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Frequency estimation methods in Python

A few simple frequency estimation methods in Python

See also https://github.com/endolith/waveform-analyzer/blob/master/frequency_estimator.py, which is mostly the same thing, maybe more up-to-date. I need to keep them both in sync with each other or delete one.

These are the methods that everyone recommends when someone asks about frequency estimation or pitch detection. Such as here:

So these are my attempts at implementation. Initially I was trying to measure the frequency of long sine waves with high accuracy (to indirectly measure clock frequency), then added methods for other types of signals later.

None of them work well in all situations, these are "offline", not real-time, and I am sure there are much better methods "in the literature", but here is some sample code for the simple methods at least.

Count zero-crossings, divide average period by time to get frequency

  • Works well for long low-noise sines, square, triangle, etc.
  • Supposedly this is how cheap guitar tuners work
  • Using interpolation to find a "truer" zero-crossing gives better accuracy
  • Pro: Fast
  • Pro: Accurate (increasing with signal length)
  • Con: Doesn't work if there are multiple zero crossings per cycle, low-frequency baseline shift, noise, etc.

Do FFT and find the peak

  • Using parabolic interpolation to find a truer peak gives better accuracy
  • Accuracy also increases with signal/FFT length
  • Con: Doesn't find the right value if harmonics are stronger than fundamental, which is common. Better method would try to be smarter about identifying the fundamental, like template matching using the "two-way mismatch" (TWM) algorithm.
  • Pro: Accurate, usually even more so than zero crossing counter (1000.000004 Hz for 1000 Hz, for instance). Due to parabolic interpolation being a very good fit for windowed log FFT peaks?

Do autocorrelation and find the peak

  • Pro: Best method for finding the true fundamental of any repetitive wave, even with weak or missing fundamental (finds GCD of all harmonics present)
  • Con: Inaccurate result if waveform isn't perfectly repeating, like inharmonic musical instruments (piano, guitar, ...), however:
  • Pro: This inaccurate result more closely matches the pitch that humans perceive :)
  • Con: Not as accurate as other methods for precise measurement of sine waves
  • Con: This implementation has trouble with finding the true peak

