Packages to install from PyPI: numpy, scipy, matplotlib, soundfile.

import numpy as np
from scipy.signal import ShortTimeFFT, find_peaks
from scipy.signal.windows import hann
from scipy.optimize import fsolve
from scipy.stats import linregress
import matplotlib.pyplot as plt
from matplotlib.colors import LogNorm
import soundfile

There is a nice song by 影虎。called Time to beat the odds:

At 1:18, there is an interesting buildup kickroll. This effect is created by mixing a steady 32nd note kickroll with a gradually-accelerating buildup kickroll. The gradually-accelerating part is not quantized to dyadic note values, so it is pretty hard to analyze by ear.

One can obtain a clearer sample of this kickroll by consulting the file Dr_KICKroll00.ogg from the BMS. The waveform of the sample looks like this:

samples, fs = soundfile.read("Kagetora_Time-to-beat-the-odds_bofet/Dr_KICKroll00.ogg")
fs /= 1000 # Use ms and kHz
if samples.ndim == 2:
	samples = samples.mean(axis=1)
N = len(samples)

plt.figure(figsize=(10, 6))
plt.plot(np.arange(N) / fs, samples, linewidth=0.5)
plt.xlabel('Time (ms)')
plt.ylabel('Amplitude')
plt.xlim(0, N/fs)
plt.ylim(-1, 1)

There is clearly a repeating pattern that is occurring at an gradually-increasing frequency. Each occurrence of the pattern is a kick. The task is to find the regularity of the kicks.

We can see that, at the start of each kick, the waveform is oscillating more rapidly than in later times of the kick. Therefore, a natural idea is to capture this high-frequency oscillation feature using the short-time Fourier transform (STFT), and we would expect to see some peaks at the high-frequency region of the spectrogram of the waveform at each kick occurrence. SciPy provides a convenient function for STFT, and then we can plot the spectrogram of the waveform:

hop = 32
SFT = ShortTimeFFT(hann(hop), hop, fs)
spectrogram = SFT.spectrogram(samples)

plt.figure(figsize=(10, 6))
im = plt.imshow(
	spectrogram,
	aspect='auto',
	origin='lower',
	extent=SFT.extent(N),
	norm=LogNorm(vmin=0.001, vmax=np.max(spectrogram)),
)
plt.xlabel('Time (ms)')
plt.ylabel('Frequency (kHz)')
plt.colorbar(im, label='Magnitude')

We can see that there are clearly some peaks at the high-frequency region of the spectrogram. Pick out the peaks:

plt.figure(figsize=(10, 6))
data = spectrogram[14:, :].mean(axis=0)
peaks, peak_properties = find_peaks(data, height=0.004, distance=7)
peaks = peaks * hop / fs
plt.xlabel('Time (ms)')
plt.ylabel('Average magnitude for high frequencies')
plt.plot(SFT.t(N), data)
plt.plot(peaks, peak_properties['peak_heights'], "o")

To check that those peaks make sense, we can mark the positions of the peaks on the waveform to see if they are indeed approximately at the start of each kick:

plt.figure(figsize=(10, 6))
plt.plot(np.arange(N) / fs, samples)
for peak in peaks:
	plt.axvline(peak, color='red')
plt.xlabel('Time (ms)')
plt.ylabel('Amplitude')
plt.xlim(0, 200)
plt.ylim(-1, 1)

We can see that although they are not perfectly at the start of each kick, the peaks are indeed approximately at the start of each kick. Now, plot the time intervals between each pair of consecutive peaks. If the plot is in log-scale, one can see that they approximately form a straight line, which means that the time intervals decay geometrically. We can use a simple linear regression to fit the data.

fig, ax = plt.subplots(figsize=(10, 6))
ax.set_yscale('log')
intervals = np.diff(peaks)
indices = np.arange(len(intervals))
lin_res = linregress(indices, np.log(intervals))
ax.plot(indices, intervals, 'o')
ax.plot(indices, np.exp(lin_res.intercept + lin_res.slope * indices))
ax.set_xlabel('Kick')
ax.set_ylabel('Interval (ms)')

With the linear regression, we can conclude that the decay ratio is $0.9762$.

