Mikael Olofsson wrote: > 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.
You're right: standard Python's math library missing the function arctan2. But you can get it from numpy or numarray. In case of doubt just compute the residuum with different possibilities . > 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 -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
