The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (2024)

50Harmonics

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (2)

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (3)

50–1Musical tones

Pythagoras is saidto have discovered the fact that two similar strings under the sametension and differing only in length, when sounded together give aneffect that is pleasant to the ear if the lengths of the stringsare in the ratio of two small integers. If the lengths are as one is totwo, they then correspond to the octave in music. If the lengths are astwo is to three, they correspond to the interval between $C$ and $G$,which is called a fifth. These intervals are generally accepted as“pleasant” sounding chords.

Pythagoras was so impressed bythis discovery that he made it the basis of a school—Pythagoreans theywere called—which held mystic beliefs in the great powers of numbers.It was believed that something similar would be found out about theplanets—or “spheres.” We sometimes hear the expression: “the musicof the spheres.” The idea was that there would be some numericalrelationships between the orbits of the planets or between other thingsin nature. People usually think that this is just a kind of superstitionheld by the Greeks. But is it so different from our own scientificinterest in quantitative relationships?Pythagoras’ discovery wasthe first example, outside geometry,of any numerical relationship in nature. It must have been verysurprising to suddenly discover that there was a fact of naturethat involved a simple numerical relationship. Simple measurements oflengths gave a prediction about something which had no apparentconnection to geometry—the production of pleasant sounds. Thisdiscovery led to the extension that perhaps a good tool forunderstanding nature would be arithmetic and mathematical analysis. Theresults of modern science justify that point of view.

Pythagoras could only have madehis discovery by making an experimental observation. Yet this importantaspect does not seem to have impressed him. If it had, physics mighthave had a much earlier start. (It is always easy to look back at whatsomeone else has done and to decide what he should have done!)

We might remark on a third aspect of this very interesting discovery:that the discovery had to do with two notes that sound pleasantto the ear. We may question whether we are any better off thanPythagoras in understandingwhy only certain sounds are pleasant to our ear. The generaltheory of aesthetics is probably no further advanced now than in thetime of Pythagoras. In this onediscovery of the Greeks, there are the three aspects: experiment,mathematical relationships, and aesthetics. Physics has made greatprogress on only the first two parts. This chapter will deal with ourpresent-day understanding of the discovery ofPythagoras.

Among the sounds that we hear, there is one kind that we callnoise. Noise corresponds to a sort of irregular vibration ofthe eardrum that is produced by the irregular vibration of some objectin the neighborhood. If we make a diagram to indicate the pressure ofthe air on the eardrum (and, therefore, the displacement of the drum)as a function of time, the graph which corresponds to a noise mightlook like that shown in Fig.50–1(a). (Such a noisemight correspond roughly to the sound of a stamped foot.) The sound ofmusic has a different character. Music is characterized bythepresence of more-or-less sustained tones—or musical“notes.” (Musical instruments may make noises as well!) The tone maylast for a relatively short time, as when a key is pressed on a piano,or it may be sustained almost indefinitely, as when a flute playerholds a long note.

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (4)The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (5)

Fig. 50–1.Pressure as a function of time for (a)a noise, and (b)amusical tone.

What is the special character of a musical note from the point of viewof the pressure in the air? A musical note differs from a noise inthat there is a periodicity in its graph. There is some uneven shapeto the variation of the air pressure with time, and the shape repeatsitself over and over again. An example of a pressure-time functionthat would correspond to a musical note is shown inFig.50–1(b).

Musicians will usually speak of a musical tone in terms of threecharacteristics: the loudness, the pitch, and the “quality.” The“loudness” is found to correspond to the magnitude of the pressurechanges. The “pitch” corresponds to the period of time for onerepetition of the basic pressure function. (“Low” notes have longerperiods than “high” notes.) The “quality” of a tone has to do withthe differences we may still be able to hear between two notes of thesame loudness and pitch. An oboe, a violin, or a soprano are stilldistinguishable even when they sound notes of the same pitch. Thequality has to do with the structure of the repeating pattern.

Let us consider, for a moment, the sound produced by a vibratingstring. If we pluck the string, by pulling it to one side andreleasing it, the subsequent motion will be determined by the motionsof the waves we have produced. We know that these waves will travel inboth directions, and will be reflected at the ends. They will sloshback and forth for a long time. No matter how complicated the wave is,however, it will repeat itself. The period of repetition is just thetime$T$ required for the wave to travel two full lengths of thestring. For that is just the time required for any wave, once started,to reflect off each end and return to its starting position, and beproceeding in the original direction. The time is the same for waveswhich start out in either direction. Each point on the string will,then, return to its starting position after one period, and again oneperiod later, etc. The sound wave produced must also have the samerepetition. We see why a plucked string produces a musical tone.

