How to plot intermediate variables of a function used by ode45 solver

Sir, I am calling ode45 for integrating a diff equation say x'=(1/(H))*(2x)

Main Program:

x0=0.1; final_time=1; [t,x]=ode45(@myfunc,[0,final_time],x0);% ODE Call plot(t(:,1),x(:,1))% plotting the state variable main ends

Function Program:

function dv=myfunc(t,x) H=1.0; Te=2*x(1);% intermediate calculations/variables for my differential equations %The Differential Equation dv=[(1/(H))*(Te)];

But i want to plot the intermediate variables say Te in this case.But ode returns t,x alone.How to plot the intermediate variables that was used in the integration process.

Answer 1

[Jan]

You can prepare your function such, that it replies the wanted value according to a special trigger method and accepts vectors as input also:

function dv = myfunc(t, x, flag)
H  = 1.0;
Te = 2 * x(:, 1)
dv = Te ./ H;      % Btw, No addition square brackets

if nargin == 3
  dv = Te;
end

Now run the integration at first:

[t,x] = ode45(@myfunc,[0,final_time],x0);

and obtain the internal variable afterwards:

Te = myFunc(t, x, 'flag');

[Thayumanavan]

Sir, Thanks for the immediate response.I tried your code and got the output plot of Te .

Inference:

In workspace i am not getting the array of Te values for all time instants.Getting a value of 4.5 (1x 1 Element)only Workspace is having the last value of Te alone ie. at t=10 if tspan=[0 10]. But the vector of state variable x is coming in workspace.(1057x1) entries.The time response of x is not constant(from the plot of (t,x(:,1))). But the The time response of Te seems to be constant with respect to time.I am getting the plot as, with respect to time (0 to 10) a line at 4.5 .so i guess it is plotting t with respect to the last value of Te alone say Te=4.5 (at t=10).(If x is time varying means Te should also vary with respect to time) .I tried giving Te as global variable also.But in vein.Please help me .

[Jan]

I do not understand your comment. Te does not depend on the time, but is simply twice as large as x in your case. Therefore I cannot understand how you get results, which "Te seems to be constant with respect to time".

Please post a copy of the code you use.

[Thayumanavan]

clear all clc; % Time Tspan =[0 10]; X0=0.7; options=odeset('RelTol',1e-3,'AbsTol',1e-6); %solving the differential Equation [t,x,s]=ode45(@myfunc,Tspan,X0,options);

%Plotting the figure

subplot(2,1,1) plot(t(:,1),x(:,1)) xlabel('t'); ylabel('x'); title('Time Response ') grid on

subplot(2,1,2) IntVar= dvdt(t, x, 'flag'); plot(t(:,1),x(1,:)) xlabel('t'); ylabel('Te') grid on

function dv=myfunc(t,x,flag) H = 1.0; Te = 2 * x(:, 1)

%The Diffenrential Equation 

       dv=Te./H;          

 if nargin == 3
  dv = Te;
end

This is the code ,what i am expecting is t=0 to 10 ,x is changing means Te should also change.But Te seems to be constant.You try the code sir It is plotting the last value of Te(0.999) wrt time.Hope you understand.

[Thayumanavan]

Sir ,I got answer for the problem that i asked help.But for simplicity i gave that differential equation.But my code is little bit different with many variables.I am attaching the main program and my function.Please help me to plot Te and Vpcc intermediate variable.

Main Code: clear all clc;

% Time Tspan =[0 10]; X0=[0.7; 1.0151; 157 ];

 options=odeset('RelTol',1e-3,'AbsTol',1e-6);

%solving the differential Equation [t,x]=ode45(@dvdt,Tspan,X0,options);

%Plotting the figures figure subplot(5,1,1) plot(t(:,1),x(:,1)) xlabel('t'); ylabel('Eprime'); title('Dynamic Wind Time Response ') grid on

subplot(5,1,2) plot(t(:,1),x(:,2)) xlabel('t'); ylabel('Delta') grid on

wradsec=x(:,3); wmech=(60*wradsec)/(2*pi); wpu=wmech/1500;

subplot(5,1,3) plot(t(:,1),wradsec) xlabel('t'); ylabel('Omega') grid on

% Plotting of Intermediate Variables as per Your Idea subplot(5,1,4) op= dvdt(t, x, 'flag'); plot(t(:,1),op(1,:)) xlabel('t'); ylabel('Vterminal') grid on

subplot(5,1,5) plot(t(:,1),op(2,:)) xlabel('t'); ylabel('ElectromagneticTorque') grid on

Function code: function dv=dvdt(t,x,flag)

