How do I scale an FFT-based cross-correlation such that its peak is equal to Pearson's rho

Question

Description of the problem

FFT can be used to compute cross-correlation between two signals or images. To determine the delay or lag between two signals A and B, it suffices to locate the peak of: IFFT(FFT(A)*conjugate(FFT(B)))

However, the amplitude of the peak is related to the amplitude of the frequency spectra of the individual signals. Thus to determine the Pearson correlation (rho), the amplitude of this peak must be scaled by the total energy in the two signals.

One way to do this is to normalize by the geometric mean of the individual autocorrelations. This gives a reasonable approximation of rho, especially when the delay between samples is small, but not the exact value.

I thought the reason for this error was that the Pearson correlation is only defined for the overlapping portions of the signal, whereas the normalization factor (the geometric mean of the two autocorrelation peaks) includes contributions from the non-overlapping portions. I considered two approaches for fixing this and producing an exact value for rho via FFT. In the first (called rho_exact_1 below), I trimmed the samples down to their overlapping portions and computed the normalization factor from those. In the second (called rho_exact_2 below), I computed the fraction of the measurements contained in the overlapping portion of the signals and multiplied the full-autocorrelation-normalization factor by that fraction.

Neither works! The figure below shows plots of the three approaches for calculating Pearson's rho using DFT-based cross-correlation. Only the region of the cross-correlation peak is shown. Each estimate is close to the correct value of 1.0, but not equal to it.

The code I used to perform the calculations is below. I used a simple sine wave as an example signal. I noticed that if I use a square-wave (w/ duty cycle not necessarily 50%) the approaches' errors change.

Can somebody explain what's going on?

A working example

import numpy as np
from matplotlib import pyplot as plt

# make a time vector w/ 256 points
# and a source signal
N_cycles = 10.0
N_points = 256.0

t = np.arange(0,N_cycles*np.pi,np.pi*N_cycles/N_points)
signal = np.sin(t)

use_rect = False
if use_rect:
    threshold = -0.75
    signal[np.where(signal>=threshold)]=1.0
    signal[np.where(signal<threshold)]=-1.0

# normalize the signal (not technically
# necessary for this example, but required
# for measuring correlation of physically
# different signals)
signal = signal/signal.std()

# generate two samples of the signal
# with a temporal offset:
N = 128
offset = 5
sample_1 = signal[:N]
sample_2 = signal[offset:N+offset]

# determine the offset through cross-
# correlation
xc_num = np.abs(np.fft.ifft(np.fft.fft(sample_1)*np.fft.fft(sample_2).conjugate()))
offset_estimate = np.argmax(xc_num)
if offset_estimate>N//2:
    offset_estimate = offset_estimate - N


# for an approximate estimate of Pearson's
# correlation, we normalize by the RMS
# of individual autocorrelations:
autocorrelation_1 = np.abs(np.fft.ifft(np.fft.fft(sample_1)*np.fft.fft(sample_1).conjugate()))
autocorrelation_2 = np.abs(np.fft.ifft(np.fft.fft(sample_2)*np.fft.fft(sample_2).conjugate()))
xc_denom_approx = np.sqrt(np.max(autocorrelation_1))*np.sqrt(np.max(autocorrelation_2))
rho_approx = xc_num/xc_denom_approx
print 'rho_approx',np.max(rho_approx)

# this is an approximation because we've
# included autocorrelation of the whole samples
# instead of just the overlapping portion;
# using cropped versions of the samples should
# yield the correct correlation:
sample_1_cropped = sample_1[offset:]
sample_2_cropped = sample_2[:-offset]

# these should be identical vectors:
assert np.all(sample_1_cropped==sample_2_cropped)

# compute autocorrelations of cropped samples
# and corresponding value for rho
autocorrelation_1_cropped = np.abs(np.fft.ifft(np.fft.fft(sample_1_cropped)*np.fft.fft(sample_1_cropped).conjugate()))
autocorrelation_2_cropped = np.abs(np.fft.ifft(np.fft.fft(sample_2_cropped)*np.fft.fft(sample_2_cropped).conjugate()))
xc_denom_exact_1 = np.sqrt(np.max(autocorrelation_1_cropped))*np.sqrt(np.max(autocorrelation_2_cropped))
rho_exact_1 = xc_num/xc_denom_exact_1
print 'rho_exact_1',np.max(rho_exact_1)

