Transfer function from non linear ode
Mostra commenti meno recenti
Functions seem to be fine in laplace domain. Don't know how to obtain numerator and denominator for a single transfer function for requency response?
Thanks :)
clc
clear
syms a b c d y(t) Y(s) x(t) X(s) s t E(t)
assume([a b c d] > 0);
%assume(X(s) ~=0)
E(t) = 5*sin(t) ;
a = 1;
b = 2;
c = 3;
d = 4;
yp = diff(y,t);
ypp = diff(y,t,2);
eqn = (y*a)*ypp*E + b*yp^2 + c*yp + d == 0 ; % the ode
V = odeToVectorField(eqn); % to get two first order linear odes
M = matlabFunction(V,{'t','Y'});
eqn1 = V(1);
eqn2 = V(2);
eqn1LT = laplace(eqn1,t,s);
eqn2LT = laplace(eqn2,t,s);
eqn1LT = subs(eqn1LT,{laplace(y,t,s), subs(diff(y(t),t),t,0),laplace(x(t),t,s),y(0)},{Y(s), 0, X(s), 0})
eqn2LT = subs(eqn2LT,{laplace(y,t,s), subs(diff(y(t),t),t,0),laplace(x(t),t,s),y(0)},{Y(s), 0, X(s), 0})
1 Commento
Star Strider
il 18 Lug 2022
‘Functions seem to be fine in laplace domain.’
They are not, really. If I remember correctly, Laplace transforms are only defined on linear differential equations with constant coefficients. Yours are nonlinear.
Probably the only way to approach this is to integrate it numerically with the input defined at specific times (use the ‘tspan’ input to define that as a vector) with the integrated result as the output.
Risposte (0)
Categorie
Scopri di più su Numerical Integration and Differential Equations in Centro assistenza e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

