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Homework Help: Convolution shortcut?

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Find (use shortcut):

    x(t) = 2e-4tu(t) * e2tu(t) * t2σ(t - 2)

    2. Relevant equations

    Convolution properties:

    # "shape of Y (output) is different from x1, x2"

    # x1 * x2 = x2 * x1

    # x1 * (x2 + x3) = (x1 * x2) + (x1 * x3)

    # x1(t) = * x2(t) * x3(t) * .....

    # step * ramp = parabolic function

    # eatu(t) * ebtu(t) = (1/a-b)[eat - ebt]u(t)

    3. The attempt at a solution

    2e-4tu(t) * e2tu(t) * t2σ(t - 2)

    just doing

    2e-4tu(t) * e2tu(t) for now, and using the last property listed:

    = 2(1/(2+4))[e-4t - e2t]u(t)

    = (1/3)[e-4t - e2t]u(t)

    I am unsure how to go about doing convolution with the * t2σ(t - 2) at this point, please help. Thanks!
    Last edited: Sep 10, 2014
  2. jcsd
  3. Sep 10, 2014 #2

    rude man

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    What is σ(t)? I've never seen that function before.

    What are you trying to find? You already have x(t). Simplifying?
  4. Sep 10, 2014 #3
    It's a impulse function supposedly (and sadly I don't seem to have any property listed where it's involved). And yes, trying to simplify and do convolution with all of this.
  5. Sep 10, 2014 #4

    rude man

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    OK, I guess σ(t) is what the rest of us call δ(t)?

    I suppose you can convolve the first & 2nd terms, the convolve the result with the third.

    So why not write the formal convolution integral for the first two terms for a starter.

    I'm afraid I don't know any 'shortcut' here.
  6. Sep 11, 2014 #5
    Hah, yes, δ(t). (Of course I forgot the 'δ' symbol was here and my poor eyesight never saw it.)


    (f * g)(t) = 0∫tf(t - τ)g(τ)dτ then, is this it?

    So if f = 2e-4tu(t) and g = e2tu(t). So it's something like

    = 0∫t[2e-4(t - τ)u(t - τ)eu(τ)]dτ

    I had to change all the function variables (including inside the 'u(t)') to τ also right? Do I have to also end up integratring the 'u(τ)'? If so, how?
  7. Sep 11, 2014 #6

    rude man

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    Permit me to write T = tau, it's a lot easier typing ...

    So, proceeding from that integral, with f(t-T) = exp{2(t - T)}U(t - T) and g(T) = 2exp{-4T}U(T), I would change t - T to -(T - t) everywhere in the integrand. Now, graph U(T) and -U(T - t) vs. T and multiply them. Then get rid of the two U functions in the integrand by suitable choice of the upper and lower limits of integration. Then, perform the integration.

    The second convolution is done the same way. Remember the sampling property of the delta function: ∫f(x)δ(x - a)dx with upper limit +∞ and lower limit -∞ = f(a).
  8. Sep 11, 2014 #7


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    use laplace!!!
    convolution in the time domain becomes multiplication in the frequency domain.
  9. Sep 11, 2014 #8

    rude man

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    Laplace is limited to t > 0. This is not specified here. But you could use Fourier. I have looked at that but found it harder to do the inversion integral back to the time domain than just plowing ahead solely in the time domain.

    BTW there is also the double-sided Laplace transform which no one uses.
  10. Sep 13, 2014 #9
    Sorry for getting back to this late

    So you have the convolution for the first two terms as:

    = 0t[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

    Does changing it to -(T - t) help simply it? It would appear as this:

    = 0t[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT

    Can you explain this? Graph any unit step function?

    In the class this convolution problem actually came before we formally covered Laplace / Fourier (which I will also need help on later).
  11. Sep 13, 2014 #10

    rude man

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    I just noticed you meant the lower limit of integration to be 0. It should be -∞. (BTW the upper limit should also be ∞. You can make the upper limit t in this case, it makes no difference here). Otherwise, good.
    Yes, it clarifies the next step.
    Yes I can. I goofed! :redface:
    I should have said "graph U(T) and U[-(T-t)] vs. T".

