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Math Help - differential eq to transfer function

  1. #1
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    differential eq to transfer function

    y''(t)-2*y'(t)-8*y(t) = 3*x'(t) - 2*x(t)

    I know how to do simpler ones, but not with the right side like it is.

    then also, If possible, with x(t) = u(t) (step function) , and all intial cond. =0 I need the zero state response. (using inverse laplace transform)

    ideally i would like to know how to do it by hand, and the proper syntax for matlab if possible.. if im asking for too much, just do what you can.
    thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by causalitist View Post
    transfer function of this?

    y''(t)-2*y'(t)-8*y(t) = 3*x'(t) - 2*x(t)

    I know how to do simpler ones, but not with the right side like it is.

    secondarily,the zero state response. with x(t) = u(t) (step function) , and all intial cond. =0 (using inverse laplace transform)
    Writing this in terma of the differential operator:

    (D^2-2D-8)y(t)=(3D-2)x(t)

    Take the LT (with implicit assumptions about initial values etc):

    (s^2-2s-8)Y(s)=(3s-2)X(s)

    and assuming x the input and y the output:

     <br />
Y(s)=H(s)X(s)=\frac{3s-2}{s^2-2s-8}X(s)<br />

    CB
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  3. #3
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    whew, I got it right. thanks, textbooks these days are complete garbage, too much filler. your response just tought me that section of the book lol
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  4. #4
    MHF Contributor chisigma's Avatar
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    With simple steps You obtain...

    H(s)= \frac{Y(s)}{X(s)} = \frac{3s-2}{s^{2} -2s -8} = \frac{a}{s+2} + \frac{b}{s-4} (1)

    ... then find a and b from (1) and finally, because is  X(s) = \frac{1}{s}, is...

    Y(s)= \frac{3s-2}{s\cdot (s^{2} -2s -8)} = \frac{a}{s\cdot (s+2)} + \frac{b}{s\cdot (s-4)} (2)

    Kind regards

    \chi \sigma
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