Error using + Matrix dimensions must agree.

close all

clear all

clc

G0=300; %initial positions

I0=2.5;

q=1/1;

A=90;

D(1)=A;

Gd=80;

Z(1)=0;

Ib=2.5;

Comm=0;

x(:,1)=[G0 0 20]'; %states initial values

xob(:,1)=[G0 0 20]';

lam(:,1)=[0 0 0]';

% patient 1

p1=0;

p2=0.0142;

p3=154*10^-5;

n=0.2814;

Pn=[p1 p2 p3 n];

%%% patient 2

%p1=0;

%p2=0.0072;

%p3=2.16*10^-5;

%n=0.2865;

Pn=[p1 p2 p3 n];

Comande(1)=10;

i=1;

B=0.05;

t(1)=0;

h=1/6;%0.005;1/6

% ODE solving

% opt1=odeset('RelTol',1e-10,'AbsTol',1e-20,'NormControl','off');

%[T,X] = ode45(@(t,x) glucoz(t,x,Pn,Gb,Comm),Ts,x0);

%V(1)=0;

YY3(1)=0;

DD(1)=0;

a=1;

while t<18000

D(i+1)=A*exp(-B*i*h);

k1=h*glucoz(x(:,i),Pn,Gd,Comande(i),D(i));

k2=h*glucoz(x(:,i)+k1'/2,Pn,Gd,Comande(i),D(i)); %the error in the + signe

k3=h*glucoz(x(:,i)+k2'/2,Pn,Gd,Comande(i),D(i));

k4=h*glucoz(x(:,i)+k3',Pn,Gd,Comande(i),D(i));

x(:,i+1)=x(:,i)+(1/6)*(k1+2*k2+2*k3+k4);

DD(i+1)=-A*B*exp(-B*i*h);

D2D(i+1)=A*B^2*exp(-B*i*h);

Food=0;

if i*h>Food

D(i+1)=2*A*exp(-B*(i*h-Food));

DD(i+1)=-A*B*exp(-B*(i*h-Food));

D2D(i+1)=A*B^2*exp(-B*(i*h-Food));

else

D(i+1)=D(i+1);

DD(i+1)=DD(i+1);

D2D(i+1)=D2D(i+1);

end

%%% tracking error %%%%%%%%%

e= x(1,i+1)-Gd;%0.1*sign(x(3,i+1)-Ib);

%%%%%%%%%%%%%%% First derivative %%%%%

edot=-x(1,i+1)*x(2,i+1)+D(i);

%%%%%%%%%%%% second derivative %%%%%%%%%%%

e2dot=-x(1,i+1)*x(2,i+1)^2+D(i+1)*x(2,i+1)+p2*x(1,i+1)*x(2,i+1)-p3*(x(3,i+1)-Ib)*x(1,i+1)+DD(i+1);

%%%%%%%%%%%%%%% Third Derivative %%%%%%%%%%%%%

%f2=[D(i+1)-x(1,i+1)*x(2,i+1)]*x(2,i+1)^2-2*p3*x(1,i+1)*x(2,i+1)*x(3,i+1)-3*p3*Ib*x(1,i+1)*x(2,i+1)-3*p2*x(1,i+1)*x(2,i+1)^2+x(2,i+1)*DD(i+1)+p3*x(3,i+1)*D(i+1)+p2*p3*x(1,i+1)*x(3,i+1)-p2*p3*Ib*x(1,i+1)-p2^2*x(1,i+1)*x(2,i+1)+D2D(i+1)+p3*x(3,i+1)*D(i+1)+p3*n*(x(3,i+1)-Ib)*x(2,i+1);

F=[D(i+1)-x(1,i+1)*x(2,i+1)]*[p2*x(2,i+1)-p3*(x(3,i+1)-Ib)+x(2,i+1)^2]-[p3*(x(3,i+1)-Ib)-p2*x(2,i+1)]*(D(i+1)+p2*x(1,i+1)-2*x(1,i+1)*x(2,i+1))+DD(i+1)*x(2,i+1)+D2D(i+1)+p3*n*x(1,i+1)*(x(3,i+1)-Ib);

%e3dot=F-p3*x(1,i+1)*U;

%%%%%%%%%%%%%%%%%%%%%%%%%%%% sliding surface %%%%%%%%%%

lambda1=0.01;%0.01;%0.01

lambda2=0.0001;%0.0001;%0.0001

QQ=.000;

KK=0.5;%0.5; 0.1 with Q=0.001, Patien 1

S=(q^2*e2dot+q*lambda1*edot+lambda2*e);%+0.001*sign(x(3,i+1)-Ib);KK=0.5;QQ=0.1;

surface(i)=S;

if KK>0

KK=KK+h*(abs(S)*sign(abs(S)-0.0001));

else

KK=KK;

end

%KK=0.9;

%KK=abs(p3*x(1,i+1)*10*F/(F^2+0.1));

YY=1.7159*[exp(a*S)-1]/[exp(a*S)+1];

g(i)=p3*x(1,i+1)*q^3;

GG=x(1,i+1)*p3;

U=((q^3)*F+(q^2)*lambda1*e2dot+q*lambda2*edot+KK*YY+QQ*S)/GG;

%___________________________________________________________

Comande(i+1)=U;

X=x(2,i+1);

I=x(3,i+1);

t(i+1)=i*h;

temps(i)=i*h;

i=i+1;

end

% Output

%torque inputs computation from the 7th,8th states inside ODE

%tt=0:(T(end)/(length(Comm)-1)):T(end);

%theta1 error plot

figure(1)

plot(t/60,x(1,:),'k')%,t/60,xob(1,:))

grid

title('G')

ylabel('error ')

xlabel('time (min)')

figure(2)

plot(t/60,Comande),

title('Commande')

figure(3)

plot(t/60,x(3,:),'k')

grid

title('X3')

ylabel('Insuline in plasma')

xlabel('time (min)')

figure(4)

plot(t/60,x(2,:),'k')

grid

title('X2')

ylabel('insuline effect')

xlabel('time (min)')

%theta2 error plot

figure(5)

plot(temps/60,surface)

grid

title('surfaces')

ylabel('insuline effect')

xlabel('time (min)')

%theta2 error plot

Answer 1

[Are Mjaavatten]

In line 43, x(:,i) is a column vector, while k1' is a row vector. Summing a column vector and a row vector is not allowed in older versions of Matlab. In Matlab 2014b:

>> a = [1;2;3];b = [10,20,30];
>> a + b
Error using  + 
Matrix dimensions must agree. 

Your code runs without error in version 2020a. Here vector a is added to each element in b, resulting in a 3 by 3 array:

>> a = [1;2;3];b = [10,20,30];
>> a+b
ans =
    11    21    31
    12    22    32
    13    23    33

But this is proably not what you want. Your glucoz function expects a three element vector, so you should simply use the column vector k1 instead of the transpose k1'.