50–2The Fourier series

We have discussed in the preceding chapter another way of looking atthe motion of a vibrating system. We have seen that a string hasvarious natural modes of oscillation, and that any particular kind ofvibration that may be set up by the starting conditions can be thoughtof as a combination—in suitable proportions—of several of thenatural modes, oscillating together. For a string we found that thenormal modes of oscillation had the frequencies $\omega_0$,$2\omega_0$,$3\omega_0$,…The most general motion of aplucked string, therefore, is composed of the sum of a sinusoidaloscillation at the fundamental frequency$\omega_0$, another at thesecond harmonic frequency$2\omega_0$, another at the thirdharmonic$3\omega_0$, etc. Now the fundamental mode repeats itself everyperiod$T_1 = 2\pi/\omega_0$. The second harmonic mode repeats itselfevery$T_2 = 2\pi/2\omega_0$. It also repeats itself every$T_1 =2T_2$, after two of its periods. Similarly, the third harmonicmode repeats itself after a time$T_1$ which is $3$of its periods. Wesee again why a plucked string repeats its whole pattern with aperiodicity of$T_1$. It produces a musical tone.

We have been talking about the motion of the string. But thesound, which is the motion of the air, is produced by themotion of the string, so its vibrations too must be composed of thesame harmonics—though we are no longer thinking about the normalmodes of the air. Also, the relative strength of the harmonics may bedifferent in the air than in the string, particularly if the string is“coupled” to the air via a sounding board. The efficiency of thecoupling to the air is different for different harmonics.

If we let $f(t)$ represent the air pressure as a function of time fora musical tone [such as that in Fig.50–1(b)], then weexpect that $f(t)$ can be written as the sum of a number of simpleharmonic functions of time—like$\cos\omega t$—for each of thevarious harmonic frequencies. If the period of the vibration is$T$,the fundamental angular frequency will be$\omega = 2\pi/T$, and theharmonics will be $2\omega$, $3\omega$, etc.

There is one slight complication. For each frequency we may expectthat the starting phases will not necessarily be the same for allfrequencies. We should, therefore, use functions like$\cos\,(\omega t+ \phi)$. It is, however, simpler to use instead both the sine andcosine functions for each frequency. We recall that\begin{equation}\label{Eq:I:50:1}\cos\,(\omega t + \phi) = (\cos\phi\cos\omega t - \sin\phi\sin\omega t)\end{equation}and since $\phi$ is a constant, any sinusoidal oscillation atthe frequency$\omega$ can be written as the sum of a termwith$\cos\omega t$ and another term with$\sin\omega t$.

We conclude, then, that any function$f(t)$ that is periodicwith the period$T$ can be written mathematically as\begin{alignat}{4}f(t) &= a_0\notag\\[.5ex]&\quad\;+\;a_1\cos&&\omega t &&\;+\;b_1\sin&&\omega t\notag\\[.65ex]&\quad\;+\;a_2\cos2&&\omega t &&\;+\;b_2\sin2&&\omega t\notag\\[.65ex]&\quad\;+\;a_3\cos3&&\omega t &&\;+\;b_3\sin3&&\omega t\notag\\[.5ex]\label{Eq:I:50:2}&\quad\;+\;\dotsb && &&\;+\;\dotsb\end{alignat}where $\omega = 2\pi/T$ and the $a$’s and$b$’s are numericalconstants which tell us how much of each component oscillation ispresent in the oscillation$f(t)$. We have added the“zero-frequency” term$a_0$ so that our formula will be completelygeneral, although it is usually zero for a musical tone. It representsa shift of the average value (that is, the “zero” level) of thesound pressure. With it our formula can take care of any case. Theequality of Eq.(50.2) is represented schematically inFig.50–2. (The amplitudes, $a_n$ and$b_n$, of the harmonicfunctions must be suitably chosen. They are shown schematically andwithout any particular scale in the figure.) Theseries(50.2) is called the Fourier series for$f(t)$.

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (6)

Fig. 50–2.Any periodic function$f(t)$ is equal to a sum of simpleharmonic functions.

We have said that any periodic function can be made up in thisway. We should correct that and say that any sound wave, or anyfunction we ordinarily encounter in physics, can be made up of such asum. The mathematicians can invent functions which cannot be made upof simple harmonic functions—for instance, a function that has a“reverse twist” so that it has two values for some values of$t$! Weneed not worry about such functions here.

50–3Quality and consonance

Now we are able to describe what it is that determines the “quality”of a musical tone. It is the relative amounts of the variousharmonics—the values of the $a$’s and$b$’s. A tone with only thefirst harmonic is a “pure” tone. A tone with many strong harmonicsis a “rich” tone. A violin produces a different proportion ofharmonics than does an oboe.

We can “manufacture” various musical tones if we connect several“oscillators” to a loudspeaker. (An oscillator usually produces anearly pure simple harmonic function.) We should choose thefrequencies of the oscillators to be $\omega$,$2\omega$,$3\omega$,etc. Then by adjusting the volume control on each oscillator, we canadd in any amount we wish of each harmonic—thereby producing tonesof different quality. An electric organ works in much this way. The“keys” select the frequency of the fundamental oscillator and the“stops” are switches that control the relative proportions of theharmonics. By throwing these switches, the organ can be made to soundlike a flute, or an oboe, or a violin.

