About Laplace transform filters in Verilog A/MS

Hi,there.
When modelling the opamp by verilog A/MS, people usually use Laplace transform filters to approximate the small signal behaviors like the code fragment below:


But I dont know the exactly meaning of the laplace transform filters function. From the verilog AMS LRM，it says that laplace_zp() implements the zero-pole form of the Laplace transform filter. The general form is:



And it implements the transfer function H(s). Then the result of the first code is V(in)/V(out) = H(s) , Vin(s)/Vout(s) = H(s), or V(in)/V(in) = IL( H(s) )? where IL( H(s) ) represents the inverse laplace transform of H(s).

The first code is used for modelling the frequency response of opamp, then does it take effects when we perform a transient analysis,or it only has effects in frequency response? Please help me understand it, any comment will be appreciated.


Regards
urian

Hi urian

providing a transfer function in the Laplace domain simply means that there is a linear differential equation relating input and output variable. The only freedom left after assigning the equation are just the initial values (initial conditions) of those variables. The description is accurate for small signal analysis but can be run in transient as well. The result will be a linearized version of the real system neglecting all large signal effects e.g. saturation

Check out **broken link removed**

Hi,dgnani, thx for commenting!

I know exactly what the Laplace transfer function means and it is evaluated in s domain. However, I dont understand what the exactly usage of Laplace transform filters. Why we put V(in) as the expr argument? And what's the result of the first code? Is it Vout(s)=Vin(s)*H(s)? or IL(Vout(s))?
If there is another contribution statement such as V(out) <= Vout_CM, does the first code takes effect together with it as a large signal superposed with a small signal quantity?

Regards
urian