HELP: looping a function and plot (ode23)

Hi, so the question for my exercise is : Write a MATLAB code to solve the differential equation for 𝑣𝑜𝑢𝑡 using function ode23() and plot the output. hint: to get the output waveform, plot(t, y(:,1))

The graph output was in subplot i believe. and the equation of each graph are the same except for the value of R. I tried doing the code with my little matlab knowledge and after searching the mathwork, i've tried to mix and match some coding that i think were relevent and make sense for my question. Here is my code;

script:

function ode23()
    function myout=fun(t,y,R)
        Vi=5; C=0.1; L=0.1;
        if nargin < 0.5 || isempty(R) %i dont really know about this one
           R = 0.5; % Your default value
        end
        myout=[(Vi-R*C*y(2)-y(1))/(L*C)];
    end
end

command windows:

tspan=[0, 10]; ohm='x03A9';
y0=[0];

[t_r1, y_r1]=ode23(@(t,y) fun(t,y,0.5), tspan, y0);
plot(t_r1, y_r1(:,1))
[t_r2, y_r2]=ode23(@(t,y) fun(t,y,2), tspan, y0);
plot(t_r2, y_r2(:,1))
[t_r3, y_r3]=ode23(@(t,y) fun(t,y,4), tspan, y0);
plot(t_r2, y_r2(:,1))

subplot(3,1,1)
plot(t_r1, y_r1(:,1))
ylabel('Vout'), xlabel('Time(s)')
title('Underdamped response of R=0.5''ohm')

subplot(3,1,2)
plot(t_r2, y_r2(:,1))
ylabel('Vout'), xlabel('Time(s)')
title('Critically damped response of R=2''ohm')

subplot(3,1,3)
plot(t_r3, y_r3(:,1))
ylabel('Vout'), xlabel('Time(s)')
title('Over damped response of R=0.5''ohm')

If anyone can help me, that would be great. Thanks in advance.

[Steven Lord]

function ode23()

Please don't name your function ode23. That name already has a meaning in MATLAB.

[Hade.R]

yes, thanks alot! i just realized that ode23 is already a function so i misunderstood when the question say 'using function ode23()'. i thought the question want me to name the function as ode23..

Answer 1

[Stephan]

tspan1 = [0, 4];
tspan2 = [4, 10];
y0 = [0, 0];

[t1_r1, y1_r1] = ode23(@(t,y) fun(t,y,0.5,5), tspan1, y0);
[t2_r1, y2_r1] = ode23(@(t,y) fun(t,y,0.5,0), tspan2, y1_r1(end,:));

[t1_r2, y1_r2] = ode23(@(t,y) fun(t,y,2,5), tspan1, y0);
[t2_r2, y2_r2] = ode23(@(t,y) fun(t,y,2,0), tspan2, y1_r2(end,:));

[t1_r3, y1_r3] = ode23(@(t,y) fun(t,y,4,5), tspan1, y0);
[t2_r3, y2_r3] = ode23(@(t,y) fun(t,y,4,0), tspan2, y1_r3(end,:));

subplot(3,1,1)
plot([t1_r1; t2_r1], [y1_r1(:,1); y2_r1(:,1)])
ylabel('Vout'), xlabel('Time(s)')
title('Underdamped response of R=0.5''ohm')
ylim([-5 10]);

subplot(3,1,2)
plot([t1_r2; t2_r2], [y1_r2(:,1); y2_r2(:,1)])
ylabel('Vout'), xlabel('Time(s)')
title('Critically damped response of R=2''ohm')
ylim([0 6])
% 
subplot(3,1,3)
plot([t1_r3; t2_r3], [y1_r3(:,1); y2_r3(:,1)])
ylabel('Vout'), xlabel('Time(s)')
title('Over damped response of R=0.5''ohm')
ylim([0 6]);

function myout=fun(~,y,R,Vi)
C=0.1;
L=0.1;
myout = zeros(2,1);
myout(1)= y(2);
myout(2)=(Vi-R*C*y(2)-y(1))/(L*C);
end

[madhan ravi]

Vin is a step input signal!

[Stephan]

See my edited code - it is always helpful to know all the details regarding a question...

[Hade.R]

thanks a lot Sephen! this time its work! but can you explain what happened in your code? may i know why do you use two tspan? and what about the hint: to get the output waveform, plot(t, y(:,1))? what does madhan ravi means by step input signal?

[Stephan]

I will try:

• As stated by Stephen Lord dont name your own functions like inbuilt functions in Matlab are named.
• As Madhan stated the function is not continuos due to the stepped Vin signal. That means we have to perform 2 integration steps, the second one starting at the falling step signal by t=4.
• That also means we need 2 tspans and the initial values for the second step are the last values of the first step.
• Since your ode is second order, the function has to be rewritten as a system of 2 first order ode's. This is what happens.
• Because your provided code (and the bugs in it...) did only provide one initial value, i first thought we talk about one first order ode and you ran wrong by using y(1) and y(2). When you provided the whole information, i understood which changes will be needed to get it run.
• Last but not least i was on my way home in the meantime, so i could not fix it quickly.

BTW: If this helped you, dont miss to accept useful answers.

[Hade.R]

thanks a lot Stephen! you really help me. i understand much better with your explaination. Thanks also for your time and i already accept the answer :)