清空筛选条件
清空筛选条件

why its giving NaN (Not a Number)

3 次查看（过去 30 天）
Ahmad Sheikh
Ahmad Sheikh 2015-6-26
评论： Ahmad Sheikh 2015-6-26
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.......

请先登录，再回答此问题。

回答（1 个）

Walter Roberson
Walter Roberson 2015-6-26
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 个评论
Ahmad Sheikh
Ahmad Sheikh 2015-6-26
Thanks for your answer but i have changed the initial conditions to
[1 12 13 1 1]
but still giving an error.....

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

尚未输入任何标签。

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by