It is interesting that to produce such “artificial” tones we needonly one oscillator for each frequency—we do not need separateoscillators for the sine and cosine components. The ear is not verysensitive to the relative phases of the harmonics. It pays attentionmainly to the total of the sine and cosine parts of eachfrequency. Our analysis is more accurate than is necessary to explainthe subjective aspect of music. The response of a microphone orother physical instrument does depend on the phases, however, and ourcomplete analysis may be needed to treat such cases.

The “quality” of a spoken sound also determines the vowel soundsthat we recognize in speech. The shape of the mouth determines thefrequencies of the natural modes of vibration of the air in themouth. Some of these modes are set into vibration by the sound wavesfrom the vocal chords. In this way, the amplitudes of some of theharmonics of the sound are increased with respect to others. When wechange the shape of our mouth, harmonics of different frequencies aregiven preference. These effects account for the difference between an“e–e–e” sound and an “a–a–a” sound.

We all know that a particular vowel sound—say “e–e–e”—still“sounds like” the same vowel whether we say (or sing) it at a highor a low pitch. From the mechanism we describe, we would expect thatparticular frequencies are emphasized when we shape our mouthfor an “e–e–e,” and that they do not change as we changethe pitch of our voice. So the relation of the important harmonics tothe fundamental—that is, the “quality”—changes as we changepitch. Apparently the mechanism by which we recognize speech is notbased on specific harmonic relationships.

What should we say now about Pythagoras’ discovery? We understand thattwo similar strings with lengths in the ratio of $2$ to$3$ will havefundamental frequencies in the ratio $3$ to$2$. But why should they“sound pleasant” together? Perhaps we should take our clue from thefrequencies of the harmonics. The second harmonic of the lower shorterstring will have the same frequency as the third harmonic of thelonger string. (It is easy to show—or to believe—that a pluckedstring produces strongly the several lowest harmonics.)

Perhaps we should make the following rules. Notes sound consonant whenthey have harmonics with the same frequency. Notes sound dissonant iftheir upper harmonics have frequencies near to each other but farenough apart that there are rapid beats between the two. Why beats donot sound pleasant, and why unison of the upper harmonics does soundpleasant, is something that we do not know how to define ordescribe. We cannot say from this knowledge of what soundsgood, what ought, for example, to smell good. In other words,our understanding of it is not anything more general than thestatement that when they are in unison they sound good. It does notpermit us to deduce anything more than the properties of concordancein music.

It is easy to check on the harmonic relationships we have described bysome simple experiments with a piano. Let us label the $3$successiveC’s near the middle of the keyboard by C,C$'$, andC$''$, and the G’sjust above by G,G$'$, andG$''$. Then the fundamentals will haverelative frequencies as follows:\begin{alignat*}{4}&\text{C}&&–2&&\quad\text{G}&&–\phantom{1}3\\[1ex]&\text{C}'&&–4&&\quad\text{G}'&&–\phantom{1}6\\[1ex]&\text{C}''&&–8&&\quad\text{G}''&&–12\end{alignat*}These harmonic relationships can be demonstrated in the following way:Suppose we press C$'$ slowly—so that it does not sound but wecause the damper to be lifted. If we then sound C, it will produce itsown fundamental and some second harmonic. The second harmonicwill set the strings of C$'$ into vibration. If we now release C(keeping C$'$ pressed) the damper will stop the vibration of the Cstrings, and we can hear (softly) the note C$'$ as it dies away. In asimilar way, the third harmonic of C can cause a vibration ofG$'$. Orthe sixth ofC (now getting much weaker) can set up a vibration in thefundamental ofG$''$.

A somewhat different result is obtained if we press G quietly and thensound C$'$. The third harmonic ofC$'$ will correspond to the fourthharmonic ofG, so only the fourth harmonic ofG will beexcited. We can hear (if we listen closely) the sound ofG$''$, whichis two octaves above the G we have pressed! It is easy to think upmany more combinations for this game.

We may remark in passing that the major scale can be defined just bythe condition that the three major chords (F–A–C); (C–E–G); and(G–B–D) each represent tone sequences with the frequencyratio$(4:5:6)$. These ratios—plus the fact that an octave (C–C$'$,B–B$'$, etc.) has the ratio$1:2$—determine the whole scale forthe “ideal” case, or for what is called “just intonation.”Keyboard instruments like the piano are not usually tuned inthis manner, but a little “fudging” is done so that the frequenciesare approximately correct for all possible starting tones. Forthis tuning, which is called “tempered,” the octave (still $1:2$) isdivided into $12$equal intervals for which the frequency ratiois$(2)^{1/12}$. A fifth no longer has the frequency ratio$3/2$, but$2^{7/12} = 1.499$, which is apparently close enough for most ears.

We have stated a rule for consonance in terms of the coincidence ofharmonics. Is this coincidence perhaps the reason that twonotes are consonant? One worker has claimed that two puretones—tones carefully manufactured to be free of harmonics—do notgive the sensations of consonance or dissonance as the relativefrequencies are placed at or near the expected ratios. (Suchexperiments are difficult because it is difficult to manufacture puretones, for reasons that we shall see later.) We cannot still becertain whether the ear is matching harmonics or doing arithmetic whenwe decide that we like a sound.