# alternatively we could try to use the
# whole sample autocorrelations and just
# scale by the number of pixels used to
# compute the numerator:
scaling_factor = float(len(sample_1_cropped))/float(len(sample_1))
rho_exact_2 = xc_num/(xc_denom_approx*scaling_factor)
print 'rho_exact_2',np.max(rho_exact_2)

# finally a sanity check: is rho actually 1.0
# for the two signals:
rho_corrcoef = np.corrcoef(sample_1_cropped,sample_2_cropped)[0,1]
print 'rho_corrcoef',rho_corrcoef

x = np.arange(len(rho_approx))
plt.plot(x,rho_approx,label='FFT rho_approx')
plt.plot(x,rho_exact_1,label='FFT rho_exact_1')
plt.plot(x,rho_exact_2,label='FFT rho_exact_2')
plt.plot(x,np.ones(len(x))*rho_corrcoef,'k--',label='Pearson rho')
plt.legend()
plt.ylim((.75,1.25))
plt.xlim((0,20))
plt.show()

Answer 1

The normalised cross correlation between two N-periodic discrete signals F and G is defined as:

Since the numerator is a dot product between two vectors (F and G_x) and the denominator is the product of the norm of these two vectors, the scalar r_x must indeed lie between -1 and +1 and it is the cosinus of the angle between the vectors (See there). If the vector F and G_x are aligned, then r_x=1. If r_x=1, then the vector F and G_x are aligned due to the triangular inequality. To ensure these properties, the vectors at the numerator must match those at the denominator.

All numerators can be computed at once by using the Discrete Fourier Transform. Indeed, that transform turns the convolution into pointwise products in the Fourier space. Here is why the different estimated normalized cross correlations are not 1 in the tests you performed.

1. For the first test "approx", sample_1 and sample_2 are both extracted from a periodic signal. Both are of the same length, but the length is not a multiple of the period as it is 2.5 periods (5pi) (figure below). As a result, since the dft performs the correlation as if they where periodic signals, it is found that sample_1 and sample_2 are not perfectly correlated and r_x<1.

   

2. For the second test rho_exact_1, the convolution is performed on signals of length N=128, but the norms at the denominator are computed on truncated vectors of size N-offset=128-5. As a result, the properties of r_x are lost. In addition, it must be noticed that the proposed convolution and norms are not normalized: the computed norms and convolution product are globally proportionnal to the number of points of the considered vectors. As a result, the norms of the truncated vectors are slightly lower compared to the previous case and r_x increases: values larger that 1 are likely encountered as the offset increases.

3. For the third test rho_exact_2, a scaling factor is introduced to try to correct the first test: the properties of r_x are also lost and values larger than one can be encountered as the scaling factor is larger than one.

Nevertheless, the function corrcoef() of numpy actually computes a r_x equal to 1 for the truncated signals. Indeed, these signals are perfectly identical! The same result can be obtained using DFTs:

xc_num_cropped = np.abs(np.fft.ifft(np.fft.fft(sample_1_cropped)*np.fft.fft(sample_2_cropped).conjugate()))
autocorrelation_1_cropped = np.abs(np.fft.ifft(np.fft.fft(sample_1_cropped)*np.fft.fft(sample_1_cropped).conjugate()))
autocorrelation_2_cropped = np.abs(np.fft.ifft(np.fft.fft(sample_2_cropped)*np.fft.fft(sample_2_cropped).conjugate()))
xc_denom_exact_11 = np.sqrt(np.max(autocorrelation_1_cropped))*np.sqrt(np.max(autocorrelation_2_cropped))
rho_exact_11 = xc_num_cropped/xc_denom_exact_11
print 'rho_exact_11',np.max(rho_exact_11)  

To provide the user with a significant value for r_x, you can stick to the value provided by the first test, which can be lower than one for identical periodic signals if the length of the frame is not a multiple of the period. To correct this drawback, the estimated offset can also be retreived and used to build two cropped signals of the same length. The whole correlation procedure must be re-run to get a new value for r_x, which will not be plaged by the fact that the length of the cropped frame is not a multiple of the period.

Lastly, if the DFT is a very efficient way to compute the convolution at the numerator for all values of x at once, the denominator can be efficiently computed as 2-norms of vector, using numpy.linalg.norm. Since the argmax(r_x) for the cropped signals will likely be zero if the first correlation was successful, it could be sufficient to compute r_0 using a dot product `sample_1_cropped.dot(sample_2_cropped).