# Thread: differential eq to transfer function

1. ## 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.

2. Originally Posted by causalitist
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:

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

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

$\displaystyle (s^2-2s-8)Y(s)=(3s-2)X(s)$

and assuming $\displaystyle x$ the input and $\displaystyle y$ the output:

$\displaystyle Y(s)=H(s)X(s)=\frac{3s-2}{s^2-2s-8}X(s)$

CB

3. 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

4. With simple steps You obtain...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$