Beat (acoustics)/Exact phase beats

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Beats can be heard in a musical interval if the two pitches are accompanied by a matching pair of harmonics with nearly the same frequency. A musical interval can display amplitude beats both pitches are accompanied by a pair of harmonicsThe calculation of a the phase beat frequency ${}^{1}\!f_{B}^{\phi }\equiv 1/T_{B},$ of a detuned just fifth (P5) is based on figure 1 (to the right.) The time axis is oriented so that the reader can scroll downward through time. Sound waves are pressure waves, and the convention in this figure is the pressure increases to the right. All the signals involve sums of sinusoidals of the form $\cos \omega t$, so that a local maximum always occurs at $t=0.$ The representation of sunusoidals as triangular waves facilitates the determination of events where the two waves are exactly aligned to form maxima. Also, there is a binary nature to the way nerve cells in the brain communicate, and for that reason, the discontinuous nature of triangular waves might better resemble how the brain processes music. When plotting waveforms, it is convenient to express time using two different units. Figure  1 shows how $T_{0}$ is subdivided into six units of $T_{x},$ and both time units are defined in Table 1. It is helpful to view time in base-6, for example, as $nT_{0}+mT_{x}.$

Table 1

$1<p/q<2$

$f_{x}>f_{p}>f_{q}>f_{0}$

$T_{x}<T_{q}<T_{p}<T_{0}$

$T_{q}=pT_{x}$	$T_{p}=qT_{x}$	$qT_{p}=pT_{q}$
$pf_{q}=f_{x}$	$qf_{p}=f_{x}$	$qf_{p}=pf_{q}$
$T_{0}=pqT_{x}\quad \quad \quad pqf_{0}=f_{x}$
$f_{p}=pf_{0}=f_{x}/q\quad \quad \quad f_{q}=qf_{0}=f_{x}/p$

Figure 1: Proof of beat formula for detuned fifth      

Detuned just fifth musical interval

Figure 2. For a just interval with $p/q$ as the frequency ratio, $T_{0}$ is the periodicity of the combined signals, and $T_{0}=pqT_{x}.$

Table 2

$n$	${\frac {T_{B}}{T_{x}}}$	$P$	$Q$	${\frac {P}{Q}}$	${\text{cents}}$
0	6	3	2	1.500	0.0
1	9	4	3	1.333	203.9
2	15	7	5	1.400	119.4
3	21	10	7	1.429	84.5
4	27	13	9	1.444	65.3
5	33	16	11	1.455	53.3
6	39	19	13	1.462	45.0
7	45	22	15	1.467	38.9
8	51	25	17	1.471	34.3
9	57	28	19	1.474	30.6
10	63	31	21	1.476	27.7
50	303	151	101	1.495	5.7
100	603	301	201	1.498	2.9
150	903	451	301	1.498	1.9
200	1203	601	401	1.499	1.4
250	1503	751	501	1.499	1.2
300	1803	901	601	1.499	1.0

Fifth (P5): 3/2 ratio

We begin our discussion with the perfect fifth. The periodicity is exactly $T_{0},$ which can be identified by the alignment of the two peaks at $t=T_{0}=6T_{X}.$ The extension of the graph beyond this alignment is not necessary beyond this point. But the extension of time beyond $T_{0}$in the first facilitates the reading of the graphs that follow.

Fourth (P4): 4/3 ratio (n=1)

The fourth is the first element in the sequence and the second column in figure 1. It seems strange to think of the fourth as a fifth detuned by two half-tones. But the beauty of a well organized sequence is that each element has something in common, and we can see a hint of the beat pattern in this first element. Everything that makes 301/201 an interesting ratio for studying a detuned version of 3/2, is true for 4/3.

