- State the Dirichlet conditions under which a Fourier series exists.
- Compute the coefficients \(a_{0}\), \(a_{n}\) and \(b_{n}\) of the trigonometric Fourier series.
- Exploit even, odd and half-wave symmetry to eliminate coefficients before integrating.
- Draw the amplitude and phase spectra of a periodic signal.
- Convert between the trigonometric and exponential forms.
- Analyse a circuit driven by a non-sinusoidal periodic source using harmonic superposition.
- Compute the RMS value and average power of a non-sinusoidal waveform.
- Define and calculate total harmonic distortion.
- Explain the role of harmonics in power systems and the operation of a spectrum analyser.
Introduction
Chapter 9 opened with three reasons for studying sinusoids, and the second was left as a promise: any periodic signal whatever can be decomposed into a sum of sinusoids. This chapter redeems it.
The claim is due to Jean-Baptiste Joseph Fourier, who in 1822 asserted that any periodic function — a square wave, a sawtooth, the jagged output of a rectifier — can be represented as an infinite sum of sines and cosines at integer multiples of its fundamental frequency. The assertion was met with scepticism by contemporaries who could not accept that a discontinuous function might equal a sum of perfectly smooth ones. Fourier was right.
The consequence for circuit analysis is immediate and enormous. A linear circuit obeys superposition, so its response to a sum of sinusoids is the sum of its responses to each. If a periodic input can be written as a sum of sinusoids, then the response to any periodic input follows from ordinary phasor analysis applied once per harmonic — the multi-frequency superposition of Section 10.8, now with infinitely many frequencies but each one handled identically.
Nothing new is needed to use the result. The work of this chapter is in obtaining the decomposition and in learning what it reveals: the spectrum, the distortion, and why harmonics are a persistent nuisance in power systems.
The Trigonometric Fourier Series
Fourier's theorem states that a periodic function \(f(t)\) of period \(T\) can be expressed as
where \(\omega_{0} = 2\pi/T\) is the fundamental angular frequency. The terms have names:
\(a_{0}\) is the DC component — the average value of \(f(t)\) over one period.
The \(n = 1\) term is the fundamental, at the same frequency as \(f(t)\) itself.
Terms with \(n \ge 2\) are the harmonics: the second harmonic at \(2\omega_{0}\), the third at \(3\omega_{0}\), and so on.
Combining each cosine–sine pair into a single cosine, as in Section 8.4, gives the amplitude–phase form:
This form is the more useful for circuit work, since each term is now a phasor waiting to be used.
Not every periodic function has a Fourier series. The sufficient conditions, due to Dirichlet, are that within any one period \(f(t)\) must be:
single-valued everywhere;
possessed of a finite number of finite discontinuities;
possessed of a finite number of maxima and minima;
absolutely integrable, so that \(\int_{t_{0}}^{t_{0}+T}\left|f(t)\right|dt < \infty\).
Every waveform arising in circuit analysis satisfies these comfortably — square waves, sawtooths, rectified sinusoids and pulse trains all qualify. The conditions exclude only pathological constructions, so in practice they need checking rarely.
At a discontinuity the series converges to the mean of the values on either side, not to either one. A square wave stepping from 0 to 1 has a series that equals exactly 0.5 at the jump.
Computing the Coefficients
The coefficients are extracted by exploiting the orthogonality of sinusoids: over a full period, the integral of the product of any two different harmonics is zero.
So multiplying the series by \(\cos m\omega_{0}t\) and integrating over one period annihilates every term except the one we want. This gives the three formulas:
Note the factor. The DC term carries \(1/T\) while the harmonic terms carry \(2/T\). The asymmetry is a frequent source of error, and it arises because \(\int_{0}^{T}\cos^{2}n\omega_{0}t\,dt = T/2\) rather than \(T\).
The integration may be taken over any full period. Choosing limits that make the algebra simplest — often \(-T/2\) to \(T/2\) for a symmetric waveform — is worth a moment's thought before starting.
