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.


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 = 
    {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]
  • $\begingroup$ 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

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};


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.


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