Problem with Laplace transformation

Discussion in 'Physics & Math' started by orange, Sep 26, 2005.

  1. orange Registered Senior Member

    Messages:
    207
    Hello everyone! I have a math problem that I can't solve, and I would greatly appreciate it if anyone took their time to help me out.

    The problem is the following:

    A dynamic system is described by the differential equation y''(t) + y(t) = x(t). What is the transfer function and the impulse response? If the outsignal
    y(t) for insignal x(t) = H(t) + delta(t-2) under the terms y(0) = y'(0) = 0.

    _______________________________
    This is the way I'm trying to solve it:

    The first thing I do is to Laplace transform the whole equation

    y''(t) + y(t) = H(t) + delta(t-2),

    which becomes

    Y(s^2 + 1) = 1/s + e^-2s

    Rewrite the equation

    Y = (1/s + e^-2s) / (s^2 + 1)

    Simplified to

    Y = 1/s(s^2 + 1) + e^-2s/(s^2 + 1)

    My problem arrises when I'm about to inverse transform these equations to find y(t).

    The first part,

    Laplace [1/s(s^2 + 1)] = integral[sin t] = -cos t.

    And the second part,

    Laplace [e^-2s/(s^2 + 1)] = sin (t-2) H(t-2)

    The whole equation,

    y = -cos t + sin(t-2) * h(t-2)

    is not defined for y(0), so I have to have made a mistake earlier on, but I have no idea what it is. I've gone through it a number of times, and still come up with the same result.

    Thanks for any help!
     
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  3. Physics Monkey Snow Monkey and Physicist Registered Senior Member

    Messages:
    869
    I assume H(t) is the Heaviside step function.

    Your mistake is in the step where you conclude that L^(-1){1/s(s^2+1)} = - cos(t), when in fact L^(-1){1/s(s^2+1)} = int^1_0 sin(t) dt = - cos(t) |^1_0 = 1 - cos(t) (and even more correctly (1 - cos(t)) H(t) ). The other inversion is correct, that is L^(-1){e^-2s/(s^2 + 1)} = sin(t-2) H(t-2).

    Putting it all together gives
    y(t) = (1 - cos(t)) H(t) + sin(t-2) H(t-2) which you can check satisfies y(0) = 0, y'(0) = 0, and the differential equation itself.

    Does that help?
     
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  5. orange Registered Senior Member

    Messages:
    207
    Thanks Physics Monkey, I recalculated and found the same inverse transform as you did, 1 - cos t.

    As for sin(t-2) H(t-2), is it defined at y(0)? That is, is sin(t-2) H(t-2) considered zero since H(t-2) is zero at t=0? In that case, what about t=5 where Heaviside is 1, but sin (3) isn't defined? Urr, weirdly asked question maybe, but it's late. Thanks again.

    Orange
     
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  7. Physics Monkey Snow Monkey and Physicist Registered Senior Member

    Messages:
    869
    Not quite sure what you mean, orange. Isn't sin(t) defined for all t?
     
  8. orange Registered Senior Member

    Messages:
    207
    Guess so.

    Please Register or Log in to view the hidden image!

    I was thinking of + 1 = sin x. That must have put me down on the incompetence scale, hehe.
     

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