50–4The Fourier coefficients

Let us return now to the idea that any note—that is, aperiodic sound—can be represented by a suitable combinationof harmonics. We would like to show how we can find out what amount ofeach harmonic is required. It is, of course, easy to compute$f(t)$,using Eq.(50.2), if we are given all thecoefficients $a$ and$b$. The question now is, if we are given$f(t)$how can we know what the coefficients of the various harmonic termsshould be? (It is easy to make a cake from a recipe; but can we writedown the recipe if we are given a cake?)

Fourier discovered that it wasnot really very difficult. The term$a_0$ is certainly easy. We havealready said that it is just the average value of$f(t)$ over one period(from $t = 0$ to$t = T$). We can easily see that this is indeed so. Theaverage value of a sine or cosine function over one period is zero. Overtwo, or three, or any whole number of periods, it is also zero. So theaverage value of all of the terms on the right-hand side ofEq.(50.2) is zero, except for$a_0$. (Recall that we mustchoose$\omega = 2\pi/T$.)

Now the average of a sum is the sum of the averages. So the averageof$f(t)$ is just the average of$a_0$. But $a_0$ is a constant, soits average is just the same as its value. Recalling the definition ofan average, we have\begin{equation}\label{Eq:I:50:3}a_0 = \frac{1}{T}\int_0^Tf(t)\,dt.\end{equation}

The other coefficients are only a little more difficult. To find them wecan use a trick discovered byFourier.Suppose we multiply both sides of Eq.(50.2)by some harmonic function—say by$\cos7\omega t$. We have then\begin{alignat}{2}f(t)\cdot\cos7\omega t &= a_0\cdot\cos7\omega t\notag\\[.5ex]&\quad+\;a_1\cos\hphantom{1}\omega t\cdot\cos7\omega t &&\;+\;b_1\sin\hphantom{1}\omega t\cdot\cos7\omega t\notag\\[.65ex]&\quad+\;a_2\cos2\omega t\cdot\cos7\omega t &&\;+\;b_2\sin2\omega t\cdot\cos7\omega t\notag\\[.65ex]&\quad+\;\dotsb &&\;+\; \dotsb\notag\\[.65ex]&\quad+\;a_7\cos7\omega t\cdot\cos7\omega t &&\;+\;b_7\sin7\omega t\cdot\cos7\omega t\notag\\[.5ex]\label{Eq:I:50:4}&\quad+\;\dotsb &&\;+\; \dotsb\end{alignat}\begin{alignat}{2}f(t)\cdot\cos7\omega t &= a_0\cdot\cos7\omega t\notag\\[.75ex]&\quad+\;a_1\cos\omega t \cdot\cos7\omega t\notag\\&\qquad\qquad+\;b_1\sin\omega t\cdot\cos7\omega t&\notag\\[.75ex]&\quad+\;a_2\cos2\omega t\cdot\cos7\omega t \notag\\&\qquad\qquad+\;b_2\sin2\omega t\cdot\cos7\omega t\notag\\[1ex]&\quad+\quad\dotsb\notag\\[2ex]&\quad+\;a_7\cos7\omega t\cdot\cos7\omega t\notag\\&\qquad\qquad+\;b_7\sin7\omega t\cdot\cos7\omega t\notag\\[.75ex]\label{Eq:I:50:4}&\quad+\quad\dotsb\end{alignat}Now let us average both sides. The average of$a_0\cos7\omegat$ over the time$T$ is proportional to the average of a cosine over$7$whole periods. But that is just zero. The average of almostall of the rest of the terms is also zero. Let us look at the$a_1$term. We know, in general, that\begin{equation}\label{Eq:I:50:5}\cos A\cos B = \tfrac{1}{2}\cos\,(A + B) + \tfrac{1}{2}\cos\,(A - B).\end{equation}The $a_1$ term becomes\begin{equation}\label{Eq:I:50:6}\tfrac{1}{2}a_1(\cos8\omega t + \cos6\omega t).\end{equation}We thus have two cosine terms, one with $8$full periods in$T$ andthe other with$6$. They both average to zero. The average ofthe $a_1$term is therefore zero.

For the $a_2$term, we would find $a_2\cos9\omega t$and$a_2\cos5\omega t$, each of which also averages to zero. For the$a_9$term, we would find $\cos16\omega t$ and$\cos\,(-2\omega t)$. But$\cos\,(-2\omega t)$ is the same as$\cos2\omega t$, so both of thesehave zero averages. It is clear that all of the $a$terms willhave a zero average except one. And that one is the$a_7$term. For this one we have\begin{equation}\label{Eq:I:50:7}\tfrac{1}{2}a_7(\cos14\omega t + \cos0).\end{equation}The cosine of zero is one, and its average, of course, is one. So wehave the result that the average of all of the $a$terms ofEq.(50.4) equals$\tfrac{1}{2}a_7$.

