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:
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]
Solve[rep == 0, s]
gives the zeros of the transfert function, not the poles. $\endgroup$