MATLAB: How do i represent the following equation in matlab

bearingfaultswitch

Best Answer

%This is OK, but there is more smarter way to write this, hope some expert will answer

        if j==k && phi_s>0 && phi_s<del_phi_do
            b=1;
        elseif j==k && phi_s>pi && phi_s<del_phi_di+pi
           b=(cdr+cdi)/(cdr-cdo);
        else
            b=0;
        end

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I don't have time to debug your code but for what it's worth, attached is my heart drawing demo.

MATLAB: Solving non linear equation using fsolve

You are not passing theta_n and other parameters to root2d(). One fix is to make root2d a Nested Function. Also, you must choose a different initial point. phi0 = [10,10] worked for me.

function solve_it()
    beta_a = 10;
    alpha_n = 10 ;
    eta= 18;
    i = 18;
    theta_i = 10;
    theta_n = 10 ;
    
    fun = @root2d;
    phi0 = [10,10];
    phi = fsolve(fun,phi0);
    phi_n = phi(1),
    phi_i = phi(2),
    function F = root2d(phi)
    F = zeros(2,1);
    F(1) = (pi*sin((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2) - (pi*cos((pi*phi(1))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((phi(1)*pi)/180)^2*(cos((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180) + sin((pi*phi(2))/180)*tan((pi*theta_i)/180)));
    F(2) = ((cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180))*((pi*cos((pi*(phi(1) + theta_n))/180)*sin((pi*phi(2))/180))/180 - (pi*cos((pi*phi(2))/180)*tan((pi*theta_i)/180))/180))/(sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2);
    end
end

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