Calculate harmonic product spectrum and find the peak

  • Pro: Good at finding the true fundamental even if weak or missing
from __future__ import division
from numpy.fft import rfft
from numpy import argmax, mean, diff, log, nonzero
from scipy.signal import blackmanharris, correlate
from time import time
import sys
try:
import soundfile as sf
except ImportError:
from scikits.audiolab import flacread
from parabolic import parabolic
def freq_from_crossings(sig, fs):
"""
Estimate frequency by counting zero crossings
"""
# Find all indices right before a rising-edge zero crossing
indices = nonzero((sig[1:] >= 0) & (sig[:-1] < 0))[0]
# Naive (Measures 1000.185 Hz for 1000 Hz, for instance)
# crossings = indices
# More accurate, using linear interpolation to find intersample
# zero-crossings (Measures 1000.000129 Hz for 1000 Hz, for instance)
crossings = [i - sig[i] / (sig[i+1] - sig[i]) for i in indices]
# Some other interpolation based on neighboring points might be better.
# Spline, cubic, whatever
return fs / mean(diff(crossings))
def freq_from_fft(sig, fs):
"""
Estimate frequency from peak of FFT
"""
# Compute Fourier transform of windowed signal
windowed = sig * blackmanharris(len(sig))
f = rfft(windowed)
# Find the peak and interpolate to get a more accurate peak
i = argmax(abs(f)) # Just use this for less-accurate, naive version
true_i = parabolic(log(abs(f)), i)[0]
# Convert to equivalent frequency
return fs * true_i / len(windowed)
def freq_from_autocorr(sig, fs):
"""
Estimate frequency using autocorrelation
"""
# Calculate autocorrelation and throw away the negative lags
corr = correlate(sig, sig, mode='full')
corr = corr[len(corr)//2:]
# Find the first low point
d = diff(corr)
start = nonzero(d > 0)[0][0]
# Find the next peak after the low point (other than 0 lag). This bit is
# not reliable for long signals, due to the desired peak occurring between
# samples, and other peaks appearing higher.
# Should use a weighting function to de-emphasize the peaks at longer lags.
peak = argmax(corr[start:]) + start
px, py = parabolic(corr, peak)
return fs / px
def freq_from_HPS(sig, fs):
"""
Estimate frequency using harmonic product spectrum (HPS)
"""
windowed = sig * blackmanharris(len(sig))
from pylab import subplot, plot, log, copy, show
# harmonic product spectrum:
c = abs(rfft(windowed))
maxharms = 8
subplot(maxharms, 1, 1)
plot(log(c))
for x in range(2, maxharms):
a = copy(c[::x]) # Should average or maximum instead of decimating
# max(c[::x],c[1::x],c[2::x],...)
c = c[:len(a)]
i = argmax(abs(c))
true_i = parabolic(abs(c), i)[0]
print('Pass %d: %f Hz' % (x, fs * true_i / len(windowed)))
c *= a
subplot(maxharms, 1, x)
plot(log(c))
show()
filename = sys.argv[1]
print('Reading file "%s"\n' % filename)
try:
signal, fs = sf.read(filename)
except NameError:
signal, fs, enc = flacread(filename)
print('Calculating frequency from FFT:', end=' ')
start_time = time()
print('%f Hz' % freq_from_fft(signal, fs))
print('Time elapsed: %.3f s\n' % (time() - start_time))
print('Calculating frequency from zero crossings:', end=' ')
start_time = time()
print('%f Hz' % freq_from_crossings(signal, fs))
print('Time elapsed: %.3f s\n' % (time() - start_time))
print('Calculating frequency from autocorrelation:', end=' ')
start_time = time()
print('%f Hz' % freq_from_autocorr(signal, fs))
print('Time elapsed: %.3f s\n' % (time() - start_time))
print('Calculating frequency from harmonic product spectrum:')
start_time = time()
freq_from_HPS(signal, fs)
print('Time elapsed: %.3f s\n' % (time() - start_time))

Motivation:

Note that the Gaussian window transform magnitude is precisely a parabola on a dB scale. As a result, quadratic spectral peak interpolation is exact under the Gaussian window. Of course, we must somehow remove the infinitely long tails of the Gaussian window in practice, but this does not cause much deviation from a parabola

Apparently this produces error if the peak is not exactly at the edge or center of a bin. Interpolating by zero-padding before the FFT does not produce this kind of error, but is more computationally expensive. So a good trade-off is to do some zero-padding interpolation and then follow with parabolic interpolation:

While we could choose our zero-padding factor large enough to yield any desired degree of accuracy in peak frequency measurements, it is more efficient in practice to combine zero-padding with parabolic interpolation (or some other simple, low-order interpolator). In such hybrid schemes, the zero-padding is simply chosen large enough so that the bias due to parabolic interpolation is negligible.

Here's a Matlab function that does the same thing.

Actually this function could be done with the polyfit() function, instead, but it requires more steps:

f = [2, 3, 1, 6, 4, 2, 3, 1]

In [8]: parabolic(f, argmax(f))
Out[8]: (3.2142857142857144, 6.1607142857142856)

In [9]: a, b, c = polyfit([2,3,4],[1,6,4],2)
In [10]: x = -0.5*b/a

In [11]: x
Out[11]: 3.2142857142857695

In [12]: a*x**2 + b*x + c
Out[12]: 6.1607142857143025

parabolic() could be updated to use this, and then it will work for non-uniform sampled input as well.