    Can you graph U(x) vs. x?
    U(x - a) vs. x?
    U(-x) vs. x?
    U[-(x-a)} vs. x?
    You can do them all just by remembering
    U(ζ) = 0, ζ < 0
    U(ζ) = 1, ζ > 0.
  12. Sep 15, 2014 #11
    Should it be -∞ to +∞ for the whole property or just for this problem? I don't think I quite understand that. I'll take your word for it though, so I think it would now be:

    -∞[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

    Isn't this just the unit step graph property? Or what's the name or term for this if it isn't? What would be 'ζ' (if this is what you are basing the graph on)? I still can not see how this is supposed to help unless you are trying to hint at doing graphical convolution.
  13. Sep 15, 2014 #12

    rude man

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    That is the general definition for the convolution integral. Often, depending on the problem, -∞ can be replaced by 0 and +∞ by t, but it's always correct to use the infinite + and - limits. Just keep in mind that if the integrand is identically zero below and/or above certain values then the infinite limits can be replaced by those respective values. This is the key idea I'm trying to point out in doing the U graphs. U(T)U{-(T-t)} has zero value below and above certain values. What are those values?
    No, I don't want you to do any graphical convolution. I am suggesting that by graphing the two U functions and multiplying them graphically you will know what the upper & lower limits of integration are for evaluating the actual convolution integral.

    Maybe you can figure out the limits of integration by looking at U(T)U{-(T-t)} some other way, but doing it graphically is real easy.

    ζ was just a dummy variable.
  14. Sep 17, 2014 #13
    This will sound real bad, but I'm afraid I haven't really heard of graphical multiplication before.

    Are you saying you want me to graph exp(-2(T - t)) and 2exp(-4T)?

    Also my bad the integral in my last post was supposed to now be this:

    -∞[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT
  15. Sep 17, 2014 #14

    rude man

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    No, I want you to graph U(T) and -U{(T-t)}, then multiply them into one graph.
    Please start with graphig these two functions separately, we can't go on until you do.
    That's fine, that was correct also but this way the integration becomes more apparent, with t = constant in the integration.
  16. Sep 18, 2014 #15
    Okay, I'm not sure if these are correct I don't really know what T is:

    https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg [Broken]

    https://imagizer.imageshack.us/v2/547x297q90/673/owhtZD.jpg [Broken]
    Last edited by a moderator: May 6, 2017
  17. Sep 18, 2014 #16

    rude man

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    T is your horizontal axis.

    You got U(T) right.
    You got -U{(T-t)} wrong. You graphed -U(T) instead.
    When is -U(x) = 0? = 1? Now let x = T - t. t is a constant in this. T is your variable. Try again ...
  18. Sep 19, 2014 #17
    So it's shifted then, is that what you are saying?:

    https://imagizer.imageshack.us/v2/547x297q90/537/vHAo1X.jpg [Broken]
    Last edited by a moderator: May 6, 2017
  19. Sep 19, 2014 #18

    rude man

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    Sorry, I keep saying -U when I mean U(-).
    So you want to graph U{-(T - t)}.
    Change (T-t) to t on the horizontal (T) axis and you'd be there.
    Now, multiply the two graphs into one new graph.
    It's real easy if you understand
    0 x 0 = 0
    0 x 1 = 0
    1 x 0 = 0
    1 x 1 = 1
  20. Sep 21, 2014 #19
    Sorry for getting back to this super late again (site moved and I had some comp problems).

    Does it mean the first graph (u(t)) is 0 and the second graph (U(-1(T - t))) is 1? I can not really see yet as to how these help with convolution integrals?
  21. Sep 23, 2014 #20
    Bump, also forgot to post the graph, does multiplying the two graphs mean it will be the same as u(T) if u(T) is 0?

    https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg [Broken]
    Last edited by a moderator: May 7, 2017
  22. Sep 23, 2014 #21

    rude man

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    No. Graph the product U(T)*U{-(T-t)} as you go along the T axis from T = - infinity to T = + infinity. (You can actually go from a small negative T to a bit above T=t).

    Those two U's appear in your convolution ingtegral! Look again at post 13.
  23. Sep 25, 2014 #22
    Okay, so like this then?:

    https://imagizer.imageshack.us/v2/479x260q90/674/Smf0nJ.jpg [Broken]

    I'm guessing now you can say that the limit goes from 0 to t or something similar then, right?
    Last edited by a moderator: May 7, 2017
  24. Sep 25, 2014 #23

    rude man

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    Congratulations! You have scaled the heights.

    Now, last big step for this first convolution: remove U(T)U{-(T-t)} from your convolution integral by suitably setting the upper and lower limits of integration.
    Last edited by a moderator: May 7, 2017
  25. Sep 27, 2014 #24
    0t[exp(-2(T - t))][2exp(-4T)]dT

    and now here I may need to review all the math down from here but I think it goes like this:

    = 20t[exp(-2(T - t))][exp(-4T)]dT

    = 20t[exp(-6T - 2t)]dT

    = 2[exp(-6T - 2t)]/-6

    = -3[exp(-6T - 2t)] | t0

    = -3[exp(-8t)] ?
  26. Sep 27, 2014 #25

    rude man

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    Great step!
    Check this - small error made. I also suggest since t is a constant here taking the t term outside the integral sign.

    Review your freshman calculus if necessary; you'll sink if you don't get this firmed up.
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