Why is the envelope curve erroneous at the beginning?

Question

I implemented a function that gives the envelope curve of discrete values. I think there might be an error as when I tested it over a date I will provide you with at the bottom of the post, I get dissimilarities between the real data points ant the envelope curve as sown in this figure example of the code applied to a given data

from scipy.interpolate import interp1d
import numpy as np
import matplotlib.pyplot as plt

def enveloppe(s):
    u_x = [0,]        
    u_y = [s[0],]
    q_u = np.zeros(s.shape)
    for k in xrange(1,len(s)-1):
        if (np.sign(s[k]-s[k-1])==1) and (np.sign(s[k]-s[k+1])==1):
            u_x.append(k)
            u_y.append(s[k])
    u_x.append(len(s)-1)
    u_y.append(s[-1]) 
    u_p = interp1d(u_x,u_y, kind = 'cubic',bounds_error = False, fill_value=0.0)
    #Evaluate each model over the domain of (s)
    for k in xrange(0,len(s)):
        q_u[k] = u_p(k)
    return q_u   

fig, ax = plt.subplots()
ax.plot(S, '-o', label = 'magnitude')
ax.plot(envelope(S), '-o', label = 'enveloppe magnitude')
ax.legend()

Data S : array([  9.12348621e-11,   6.69568658e-10,   6.55973768e-09,
         1.26822485e-06,   4.50553316e-09,   5.06526113e-07,
         2.96728433e-09,   2.36088205e-07,   1.90802318e-09,
         1.15867354e-07,   1.18504790e-09,   5.72888034e-08,
         6.98672478e-10,   2.75361324e-08,   3.82391643e-10,
         1.25393143e-08,   1.96697343e-10,   5.96979943e-09,
         1.27009013e-10,   4.46365555e-09,   1.31769958e-10,
         4.42024233e-09,   1.42514400e-10,   4.17757107e-09,
         1.41640360e-10,   3.65170558e-09,   1.29784598e-10,
         2.99790514e-09,   1.11732461e-10])

Why do you interpolate? Cubic spline interpolation is causing the effect that you observe. — Dima Chubarov, Aug 08 '17 at 13:29
would you please suggest how to do it without interpolation ? — Tassou, Aug 08 '17 at 13:37
What about linear interpolation, which does not overshoot? Or if this is in a signal processing background check out [analytic signal](https://en.wikipedia.org/wiki/Analytic_signal#Envelope_and_instantaneous_phase) and [hilbert transform](https://en.wikipedia.org/wiki/Hilbert_transform). — MB-F, Aug 08 '17 at 13:44
could you provide me with what I can change in my code so there is no overshooting in the interpolation? — Tassou, Aug 08 '17 at 14:04

score 1 · Accepted Answer · answered Aug 08 '17 at 14:44

I would make two modifications to your enveloppe function to get a more monotone output

The idea is to avoid implicit addition of the left and right ends to the list of peaks that is used to construct the envelope

def enveloppe(s):
    u_x = [] # do not add 0
    u_y = []
    q_u = np.zeros(s.shape)
    for k in range(1,len(s)-1):
        if (np.sign(s[k]-s[k-1])==1) and (np.sign(s[k]-s[k+1])==1):
            u_x.append(k)
            u_y.append(s[k])
    print(u_x)
    u_p = interp1d(u_x,u_y, kind = 'cubic',
              bounds_error = False, 
              fill_value="extrapolate") # use fill_value="extrapolate"
    for k in range(0,len(s)):
        q_u[k] = u_p(k)
    return q_u

Why is the envelope curve erroneous at the beginning?

1 Answers1