Transfer function from non linear ode

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})
eqn1LT = 
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})
eqn2LT = 

1 Comment

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.

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R2021b

Asked:

on 18 Jul 2022

Commented:

on 18 Jul 2022

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