# Math Help - Differential equation from transfer function

1. ## Differential equation from transfer function

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

1/s^2 + 2s +7

2. Originally Posted by moisesbr
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.

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

So:

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

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

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

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

CB

3. 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)

4. Originally Posted by moisesbr
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 $kf(t)$ is $kF(t)$ and vice versa, where $f$ and $F$ are a LT pair:

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

gives:

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

Take the inverse LT:

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

CB

5. 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)

6. Originally Posted by moisesbr
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 s³ will give you:

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

Writing the DE is not so difficult now.

Coomast