To create the fourth, we begin with the first column's wave at $T_{0}+2T_{x}=8T_{x}.$ This is a local maximum ("peak") associated with the short period p-wave's period (shown in black). Now we "stretch" this p-wave so that it's peak aligns with the peak of the (orange) q-wave's peak at $T_{0}+3T_{x}=9T_{x}.$ By realigning the peaks that were initially aligned at $t=0,$ we establish the periodicity for sum of the two waves to be $T_{B}=9T_{x}.$ This is consistent with,

  $T_{B}^{n}=(6n+3)T_{x}\,,$Y

where $n=1$. We will later confirm that this formula describes the periodicity for all the waveforms depicted in figure 1 and table 2. This stretch lengthens the period of the p-wave from $T_{p}=2T_{x}$ to a period we shall call ${\widetilde {T}}_{p}.$ To find this stretched period, multiply $T_{p}$ by the factor by which the p-wave was stretched, namely from $8T_{x}$ to $9T_{x}$. Hence, ${\widetilde {T}}_{p}^{1}=2(9/8)T_{x}=(9/4)T_{x},$ which is consistent with,

  ${\widetilde {T}}_{p}^{n}={\frac {6n+3}{3n+1}}T_{x}\,,$Y

for $n=1.$ We now turn our attention to the ratios that define our sequence of detuned fifths. It is easy to guess the formula for a sequence that starts with 4/3 and converges to 3/2:

  $P_{n}=4+3(n-1)=3n+1$Y

  $Q_{n}=3+2(n-1)=2n+1$Y

Defining P and Q

The definition of $P_{n}$ and $Q_{n}$ follows from two properties about the graphs in figure 1: (1) the period of the orange q-wave is $T_{q}=3T_{x}$, and (2) The period of the black p-wave adjusted from $T_{p}=2T_{x}$ to ${\widetilde {T}}_{p}$, so that:

  ${\frac {{\widetilde {T}}_{p}}{T_{q}}}={\frac {Q_{n}}{P_{n}}}\,.$

Proof by induction

Four formulas have been demonstrated for $n=1$ and postulated to be true for all $n>1$. Proof by induction requires two steps: First we prove something is true for $n=1$. Then we assume it true for $n$ and use that assumption to prove it true for $n+1.$ We don't really need a "proof" regarding the values of $P_{n}$ and $Q_{n}$ because they obviously converge to the desired limit:

  $\lim _{n\to \infty }{\frac {P_{n}}{Q_{n}}}=\lim _{n\to \infty }{\frac {4+3(n-1)}{3+2(n-1)}}={\frac {3}{2}}$Y

This equations tells us how many wavelengths of each pitch must be added to create the $(n+1)^{\text{th}}$ interval. We need 3 cycles of the p-wave and 2 of the q-wave. Since the p-wave must be stretched, we must use q-wave to calculate the next alignment of the peaks: Two cycles of the q-wave correspond to a time interval of $6T_{x}$. Hence, we must compare adding $6T_{x}$ to $T_{B}^{n}$ with evaluating $T_{B}^{n+1}$ using our postulated formula.

For simplicity, we temporarily set $T_{x}=1$ and do our comparison: Given that $T_{B}^{n}=6n+3$, does adding 6 recover our postulated formula for $T_{B}^{n+1}?$

  $\underbrace {T_{B}^{n}+6} _{(6n+3)+6}=\underbrace {T_{B}^{n+1}} _{6(n+1)+3}$ Yes it does!Y

Continuing to measure time in units of $T_{x}$, we now seek the factor by which the p-wave is stretched. Now the logic gets so contorted that I must confess that I only believe it because the spreadsheet that generated table 2 tells me so. Define $\Delta t$ as the length the p-wave before it was stretched. It is important to note that these are total lengths, measured from the graph at $t=0$ to the time corresponding to the tail of the arrow pointing to the peak near $nT_{0}$ where each graph terminates. We have to carefully define what is meant by "before" the stretch. Are we stretching the p-wave from its length when it was the exactly tuned just fifth in the first column? Or is the $n^{\text{th}}$ iteration made from the previous stretch at $n-1\,?$ No answer will be given here because that would make a good homework or prelim exam question. So here goes the algebra: Each iteration adds $T_{0}$ units of time to the periodicity. As shown in figure 1, the each p-wave measures from $t=0$ to $nT_{0}+2T_{x}$ before the stretch. After the stretch, the peak's new location is at $nT_{0}+3T_{x}$. This defines the factor by which $T_{p}=2T_{x}$ is stretched to its new length, which is defined as ${\widetilde {T}}_{p}:$

