使用边界时如何优化我的鼻窦拟合？

Question

i'm trying to do a sinus-fit on a given part of a curve. There a two restrictions. First the offset of my fitted sinus-curve should be 0, second the amplitude of my fitted sinus-curve should be identically to the minimum value of my original data.

When i'm using my code below, the fitting looks like in the picture i added ( 1 ). In my opinion the period of the sinus-function should be higher. The fitted curve only matches at the minimum with my original data, the fitting-curve isn't wide enough. When i don't use the bounds for c and A my fitting looks good ( 2 ). What am i doing wrong? Is there a way to modify the fitting so that the sinus-curves fits better when using the bounds for A and c?

Fitting without bounds

Fitting with bounds

Edit: Something thing I mentioned is, that the fitting extremely depends on the start value of the amplitude (ff_guess). When I manually change it to X (eg. 10 or 30), than the fitted sinus curve always shows an amplitude near to X (10.3 or 32.5). Is there any setting I haven't considered yet, that prevents the optimizer from varying the amplitude value?

import matplotlib.pyplot as plt
from matplotlib.ticker import FormatStrFormatter
from matplotlib.pyplot import figure, rcParams
import numpy as np
from scipy.optimize import curve_fit


#Time
t = [313.544, 313.545, 313.546, 313.547, 313.548, 313.549, 313.55,  313.551, 313.552, 313.553, 313.554, 313.555, 313.556, 313.557, 313.558, 313.559, 313.56,  313.561, 313.562, 313.563, 313.564, 313.565, 313.566, 313.567,]

#y-Values
s = [0.911188, -0.43135, -1.80997, -3.27816, -4.85784, -6.59428, -8.2214, -9.53617, -10.6892, -11.6003, -12.0844, -12.0524, -11.9749, -11.4891, -10.6131, -9.49873, -8.1154, -6.41442, -5.09357, -3.99165, -2.72991, -1.71446, -0.56306, 0.440741]
   
    
#fourier frequency
ff = np.fft.fftfreq(len(t), (t[1]-t[0]))                               
#fourier amplitude
fa = abs(np.fft.fft(s, len(t)))                                            
#Position of maximum Amplitude 
pos_amax = np.argmax(fa[1:])+1                                        
#Frequency at maximum Amplitude (w/2pi)
ff_max = abs(ff[pos_amax])                                             
ff_guess = ff_max   
T_guess = 1000/ff_max 
#A_guess = np.std(s) *2. **0.5                                          
A_guess = min(s)
#c_guess = np.mean(s)
c_guess = 0
#First Guess for all paramters                           
f_guess = np.array([A_guess, 2*np.pi*ff_guess, 0., c_guess])  
#Sinus_Curve
def sin_func(t, A, w, phi, c):
   return A * np.sin(w*t + phi) + c    
#Defining Bounds for A and c
c_bound = 0.1
A_bound = min(s)
#Bounds Array for curve_fit
param_bounds=([1.01*A_bound, -np.inf, -np.inf, -1*c_bound],[0.99*A_bound, np.inf, np.inf, c_bound])
popt, pcov = curve_fit(sin_func, t, s, p0=f_guess, bounds=param_bounds, maxfev=10000000)
#popt, pcov = curve_fit(sin_func, t, s, p0=f_guess, maxfev=10000000)
#
A, w, phi, c = popt 
f = w/(2.*np.pi)
T = 1000/f

t = np.array(t)  
s = np.array(s)

plt.figure(1)
#Generate Sinus Function
s_fit = A * np.sin(w*t + phi) + c
#Plotting
rcParams['figure.figsize'] =10, 5
fig, ax = plt.subplots()  
plt.plot(t, s, "b", label="Original")     
plt.plot(t, s_fit, 'k--', label="Fitting")  
ytitle='ytitle'
xtitle='xtitle'
ax.set(xlabel=xtitle, ylabel=ytitle)
ax.xaxis.set_major_formatter(FormatStrFormatter('%.2f'))               
ax.yaxis.set_major_formatter(FormatStrFormatter('%.2f'))  
ax.grid()
#Sidetext
ausgabe = ("Sinus-Fit \nAmplitude = {:.2f} m/s^2 \nPeriode = {:.2f} ms \nOffset = {:.2f} m/s^2".format(A, abs(T), c))
plt.text(0.795, 0.7, ausgabe, family="sans-serif", fontsize=10, ha='left', va='top', transform=fig.transFigure)  
box = ax.get_position()
ax.set_position([box.x0, box.y0, box.width * 0.85, box.height]) 
plt.show()

Answer 1

If the amplitude is fixed and offset is supposedly zero, why fitting them in the first place. Moreover, there is no need for fft to estimate parameters, as there is a simple linear method.

Looks like this:

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import cumtrapz


#Time
t = np.array([
    313.544, 313.545, 313.546, 313.547, 313.548, 313.549, 313.55,
    313.551, 313.552, 313.553, 313.554, 313.555, 313.556, 313.557,
    313.558, 313.559, 313.56,  313.561, 313.562, 313.563, 313.564,
    313.565, 313.566, 313.567
])
# ~t -= 313.544

#y-Values
s = np.array([
    0.911188, -0.43135, -1.80997, -3.27816, -4.85784, -6.59428,
    -8.2214, -9.53617, -10.6892, -11.6003, -12.0844, -12.0524,
    -11.9749, -11.4891, -10.6131, -9.49873, -8.1154, -6.41442,
    -5.09357, -3.99165, -2.72991, -1.71446, -0.56306, 0.440741
])

amp = np.max( np.abs( s ) )

def sine(t, w, f ):
    return amp * np.sin( w * t + f )

Sy = cumtrapz( s, x=t, initial = 0 )
SSy = cumtrapz( Sy, x=t, initial = 0 )
VMXT = np.array( [ s, t, np.ones( len( t ) ) ] )
VMX = np.transpose( VMXT )
A = np.dot( VMXT, VMX )
SV = np.dot( VMXT, SSy )
AI = np.linalg.inv( A )
alpha = np.dot( AI , SV )
wstart = np.sqrt( -1 / alpha[0] )


VMXT = np.array( [ np.sin( wstart * t ), np.cos( wstart * t ) ] )
VMX = np.transpose( VMXT )
A = np.dot( VMXT, VMX )
SV = np.dot( VMXT, s )
AI = np.linalg.inv( A )
alpha = np.dot( AI , SV )
phistart = np.arctan2( alpha[1], alpha[0] )


op, cv = curve_fit( sine, t, s, p0=( wstart, phistart ) )
print( op )
yth = sine( t, *op )
fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1 )
ax.plot( t, s )
ax.plot( t, yth )
plt.show()

Works just fine.