Problem. Find the Fourier series of a periodic pulse train that equals 1 for \(0 < t < 1\) and 0 for \(1 < t < 2\), with period \(T = 2\).
Solution. The fundamental frequency is \(\omega_{0} = 2\pi/T = \pi\).
The DC term is the average, which by inspection is one half — the waveform is at 1 for half the period:
The cosine terms:
since \(\sin n\pi = 0\) for every integer \(n\). The sine terms:
Now \(\cos n\pi\) is \(-1\) for odd \(n\) and \(+1\) for even, so
Two checks. At \(t = 0.5\), all the sine terms give \(\sin(n\pi/2)\) which alternates \(+1, -1, +1\), and the series becomes \(\tfrac{1}{2}+\tfrac{2}{\pi}\left(1-\tfrac{1}{3}+\tfrac{1}{5}-\cdots\right) = \tfrac{1}{2}+\tfrac{2}{\pi}\cdot\tfrac{\pi}{4} = 1\) \(\checkmark\). At the discontinuity \(t = 0\) every sine vanishes, leaving \(f = 0.5\) — the mean of 0 and 1, exactly as the Dirichlet condition predicts.
Note that only odd harmonics appear. Section 17.4 explains why.
Problem. Find the Fourier series of the sawtooth \(f(t) = t\) for \(0 < t < 1\), repeating with period \(T = 1\).
Solution. Here \(\omega_{0} = 2\pi\).
Both harmonic coefficients require integration by parts. For the cosine terms:
At integer \(n\) we have \(\sin 2n\pi = 0\) and \(\cos 2n\pi = 1\), so the bracket gives \(1/(2n\pi)^{2}\) at both limits and
For the sine terms:
Compare the two examples. The pulse train's harmonics fall off as \(1/n\) but only odd ones appear; the sawtooth's also fall as \(1/n\) but all harmonics are present. The rate of decay reflects the waveform's smoothness — a discontinuous waveform gives \(1/n\), a continuous one with discontinuous slope gives \(1/n^{2}\), and so on. Sharp edges require high harmonics.
Symmetry Considerations
Computing every coefficient by integration is laborious, and much of the labour can be avoided. A waveform's symmetry determines in advance which coefficients vanish.
Even symmetry — \(f(-t) = f(t)\), so the waveform is a mirror image about the vertical axis. A cosine is even.
The series contains only cosines and a DC term. Note the halved range of integration and the doubled factor.
Odd symmetry — \(f(-t) = -f(t)\), so the waveform is antisymmetric about the origin. A sine is odd.
The series contains only sines. An odd function necessarily has zero average, which is why \(a_{0}\) vanishes too.
Half-wave symmetry — \(f\left(t-T/2\right) = -f(t)\), so each half-period is the inverted image of the other.
Only odd harmonics survive. This explains Example 17.1: shifting that pulse train down by its mean of 0.5 gives a waveform with half-wave symmetry, and only odd harmonics appeared.
| Symmetry | Condition | Vanishing coefficients | Series contains |
|---|---|---|---|
| Even | \(f(-t) = f(t)\) | all \(b_{n}\) | DC and cosines |
| Odd | \(f(-t) = -f(t)\) | \(a_{0}\) and all \(a_{n}\) | sines only |
| Half-wave | \(f\left(t-\tfrac{T}{2}\right) = -f(t)\) | \(a_{0}\), and all even \(n\) | odd harmonics only |
Symmetries can combine. A waveform that is both odd and half-wave symmetric has only odd sine terms — three quarters of the coefficients are known to be zero before any integration is attempted. Always inspect the waveform first.
One practical note: symmetry depends on where the time origin is placed. Shifting a waveform can create or destroy even or odd symmetry, and choosing the origin to make the waveform symmetric is a legitimate and worthwhile simplification.