The $b$terms are even easier. When we multiply by any cosine termlike$\cos n\omega t$, we can show by the same method that allof the $b$terms have the average value zero.

We see that Fourier’s“trick” has acted like a sieve. When we multiply by$\cos7\omega t$and average, all terms drop out except$a_7$, and we find that\begin{equation}\label{Eq:I:50:8}\operatorname{Average}\,[f(t)\cdot\cos7\omega t]=a_7/2,\end{equation}or\begin{equation}\label{Eq:I:50:9}a_7 = \frac{2}{T}\int_0^Tf(t)\cdot\cos7\omega t\,dt.\end{equation}

We shall leave it for the reader to show that the coefficient$b_7$can be obtained by multiplying Eq.(50.2) by$\sin7\omegat$ and averaging both sides. The result is\begin{equation}\label{Eq:I:50:10}b_7 = \frac{2}{T}\int_0^Tf(t)\cdot\sin7\omega t\,dt.\end{equation}

Now what is true for$7$ we expect is true for any integer. So we cansummarize our proof and result in the following more elegantmathematical form. If $m$ and$n$ are integers other than zero, and if$\omega = 2\pi/T$, then\begin{align}\label{Eq:I:50:11}&\text{I.}\quad\int_0^T\sin n\omega t\cos m\omega t\,dt = 0.\\[1ex]% ebook break&\left.\hspace{-2mm}\begin{alignedat}{3}&\text{II.}\quad\int_0^T\cos n\omega t \cos m\omega t\,dt ={}\\[1ex]&\text{III.}\quad\int_0^T\sin n\omega t \sin m\omega t\,dt ={}\end{alignedat}\label{Eq:I:50:12}\right\}\;\begin{cases}0 & \kern{-1ex}\text{if $n \neq m$}.\\[1ex]T/2 & \kern{-1ex}\text{if $n = m$}.\end{cases}\\[1ex]% ebook break\label{Eq:I:50:13}&\text{IV.}\quadf(t) = a_0 + \sum_{n = 1}^\infty a_n\cos n\omega t +\sum_{n = 1}^\infty b_n\sin n\omega t.\\[1ex]\label{Eq:I:50:14}&\text{V.}\quada_0 = \frac{1}{T}\int_0^Tf(t)\,dt.\\[1.5ex]\label{Eq:I:50:15}&\text{}\qquada_n = \frac{2}{T}\int_0^Tf(t)\cdot\cos n\omega t\,dt.\\[1.5ex]\label{Eq:I:50:16}&\text{}\qquadb_n = \frac{2}{T}\int_0^Tf(t)\cdot\sin n\omega t\,dt.\end{align}\begin{align}\label{Eq:I:50:11}&\text{I.}\;\int_0^T\!\sin n\omega t\cos m\omega t\,dt = 0.\\[1ex]% ebook break&\left.\hspace{-2mm}\begin{alignedat}{3}&\text{II.}\;\int_0^T\!\cos n\omega t \cos m\omega t\,dt ={}\\[1ex]&\text{III.}\;\int_0^T\!\sin n\omega t \sin m\omega t\,dt ={}\end{alignedat}\right\}\notag\\\label{Eq:I:50:12}&\kern{6em}\begin{cases}0 & \kern{-1ex}\text{if $n \neq m$}.\\[1ex]T/2 & \kern{-1ex}\text{if $n = m$}.\end{cases}\\[1ex]% ebook break&\text{IV.}\;f(t) = a_0 + \sum_{n = 1}^\infty a_n\cos n\omega t\notag\\\label{Eq:I:50:13}&\kern{5.5em}+\sum_{n = 1}^\infty b_n\sin n\omega t.\\[1ex]\label{Eq:I:50:14}&\text{V.}\;a_0 = \frac{1}{T}\int_0^T\!f(t)\,dt.\\[1.5ex]\label{Eq:I:50:15}&\text{}\quada_n = \frac{2}{T}\int_0^T\!f(t)\cdot\cos n\omega t\,dt.\\[1.5ex]\label{Eq:I:50:16}&\text{}\quadb_n = \frac{2}{T}\int_0^T\!f(t)\cdot\sin n\omega t\,dt.\end{align}

In earlier chapters it was convenient to use the exponential notationfor representing simple harmonic motion. Instead of$\cos\omega t$ weused$\FLPRe e^{i\omega t}$, the real part of the exponentialfunction. We have used cosine and sine functions in this chapterbecause it made the derivations perhaps a little clearer. Our finalresult of Eq.(50.13) can, however, be written in thecompact form\begin{equation}\label{Eq:I:50:17}f(t) = \FLPRe\sum_{n = 0}^\infty\hat{a}_ne^{in\omega t},\end{equation}where $\hat{a}_n$ is the complex number$a_n - ib_n$ (with $b_0 =0$). If we wish to use the same notation throughout, we can write also\begin{equation}\label{Eq:I:50:18}\hat{a}_n = \frac{2}{T}\int_0^Tf(t)e^{-in\omega t}\,dt\quad(n \geq 1).\end{equation}

