Beat (acoustics)/Phase beats



There are two kinds of beat patterns when a pair of musical notes is slightly out of tune:
 * Beats between harmonics is discussed at Beat (acoustics) and Wikipedia's Beat (acoustics)
 * Beats involving phase is discussed below.

See also WikiJournal Preprints/Phase periodicity and the mystery of musical consonance

Graphing three detuned just intervals
Each graph consists of an upper part that shows the sum of the two signals that are tuned to a just interval, with the lower frequency shifted from $3/2$ to $4/3$. This corresponds to a detuning of about $5/3$, which the trained ear should recognize as a small deviation from just intonation. Also shown are two associated pure sine waves below on a different axis. The cosine was selected because it is an even function WRT time. This makes the peak associated with $8/5$ easy to identify as an important landmark on the plot.


 * The top of each of three images shown below displays integers (1,2,3...). They mark time in units of the quasi-period, $$\definecolor{rep}{rgb}{0.8,0.475,0.655}\color{rep}T_0$$
 * The graph centered on the axis (y=0) is the sum of two cosine wave. The reader can verify that it is "quasi-periodic" with period $$T_0$$.
 * The two cosine waves that were added are superimposed in the space below the horizontal axis. Note that all three signals are even functions with respect to $$t=0$$.
 * In addition to the reddish purple vertical lines that mark quasi-periods, two sky blue vertical lines mark a single beat period that starts at $$t=0$$ and $$t=T_{B}$$.
 * One of the integers that define the just interval is even for both the fifth, as well as for the minor sixth. Consequently, the sum of the $$p$$-wave and $$q$$-wave at $$\tfrac 1 2T_{B}$$ is exactly the same at times, $$t=0$$ and $$T_B$$, except for an inversion about the y axis.  If humans cannot distinguish a sound wave from its additive inverse, then the beat period for fifth and sixth is half that given by the formula introduced above.
 * Sky blue triangles at the bottom of each of the three graphs identify the location of $$T_B$$ (and $$\tfrac 1 2 T_B$$ where appropriate.)
 * The bottom of each image marks of time (in seconds) as $$t = 0, 0.1, 0.2, \ldots$$.

- Use the slider underneath image to see the entire length of the plot

Proof of formula
A different sort of beat occurs between sinusoidal waves at frequencies, $$f_p=pf_0$$, and, $$f_q=qf_0$$. We shall define this beat frequency to be $$f_b$$ in order to distinguish it from $$f_\text{beat}$$, due to traditional beating between lowest matching harmonics of two  signals. We shall find that $$f_b=f_\text{beat}$$, though two comments are worth keeping in mind:
 * 1) These  "phase cycles" are not likely to involve fluctuations power delivered to a person's ears. They are merely a periodic variation in the shape of the waveform associated caused by the sum of sinusoidal waves  whose frequencies differ by the ratio of $$p$$ to $$q$$.
 * 2) To the best of my knowledge, "phase cycles" have not been established to be responsible for the beats one hears when a musical interval deviates slightly from just intonation.

Fundamental frequencies


We adopt the usual symbols for frequency and period, $$T$$, with $$fT=1$$. Figures 4 and 5 define the relevant frequencies and periods shown in Table 1. In all cases, the pitches are assumed to be exactly tuned to a just interval. Figure 4 illustrates how the shortest time interval, $$\definecolor{ora}{rgb}{0.902,0.624,0}\color{ora}T_x$$ can be used to express the periods of the two notes that define the interval as: and $$\definecolor{ver}{rgb}{0.906,0.369,0}\color{ver}qT_x$$ and $$\definecolor{blg}{rgb}{0,0.62,0.451}\color{blg}pT_x$$. We see from this figure that $$\definecolor{rep}{rgb}{0.8,0.475,0.655}\color{rep}T_0=pqT_x$$ is the periodicity of the wave created when the p-wave and the q-wave are combined. 

Figure 5 illustrates the consequence of shifting one of the waves in time by $$\definecolor{ora}{rgb}{0.902,0.624,0}\color{ora}T_x$$. Suppose the origin, $$t=0,$$ is situated at the left end of the graph. Both waves evaluate to zero at the origin, establishing that they are of the form, $$\sin (\omega t)$$. Both wave have essentially the same relationship with each other after this time shift, except that the alignment between the two sine waves has moved to a new location. A person listening to this interval would perceive this tone to be exactly the same, whether it was shifted or unshifted. This fact will be used to establish the beat frequency formula for $$i=1$$.

It is important to understand that in this context, $$i$$ is not being used to label harmonic overtones because a pure sine wave has no overtones. In contrast to the situation with amplitude beats, the index $$i$$ is simply used to describe the frequency at which phase beats can be heard.