Problem. Find the Fourier series of the square wave that equals \(+1\) for \(0 < t < 1\) and \(-1\) for \(1 < t < 2\), with \(T = 2\).
Solution. Inspect first. The waveform is odd about the origin, so \(a_{0} = 0\) and every \(a_{n} = 0\). It also has half-wave symmetry, so only odd harmonics survive. Only odd-\(n\) sine terms need computing — one integral instead of three.
The fundamental has amplitude \(4/\pi = 1.273\) — larger than the square wave's own amplitude of 1. This is not an error: the higher harmonics subtract near the peak to flatten the top, and the fundamental must overshoot for the sum to come out right.
The Gibbs phenomenon. Truncating the series after a finite number of terms produces an overshoot of about 9 % just either side of each discontinuity. Adding more terms narrows the overshoot but never reduces its height — it converges to 8.95 % no matter how many terms are taken. This is why a square wave passed through a band-limited channel shows ringing at its edges, and why an oscilloscope's step response has a small overshoot that no amount of bandwidth removes entirely.
Amplitude and Phase Spectra
The Fourier coefficients are naturally displayed as a plot against frequency. Since only discrete frequencies \(n\omega_{0}\) are present, the plot consists of isolated lines rather than a continuous curve — hence line spectrum.
The amplitude spectrum plots \(A_{n}\) against \(n\omega_{0}\).
The phase spectrum plots \(\phi_{n}\) against \(n\omega_{0}\).
Together they contain exactly the same information as the time-domain waveform, and either representation determines the other completely.
Reading a spectrum is a skill worth developing. Widely spaced lines mean a low fundamental frequency; slowly decaying amplitudes mean sharp edges in the time domain; missing even harmonics mean half-wave symmetry; a line at zero frequency means a DC offset. An instrument that displays this plot directly — a spectrum analyser — is among the most useful pieces of laboratory equipment there is.
The Exponential Fourier Series
Euler's identity allows the sines and cosines to be replaced by complex exponentials, giving a form that is more compact and often easier to manipulate.
Substituting and collecting terms produces
Three features distinguish this form. The sum runs over all integers, positive and negative, so negative frequencies appear — a mathematical convenience with no physical counterpart, since each negative-frequency term pairs with its positive twin to give a real result. There is a single formula for all coefficients, with no special case for \(n = 0\). And the coefficients are complex, carrying amplitude and phase together.
The connection to the trigonometric form:
so the exponential spectrum is two-sided, with each line of the one-sided amplitude spectrum split into two lines of half the height at \(\pm n\omega_{0}\). Since \(c_{-n} = c_{n}^{*}\), the amplitude spectrum is even and the phase spectrum odd — a property of every real signal.
The exponential form is preferred in signal processing and is the natural stepping stone to the Fourier transform, in which the period is allowed to become infinite and the discrete spectrum becomes continuous.
Problem. Find the exponential Fourier coefficients of the square wave of Example 17.3, and relate them to the trigonometric coefficients already found.
Solution. From Example 17.3, \(a_{n} = 0\) and \(b_{n} = 4/n\pi\) for odd \(n\). Using the conversion:
Checking the conjugate property for \(n = 1\): \(c_{1} = -j2/\pi\) and \(c_{-1} = -j2/(-\pi) = +j2/\pi = c_{1}^{*}\) \(\checkmark\)
The magnitudes are
each exactly half the corresponding \(A_{n}\) from the one-sided spectrum, as expected. The energy is not lost — it is shared between the lines at \(+n\omega_{0}\) and \(-n\omega_{0}\).
That every \(c_{n}\) is purely imaginary reflects the waveform's odd symmetry: an odd real function has purely imaginary Fourier coefficients, just as an even one has purely real ones.
Circuit Applications
Here the machinery pays off. A circuit driven by a periodic non-sinusoidal source is analysed by harmonic superposition, which is the multi-frequency superposition of Section 10.8 applied term by term.
