# differential eq to transfer function

• Mar 30th 2010, 10:59 AM
causalitist
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.
• Mar 30th 2010, 11:00 PM
CaptainBlack
Quote:

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:

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

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

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

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

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

CB
• Mar 31st 2010, 09:36 AM
causalitist
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
• Apr 4th 2010, 05:25 AM
chisigma
With simple steps You obtain...

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

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

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

Kind regards

$\chi$ $\sigma$