Search code examples
pythonmatplotlibsignal-processingcross-correlationspectral-density

How to use the cross-spectral density to calculate the phase shift of two related signals


I've two signals, from which I expect that one is responding on the other, but with a certain phase shift.

Now I would like to calculate the coherence or the normalized cross spectral density to estimate if there is any causality between the input and output to find out on which frequencies this coherence appear.

See for example this image (from here) which seems to have high coherence at the frequency 10: enter image description here

Now I know that I can calculate the phase shift of two signals using the cross-correlation, but how can I use the coherence (of frequency 10) to calculate the phase shift?

Code for image:

"""
Compute the coherence of two signals
"""
import numpy as np
import matplotlib.pyplot as plt

# make a little extra space between the subplots
plt.subplots_adjust(wspace=0.5)

nfft = 256
dt = 0.01
t = np.arange(0, 30, dt)
nse1 = np.random.randn(len(t))                 # white noise 1
nse2 = np.random.randn(len(t))                 # white noise 2
r = np.exp(-t/0.05)

cnse1 = np.convolve(nse1, r, mode='same')*dt   # colored noise 1
cnse2 = np.convolve(nse2, r, mode='same')*dt   # colored noise 2

# two signals with a coherent part and a random part
s1 = 0.01*np.sin(2*np.pi*10*t) + cnse1
s2 = 0.01*np.sin(2*np.pi*10*t) + cnse2

plt.subplot(211)
plt.plot(t, s1, 'b-', t, s2, 'g-')
plt.xlim(0,5)
plt.xlabel('time')
plt.ylabel('s1 and s2')
plt.grid(True)

plt.subplot(212)
cxy, f = plt.cohere(s1, s2, nfft, 1./dt)
plt.ylabel('coherence')
plt.show()

.
.
EDIT:

For what it's worth, I've add an answer, maybe it's right, maybe it's wrong. I'm not sure..


Solution

  • Let me try to answer my own question and maybe one day it might be useful to others or function as a starting point for a (new) discussion:

    Firstly calculate the power spectral densities of both the signals,

    subplot(121)
    psd(s1, nfft, 1/dt)
    plt.title('signal1')
    
    subplot(122)
    psd(s2, nfft, 1/dt)
    plt.title('signal2')
    
    plt.tight_layout()
    show()
    

    resulting in:enter image description here

    Secondly calculate the cross-spectral density, which is Fourier transform of the cross-correlation function:

    csdxy, fcsd = plt.csd(s1, s2, nfft, 1./dt)
    plt.ylabel('CSD (db)')
    plt.title('cross spectral density between signal 1 and 2')
    plt.tight_layout()
    show()
    

    Which gives:

    enter image description here

    Than using the cross-spectral density we can calculate the phase and we can calculate the coherence (which will destroy the phase). Now we can combine the coherence and the peaks that rise above the 95% confidence level

    # coherence
    cxy, fcoh = cohere(s1, s2, nfft, 1./dt)
    
    # calculate 95% confidence level
    edof = (len(s1)/(nfft/2)) * cxy.mean() # equivalent degrees of freedom: (length(timeseries)/windowhalfwidth)*mean_coherence
    gamma95 = 1.-(0.05)**(1./(edof-1.))
    conf95 = np.where(cxy>gamma95)
    print 'gamma95',gamma95, 'edof',edof
    
    # Plot twin plot
    fig, ax1 = plt.subplots()
    # plot on ax1 the coherence
    ax1.plot(fcoh, cxy, 'b-')
    ax1.set_xlabel('Frequency (hr-1)')
    ax1.set_ylim([0,1])
    # Make the y-axis label and tick labels match the line color.
    ax1.set_ylabel('Coherence', color='b')
    for tl in ax1.get_yticklabels():
        tl.set_color('b')
    
    # plot on ax2 the phase
    ax2 = ax1.twinx()
    ax2.plot(fcoh[conf95], phase[conf95], 'r.')
    ax2.set_ylabel('Phase (degrees)', color='r')
    ax2.set_ylim([-200,200])
    ax2.set_yticklabels([-180,-135,-90,-45,0,45,90,135,180])
    
    for tl in ax2.get_yticklabels():
        tl.set_color('r')
    
    ax1.grid(True)
    #ax2.grid(True)
    fig.suptitle('Coherence and phase (>95%) between signal 1 and 2', fontsize='12')
    plt.show()
    

    result in:

    enter image description here

    To sum up: the phase of the most coherent peak is ~1 degrees (s1 leads s2) at a 10 min period (assuming dt is a minute measurement) -> (10**-1)/dt

    But a specialist signal processing might correct me, because I'm like 60% sure if I've done it right