If we had guessed that the time intervals decay geometrically and that the first kick has the length of a 32nd note, we could calculate the decay ratio $a$ by solving the equation $$\fr18\p{1-a^{59}}=4\p{1-a},$$ where $1/8$ is the note value of a 32nd note, and $4$ is the total note value of this kickroll (a measure with $4$ quarter notes), and $59$ is the total number of kicks (obtained by counting the patterns in the waveform). Solving this equation numerically gives us $a=0.9764$, which is pretty close to the value we got before.

First, let’s define what is a note.

A note is an element of the countable set $N$. Here exists a bijection from $\mathbb Z$ to $N$ denoted as $k\mapsto \nu_k$. Therefore, all notes form a sequence $$\dots,\nu_{-2},\nu_{-1},\nu_0,\nu_1,\nu_2,\dots$$

Our goal is to define a frequency assignment, which is a mapping $f:N\rightarrow\mathbb R^+$, whose meaning is the frequency (in Hz) of the sound of a note.

It is a natural idea to define a sequence $$f_k\coloneqq f\!\left(\nu_k\right).$$

It makes sense that the sequence is in strictly increasing order.

The octaves

Now, let’s think about a musical interval. At this stage, a musical interval can be defined as an unordered pair of notes. After long time of experimenting, people find that they tend to think a musical interval extremely harmonic if it consists of such two notes that the frequency of one of them doubles that of the other.

In other words, the musical interval of $\nu_a$ and $\nu_b$ is extremely harmonic if $f_b=2f_a$. In fact, it is so harmonic that if the two notes are played simultaneously, a person tend to think there is only one note being played:

          
            motif = c4 c g g a a g2

            piano:

            V1: o4 motif motif

            V2: o5 motif

          
        

The audio above is an illustration for octave intervals. As can be heard, the first part of the audio is played in octave intervals while the second part is played in single notes.

Considering that, here comes an amazing idea by which we can kind of make the sequence $\left\{f_k\right\}$ seem “periodic”. Let the “period” be denoted as $n$. Because it is not virtually periodic, we tend to call it an octave instead of a period. After that, the constant $n$ is the length of an octave.

What on earth is an octave defined? It is $$\forall k:f_{k+n}=2f_k.$$ Why do we say an octave is like a period? It is because according to the explanation above, corresponding notes in different octaves sound so harmonic that a person almost think they are the same. In this way, for some questions, we only need to consider a single octave instead of all notes. Let’s define the base octave, notes of which can generate all other notes by multiplying the frequency by a power of 2: $$O_0\coloneqq\left\{\nu_k\,\middle|\,k\in T\right\},$$ so we can say the frequency assignment has $n$ different tones. A tone is an integer in $$T\coloneqq\left[0,n\right)\cap\mathbb Z$$ representing where a note is in an octave.

We can thus define a sequence of octaves $$O_m\coloneqq\left\{\nu_{k+mn}\,\middle|\,\nu_k\in O_0\right\}.$$

In fact, any octave can be the base octave. They are all the same. For any $m$, all notes can be generated by notes in $O_m$. $$\begin{array}{|c|cccccc|} \hline T & \cdots & O_{-1} & O_0 & O_1 & O_2 & \cdots\\ \hline 0&\cdots& \nu_{-n}&\nu_0& \nu_l & \nu_{2n}&\cdots\\ 1&\cdots&\nu_{-n+1}&\nu_1&\nu_{n+1}&\nu_{2n+1}&\cdots\\ 2&\cdots&\nu_{-n+2}&\nu_2&\nu_{n+2}&\nu_{2n+2}&\cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ n-1&\cdots&\nu_{-1}&\nu_{n-1}&\nu_{2n-1}&\nu_{3n-1}&\cdots\\ \hline \end{array}$$ If we define $p_k\coloneqq\log_2f_k$, it is interesting to see that $$\forall k:p_{k+n}=p_k+1.$$ From that, we can naturally think an excellent frequency assignment be defined as $$p_k\coloneqq p_0+\frac kn,$$ $$(1)$$ which is an elegant arithmetic progression.

The octave group

Note that here “group” is the group concept in algebra.

Let $O_0$ be isomorphic to the additive group of $\mathbb Z/n\mathbb Z$, the integers modulo $n$, under the isomorphism $k\mapsto\nu_k$, which means to make $O_0$ a cyclic group of order $n$. I call this group an octave group.

