# Thread: Inverse of this transfer function

1. ## 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.

2. 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 $\displaystyle s^2 LC + s RC +1$ depends on which from 3 types will you use

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

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 ... )

3. okay this what i have

for type :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

so if you need that or any another, just say