2
$\begingroup$

The mathematical model is

$$V'=\frac{I}{C}, V(0)=0$$ and $$I'=\frac{-V}{L}-\frac{RI}{L}+\frac{V_β}{L}, I(0)=0$$ With I.C $V_β=5 V, R=0.1Ω, L=4*10^{-9}H and C=0.5*10^{-9}F $

I have calculated and I have found that the differential has one equilibrium point $(V_eq,I_eq)=(5,0)$.

Now I want to find The characteristic equation for the eigenvalues of the Jacobian matrix.

I use the following code but no result! Where have i done the mistake?

vb = 5; r = 0.1; l = 4*Power[10, -9]; c = 0.5*Power[10, -9];
f = i/c;
g = (-v/l) - (r*i/l) + vb/l;
deq1 = v'[t] == f
deq2 = i'[t] == g
equilibrio = Solve[{f == 0, g == 0}, {v[t], i[t]}]
MatrixForm[jacobian = D[{f, g}, {{v[t], t[t]}}] // Simplify]
Simplify[MatrixForm[jacobian] /. equilibrio]
$\endgroup$
2

1 Answer 1

3
$\begingroup$

Based on the answer I linked to in the comment above, this may be what you want:

Clear[i, v]

vb = 5; r = 0.1; l = 4*Power[10, -9]; c = 0.5*Power[10, -9];

eqns = {i/c , -v/l - r*i/l + vb/l}

(* {2.*10^9 i, 1250000000 - 2.5*10^7 i - 250000000 v} *)

eq = Solve[eqns == {0, 0}, {v, i}]

(* {{v -> 5., i -> 0.}} *)

eq[[1]]

(* {v -> 5., i -> 0.} *)

j = D[eqns, {{i, v}}]

(* {{2.*10^9, 0}, {-2.5*10^7, -250000000}} *)

j /. eq[[1]]

(* {{2.*10^9, 0}, {-2.5*10^7, -250000000}} *)

CharacteristicPolynomial[j /. eq[[1]], \[Lambda]]

(* -5.*10^17 - 1.75*10^9 \[Lambda] + \[Lambda]^2 *)

Eigenvalues[ j /. eq[[1]] ]

(* {2.*10^9, -2.5*10^8} *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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