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
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
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)