Sine Wave Curve Fit Question

[Mikael Olofsson]

Helmut Jarausch wrote:
> Your model is  A*sin(omega*t+alpha)  where  A and alpha are sought.
> Let T=(t_1,...,t_N)' and  Y=(y_1,..,y_N)'  your measurements (t_i,y_i)
> ( ' denotes transposition )
> 
> First, A*sin(omega*t+alpha) =
>   A*cos(alpha)*sin(omega*t) + A*sin(alpha)*cos(omega*t) =
>   B*sin(omega*t) + D*cos(omega*t)
> 
> by setting  B=A*cos(alpha)  and  D=A*sin(alpha)
> 
> Once, you have  B and D,  tan(alpha)= D/B   A=sqrt(B^2+D^2)

This is all very true, but the equation tan(alpha)=D/B may fool you. 
This may lead you to believe that alpha=arctan(D/B) is a solution, which 
is not always the case. The point (B,D) may be in any of the four 
quadrants of the plane. Assuming B!=0, the solutions to this equation 
fall into the two classes

     alpha = arctan(D/B) + 2*k*pi

and

     alpha = arctan(D/B) + (2*k+1)*pi,

where k is an integer. The sign of B tells you which class gives you the 
solution. If B is positive, the solutions are those in the first class. 
If B is negative, the solutions are instead those in the second class. 
Whithin the correct class, you may of course choose any alternative.

Then we have the case B=0. Then the sign of D determines alpha. If D is 
positive, we have alpha=pi/2, and if D is negative, we have alpha=-pi/2.

Last if both B and D are zero, any alpha will do.

/MiO