How do I solve these systems of equations? Keep getting empty matrices.

So here are the two equations:

(1−e^φ )*P = (1−e^β )*G + (e^β)*(1−e^α)*W

α = φ −β

Where φ = 90 and P,G and W are 1x2 matrices of [2,1], [0,0] and [1.3822, 1.3822] respectively.

Here is my code:

function y = RR2CrankAng(P12,G,W1,theta12)
syms beta12 alpha12
eqn1 = ((1-exp(theta12))*P12) == ((1-exp(beta12))*G)+((exp(beta12))*(1-exp(alpha12))*W1);
eqn2 = alpha12 == theta12 - beta12;
sol = solve([eqn1, eqn2], [beta12, alpha12]);
xSol = sol.beta12;
ySol = sol.alpha12;
y = [xSol,ySol];
end

calling it

CA = RR2CrankAng(P12,G,W1,theta12)

CA =

Empty sym: 0-by-2

What is happening here?

Answer 1

[Roger Stafford]

If you specify theta12 as a known, as you have done, you have in effect three equations to solve and only two unknowns. I think ‘solve’ will object to that. In any case, there is a way of solving these without the use of ‘solve’. Consider x = exp(beta) as one unknown and y = exp(phi) as the other unknown. Then you have two linear equations

   (1-y)*P(1) = (1-x)*G(1)+(x-y)*W(1)
   (1-y)*P(2) = (1-x)*G(2)+(x-y)*W(2)

which can easily be solved for x and y using the backslash operator. From the values of x and y, you can get alpha and beta using:

   beta = log(x)
   alpha = log(y)-beta

[S R]

Sorry I'm still learning Matlab. I've read the backlash operator documentation and I still don't understand how to use it in this context. Maybe provide me with an example?

I see what you did breaking up the matrix into its two parameters, I just tried that and its still giving me 0x1 empty matrices for me answers.

Also in the third term of the eqn, I don't understand how you got

    (e^β)*(1−e^α)*W ==> (x-y)*W(1)

from my original eqn.

[Roger Stafford]

The equations

   (1-y)*P(1) = (1-x)*G(1)+(x-y)*W(1)
   (1-y)*P(2) = (1-x)*G(2)+(x-y)*W(2)

can be translated to

   (G(1)-W(1))*x + (W(1)-P(1))*y = G(1)-P(1)
   (G(2)-W(2))*x + (W(2)-P(2))*y = G(2)-P(2)

and expressing this as a problem in matlab gives the matrix equation

   [G(:)-W(:),W(:)-P(:)]*XY = [G(:)-P(:)]

where XY is the unknown column matrix [x;y]. Its solution is

   XY = [G(:)-W(:),W(:)-P(:)]\[G(:)-P(:)] = [1;1]

for which x = 1 and y = 1. This corresponds to the trivial solution β = 0, φ = 0, and α = 0. Somehow I doubt if you expected this as an answer, but if you go back and substitute those values in, you will see that it is actually valid.

As to your question about

(e^β)*(1−e^α)*W ==> (x-y)*W(1)

this is equivalent to

   (e^β-e^(α+β))*W = (e^β-e^φ)*W = (x-y)*W

I think I have already explained why I think you are getting empty solutions from 'solve'. You are presenting it with more equations than unknowns, which 'solve' is unhappy with.