We now know how to “analyze” a periodic wave into its harmoniccomponents. The procedure is called Fourieranalysis, and the separate terms are calledFourier components. We have notshown, however, that once we find all of the Fouriercomponents and add them together, we doindeed get back our $f(t)$. The mathematicians have shown, for a wideclass of functions, in fact for all that are of interest tophysicists, that if we can do the integrals we will get back$f(t)$. There is one minor exception. If the function $f(t)$ isdiscontinuous, i.e., if it jumps suddenly from one value to another,the Fourier sum will give a value at thebreakpoint halfway between the upper and lower values at thediscontinuity. So if we have the strange function $f(t) = 0$, $0 \leqt < t_0$, and $f(t) = 1$ for $t_0 \leq t \leq T$, the Fouriersum will give the right value everywhereexcept at $t_0$, where it will have the value $\tfrac{1}{2}$instead of $1$. It is rather unphysical anyway to insist that afunction should be zero up to $t_0$, but $1$ right at$t_0$. So perhaps we should make the “rule” for physicists that anydiscontinuous function (which can only be a simplification of areal physical function) should be defined with halfway valuesat the discontinuities. Then any such function—with any finitenumber of such jumps—as well as all other physically interestingfunctions, are given correctly by the Fourier sum.

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (7)

Fig. 50–3.Square-wave function.

As an exercise, we suggest that the reader determine the Fourierseries for the function shown inFig.50–3. Since the function cannot be written in anexplicit algebraic form, you will not be able to do the integrals fromzero to$T$ in the usual way. The integrals are easy, however, if weseparate them into two parts: the integral from zero to$T/2$ (overwhich $f(t) = 1$) and the integral from $T/2$ to$T$ (over which $f(t) =-1$). The result should be\begin{equation}\label{Eq:I:50:19}f(t) = \frac{4}{\pi}(\sin\omega t + \tfrac{1}{3}\sin3\omega t +\tfrac{1}{5}\sin5\omega t + \dotsb),\end{equation}where $\omega = 2\pi/T$. We thus find that our square wave (with theparticular phase chosen) has only odd harmonics, and their amplitudesare in inverse proportion to their frequencies.

Let us check that Eq.(50.19) does indeed give us back$f(t)$ for some value of$t$. Let us choose $t = T/4$, or $\omega t =\pi/2$. We have\begin{align}\label{Eq:I:50:20}f(t) &= \frac{4}{\pi}\biggl(\sin\frac{\pi}{2} +\frac{1}{3}\sin\frac{3\pi}{2} + \frac{1}{5}\sin\frac{5\pi}{2} +\dotsb\biggr)\\[1.5ex]\label{Eq:I:50:21}&= \frac{4}{\pi}\biggl(1 - \frac{1}{3} + \frac{1}{5} -\frac{1}{7}\pm\dotsb\biggr).\end{align}The series1 has the value$\pi/4$, and we find that$f(t) = 1$.

50–5The energy theorem

The energy in a wave is proportional to the square of itsamplitude. For a wave of complex shape, the energy in one period willbe proportional to$\int_0^Tf^2(t)\,dt$. We can also relate thisenergy to the Fourier coefficients. Wewrite\begin{equation}\label{Eq:I:50:22}\int_0^Tf^2(t)\,dt =\int_0^T\biggl[a_0 + \sum_{n = 1}^\inftya_n\cos n\omega t +\sum_{n = 1}^\inftyb_n\sin n\omega t\biggr]^2\,dt.\end{equation}\begin{gather}\label{Eq:I:50:22}\int_0^Tf^2(t)\,dt =\\\int_0^T\biggl[a_0 + \sum_{n = 1}^\inftya_n\cos n\omega t +\sum_{n = 1}^\inftyb_n\sin n\omega t\biggr]^2\,dt.\notag\end{gather}When we expand the square of the bracketed term we will get all possible crossterms, such as$a_5\cos5\omega t\cdot a_7\cos7\omega t$ and $a_5\cos5\omega t\cdot b_7\sin7\omega t$. We have shown above,however, [Eqs. (50.11) and(50.12)] that theintegrals of all such terms over one period is zero. We have left only thesquare terms like$a_5^2\cos^2 5\omega t$. The integral of any cosine squared orsine squared over one period is equal to$T/2$, so we get\begin{align}\int_0^Tf^2(t)\,dt &= Ta_0^2 + \frac{T}{2}\,(a_1^2 + a_2^2 + \dotsb + b_1^2 + b_2^2 + \dotsb)\notag\\[.5ex]\label{Eq:I:50:23}&= Ta_0^2 + \frac{T}{2}\sum_{n = 1}^\infty(a_n^2 + b_n^2).\end{align}\begin{gather}\label{Eq:I:50:23}\int_0^Tf^2(t)\,dt =\\[.5ex]Ta_0^2 + \frac{T}{2}\,(a_1^2 + a_2^2 + \dotsb + b_1^2 + b_2^2 + \dotsb)= \notag\\[.5ex]Ta_0^2 + \frac{T}{2}\sum_{n = 1}^\infty(a_n^2 + b_n^2).\notag\end{gather}This equation is called the “energy theorem,” and says that the total energyin a wave is just the sum of the energies in all of the Fourier components. Forexample, applying this theorem to the series(50.19), since $[f(t)]^2= 1$ we get\begin{equation*}T = \frac{T}{2}\cdot\biggl(\frac{4}{\pi}\biggr)^2\biggl(1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dotsb\biggr),\end{equation*}so we learn that the sum of the squares of the reciprocals of the odd integersis$\pi^2/8$. In a similar way, by first obtaining the Fourier series for thefunction$f(t)=(t-T/2)^2$ and using the energy theorem, we can prove that $1 +1/2^4 + 1/3^4 + \dotsb$ is$\pi^4/90$, a result we needed inChapter45.

