MATLAB: Matlab computing a wrong value

tan function

I am trying to do a simple calculation but matlab is outputting a different value than from what it should be, any ideas why?

 R=0.5;
 Cv=0.4;
 B1= 60;
 R= Cv/2*(tan(B1) + tan(B2));
 [B2] = vpasolve([R], [B2]);
 disp(B2);

You are supposed to find that B2 should be around 37.5 but matlab displays a value of -0.3097 so why is matlab getting a different answer?

Best Answer

Note that the tan function uses radians, NOT degrees. So you use tan(B1), where B1 is clearly a number written in degrees.

You can either use the function tand, which works in degrees, or you can convert degrees to radians.

Next, first, you set R = 0.5, then you set it equal to Cv/2*(tan(B1)+tan(B2)). This is wrong, since you wish to SOLVE for that.

syms B2
R=0.5;
Cv=0.4;
B1= 60;
B2 = solve(R == Cv/2*(tan(B1*pi/180) + tan(B2*pi/180)))

Note the use of == in there, NOT = as you had done.

B2 =
-(180*atan(3^(½) - 5/2))/pi
vpa(B2)
ans =
37.522431511944199990343681864777

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The Symbolic Math Toolbox is not the best way to solve iterative problems.

I coded your equations as anonymous functions and gave them to the Optimization Toolbox fsolve function that works best for problems like these. It works, but the ‘SWR’ value is negative (and since I believe that stands for ‘standing wave ratio’ on a transmission line, absolutely does not make sense, at least in that context).

Look over your equations to be sure they are set up correctly:

Cp=0.5;
% Cv=0.3;
Cvv = [0.3:0.1:101];
R = @(A1,B2,Cv) Cv.^2.*((1./Cv-tan(A1)).^2 - (tan(B2)).^2)./2./(1-Cv.*tan(A1) - Cv.*tan(B2)); %%Eq 1
Cp = @(A1,B2,Cv) 1 - (tan(B2)./(1./Cv - tan(A1))).^2; %%Eq2
% MAPPING: A1 = p(1),    B2 = p(2)
Eqn = @(p,Cv) [R(p(1),p(2),Cv), Cp(p(1),p(2),Cv)];
for k1 = 1:length(Cvv)
    Cv = Cvv(k1);
    [P(:,k1),Fval] = fsolve(@(p)Eqn(p,Cv), [-145; 23]);
    SWR(k1) = 1 - Cv*(tan(P(1,k1)) + tan(P(2,k1)));
end
figure(1)
plot(Cvv, SWR)
grid
xlabel('C_v')
ylabel('SWR')

I did the symbolic solution first with the initial value of ‘Cv’ (0.3), and used that solution ‘[-145,23]’ for the initial estimates in my loop. If you believe they should be other values, change them, run the loop, and see what works best. Since your equations are periodic, the ‘correct’ initial parameter estimates are important.

Best Answer

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