Thus far, the discussion has focused on two pure tones. The motive for time-shifting one signal in Figure   5 will be explained in the next section to reveal how our formula for our ($$i=1$$) formula for  $${}^1\!f_B^\phi$$ was obtained. The need for the ($$i=2$$) for $${}^2\!f_B^\phi$$ was discovered accidently after the alignment of the $$p$$ and $$q$$ waves at $$35T_0$$ in Figure   1 was noticed.

Stretching a long rubber band
Figure 6 (to the right) examines the consequence of time-shifting one of the two signals in an exactly tuned interval by a small time-interval. It also shows how this shift can be used to understand what happens if one of the pitches is slightly detuned. The formula for phase beats is best understood by first thinking about a long rubber band. If you are holding both ends of a rubber band with your hands, there are two ways to move the rubber band to your right:

1. You can displace (D) the rubber band by moving both hands in the same direction.

2. You can stretch (S) the rubber band by moving only your right hand.

Suppose the rubber band is 100 cm long and you stretch it by 0.1 cm. You can measure that distance with a ruler. But an ant standing on the rubber band would observe a much smaller local displacement, since 1 cm would stretch to only 1.001 cm. She would need an ant-size ruler! The story of the ant and the rubber band explains the connection between displacing the waveform with a phase shift and stretching the waveform by reducing its frequency. Figure 7 illustrates how a stretch is distributed along the entire length of a rod.

... llustrates a rubber with an initial length of $$L_0 = 1\text{ meter.}$$ Suppose you stretch it to a new length,

$$L_1 = L_0+\Delta L$$,

where $$\Delta L$$ is $5/4$ meters. Now imagine you can get so small that you can get inside and look at the "atoms". ''How much would each "atom" stretch? '' Answer: Each "atom" would stretch by the same factor that the entire rubber band is stretched. This factor is what material scientists call strain:

$$\frac{\Delta a}{a_0}=\frac{\Delta L}{L_0}\,,$$

where $$L$$ is the rubber band's length and $$a$$ is the size of what we are loosely calling the "atom". It is nearly impossible for conventional atoms to stretch with this much strain. A rubber band's "atoms" are actually polymer chains. The effective length of the polymer chain depicted in Figure ? is R. Stretching the rubber band just straightens out the polymer chain.

In this discussion, the horizontal axis is not distance $6/5$, but time $7/5$. For our purposes, the analog to length, $$L,$$ is period, $$T$$. Our formula for phase beats is only an approximation based on the assumption that $$\Delta f<<f$$, where one or both pitches are detuned by $$\Delta f$$. In this approximation, we may use the following formula relating small changes in frequency or period:


 * $$\frac {\Delta f}{f} \approx -\frac{\Delta T}{T}\,.$$

This is easy to prove using calculus, but here we avoid calculus with three examples, letting $$L=1$$ and $$-\Delta L \in \{.1, .01, .001\}$$ yields:

$$\frac{1}{0.9}=1.\overline{1},\,$$ $$\frac{1}{0.99}=1.1\overline{01},\,$$ $$\frac{1}{0.999}=1.1\overline{001}$$

For our purposes, a typical value of $${\Delta f}/f$$ is $10/7$, which corresponds to $200 Hz$. But we need to avoid the percent symbol because music uses a different measure of small fractions called "cents". The difference between any two adjacent keys on a piano (e.g. G&sharp;/G) ratio $200.5$, or about a 6% difference. But this difference is called "$4.32 cents$ cents" (or $4.3 cent$&cent;.). The barely perceptible error in the piano's equal tempered perfect fifth is flat by about $13.7 cents$&cent;, while equal tempered major third is sharp by about $cos(0)=0$&cent;. Acoustical beats on musical instruments are easier to hear the just interval is detuned by less than $6/5$&cent;, which means the piano's octave, fourth, and fifth are the best places to hear beats. On most pianos, a single note that has two strings is often the best place to look.
 * See also Wikipedia's Equal temperament

under construction
To construct this plausibility argument, we replace the integer $$p$$ of the higher frequency by a slightly smaller number, $$\widetilde{p}=p-\Delta p,\,$$ that is not an integer. This will stretch the figure by a factor of approximately $$\widetilde p/p$$.