50–6Nonlinear responses

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (8)The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (9)

Fig. 50–4.Linear and nonlinear responses.

Finally, in the theory of harmonics there is an important phenomenonwhich should be remarked upon because of its practicalimportance—that of nonlinear effects. In all the systems that wehave been considering so far, we have supposed that everything waslinear, that the responses to forces, say the displacements or theaccelerations, were always proportional to the forces. Or that thecurrents in the circuits were proportional to the voltages, and soon. We now wish to consider cases where there is not a strictproportionality. We think, at the moment, of some device in which theresponse, which we will call$x_{\text{out}}$ at the time$t$, isdetermined by the input$x_{\text{in}}$ at the time$t$. For example,$x_{\text{in}}$ might be the force and$x_{\text{out}}$ might be thedisplacement. Or $x_{\text{in}}$ might be the currentand$x_{\text{out}}$ the voltage. If the device is linear, we would have\begin{equation}\label{Eq:I:50:24}x_{\text{out}}(t) = Kx_{\text{in}}(t),\end{equation}where $K$ is a constant independent of$t$ andof$x_{\text{in}}$. Suppose, however, that the device is nearly, but notexactly, linear, so that we can write\begin{equation}\label{Eq:I:50:25}x_{\text{out}}(t) = K[x_{\text{in}}(t) + \epsilon x_{\text{in}}^2(t)],\end{equation}where $\epsilon$ is small in comparison with unity. Such linear andnonlinear responses are shown in the graphs of Fig.50–4.

The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (10)

Fig. 50–5.The response of a nonlinear device to the input$\cos\omegat$. A linear response is shown for comparison.

Nonlinear responses have several important practical consequences. Weshall discuss some of them now. First we consider what happens if weapply a pure tone at the input. We let$x_{\text{in}} = \cos\omegat$. If we plot$x_{\text{out}}$ as a function of time we get the solidcurve shown in Fig.50–5. The dashed curve gives, forcomparison, the response of a linear system. We see that the output isno longer a cosine function. It is more peaked at the top and flatterat the bottom. We say that the output is distorted. We know,however, that such a wave is no longer a pure tone, that it will haveharmonics. We can find what the harmonics are. Using$x_{\text{in}} =\cos\omega t$ with Eq.(50.25), we have\begin{equation}\label{Eq:I:50:26}x_{\text{out}}(t) = K(\cos\omega t + \epsilon\cos^2\omega t).\end{equation}From the equality$\cos^2\theta = \tfrac{1}{2}(1 + \cos2\theta)$, wehave\begin{equation}\label{Eq:I:50:27}x_{\text{out}}(t) = K\Bigl(\cos\omega t + \frac{\epsilon}{2} +\frac{\epsilon}{2}\cos2\omega t\Bigr).\end{equation}The output has not only a component at the fundamental frequency, thatwas present at the input, but also has some of its secondharmonic. There has also appeared at the output a constantterm$K(\epsilon/2)$, which corresponds to the shift of the average value,shown in Fig.50–5. The process of producing a shift ofthe average value is called rectification.

A nonlinear response will rectify and will produce harmonics of thefrequencies at its input. Although the nonlinearity we assumedproduced only second harmonics, nonlinearities of higher order—thosewhich have terms like $x_{\text{in}}^3$ and$x_{\text{in}}^4$, forexample—will produce harmonics higher than the second.

Another effect which results from a nonlinear response ismodulation. If our input function contains two (or more) puretones, the output will have not only their harmonics, but still otherfrequency components. Let$x_{\text{in}} = A\cos\omega_1t +B\cos\omega_2t$, where now $\omega_1$ and$\omega_2$ are notintended to be in a harmonic relation. In addition to the linear term(which is $K$times the input) we shall have a component in the outputgiven by\begin{align}\label{Eq:I:50:28}x_{\text{out}} &= K\epsilon(A\cos\omega_1t + B\cos\omega_2t)^2\\[.5ex]\label{Eq:I:50:29}&= K\epsilon(A^2\cos^2\omega_1t + B^2\cos^2\omega_2t +2AB\cos\omega_1t\cos\omega_2t).\end{align}\begin{align}\label{Eq:I:50:28}x_{\text{out}} &= K\epsilon(A\cos\omega_1t + B\,\cos\omega_2t)^2&\\[1ex]&= K\epsilon(A^2\cos^2\omega_1t + B^2\cos^2\omega_2t \notag\\[.5ex]\label{Eq:I:50:29}&\phantom{=K\epsilon}+\,2AB\cos\omega_1t\cos\omega_2t).\end{align}The first two terms in the parentheses of Eq.(50.29) arejust those which gave the constant terms and second harmonic terms wefound above. The last term is new.

