how to solve this equation?

clc

clear

syms Th1 Th2 Th3 Th4 Rs Ld Lq kexi Ts

eq1 = (-Ts*Rs + 2*Lq)/(Ts*Rs + 2*Lq);

eq2 = (Ts)/(Ts*Rs + 2*Lq);

eq3 = -(Ts*Ld)/(Ts*Rs + 2*Lq);

eq4 = -(Ts*kexi)/(Ts*Rs + 2*Lq);

d = solve(eq1==Th1,eq2==Th2,eq2==Th3,eq2==Th4,Rs,Ld,Lq,kexi);

d.Rs

d.kexi

d.Ld

d.Lq

ans =

command:

Empty sym: 0-by-1

ans =

Empty sym: 0-by-1

ans =

Empty sym: 0-by-1

ans =

Empty sym: 0-by-1

Answer 1

[Swatantra Mahato]

Hi

I am assuming you want to solve the equation

d = solve(eq1==Th1,eq2==Th2,eq2==Th3,eq2==Th4,Rs,Ld,Lq,kexi);

You can try using the "solve" function in the Symbolic Math Toolbox with the Parameter "ReturnConditions" set to true as below:

d = solve([eq1==Th1,eq2==Th2,eq2==Th3,eq2==Th4],[Rs,Ld,Lq,kexi],'ReturnConditions',true);

which gives

>> d

d = 

  struct with fields:

            Rs: [3×1 sym]
            Ld: [3×1 sym]
            Lq: [3×1 sym]
          kexi: [3×1 sym]
    parameters: [1×4 sym]
    conditions: [3×1 sym]

where the parameters used in the solutions and the conditions under which the respective solutions are valid are stored in the struct "d" under the fields "parameters" and "conditions" respectively.

I noticed that

d = solve([eq1==Th1,eq2==Th2,eq3==Th3,eq4==Th4],[Rs,Ld,Lq,kexi]);

seems to give a unique solution if that was what you wanted to actually solve.

You can refer the documentation for "solve" for more information:

https://www.mathworks.com/help/symbolic/solve.html#bue1fkj-5

Hope this helps

[Lu gy]

Hi

Thank you very much for providing me with the ideas to solve the problem !