Solving a RL Circuit Using MATLAB ode45

[gurhan]

I have analysed a RL circuit with freewheelind diode, switch and additional source. I have used ode solver which is called ode45 in matlab. This expression is [t, i]=ode45('inducwithE', tspan, i0'). There is three stages in that circuit and every last 'i' (current-this is the state variable) value for the previous stage becomes initial value for the following stages. But when I am using if statement for three different conditions I am always referring the expression ([t, i]=ode45('inducwithE', tspan, i0')). In that point 'i' becomes always '0'. How can I assing those initial values for second and third stages? Could anyone help me about it?
Thanks

function dcrl = inducwithE(t,i)
% function for RL circuit calculations
Vs=50;
E=30;
L=0.1;
R=1;
Z=E/L;
X = (Vs/L)-Z;
Y = i*R/L;
if t<0.7;
dcrl = X - Y;VT=Vs;
else 0.7<t<0.75;
dcrl=-Z-Y;
end
if t>0.75;
dcrl=Z-Y;
end
if t>1.4
dcrl = X - Y;
end

tspan=[0 2];
i0=0;
[t, i]=ode45('inducwithE',tspan,i0')

 

[Aaron Barnard]

You can use the same statement to solve the second and third stages. The initial value for the second stage is just the last value of the current from the first stage, and the initial value for the third stage is the last value of the current from the second stage. So you can just do:

tspan=[0.7 0.75];
i0=i(end);
[t, i]=ode45('inducwithE',tspan,i0')

tspan=[0.75 1.4];
i0=i(end);
[t, i]=ode45('inducwithE',tspan,i0')

 

[tacman]

% solving using ode45

To assign initial values for the second and third stages, you can modify the code to include the initial condition for each stage separately. For example, for the second stage, you can set the initial condition to be the final value of current from the first stage. Similarly, for the third stage, you can set the initial condition to be the final value of current from the second stage. This way, you can ensure that the initial values are correctly assigned for each stage and the calculation continues from the previous stage's final value. Here is an example of how you can modify the code:

function dcrl = inducwithE(t,i)
% function for RL circuit calculations
Vs=50;
E=30;
L=0.1;
R=1;
Z=E/L;
X = (Vs/L)-Z;
Y = i*R/L;
if t<0.7;
dcrl = X - Y;VT=Vs;
else 0.7<t<0.75;
dcrl=-Z-Y;
end
if t>0.75;
dcrl=Z-Y;
end
if t>1.4
dcrl = X - Y;
end

tspan=[0 2];
i0_1=0; % initial condition for first stage
i0_2=i(end); % final value of current from first stage as initial condition for second stage
i0_3=i(end); % final value of current from second stage as initial condition for third stage

[t, i]=ode45('inducwithE',tspan,[i0_1 i0_2 i0_3]); % solving using ode45 with multiple initial conditions

Hope this helps!