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

1/s^2 + 2s +7

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- Sep 4th 2009, 02:31 AMmoisesbrDifferential equation from transfer function
*How can I Obtain the differential equation from transfer function below ?*

1/s^2 + 2s +7

- Sep 4th 2009, 08:52 AMCaptainBlack
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 - Sep 4th 2009, 09:29 AMmoisesbr
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)

- Sep 4th 2009, 09:55 AMCaptainBlack
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 - Sep 4th 2009, 10:31 AMmoisesbr
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)

- Sep 5th 2009, 11:01 PMCoomast