MATLAB: Is it possible to make Acos return values greater than pi

acoscosMATLABsignal processingsin

I know the a, b and c and x values. I want to find Theta value. I use the command 'acos' for this, but 'acos' returns a value between [0, pi]. For example, it needs to return 240 degrees, but it retuns 120 deg. Since theta value is a phase shift, it is problematic. Any advice?

Best Answer

Hi Berke,

I wrote a clarification to your previous question which I think should resolve that issue. It looks like you have determined (or can determine) a, c and phi and wish to solve for b and theta. The first equation in your question is just the imaginary part of

a*exp(i*x) + b*exp(i*x + i*theta) = c*exp(i*x + i*phi)

Factoring out the exp(i*x) leaves

a + b*exp(i*theta) = c*exp(i*phi)

so the sum of two phasors is a third one. [ note that multiplying this equation by its complex counjugate gives the second equation in the question ]. Solving this gives

b = abs(c*exp(i*phi) - a)
theta = angle(c*exp(i*phi) - a)

Just like acos, angle has a convention that deals with 2pi ambiguity, which is -pi < angle <=pi. This means that for the case that you mentioned, angle will produce -120 degrees, not 240. That's how it is, but if you want only positive angles, then (for degrees) it would be mod(angle,360).

Related Solutions

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

MATLAB: How can i plot radiation pattern in cartesian coordinate in matlab

Dear Naveedp, here is the code you can use for plotting:

theta = -90:90;
theta_rad = degtorad(theta);
r = 1;
elevation = 0;
[x,y,z] = sph2cart(theta_rad,elevation,r);
E_theta = 1j * 30 * exp(-1j * 2 * pi) * sin(5 * pi * sin(theta)) ./ (5 * pi * sin(theta));
E_phi = 1j * 30 * exp(-1j * 2 * pi) * cos(theta) .* sin(3 * pi * sin(theta)) ./ (3 * pi * sin(theta));
figure, subplot(2,1,1), plot3(x, y, abs(E_theta)), xlabel('x'), ylabel('y'), zlabel('z'), title('E(theta) amplitude')
        subplot(2,1,2), plot3(x, y, angle(E_theta)), xlabel('x'), ylabel('y'), zlabel('z'), title('E(theta) phase')
figure, subplot(2,1,1), plot3(x, y, abs(E_phi)), xlabel('x'), ylabel('y'), zlabel('z'), title('E(phi) amplitude')
        subplot(2,1,2), plot3(x, y, angle(E_phi)), xlabel('x'), ylabel('y'), zlabel('z'), title('E(phi) phase')

Since, I did't see use of angle phi so I assumed it to be 0 and assumed value of radius to be 1. I hope it will help you

Best Answer

Related Solutions

MATLAB: Solving non linear equation using fsolve

MATLAB: How can i plot radiation pattern in cartesian coordinate in matlab

Related Question