Laplace Transform for ODE - RC Circuit

Hello guys,

I am trying to make a laplace transform from a ODE for a simple RC Circuit.

But When try to use the function subs() in order to replace my laplace transfromed variable >>> laplace(q(t), t, s) for Q(s) I don't see any change at all.

I would like to undestand why and How I can fix this.

Thanks for your helping

%For the capacitor 
syms q(t) s
%Charge (C) 
%initial condition in the beginning 
cond_q = q(0) == 0;
%ODE for capacitor in series with resistor 
ode_capacitor = diff(q, t) + P_c*q == Q_c
%Solving ODE
q = dsolve(ode_capacitor, cond_q);
% simplify(q)
% % % Calculate the derivative of 'Charge' - the current(A) 
i_c = diff(q)
limit(i_c, t, 0)
limit(i_c, t, inf)
% 
% %transformada de laplace
laplace(i_c)
a = laplace(ode_capacitor, t, s)
syms Q
a = subs(a, laplace(q, t, s), Q)

Answer 1

[Paul]

I think that last subs command is getting confused because q is defined as the solution of the ode. I'm not sure why that's a problem (or if I'm even correct about that), but you can get around this by using a different variable name as the solution of dsolve:

syms q(t) Q(s) P_c Q_c
cond_q = q(0) == 0;
ode_capacitor = diff(q(t),t) + P_c*q == Q_c;
qsol(t) = dsolve(ode_capacitor,cond_q);
i_c(t) = diff(qsol(t),t);
I_c(s) = laplace(i_c(t));

% Laplace transform of the ode
E(s) = laplace(ode_capacitor);
E(s) = subs(E(s),laplace(q(t),t,s),Q(s));
E(s) = isolate(E(s),Q(s));
Q(s) = rhs(E(s));
Q(s) = subs(Q(s),q(0),0);

% compare Q(s) with laplace(qsol(t))
[Q(s) simplify(laplace(qsol(t)))]

% verify relationship between charge and current
[s*Q(s) I_c(s)]

ans =
 
[ Q_c/(s*(P_c + s)), Q_c/(s*(P_c + s))]
 
 
ans =
 
[ Q_c/(P_c + s), Q_c/(P_c + s)]
 

[Guilherme Lima]

Hello Paul!

Thanks for your help! It did work! :)

[darova]