MATLAB: If I use multi funcions with syms variable, How to get track min value among that funcions

MATLABSymbolic Math Toolbox

clear; clc; close all;
syms seta
Rx1= cos(seta)
Rx2= sin(seta)

If I set Rx1, Rx2 funcion like that, How can I get the graph with min value among Rx1, Rx2?

If I use discrete value, I can use like this

xmin=min([R1(x(:)');R2(x(:)');])
plot(x,xmin,'.-');

but, In this case I using Symbolic Math Toolbox, I don't know how I can make it works with syms variable…

Best Answer

You need to use something like

Min = @(tworows) arrayfun( @(colidx) piecewise(tworows(1,colidx) <= tworows(2,colidx), tworows(1,colidx), tworows(2,colidx)), 1:size(tworows,2);

The code could be simplified if you were willing to use R1 and R2 as separate arguments instead of putting them together in one array.

Related Solutions

MATLAB: How to get maximum and minimum value with sym variables

Use matlabFunction() and fminbnd() to find the minimum value of that function and to find the maximum extreme use negative sign for the function and feed it to fminbnd().

 doc matlabFunction
 doc fminbnd

And as always eval() here is completely not necessary and should be avoided.

MATLAB: Is the script not working

I try

1) I think you are caluclating the jacobian matrix. To avoid mistakes I did this symbolically and the result is

>> syms r1 r2 r3 r4 r5 r6 t1 t2 t3 t4 t5 t6
>> f = [r2*cos(t2)-r3*cos(t3)+r1; r2*sin(t2)-r3*sin(t3); r6-r4*cos(t4)+r1; r5-r4*sin(t4); t4-t3]
 
f =
 
 r1 + r2*cos(t2) - r3*cos(t3)
      r2*sin(t2) - r3*sin(t3)
         r1 + r6 - r4*cos(t4)
              r5 - r4*sin(t4)
                      t4 - t3
 
>> A = jacobian(f,[r3 r4 r5 t3 t4])
 
A =
 
[ -cos(t3),        0, 0,  r3*sin(t3),           0]
[ -sin(t3),        0, 0, -r3*cos(t3),           0]
[        0, -cos(t4), 0,           0,  r4*sin(t4)]
[        0, -sin(t4), 1,           0, -r4*cos(t4)]
[        0,        0, 0,          -1,           1]

It seems that you have different signs in your variable dfdt3

2) you load your derivatives in the matrix columns. It seems to me you are working with the transpose Jacobian.

3) Never ever use inv(A)*f but rather use matrix solution A\f

All in all, I modified your code as follows

clear all; clc;
pi=4.0*atan(1.0);
%scalar knowns
r1=4.8;
r2=2.0;
r6=3.63;
t1=-pi;
t5=-pi/2;
t6=0;
%input theta 2
t2= 283*pi/180;
%guess values of scalar unknowns
r3=3.0;
r4=11.0;
r5=4.0;
t3=5.236;
t4=5.236;
%vector containing initial guesses:
x0=[r3;r4;r5;t3;t4];
x=[0;0;0;0;0];
% counter to limit iterations
counter = 0;
counterLimit = 1e3;
while abs(x-x0) >= 1e-2
    counter = counter+1;
    
    ct2=cos(t2);
    st2=sin(t2);
    ct3=cos(t3);
    st3=sin(t3);
    ct4=cos(t4);
    st4=sin(t4);
    
    %compute functions at current guessed values
    f1=r2*ct2-r3*ct3+r1;
    f2=r2*st2-r3*st3;
    f3=r6-r4*ct4+r1;
    f4=-r5-r4*st4;
    f5=t4-t3;
    
    %define vector for computed functions of guessed values
    f=[f1;f2;f3;f4;f5];
    
    %calculate partial derivatives of f w.r.t. each element of x
    dfdr3=[-ct3; 0; 0; r3*st3; 0];
    dfdr4=[-st3; 0; 0; -r3*ct3; 0];
    dfdr5=[0; -ct4; 0; 0; r4*st4];
    dfdt3=[0; -st4; 1; 0; -r4*ct4];
    dfdt4=[0; 0; 0; -1; 1];
    
    % Define the A matrix
    A= [dfdr3 dfdr4 dfdr5 dfdt3 dfdt4];
    
    %use equation (2.67) to compute the solution x
    x=-A.'\f+x0;
    r3=x(1);
    r4=x(2);
    r5=x(3);
    t3=x(4);
    t4=x(5);
end
if counter == counterLimit
    warning('Failed')
end
fprintf(' %f\n', x);

This code does provide a solution, if it is correct is up to you.

As a side note, I think it should be better coding not to use r1 r2 r3 r4 r5 r6 t1 t2 t3 t4 t5 t6 as variables, but I would rather use the vector x, and accessing with indexing (x(1), x(2), ...)

Best Answer

Related Solutions

MATLAB: How to get maximum and minimum value with sym variables

MATLAB: Is the script not working

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