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

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Sir, I want to plot the intermediate variables say Te from a function that was used by ode45 solver .But i am not getting the response exactly. 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. The problem is ,in workspace i am not getting the array of Te values for all time instants.I am getting a single last value of Te say 4.5 (1x 1 Element) 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.(say 1057x1) entries.(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 .
Main Pragram:
x0=0.1;
final_time=1;
[t,x]=ode45(@myfunc,[0,final_time],x0);% ODE Call
plot(t(:,1),x(:,1))% plotting the state varaible
Intervar=myfunc(t,x,'flag');
plot(t(:,1),Intervar)
main ends
Function :
function dv=myfunc(t,x)
H=1.0;
Te=2*x(1);% intermediate calculations/variables for my differential equations
%The Diffenrential Equation
dv=[(1/(H))*(Te)];
if nargin == 3
dv=Te; //Returns the intermediate variable Te
end

Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 1 Aug 2013
Edited: Azzi Abdelmalek on 1 Aug 2013
In your code Te represent 2*x, then why you don't just type
Te=2*x(:,2)

Jan
Jan on 1 Aug 2013
Edited: Jan on 1 Aug 2013
Please copy the code I have posted exactly:
function dv = myfunc(t, x, flag)
H = 1.0;
Te = 2 * x(:, 1); % <-- Not "Te=2*x(1);"
dv = Te ./ H; % Btw, No additional square brackets
if nargin == 3
dv = Te;
end
  1 Comment
Thayumanavan
Thayumanavan on 18 Aug 2013
Sir, my code is this:Main Program
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('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
It is giving the last value only.In workspace Te,VPCC is a 1x1 element.iT SHOULD BE AN ARRAY similar to x.I tried giving Te,VPCC Global,but in vein.Simulate the code ,Please help.

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