Always store metadata rotation as a string

And adjust the period locator accordingly.
Fitting \sin is the same mathematically, it's just conceptually more
straightforward since we're locating zero crossings anyway.

src/prep.py

@@ -80,7 +80,7 @@ def main(argv = None):
     f.check_dest_metadata({ "prep_raw_source": f.src.path,
                             "prep_sinefit_source": sinefit.path,
                             "prep_column": args.column,
-                            "prep_rotation": rotation })
+                            "prep_rotation": repr(rotation) })
 
     # Run the processing function on all data
     f.process_numpy(process, args = (client_sinefit, sinefit.path, args.column,

src/sinefit.py

@@ -98,12 +98,12 @@ def process(data, interval, args, insert_function, final):
             continue
 
         #p.plot(arange(N), this)
-        #p.plot(arange(N), A * cos(f0/fs * 2 * pi * arange(N) + phi) + C, 'g')
+        #p.plot(arange(N), A * sin(f0/fs * 2 * pi * arange(N) + phi) + C, 'g')
 
-        # Period starts when the argument of cosine is 3*pi/2 degrees,
+        # Period starts when the argument of sine is 0 degrees,
         # so we're looking for sample number:
-        #     n = (3 * pi / 2 - phi) / (f0/fs * 2 * pi)
-        zc_n = (3 * pi / 2 - phi) / (f0 / fs * 2 * pi)
+        #     n = (0 - phi) / (f0/fs * 2 * pi)
+        zc_n = (0 - phi) / (f0 / fs * 2 * pi)
         period_n = fs/f0
 
         # Add periods to make N positive
@@ -149,9 +149,9 @@ def sfit4(data, fs):
 
     Output:
       Parameters [A, f0,  phi, C] to fit the equation
-        x[n] = A * cos(f0/fs * 2 * pi * n + phi) + C
+        x[n] = A * sin(f0/fs * 2 * pi * n + phi) + C
       where n is sample number.  Or, as a function of time:
-        x(t) = A * cos(f0 * 2 * pi * t + phi) + C
+        x(t) = A * sin(f0 * 2 * pi * t + phi) + C
 
     by Jim Paris
     (Verified to match sfit4.m)
@@ -188,12 +188,11 @@ def sfit4(data, fs):
     # if something fails with the least squares fit, etc.
     try:
         # first guess for A0, B0 using 3-parameter fit (step c)
+        s = zeros(3)
         w = 2*pi*f0
-        D = c_[cos(w*t), sin(w*t), ones(N)]
-        s = linalg.lstsq(D, data)[0]
 
-        # Now iterate 6 times (step i)
-        for idx in range(6):
+        # Now iterate 7 times (step b, plus 6 iterations of step i)
+        for idx in range(7):
             D = c_[cos(w*t), sin(w*t), ones(N),
                   -s[0] * t * sin(w*t) + s[1] * t * cos(w*t) ] # eqn B.16
             s = linalg.lstsq(D, data)[0] # eqn B.18
@@ -202,7 +201,7 @@ def sfit4(data, fs):
         ## Extract results
         A = sqrt(s[0]*s[0] + s[1]*s[1]) # eqn B.21
         f0 = w / (2*pi)
-        phi = -arctan2(s[1], s[0]) # eqn B.22
+        phi = arctan2(s[0], s[1]) # eqn B.22 (flipped for sin instead of cos)
         C = s[2]
         return (A, f0, phi, C)
     except Exception as e: