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Thread: Differential equation from transfer function

  1. #1
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    Differential equation from transfer function

    How can I Obtain the differential equation from transfer function below ?

    1/s^2 + 2s +7


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  2. #2
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    Quote Originally Posted by moisesbr View Post
    How can I Obtain the differential equation from transfer function below ?

    1/s^2 + 2s +7
    This is the reverse process of the question of obtaing the LT form of the transfer function defined by a differential equation we did earlier.

    $\displaystyle H(s)=\frac{1}{s^2+2s+7}$

    So:

    $\displaystyle Y(s)= H(s)X(s)=\frac{1}{s^2+2s+7}X(s)$

    $\displaystyle (s^2+2s+7)Y(s)=X(s)$

    Taking inverse Laplace transforms (assuming $\displaystyle y(0)=0$ and $\displaystyle y'(0)=0$):

    $\displaystyle y''(t)+2y'(t)+7y(t)=x(t)$

    CB
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  3. #3
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    Thanks, your comprehension is very appreciated

    I am trying to solve a second excercice as you did, but
    I got lost in third line. I don't know where the numerator 10 should
    go from the second to 3hd line


    H(s) = 10/(s+7) * (s+8)

    y(s) = H(s) * X(s) = 10/([s+7] * [s+8]) * X(s)

    ([s+7] * [s+8]) Y(s) = X(s)






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  4. #4
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    Quote Originally Posted by moisesbr View Post
    Thanks, your comprehension is very appreciated

    I am trying to solve a second excercice as you did, but
    I got lost in third line. I don't know where the numerator 10 should
    go from the second to 3hd line


    H(s) = 10/(s+7) * (s+8)

    y(s) = H(s) * X(s) = 10/([s+7][s+8]) * X(s)

    ([s+7] * [s+8]) Y(s) = X(s)




    The LT of $\displaystyle kf(t)$ is $\displaystyle kF(t)$ and vice versa, where $\displaystyle f$ and $\displaystyle F$ are a LT pair:

    $\displaystyle Y(s) = H(s) X(s) = \frac{10}{(s+7)(s+8) }X(s)$

    gives:

    $\displaystyle (s^2+15s+56)Y(s)=10X(s)$

    Take the inverse LT:

    $\displaystyle y''(t)+15y'(t)+56y(t)=10x(t)$

    CB
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  5. #5
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    Thank you. Now I know how to move the numerator

    I have this last one

    I am stuck in line 3 because I'm in doubt if I can sum s^3 + 8s^2 and 9s or let it be

    s+2/s^3 + 8s^2 +9s +15

    y(s)= H(s) X(s) = s+2/s^3 + 8s^2 +9s +15 * X(s)

    (s^3 + 8s^2 +9s +15) * Y(s) = s+2 * X(s)

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  6. #6
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    Quote Originally Posted by moisesbr View Post
    Thank you. Now I know how to move the numerator

    I have this last one

    I am stuck in line 3 because I'm in doubt if I can sum s^3 + 8s^2 and 9s or let it be

    s+2/s^3 + 8s^2 +9s +15

    y(s)= H(s) X(s) = s+2/s^3 + 8s^2 +9s +15 * X(s)

    (s^3 + 8s^2 +9s +15) * Y(s) = s+2 * X(s)
    Using the same manner as before after multiplying with sł will give you:

    $\displaystyle s^3 \cdot Y(s)=(8s^5+(9+1)s^4+15s^3+2)\cdot X(s)$

    Writing the DE is not so difficult now.

    Coomast
    Last edited by Coomast; Sep 5th 2009 at 11:01 PM. Reason: eliminating a typo
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