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why its giving NaN (Not a Number)

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Ahmad Sheikh
Ahmad Sheikh on 26 Jun 2015
Commented: Ahmad Sheikh on 26 Jun 2015
I am working on takagi Sugeno fuzzy modeling
the example i have done is this one
function derivative_x = book_TS(t,x)
derivative_x = zeros(2,1);
z1=x(1).*(x(2).^2);
z2=(3+x(2)).*(x(1).^2);
a1=1;
a2=-1;
b1=4;
b2=0;
M1=(a1-z1)./(a1-a2)
M2=(z1-a2)./(a1-a2)
M3=(b1-z2)./(b1-b2)
M4=(z2-b2)./(b1-b2)
h1= M1.*M3
h2= M1.*M4
h3= M2.*M3
h4= M2.*M4
h=h1+h2+h3+h4
A1=[-1 1;4 -1];
A2=[-1 1;0 -1];
A3=[-1 -1;4 -1];
A4=[-1 -1;0 -1];
derivative_x(1)= -h1*x(1)+ 4*h1*x(2)-h2*x(1)-h3*x(1)+4*h3*x(2)-h4*x(1);
derivative_x(2)=h1*x(1)-h1*x(2)+h2*x(1)-h2*x(2)-h3*x(1)-h3*x(2)-h4*x(1)-h4*x(2);
when i run this function in a seperate file j get correct results
Initial_Time=0;
Final_Time=10;
[x1,x2] = meshgrid(-2:0.1:2);
[t,x] = ode45(@book_TS, [Initial_Time Final_Time], [0.5 0.5]);
figure(1);
plot(t,x(:,1));
but now i am working on some different example
function derivative_x = example_aircraft(t,x)
derivative_x = zeros(5,1);
z1=-35.137.*x(1).*x(2)-0.6678.*x(1)+0.7870.*x(2)-0.10977;
z2=-8.50255.*x(2)-0.66728-1.043.*x(3);
z3=-132.7./x(1);
a1 = 0.0269;
a2 = -0.0660;
b1 = -7.91;
b2 = -8.73;
c1 = -0.923;
c2 = -2.09;
a= a1-a2;
b= b1-b2;
c= c1-c2;
M11= (a1 - z1) ./ a;
M21= (z1 - a2) ./ a;
M12= (b1 - z2) ./ b;
M22= (z2 - b2) ./ b;
M13= (c1 - z3) ./ c;
M23= (z3 - c2) ./ c;
h1= M11 .* M12 .* M13
h2= M11 .* M12 .* M23
h3= M11 .* M22 .* M13
h4= M11 .* M22 .* M23
h5= M21 .* M12 .* M13
h6= M21 .* M12 .* M23
h7= M21 .* M22 .* M13
h8= M21 .* M22 .* M23
h=h1+h2+h3+h4+h5+h6+h7+h8
derivative_x(1)= -0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4)-0.0142*x(1)+1.47*x(2)+0.0374*x(3)-9.80*x(4);
derivative_x(2)= -0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3)-0.000368*x(1)-0.758*x(2)+0.934*x(3);
derivative_x(3)= -0.0660*x(1)-8.73*x(2)-2.09*x(3)-0.0660*x(1)-8.73*x(2)-0.923*x(3)-0.0660*x(1)-7.91*x(2)-2.09*x(3)-0.0660*x(1)-7.91*x(2)-0.923*x(3)+0.0269*x(1)-8.73*x(2)-2.09*x(3)+0.0269*x(1)-8.73*x(2)-0.923*x(3)+0.0269*x(1)-7.91*x(2)-2.09*x(3)+0.0269*x(1)-7.91*x(2)-0.923*x(3);
derivative_x(4)= 8*x(3);
derivative_x(5)= 150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4)+150*x(2)-150*x(4);
Now when i am running this one
Initial_Time=0;
Final_Time=10;
[x1,x2] = meshgrid(-2:0.1:2);
[t,x] = ode45(@example_aircraft, [Initial_Time Final_Time], [0 12 13 0 0]);
figure(1);
plot(t,x(:,1));
xlabel('time')
ylabel('state x1k')
its saying that h1 h2 h3 h4 ...h8 are not a number i-e NaN
kindly do help if any one know why its saying like this.......

Answers (1)

Walter Roberson
Walter Roberson on 26 Jun 2015
You divide by x(1) . Your initial x vector is [0 12 13 0 0] which starts with 0 so your x(1) is going to be 0. You are dividing by 0. You are dividing a positive number by that 0 so the result will be infinity. That infinity is going to "poison" the equations, and if you ever multiply the infinity by 0 you will get NaN as the result. Any arithmetic operation on the NaN is going to result in NaN.
  1 Comment
Ahmad Sheikh
Ahmad Sheikh on 26 Jun 2015
Thanks for your answer but i have changed the initial conditions to
[1 12 13 1 1]
but still giving an error.....

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