Is it possible to make Acos return values greater than pi?

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?

[Birdman]

Can you share your code?

[David Hill]

Just generate both first and second quandrant solutions and discard the one that doesn't check.

[Berke Ogulcan Parlak]

w = 5;
a=5;
b=10;
phase = degtorad(200);
t=0:0.01:5;
A = a*sin(w*t);
B = b*sin(w*t + phase);
C = A+B;
% plot(t,A); hold on; plot(t,B,'r--'); hold on; plot(t,C,'g-o');
[val_Cpeak,loc_Cpeak] = findpeaks(C);
c = max(val_Cpeak);
faz = real(acos((c^2-a^2-b^2)/(2*a*b)));

[David Hill]

So faz should be roughly equal to phase? What are you trying to do (big picture)?

[Berke Ogulcan Parlak]

What I really want to do is to give the system a sinusoidal input (A.sin (wt)) and find the frequency transfer function of the system by looking at the system's steady state output (A. | G (w) | .sin (wt + phase)). So I need to determine this phase and yes, 'faz' must be directly equal to phase.

[David Hill]

I am still a bit confused. If you are only given C (sinusoidal input) how do you know what (a) and (c) are? If you know what A and B are, then you already know the (phase). Like I said before you could try all the possible solutions of (faz) and discard those that don't result in the equation equality.

Answer 1

[David Goodmanson]

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).

[Berke Ogulcan Parlak]

Hi David,

Thank you for explaining everything very clearly. This worked really well for me!

Answer 2

[Guillaume]

"Is it possible to make Acos return values greater than pi?"

No. The function codomain is mathematically defined as [0, π]. You can define your own reciprocal of the cosinus function with a different codomain but that won't be arccos.eg:

function y = myaccosd(x)
    y = -accosd(x);
end

which would return your 240 degrees (mod 360) for an input of -0.5.

[Berke Ogulcan Parlak]

"For real values of X in the interval [-1, 1], acosd(X) returns values in the interval [0, 180]. "

How can I get 240 degrees output from this function?

[Guillaume]

"How can I get 240 degrees output from this function?"

360 - acosd(x)

[Berke Ogulcan Parlak]

What if the actual phase value is 120 degrees, then it will still get 240 degrees.

[Walter Roberson]

What information do you have that would permit you to distinguish whether 120 or 240 was the "actual" phase" ? And not (say) 600 ?