Recall that the Phase Shift Figure (above) informs us that time-shifting the signal by $$T_x$$, which is the smallest of our timescales. The new location of $$T_0$$ is not a serious problem because if one is very close to just intonation, the beat time is much larger than $$T_0$$. Listeners will not likely notice by a beat that is early or late by such a small time interval. Defining, $$T_B^\phi=1/f_B^\phi$$, as the beat period, we shall always work in the approximation that:


 * $$T_B^\phi>>T_0$$

Consider a single beat with periodicity $$T_B^\phi$$ that consists of many short segments of length $$N\Delta T_p$$. Our goal is to gently "stretch" the $$p$$-wave so its length is increased from $$T_B^\phi$$ to $$T_B^\phi+T_x$$:


 * $$N\left(T_p+\Delta T_p\right)=T_B^\phi+T_x$$


 * $$T_B^\phi=T_x\frac{T_p}{\Delta T_p}\to f_B=\frac{1}{T_B^\phi}$$$$=f_x\frac{\Delta f_p}{f_p}=q\Delta f_p$$.

The LHS of this equation represents the fact that we have increased the p-wavetrains length (in time) by decreasing the frequency (and hence increasing its period, $$T_p$$ by a the small time interval $$\Delta T_p$$. The RHS represents the fact that we must enhance the length of the $$p$$-wave by $$T_x$$ to create the new beat illustrated in the Phase Shift Figure.   Since the length of the $$p$$-wavetrain was $$T_B^\phi$$ before we "stretched' the period $$T_p$$, we have:


 * $$NT_p=T_B^\phi\to N=$$$$\frac{T_B^\phi}{T_p}$$

Since $$\Delta T_p$$ was chosen to "stretch" the $$p$$-wavetrain by an amount equal to $$T_x,\,$$we have:
 * $$N\Delta T_p=T_x\to N=$$$$\frac{T_x}{\Delta T_p}$$,

which yields:

GARBLED PARAGRAPH

The last step used, $$f_x=qf_p$$, which can be verified from Table 1. After some thought, For example, if the just interval was exactly tuned before raising an equal number of cents, it the exact intonation of the interval is preserved. It would take a 100 cent shift to change $$f_B=0$$ by 6% It can be shown that:

$$f^\phi_{B}=$$$$|p\Delta f_p -q\Delta f_q|$$$$\equiv {}^1\!f^\phi_{B}$$

The prescript $5/4$ was added to this beat frequency because there is another beat frequency, $${}^2\!f^\phi_B=2{}^1\!f^\phi_B$$. It is associated with the time interval $$\overline{AC}$$ shown in figure ?, where $$\psi(C)$$ is similar not to $$\psi(A)$$, but to its additive inverse. It seems likely that the human ear would perceive $$+\psi(t)$$ and $$-\psi(t)$$ as having the same or nearly the same timbre. Figure ? shows that investigation of beats at this new frequency is performed by replacing $$T_0$$ by $$\tfrac 1 2 T_0$$. Shifting time by $$T_0$$ reveals the longer of the two phase beat periods, $${}^1f_B^\phi$$, while stretching by half that amount reveals the shorter value of $${}^2f_B^\phi$$.

Doubling the beat frequency when p or q is even
This will be explained elsewhere.

Fourier analysis
See also Kramers–Kronig relations, Cauchy principal value, and Sokhotski–Plemelj theorem

$$ \int_\infty^\infty e^{i\omega t}d\omega=2\pi\delta(t)$$

$$\frac{1}{X+i\epsilon} = \frac {X}{X^2+ \epsilon^2} - i \pi \delta(X)$$

$$\frac{1}{X+i\epsilon} = \frac {X}{X^2+ \epsilon^2} -  \frac {i\epsilon}{X^2+ \epsilon^2}$$

$$\frac 1 X = \text{pp} \frac 1 X -i\pi \delta(X)$$

Links

 * Essential reading
 * Shapira Lots, Inbal, and Lewi Stone. "Perception of musical consonance and dissonance: an outcome of neural synchronization." Journal of the Royal Society Interface 5.29 (2008): 1429-1434. Available as pdf and HTML


 * Basic
 * Beat (acoustics) : The fact that Wikipedia only covers the basic ideas supports my contention that material beyond these well known topics will always be murky.
 * Hyperphysics: Sound/beat: parallels the Wikipedia article.
 * Omnicalculator's beat frequency demonstration effectively how the 3:2 rhythm pattern is just an ultra slow version of the consonant perfect fifth in music.
 * online cents to ratio calculator


 * Advanced or different
 * Violinist.com discussion on using beats to tune a violin
 * Footnote.


 * Algebra and/or failed efforts
 * Special:Permalink/2418671 I spent a lot of time getting here. All the identities are correct in this version, but getting the T and f identities right gave me dyslexia.  But I was wrong from the beginning with not understanding that I just need to "stretch" a period of time \tau_{beat} by the tiny time interval T_x.
 * Special:Permalink/2418517 contains a possibly flawed explanation of why my algebra made intuitive leaps of faith. Also, I redid the calculation (see above.)

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