# Inverse of this transfer function

• August 22nd 2010, 05:59 PM
algorithm
Inverse of this transfer function
Hi,

I have the transfer function for a particular configuration on an RLC circuit:

H(s) = sRC/(LC*s^2 + sRC + 1)

R, L, and C are constant. I want to transform it back to the time domain. Is there a standard transform pair I can match this to?

Thanks.
• August 22nd 2010, 06:12 PM
yeKciM
Quote:

Originally Posted by algorithm
Hi,

I have the transfer function for a particular configuration on an RLC circuit:

H(s) = sRC/(LC*s^2 + sRC + 1)

R, L, and C are constant. I want to transform it back to the time domain. Is there a standard transform pair I can match this to?

Thanks.

as a matter a fact there is but it's based on your $s^2 LC + s RC +1$ depends on which from 3 types will you use :D:D:D

is it $B^2 < 4AC$ or $B^2 > 4AC$ or $B^2 = 4AC$

i had have it here on the desk but now... lol try google it, and soon as i find it i'll edit this post... (now i'm confused... like it has just disappeared )

P.S. look for table of inverse Laplace transformations (there should be like 60 or so formulas for different types ... )
• August 27th 2010, 10:11 AM
yeKciM
okay this what i have :D:D:D

for type :

$\displaystyle \frac {mS+n}{AS^2+BS+C}$

where those with S are just any numbers :D (this is type that is what you need just put n=0, and you will get that function in time domain)

$\displaystyle \frac {RCS}{LCS^2+RCS+1} \Rightarrow m=RC ; n=0 ; A=LC; B=RC; C=1$

now you have 3 different formulas :D depending on values of your A,B,C :D

if $B^2>4AC$ than you will use :

$\displaystyle h(t)= e^{-at}(\frac {m}{A} \cosh {\omega_0 t} + \frac {n-am}{A\omega_0} \sinh{\omega_0 t})$

where $\displaystyle a=\frac {B}{2A}$ and $\displaystyle \omega_0 = \sqrt{a^2- \frac {C}{A}}$

if $B^2=4AC$ than you will use :

$\displaystyle h(t)= e^{-at}(\frac {m}{A} + \frac {n-am}{A}t)$

$\displaystyle a=\frac {B}{2A}$

if $B^2<4AC$ than you will use :

$\displaystyle h(t)= e^{-at}(\frac {m}{A} \cos {\omega_0 t} + \frac {n-am}{A\omega_0} \sin{\omega_0 t})$

where $\displaystyle a=\frac {B}{2A}$ and $\displaystyle \omega_0 = \sqrt{\frac {C}{A} - a^2}$

there are "little" different formulas if you need to do this one for example :

$\displaystyle \frac {mS^2+nS+q}{S(AS^2+BS+C)}$

so if you need that or any another, just say :D