what the problem that I get to during plotting the graph

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SAHIL SAHOO 2022-7-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1756815-what-the-problem-that-i-get-to-during-plotting-the-graph

编辑： Torsten 2022-7-11

ti = 0;

tf = 10E-4;

tspan=[ti tf];

y0=[10;10;1;1;1].*10E-2;

[T,Y]= ode45(@(t,y) rate_eq(t,y),tspan,y0);

plot(T,Y(:,1));

xlim([0 4E-4])

ylim([0 8])

xlabel('time')

ylabel('amplitude')

yyaxis right

ylabel('phase difference')

hold on

plot(T,((Y(:,5))./6.28));

hold off

function dy = rate_eq(t,y)

dy = zeros(5,1);

o = 2E5;

tc = 30E-9;

tf = 230E-6;

a1 = 0.1;

a2 = 0.1;

P1 = 0.2;

P2 = 0.2;

kc = 3E-3;

s = 0.17;

k = s.*kc;

dy(1) = ((y(3) - a1) * y(1) + k * y(2) * cos(y(5))) / tc;

dy(2) = ((y(4) - a2) * y(2) + k * y(1) * cos(y(5))) / tc;

dy(3) = (P1 - y(3) * (y(1)^2 + 1)) / tf;

dy(4) = (P2 - y(4) * (y(2)^2 + 1)) / tf;

dy(5) = o - k / tc * (y(1) / y(2) + y(2) / y(1) * sin(y(5)));

end

I have marked the circle which part is not proper waveform and I need a proper waveform with same amplitude, I need to know what is causing this, is it the constant value which I have given , initial conditions or something

and please check that if I solve the differential equation correctly or not, I mean the techinques that I apply to solve this differential equation.

these are the differential equation

y(1) = A1

y(2) = A2

y(3) = G1

y(4) = G2

y(5) = Φ

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Answer 1

VBBV 2022-7-11

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1756815-what-the-problem-that-i-get-to-during-plotting-the-graph#answer_1004155

在 MATLAB Online 中打开

ti = 0;

tf = 10E-4;

tspan=[ti tf];

y0=[2;2;1;1;1].*10E-2; % this change

[T,Y]= ode45(@(t,y) rate_eq(t,y),tspan,y0);

plot(T,Y(:,1));

xlim([0 4E-4])

ylim([0 8])

xlabel('time')

ylabel('amplitude')

yyaxis right

ylabel('phase difference')

hold on

plot(T,((Y(:,5))./6.28));

hold off

function dy = rate_eq(t,y)

dy = zeros(5,1);

o = 2E5;

tc = 30E-9;

tf = 230E-6;

a1 = 0.1;

a2 = 0.1;

P1 = 0.2;

P2 = 0.2;

kc = 3E-3;

s = 0.17;

k = s.*kc;

dy(1) = ((y(3) - a1) * y(1) + k * y(2) * cos(y(5))) / tc;

dy(2) = ((y(4) - a2) * y(2) + k * y(1) * cos(y(5))) / tc;

dy(3) = (P1 - y(3) * (y(1)^2 + 1)) / tf;

dy(4) = (P2 - y(4) * (y(2)^2 + 1)) / tf;

dy(5) = o - k / tc * (y(1) / y(2) + y(2) / y(1) * sin(y(5)));

end

Yes, the constant values given as initial conditions need to be modified to get proper waveform.

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SAHIL SAHOO 2022-7-11

I couln't found any suitable initial condition for which the waveform is proper and please can you check if my algorithm is correct for solving the differential equation that I gave you in the picture.

Torsten 2022-7-11

编辑：Torsten 2022-7-11

在 MATLAB Online 中打开

dy(5) doesn't coincide with your equation in the picture.

And we don't know whether I1 = y(1)^2 and I2 = y(2)^2.

ti = 0;

tf = 10E-4;

tspan=[ti tf];

y0=[10;10;1;1;1].*10E-2; % this change

options = odeset('RelTol',1e-8,'AbsTol',1e-8);

[T,Y]= ode45(@(t,y) rate_eq(t,y),tspan,y0,options);

plot(T,Y(:,1));

xlim([0 4E-4])

ylim([0 8])

xlabel('time')

ylabel('amplitude')

yyaxis right

ylabel('phase difference')

hold on

plot(T,((Y(:,5))./6.28));

hold off

function dy = rate_eq(t,y)

dy = zeros(5,1);

o = 2E5;

tc = 30E-9;

tf = 230E-6;

a1 = 0.1;

a2 = 0.1;

P1 = 0.2;

P2 = 0.2;

kc = 3E-3;

s = 0.17;

k = s.*kc;

dy(1) = ((y(3) - a1) * y(1) + k * y(2) * cos(y(5))) / tc;

dy(2) = ((y(4) - a2) * y(2) + k * y(1) * cos(y(5))) / tc;

dy(3) = (P1 - y(3) * (y(1)^2 + 1)) / tf;

dy(4) = (P2 - y(4) * (y(2)^2 + 1)) / tf;

dy(5) = o - k / tc * (y(1) / y(2) + y(2) / y(1)) * sin(y(5));

end

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