Alternative estimators with even lower error: How to Interpolate the Peak Location of a DFT or FFT if the Frequency of Interest is Between Bins

from __future__ import division
from numpy import polyfit, arange
def parabolic(f, x):
"""Quadratic interpolation for estimating the true position of an
inter-sample maximum when nearby samples are known.
f is a vector and x is an index for that vector.
Returns (vx, vy), the coordinates of the vertex of a parabola that goes
through point x and its two neighbors.
Example:
Defining a vector f with a local maximum at index 3 (= 6), find local
maximum if points 2, 3, and 4 actually defined a parabola.
In [3]: f = [2, 3, 1, 6, 4, 2, 3, 1]
In [4]: parabolic(f, argmax(f))
Out[4]: (3.2142857142857144, 6.1607142857142856)
"""
xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x
yv = f[x] - 1/4. * (f[x-1] - f[x+1]) * (xv - x)
return (xv, yv)
def parabolic_polyfit(f, x, n):
"""Use the built-in polyfit() function to find the peak of a parabola
f is a vector and x is an index for that vector.
n is the number of samples of the curve used to fit the parabola.
"""
a, b, c = polyfit(arange(x-n//2, x+n//2+1), f[x-n//2:x+n//2+1], 2)
xv = -0.5 * b/a
yv = a * xv**2 + b * xv + c
return (xv, yv)
if __name__ == "__main__":
from numpy import argmax
import matplotlib.pyplot as plt
y = [2, 1, 4, 8, 11, 10, 7, 3, 1, 1]
xm, ym = argmax(y), y[argmax(y)]
xp, yp = parabolic(y, argmax(y))
plot = plt.plot(y)
plt.plot(xm, ym, 'o', color='silver')
plt.plot(xp, yp, 'o', color='blue')
plt.title('silver = max, blue = estimated max')
@ted107mk
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ted107mk commented Jun 6, 2017

@CMCDragonkai I would guess the frequency sampling, or the sampling rate: the number of audio samples per second.

@mindman21
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Thanks

@mindman21
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What is a pababolic

@appetrosyan
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Code is buggy as hell. I understand the concept, and I don't understand why does Find Always go out of bounds

@endolith
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endolith commented Mar 22, 2018

@mindman21 https://gist.github.com/endolith/255291#file-parabolic-py-L6 is a function for parabolic interpolation

@appetrosyan What are you doing that causes something to go out of bounds?

@mindman21
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Can you separate base of piano? Sound of overlap

@endolith
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endolith commented Apr 3, 2018

@mindman21 These can only recognize one frequency at a time

@Master64k
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Master64k commented Aug 12, 2018

    def __barry_quinn_bin_estimator(self, sig, idx):

        """
        My trial to implement the Berry Quinn's second method
        based on http://www.dspguru.com/dsp/howtos/how-to-interpolate-fft-peak

        :param sig: the chunk of data to analyze
        :param idx: the index of the peak in the chunk
        :return: the interpolated bin location
        """

        out = sig

        k = idx

        d = pow(np.real(out[k]), 2.0) + pow(np.imag(out[k]), 2.0)
        ap = (np.real(out[k + 1]) * np.real(out[k]) + np.imag(out[k + 1]) * np.imag(out[k])) / d
        dp = -ap / (1.0 - ap)
        am = (np.real(out[k - 1]) * np.real(out[k]) + np.imag(out[k - 1]) * np.imag(out[k])) / d

        dm = am / (1.0 - am)
        d = (dp + dm) / 2 + self.__tau(dp * dp) - self.__tau(dm * dm)

        bin_location = k + d

        return bin_location

@KangarooD
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Hi! Do you know why I would get an indexerror?

xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x

IndexError: index 1921 is out of bounds for axis 0 with size 1921

@endolith
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@KangarooD Basically it means it couldn't identify the frequency. Which function are you calling?

@KangarooD
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KangarooD commented Jun 26, 2019 via email

@endolith
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@KangarooD Can you post the audio file somewhere?

@KangarooD
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KangarooD commented Jun 26, 2019 via email

@endolith
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@KangarooD That recording doesn't seem to have any repetition at all. What result are you expecting?

@monk1337
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monk1337 commented May 1, 2020

@endolith, How can I calculate the frequency estimation in time-domain instead of frequency domain?

@endolith
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@monk1337 freq_from_crossings is time-domain, and freq_from_autocorr sort of is, too.

@monk1337
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monk1337 commented Sep 23, 2020

@endolith How can I calculate center-clipped autocorrelation method to calculate the pitch, proposed in this paper https://ieeexplore.ieee.org/document/1161986

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