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alternative for subs function

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Neel
Neel 2013-11-27
评论： Neel 2013-11-27
actually i am trying to solve matrix calculations with symbols when i use subs it does not evaluate but keeps it in huge number fractions huge means in millions and millions and i have to repeat for 10000 time with different values every time i get in previous step so please tell me some alternative i used simple but its not working and the whole process is so slow that in the 6th step it takes about 6 mins to solve my code is as follows:
I am attaching the m file also please help me out its urgent
thanks in advance
clc
clear
format short
syms z1 z2 z3 z4 u z
z01 = input ('z01');
z02 = input ('z02');
z03 = input ('z03');
z04 = input ('z04');
u01 = input ('u01');
E = input ('E');
z0 = [z01;z02;z03;z04;u01;E];
z_dot = state()
J = sub(state())
s = 0.0001;
z_old = z0(2,1)
dJ = (der(z0(6,1)))
for i = 1:6
z_new = (z_old) - s*(subs(dJ,{z1,z2,z3,z4,u},{z0(1,1),z_old,z0(3,1),z0(4,1),z0(5,1)}))
z_old=z_new;
hold on
plot(i,z_old,'c*')
end
and the functions i used are as follows:
function z_dot = state()
syms z1 z2 z3 z4 u
m = 0.23;
M = 1.0731;
I = 0.0079;
l = 0.3302;
b = 5.4;
g = 9.81;
%z = [z10;z20;z30;z40;u0];
%z = subs(z,[z1,z2,z3,z4])
F = u;
z1_dot = z3;
z2_dot = z4;
z3_dot = ((I+m*l^2)*F)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) + (m*l*(I+m*l^2)*sin(z2)*(z4)^2)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) - (b*(I+m*l^2)*z3)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) + (m^2*l^2*sin(z2)*cos(z2))/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2);
z4_dot = (-m*g*l*(M+m)*sin(z2))/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) - (m*l*cos(z2)*F)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) - (m^2*l^2*sin(z2)*cos(z2)*(z4)^2)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2) - (m*l*b*cos(z2)*z3)/((M+m)*(I+m*l^2)-m^2*l^2*(cos(z2))^2);
z_dot = [z1_dot;z2_dot;z3_dot;z4_dot];
%z_dot = simple(subs(z_dot,[z1,z2,z3,z4,u],[z(1,1),z(2,1),z(3,1),z(4,1),z(5,1)]));
z_dot = simple(z_dot);
z_dot = simple(z_dot);
end
function derivativ = der(E)
format short
syms z1 z2 z3 z4 u
%E = input('E')
%z10 = input('z10')
%z20 = input('z20')
%z30 = input('z30')
%z40 = input('z40')
%u0 = input('u0')
%E = 0.01;
J2 = sub(state());
J2 = (subs(J2,{z1,z2,z3,z4,u},{z1,z2+E,z3,z4,u}));
J1 = sub(state()); J1 = (subs(J1,{z1,z2,z3,z4,u},{z1,z2,z3,z4,u}));
derivativ = ((J2-J1)/E);
end
function substitute = sub(zk)
substitute = (norm(zk))^2;
end
  1 个评论
Neel
Neel 2013-11-27
my initial inputs are
z01 = 0
z02 = 0.2
z03 = 0
z04 = 0
u01 = 0
E = 0.01

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采纳的回答

Walter Roberson
Walter Roberson 2013-11-27
You need to expect that sort of thing to happen when you do iteration of functions in rational fractions. It is needed to express the exact solution.
If you do not need the exact solution, have a look at vpa()

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