Xs=0.0754;Xr=0.0326;H=0.20; Rl=0.0083;Rcon=0.0089;Xl=0.033;Xcon=0.01789; Vdinf=1.0;Vqinf=0.0;wspu=157.0; T=0.7855;Xdash=0.1076;D=6.4165; Rt=Rl+Rcon;Xt=Xl+Xcon; xtrans=Xt+(D/Xr); detm=(Rt*Rt)+(xtrans*xtrans); J1=-xtrans;J2=-Rt;J3=-Rt;J4=xtrans; iq=(1/detm)*((J3*((x(1)*cos(x(2)))-Vqinf))+(J4*((-x(1)*sin(x(2)))-Vdinf))); id=(1/detm)*((J2*((-x(1)*sin(x(2)))-Vdinf))+(J1*((x(1)*cos(x(2)))-Vqinf))); Vq=(1/Xr)*((D*id)+(Xr*x(1)*cos(x(2)))); Vd=(-1/Xr)*((D*iq)+(Xr*x(1)*sin(x(2)))); Vdpc=(-Rcon*id)+(Xcon*iq)+Vdinf; Vqpc=(-Rcon*iq)-(Xcon*id)+Vqinf; VPCC=sqrt((Vdpc*Vdpc)+(Vqpc*Vqpc));% Intermediate Variable 1 t1=(Vq*cos(x(2)))-(Vd*sin(x(2))); t2=(Vq*sin(x(2)))+(Vd*cos(x(2))); t3=((Vd*cos(x(2)))+(Vq*sin(x(2)))); Te=(x(1)*t3/Xdash); % Intermediate Variable 2

%The Diffenrential Equation 

       dv=[ (1/T)*((-Xs*x(1)/Xdash)+(((Xs-Xdash)/Xdash)*t1)); % My 1st Differential Equation
           (x(3)-wspu)-(((Xs-Xdash)*t2) /(Xdash*T*x(1)));    %My 2nd  Differential Equation  
           (wspu/(2*H))*(Tm+Te)]; %  My 3rd Differential Equation

 if nargin == 3
  dv = [VPCC;
       Te];

end

You copy paste the code execute and suggest me to plot Te,Vpcc.Thanks

[Thayumanavan]

clear all clc; % Time Tspan =[0 10]; X0=[0.7; 1.0151; 157 ];

 options=odeset('RelTol',1e-3,'AbsTol',1e-6);

%solving the differential Equation [t,x]=ode45(@dvdt,Tspan,X0,options);

%Plotting the figures figure subplot(5,1,1) plot(t(:,1),x(:,1)) xlabel('t'); ylabel('Eprime'); title('Dynamic Wind Time Response ') grid on

subplot(5,1,2) plot(t(:,1),x(:,2)) xlabel('t'); ylabel('Delta') grid on

wradsec=x(:,3); wmech=(60*wradsec)/(2*pi); wpu=wmech/1500;

subplot(5,1,3) plot(t(:,1),wradsec) xlabel('t'); ylabel('Omega') grid on

% Plotting of Intermediate Variables as per Your Idea subplot(5,1,4) op= dvdt(t, x, 'flag'); plot(t(:,1),op(1,:)) xlabel('t'); ylabel('Vterminal') grid on

subplot(5,1,5) plot(t(:,1),op(2,:)) xlabel('t'); ylabel('ElectromagneticTorque') grid on

Function Code: function dv=dvdt(t,x,flag)

Xs=0.0754;Xr=0.0326;H=0.20; Rl=0.0083;Rcon=0.0089;Xl=0.033;Xcon=0.01789; Vdinf=1.0;Vqinf=0.0;wspu=157.0; T=0.7855;Xdash=0.1076;D=6.4165; Rt=Rl+Rcon;Xt=Xl+Xcon; xtrans=Xt+(D/Xr); detm=(Rt*Rt)+(xtrans*xtrans); J1=-xtrans;J2=-Rt;J3=-Rt;J4=xtrans; iq=(1/detm)*((J3*((x(1)*cos(x(2)))-Vqinf))+(J4*((-x(1)*sin(x(2)))-Vdinf))); id=(1/detm)*((J2*((-x(1)*sin(x(2)))-Vdinf))+(J1*((x(1)*cos(x(2)))-Vqinf))); Vq=(1/Xr)*((D*id)+(Xr*x(1)*cos(x(2)))); Vd=(-1/Xr)*((D*iq)+(Xr*x(1)*sin(x(2)))); Vdpc=(-Rcon*id)+(Xcon*iq)+Vdinf; Vqpc=(-Rcon*iq)-(Xcon*id)+Vqinf; VPCC=sqrt((Vdpc*Vdpc)+(Vqpc*Vqpc));% Intermediate Variable 1 t1=(Vq*cos(x(2)))-(Vd*sin(x(2))); t2=(Vq*sin(x(2)))+(Vd*cos(x(2))); t3=((Vd*cos(x(2)))+(Vq*sin(x(2)))); Te=(x(1)*t3/Xdash)% Intermediate Variable 2 Tm=1.0;

%The Diffenrential Equation 

       dv=[ (1/T)*((-Xs*x(1)/Xdash)+(((Xs-Xdash)/Xdash)*t1)); % My 1st Differential Equation
           (x(3)-wspu)-(((Xs-Xdash)*t2) /(Xdash*T*x(1)));    %My 2nd  Differential Equation  
           (wspu/(2*H))*(Tm+Te)]; %  My 3rd Differential Equation

 if nargin == 3
  dv = [VPCC;
       Te];

end

Little change in code .i forgot to give 1 parameter .You pls simulate and tell me what change i need to do in the code to get plot of Te,Vpcc

Answer 2

[Shashank Prasanna]

You can define an output function (outputfcn) that will be called after each iteration.

http://www.mathworks.com/help/matlab/ref/odeset.html#f92-1016858

you can specify that in the options structure. In your custom output function you can use a plot command with a hold on in order to plot all you intermediate values.

[Jan]

Even the output function has no access to the internally used variables.

[Thayumanavan]

Thanks Sir