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Math Help - Differential Equation to Transfer Function

  1. #1
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    Differential Equation to Transfer Function

    Hello,

    I have done this in a long time but is this right? I have a differential equation and I want to find the transfer function. Is that right?

    Thanks,

    r, m, and k are numbers.

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  2. #2
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    Quote Originally Posted by wqvong View Post
    Hello,

    I have done this in a long time but is this right? I have a differential equation and I want to find the transfer function. Is that right?

    Thanks,

    r, m, and k are numbers.

    Your ODE is:

    m\frac{d^2}{dt^2}x(t)+r\frac{d}{dt}x(t)+kx(t)=a_0\  cos(\omega t)

    This is of the form:

    m\frac{d^2}{dt^2}x(t)+r\frac{d}{dt}x(t)+kx(t)=y(t)

    Now take the Laplace Transform (and assume zero initial conditions):

    ms^2X(s)+rsX(s)+kX(s)=Y(s)

    So:

    X(s)=\frac{1}{ms^2+rs+k}Y(s)

    or:

    G(s)=\frac{1}{ms^2+rs+k}

    CB
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  3. #3
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    Hello,

    What happens to a0 cos(wt)? let assume that the initial condition is not zero how would i solve for it now?

    Thanks,
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  4. #4
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    Quote Originally Posted by wqvong View Post
    Hello,

    What happens to a0 cos(wt)? let assume that the initial condition is not zero how would i solve for it now?

    Thanks,
    a_0\cos(\omega t) is the forcing or input function y(t)

    The initial conditions are zero, that is how you construct a transfer function.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    a_0\cos(\omega t) is the forcing or input function y(t)

    The initial conditions are zero, that is how you construct a transfer function.

    CB
    Ok, I get that part. What happens when the initial condition isn't zero anymore? lets just say it is 1. Do i have to take the Laplace of (ao cos wt) = s / (s^2 + w^2), multiple it by the transfer function where it will equal to

    s / (s^2 + w^2) * 1 / (k + r s + r s^2) ?

    or i still use the regular transfer function 1 / (k + r s + r s^2) and plug one into that?

    Another question I have is, lets say I solve the second order differential equation and I come up with the general solution. If i take the Laplace transform of that is that similar to the transfer function? What is the differences?

    Thanks,
    Last edited by wqvong; May 13th 2010 at 11:14 AM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by wqvong View Post
    Ok, I get that part. What happens when the initial condition isn't zero anymore? lets just say it is 1. Do i have to take the Laplace of (ao cos wt) = s / (s^2 + w^2), multiple it by the transfer function where it will equal to

    s / (s^2 + w^2) * 1 / (k + r s + r s^2) ?

    or i still use the regular transfer function 1 / (k + r s + r s^2) and plug one into that?
    The system you are deriving a transfer function for is like a black box and the transfer function relates the output to the input. When you plug the device in the initial conditions are all zero by definition.

    Another question I have is, lets say I solve the second order differential equation and I come up with the general solution. If i take the Laplace transform of that is that similar to the transfer function? What is the differences?
    No, you are doing something different from characterising the behaviour of a linear system to an arbitrary input.

    CB
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