To make you have a good sense of what on earth the group looks like, the definition of its group operation can be defined as $$\nu_a\circ\nu_b\coloneqq\nu_{\left(a+b\right)\mathbin\%n},$$ $$(2)$$ where $x\mathbin\%y\coloneqq x-y\left\lfloor\frac xy\right\rfloor$.

Note that this binary operator $\circ$ can be extended to be used for the whole $N$ while the definition remains the same as Formula 2.

The octave group has a musical meaning which we should take a further look at musical intervals to find out.

The musical intervals

Taking Formula 1, people find that although it is sometimes subjective whether a musical interval sounds harmonic or not, it does not depend on where the interval is located but on how far the two notes making up the interval are.

Taking this sense, we can consider only those intervals involving $\nu_0$ because we can always translate a interval so that one of its notes is $\nu_0$. Taking this idea, we can conclude that an interval can be represented by a note $\nu_k$ because the interval is equivalent to another interval which consists of $\nu_0$ and $\nu_k$.

We can make this idea even further. Previously, I have stated that a note can always be generated by a note in $O_0$. Therefore, an interval can be represented as a note in $O_0$.

Now, let’s look back to the octave group $\left(O_0,\circ\right)$. Denote the interval $b$ as that represented by $\nu_b$. Then, translate interval $b$ to such a location that its lower note is $\nu_a$. Then, its higher note represents the same interval as $\nu_a\circ\nu_b$.

Well, why do we focus on the group? It is because we need to mention an important concept in group theory, which is “generator”. It has something to do with determining the value of $n$.

The value of $p_0$ does not matter because changing it is just a translation of the whole sequence. What really matters is the value of $n$.

The inventor of the current prevailing frequency assignment (which is the $12$-tone equal temperament shown in Formula 3) may think the generator of the group a vital thing. Actually, people think it a wonderful thing that a note representing a very harmonic interval is a generator of the octave group.

Fortunately, such a goal is achievable. People find that if $p_b=p_a+\frac7{12}$, then the interval (which is the perfect fifth interval if you know music theory) consisting of $\nu_a$ and $\nu_b$ sounds very harmonic:

          
            motif = e4 a f2 d4 g c2

            piano:

            V1: motif

            V2: (transpose 7) motif

          
        

Furthermore, as can be seen in the following table, $\nu_7$ is a generator of the group $O_0$ if $n\coloneqq12$: $$\begin{array}{|c|cccccccccccc|} \hline j&0&1&2&3&4&5&6&7&8&9&10&11 \\\hline \nu_7^{\circ j}& \nu_0&\nu_7&\nu_2&\nu_9&\nu_4&\nu_{11}&\nu_6&\nu_1& \nu_8&\nu_3&\nu_{10}&\nu_5 \\\hline \end{array}$$ Thus, wonderful! Let’s take $n\coloneqq12$.

The $12$-tone equal temperament

The $12$-tone equal temperament is the most popular frequency assignment used nowadays. It is defined as $$f_k\coloneqq16.3516\cdot 2^\frac k{12},$$ $$(3)$$ which can be derived from Formula 1 taking $$p_0\coloneqq4.03136\qquad n\coloneqq12.$$

The frequency assignment has $12$ different tones, $7$ of which have their names: $$\begin{align*} \mathrm C_m&\coloneqq\nu_{12m},\\ \mathrm D_m&\coloneqq\nu_{12m+2},\\ \mathrm E_m&\coloneqq\nu_{12m+4},\\ \mathrm F_m&\coloneqq\nu_{12m+5},\\ \mathrm G_m&\coloneqq\nu_{12m+7},\\ \mathrm A_m&\coloneqq\nu_{12m+9},\\ \mathrm B_m&\coloneqq\nu_{12m+11}. \end{align*}$$ The famous “middle C” is $\mathrm C_4$.

This notation is called the scientific pitch notation.

Note that in fact, this definition of $12$-tone equal temperament has some slight error. The accurate value for $p_0$ is $$p_0\coloneqq\log_255-\frac74$$ because it is stipulated that $f\left(\mathrm A_4\right)=440$, which is standardized as ISO 16 and known as A440.

Why $\frac 7{12}$

People think if the ratio of two frequencies is a simple rational number, then the interval of the two notes is harmonic.

$2^\frac 7{12}\approx\frac32$, which is a simple ratio. Harmonic, huh.

(Finally, as is a notice, codes appearing above are alda codes, which are used to write music.)