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    Of course not.
    I think this discussion has borne all the fruit it can. Thanks for your patience.
     
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