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RLC Circuit Equation Implementation-Runge Kutta

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OvercookedRamen
OvercookedRamen on 12 Nov 2019
Commented: darova on 20 Nov 2019
Hello, I need to convert an RLC equation to work inside the functions I have wirtten. Currently MatLab is telling me there are too many variables inside my equation for it to work. Here is the code of the function I am trying to work with
function [derriv_value] = RLC(y)
%Function that contains the derrivative value
%RLC Implementation: The equation is given by L*Q'' + R*Q' + (1/C)*Q = E(t)
% E(t) is the sum of total voltage drop across the circuit.
% R is resistance, L is indunctance, and C is capacitance. Q Is the charge
% in coloumbs. This equation is currently in a second order form.
% In most cases, the voltage in the system is known, it is usually the
% current (I) that is unknown.
% If we differentiate both sides and subsitute Q for I, we get,
% L*I'' + R*I' + (1/C)*I = E'(t)
% Since this is the more pratical case, we will use this format.
% Here are some values that could be used to simulate this.
% These above are some example values.
R = 10; % Resistance (Series RL Branch), Ohms..
L = 100e-3; % Inductance (Series RL Branch), H..
C = 50e-6; % Capacitor (Series RLC Branch), F
derriv_value = L*y(3)+R*y(2)+(1/C)*y;
end
And this function will be called in a Runge Kutta implementation file. I'll attach it. I have another file called FunctionC that has a format that matlab accepts. It works with the Runge Kutta file.

  2 Comments

OvercookedRamen
OvercookedRamen on 12 Nov 2019
Yeah I already looked over that. I have the equation converted fine on paper, the issue is trying to set up a system of 3 first order ODEs in a format that will run like the equation in FunctionC.

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Accepted Answer

Daniel M
Daniel M on 12 Nov 2019
Instead of this
derriv_value = L*y(3)+R*y(2)+(1/C)*y;
Do you mean this?
derriv_value = L*y(3)+R*y(2)+(1/C)*y(1);

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OvercookedRamen
OvercookedRamen on 12 Nov 2019
That seemed to work, that last format you posted there. I think that is what the problem was. Setting the xi values up for the function. Thanks!
OvercookedRamen
OvercookedRamen on 20 Nov 2019
How would you get this equation into a system of 3 first order ODEs?
darova
darova on 20 Nov 2019
Your equation is of second order
% L*Q'' + R*Q' + (1/C)*Q = E(t)
% Q'' = 1/L*( E(t) - R*Q' - Q/C );
f = @(t,y) [y(2)
1/L*(E - R*y(2) - y(1)/C)];
[t,y] = ode45(f,ts,y0);

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