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Is there a mistake in my calculation or in my reasoning?

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    y'' + 3y' + 2y = r(t),

    r(t) = u(t - 1) - u(t - 2),

    y(0) = y'(0) = 0.

    I need to solve this by convolution, which I know is commutative. The problem is that my calculation gives (f * g) =/= (g * f). Could someone please tell me where my mistake is?


    2. Relevant equations

    (f * g) = ∫f(τ)g(t-τ)dτ, from 0 to t

    3. The attempt at a solution
    I split the problem into 3 cases:

    First (0 < t < 1): r(t) = 0, so y(t) = 0

    Second (1 < t < 2): r(t) = 1 This is where it goes wrong.

    I get f(t) = 1, g(t) = e^(-t) - e^(-2t), then y = (f * g) = ∫[ e^-(t-τ) - e^-2(t-τ) ]dτ, from 1 to t = (1/2) - e^-(t-1) - (1/2)e^-2(t-1), which is the correct answer according to my textbook.

    But, y = (g * f) = ∫[ e^(-τ) - e^(-2τ) ]dτ, from 1 to t = -e^(-t) + (1/2)e^(-2t) + e^(-1) - (1/2)e^(-2) =/= (f * g)

    I integrated from 1 instead of 0 since r=0 for t < 1.
     
  2. jcsd
  3. Oct 20, 2015 #2

    Orodruin

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    I suggest you look over your integration boundaries for the second case. Think about what the convolution really means.
     
  4. Oct 20, 2015 #3
    I don't think I understand. I get the right answer for (g * f) if I integrate from 0 to (t - 1) instead, but I don't understand why. Could you please explain?
     
  5. Oct 20, 2015 #4

    Orodruin

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    Try writing the integrals using the actual expressions for the functions instead. Where are the heaviside functions non-zero?
     
  6. Oct 20, 2015 #5
    I have now tried that for both (f * g) and (g * f), but in both cases I get the correct answer for case 3 (t > 2), when i'm trying to solve for case 2:

    (f * g) = ∫[ e^(-τ) - e^(-2τ) ] * [ u(t - τ - 1) - u(t - τ - 2)]dτ

    I find that for case 2:

    t - τ - 1 > 0 Λ t - τ - 2 < 0

    which gives me:

    t - 2 < τ < t - 1

    So the integral becomes:

    [ -e^(-τ) + (1/2)e^(-2τ) ] from (t - 2) to (t - 1) = e^-(t - 2) - (1/2)e^-2(t - 2) - (e^-(t - 1) - (1/2)e^-2(t - 2))

    but this is correct only if t > 2, so I don't understand why I get that when I try to solve for 1 < t < 2
     
  7. Oct 20, 2015 #6

    Ray Vickson

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    I don't see why you want to use convolutions, instead of just applying standard properties of Laplace transforms to get an easily-recognized form of inverse Laplace transform. However, if you insist on using convolutions you will have
    [tex] y(t) = \int_0^t f(t - \tau) r(\tau) \, d \tau = \underbrace{\int_0^t f(t-\tau) u(\tau-1) \, d\tau}_{= I_1} -
    \underbrace{\int_0^t f(t-\tau) u(\tau-2) \, d\tau}_{= I_2} [/tex]
    For ##t \leq 1## we have ##I_1 = 0## because the ##u##-factor vanishes. For ##t > 1## we have ##I_1 = \int_1^t f(t-\tau) \, d \tau##, etc. The term ##I_2## is similar.
     
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