We can look at this new “cross term”$AB\cos\omega_1t\cos\omega_2t$in two ways. First, if the two frequencies are widely different (forexample, if $\omega_1$ is much greater than$\omega_2$) we canconsider that the cross term represents a cosine oscillation ofvarying amplitude. That is, we can think of the factors in this way:\begin{equation}\label{Eq:I:50:30}AB\cos\omega_1t\cos\omega_2t = C(t)\cos\omega_1t,\end{equation}with\begin{equation}\label{Eq:I:50:31}C(t)=AB\cos\omega_2t.\end{equation}We say that the amplitude of$\cos\omega_1t$ is modulated with thefrequency$\omega_2$.

Alternatively, we can write the cross term in another way:\begin{equation}\label{Eq:I:50:32}AB\cos\omega_1t\cos\omega_2t =\frac{AB}{2}\,[\cos\,(\omega_1 + \omega_2)t +\cos\,(\omega_1 - \omega_2)t].\end{equation}\begin{gather}\label{Eq:I:50:32}AB\cos\omega_1t\cos\omega_2t =\\\frac{AB}{2}\,[\cos\,(\omega_1 + \omega_2)t +\cos\,(\omega_1 - \omega_2)t].\notag\end{gather}We would now say that two new components have been produced,one at the sum frequency$(\omega_1 + \omega_2)$, another atthe difference frequency$(\omega_1 - \omega_2)$.

We have two different, but equivalent, ways of looking at the sameresult. In the special case that $\omega_1 \gg \omega_2$, we canrelate these two different views by remarking that since $(\omega_1 +\omega_2)$ and$(\omega_1 - \omega_2)$ are near to each other we wouldexpect to observe beats between them. But these beats have just theeffect of modulating the amplitude of the averagefrequency$\omega_1$ by one-half the differencefrequency$2\omega_2$. We see, then, why the two descriptions are equivalent.

In summary, we have found that a nonlinear response produces severaleffects: rectification, generation of harmonics, and modulation, orthe generation of components with sum and difference frequencies.

We should notice that all these effects (Eq.50.29) areproportional not only to the nonlinearity coefficient$\epsilon$, butalso to the product of two amplitudes—either $A^2$,$B^2$,or$AB$. We expect these effects to be much more important forstrong signals than for weak ones.

The effects we have been describing have many practicalapplications. First, with regard to sound, it is believed that the earis nonlinear. This is believed to account for the fact that with loudsounds we have the sensation that we hear harmonics and alsosum and difference frequencies even if the sound waves contain onlypure tones.

The components which are used in sound-reproducingequipment—amplifiers, loudspeakers, etc.—always have somenonlinearity. They produce distortions in the sound—they generateharmonics, etc.—which were not present in the original sound. Thesenew components are heard by the ear and are apparentlyobjectionable. It is for this reason that “Hi-Fi” equipment isdesigned to be as linear as possible. (Why the nonlinearities of theear are not “objectionable” in the same way, or how weeven know that the nonlinearity is in the loudspeaker ratherthan in the ear is not clear!)

Nonlinearities are quite necessary, and are, in fact,intentionally made large in certain parts of radio transmitting andreceiving equipment. In an am transmitter the “voice”signal (with frequencies of some kilocycles per second) is combinedwith the “carrier” signal (with a frequency of some megacycles persecond) in a nonlinear circuit called a modulator, to producethe modulated oscillation that is transmitted. In the receiver, thecomponents of the received signal are fed to a nonlinear circuit whichcombines the sum and difference frequencies of the modulated carrierto generate again the voice signal.

When we discussed the transmission of light, we assumed that the inducedoscillations of charges were proportional to the electric field of thelight—that the response was linear. That is indeed a very goodapproximation. It is only within the last few years that light sourceshave been devised (lasers) which produce an intensity oflight strong enough so that nonlinear effects can be observed. It is nowpossible to generate harmonics of light frequencies. When a strong redlight passes through a piece of glass, a little bit of bluelight—second harmonic—comes out!

  1. The series can be evaluated in the followingway. First we remark that $\int_0^x[dx/(1 + x^2)] = \tan^{-1}x$. Second, we expand the integrand in a series$1/(1 + x^2) = 1 - x^2+ x^4 - x^6 \pm\dotsb$ We integrate the series term by term (from zeroto$x$) to obtain$\tan^{-1} x = x - x^3/3 + x^5/5 - x^7/7 \pm\dotsb$Setting $x = 1$, we have the stated result, since $\tan^{-1}1 =\pi/4$.
The Feynman Lectures on Physics Vol. I Ch. 50: Harmonics (2024)
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