  $\underbrace {\widetilde {T_{p}}} _{\text{after}}=\underbrace {\frac {nT_{0}+3T_{x}}{nT_{0}+2T_{x}}} _{\text{factor}}\cdot \underbrace {2T_{x}} _{\text{before}}={\frac {6n+3}{3n+1}}T_{x}$Y

Relabeling beat period and frequency

It was shown above that the periodicity is $T_{B}^{n}=(6n+3)T_{x}.$ We relabel this periodicity (or beat frequency) to adhere to the convention that these are phase $(\phi )$ associated with $i=1$, meaning that this is the "first" or smallest beat frequency. This relabeling is accomplished by replacing $T_{B}^{n}$ by:

  ${}^{1}\!T_{B}^{\phi }=\left({}^{1}\!f_{B}^{\phi }\right)^{-1}$Y

With this replacement, ${}^{1}\!T_{B}^{\phi }=(6n+3)T_{x}$ implies that:

  ${}^{1}\!f_{B}^{\phi }={\frac {f_{x}}{6n+3}}={\frac {6f_{0}}{6n+3}}={\frac {2f_{0}}{2n+1}}$

Comparison of these phase beats with Helmholtz (harmonic) beats

Here, Helmholtz (or harmonic) beats are amplitude beats between matching harmonics. The pre-superscript $i=1$ can be used to identify the harmonics as the lowest possible pair. For the just fifth between $2f_{0}$ and $3f_{0}$, these $i=1$ matching harmonics have frequency $6f_{0}$.

• To learn more about this type of beating, visit Beat_(acoustics)/Phase_beats

From that aforementioned page, the lowest order $i=1$ beats in the fifth is between the third harmonic of the bottom pitch and the second harmonic of the higher pitch. On this page, only the p-wave is detuned in our sequence of just intervals. Also, the frequency is reduced, causing $\Delta f_{p}$ to be negative:

  ${}^{1}\!f_{B}^{H}=|q\Delta f_{p}|=-2\Delta f_{p}\,.$ Y

We now calculate $\Delta f_{p}$ for the just intervals in our sequence (4/3, 5/7,.... Due to the geometrical nature of our treatment of these intervals, we start with periods and convert to frequencies:

  $T_{q}=3T_{x}$ and ${\frac {{\widetilde {T}}_{p}}{T_{q}}}={\frac {Q_{n}}{P_{n}}}$ implies ${\widetilde {T}}_{p}=3T_{x}{\frac {Q_{n}}{P_{n}}}={\frac {1}{{\widetilde {f}}_{p}}}$Y

  ${\widetilde {f}}_{p}={\frac {P_{n}}{Q_{n}}}{\frac {f_{x}}{3}}={\frac {3n+1}{2n+1}}{\frac {6f_{0}}{3}}=2{\frac {3n+1}{2n+1}}f_{0}={\frac {6n+2}{2n+1}}f_{0}$Y

  $\Delta f_{p}={\widetilde {f}}_{p}-f_{p}$Y,

where ${\widetilde {f}}_{p}={\frac {6n+2}{2n+1}}$ and $f_{p}=3f_{0}={\frac {6n+3}{2n+1}}f_{0}.$ This last step sets up the LCD for subtracting to obtain:

$\Delta f_{p}=\left({\frac {6n+2}{2n+1}}-{\frac {6n+3}{2n+1}}\right)f_{0}=-{\frac {f_{0}}{2n+1}}.$ Multiply by $q=2$ to obtain the Helmholtz (harmonic) beat frequency:

  ${}^{1}\!f_{B}^{H}={\frac {2f_{0}}{2n+1}}.$

The fact that ${}^{1}\!f_{B}^{H}={}^{1}\!f_{B}^{\phi }$ verifies that both mechanism yield the same beat frequency.