2
$\begingroup$

I just started using Mathematica and I am a bit stuck. I want to compute poles of transfer function that are on edge of stability i.e. Re[s] = 0 and I want to find for which RE that happens.

This is my transfer function:

enter image description here

I know such solutions exist because I've computed them manually using slider.

I've tried

sol1 = Solve[eq == 0, s];
real = ComplexExpand[Re[st[[2]]]];
sol2 = Solve[sol1 == 0, RE]

but I don't get anything. I wrote st[[2]]] because I get 3 poles, one is strictly real and always on lhs, and other two are complex conjugate that can pass on rhs, I hoped it will speed up stuff.

Edit

Full code:

Y = s*CF*(1 - (A0*s*τ1)/((1 + s*τ1)*(1 + s*τ2)));
Yin = s*CE + 1/RE + Y;
τ1n = 2.75*10^-07;
τ2n = 7.52*10^-09;
CFn = 18.951;
CEn = 1;
rep = 
  ReplaceAll[
    Yin, 
    {CE -> CEn*10^-12, τ1 -> τ1n, τ2 -> τ2n, CF -> CFn*10^-12, A0 -> 2}]

st = Solve[rep == 0, s];
real = ComplexExpand[Re[st[[2]]]];
sol = Solve[real == 0, RE]
$\endgroup$
  • $\begingroup$ please post complete code. Including eq used. $\endgroup$ – Nasser Oct 28 '17 at 20:41
  • $\begingroup$ I've included full code into the original question. $\endgroup$ – Dominik Oct 28 '17 at 21:00
  • $\begingroup$ Something is not clear : Solve[rep == 0, s] gives the zeros of the transfert function, not the poles. $\endgroup$ – andre314 Oct 28 '17 at 23:14
  • $\begingroup$ Sorry Yin as a transfer function is taken from a script sent to me by my mentor I just wanted to test some other things. Y is admittance of a system and s*CE + 1/RE is external admittance. I want to see how stability changes depending on values of CE an RE $\endgroup$ – Dominik Oct 29 '17 at 13:44
1
$\begingroup$

Based on the equations you have, RE and CE have no effect on system stability.

rep = Yin /. {τ1 -> τ1n, τ2 -> τ2n, CF -> CFn*10^-12, A0 -> 2};

Denominator[Together[rep]]

RE (3.63636*10^6 + 1. s) (1.32979*10^8 + 1. s)

CE does not appear in the denominator, and RE is just a factor and does not affect the pole locations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.