Express the source as a Fourier series: a DC term plus a sum of sinusoids.
Solve for the DC term with \(\omega = 0\), so inductors are short circuits and capacitors open circuits.
Solve for each harmonic separately using phasors, recomputing every impedance at \(n\omega_{0}\).
Add the results in the time domain — never as phasors, since the frequencies differ.
The one point requiring care is step 3: the impedances change with every harmonic. An inductor's \(jn\omega_{0}L\) grows with \(n\), so a circuit containing inductance attenuates higher harmonics more strongly — which is exactly why a choke smooths a rectifier's output.
In practice the series is truncated after a few terms, because the harmonics decay and the circuit usually attenuates them further. Three or four terms often give better than 1 % accuracy.
Problem. A series circuit of \(R = 30~\Omega\) and \(L = 0.4~\mathrm{H}\) is driven by
Find the current \(i(t)\).
Solution. Three terms, three separate problems.
DC term (\(\omega = 0\)). The inductor is a short circuit:
Fundamental (\(\omega = 100~\mathrm{rad/s}\)).
Second harmonic (\(\omega = 200~\mathrm{rad/s}\)). The reactance doubles:
Sum in the time domain.
Observe the filtering. The source's harmonics are in the ratio \(5 : 10 : 4\), but the current's are \(0.1667 : 0.2 : 0.0468\). The second harmonic has been attenuated far more than the fundamental, and the fundamental more than the DC, because the inductive reactance rises with frequency. The output waveform is smoother than the input — the circuit is a low-pass filter, and this is precisely how a smoothing choke works.
Average Power and RMS Values
Chapter 11 defined RMS values and average power for sinusoids. Both extend to non-sinusoidal periodic waveforms, and the extension is remarkably clean.
The RMS value. Applying the definition of Section 11.4 to a Fourier series, all the cross terms integrate to zero by orthogonality, leaving only the squares:
RMS values add in quadrature. Each harmonic contributes its own RMS value squared, and the DC term contributes its square directly. This result is Parseval's theorem, and it says that the total power is the sum of the powers in the individual harmonics.
The average power. Again orthogonality eliminates all cross terms — a voltage harmonic and a current harmonic of different order contribute no average power whatever. Only matching pairs survive:
or, in RMS terms, \(P = V_{0}I_{0}+\sum V_{n,\mathrm{rms}}I_{n,\mathrm{rms}}\cos\theta_{n}\). The DC term carries no factor of one half; the harmonic terms do, because they use amplitudes.
The consequence is worth stating plainly. If a source contains only a fundamental and a load draws only third-harmonic current, no power is transferred at all — however large both may be. Power flows only between harmonics of the same order.
Problem. For the circuit of Example 17.5, find the RMS values of the source voltage and the current, and the average power delivered. Verify the power by two independent routes.
RMS voltage. The DC term contributes its square; each harmonic contributes \(A_{n}^{2}/2\):
RMS current.
Average power, harmonic by harmonic. The angle in each term is the phase difference between that harmonic's voltage and current — which here is just the impedance angle:
Verification. All the power is dissipated in the resistor, so it must also equal \(I_{\mathrm{rms}}^{2}R\):
The two routes agree exactly. Note how unevenly the power is distributed: the DC term supplies 57 % of it and the second harmonic barely 2 %, even though the second-harmonic voltage is nearly as large as the DC. The inductor's rising reactance has suppressed it.
Note also that the inductor consumes no average power at any harmonic — \(\cos 90^{\circ} = 0\) for a pure reactance, exactly as in Chapter 11. Only \(R\) dissipates.
Total Harmonic Distortion
An ideal AC supply is a pure sinusoid. Real supplies are not, and a single number is needed to say how far from ideal a waveform is. That number is the total harmonic distortion:
It is the RMS value of everything except the fundamental, expressed as a fraction of the fundamental. A pure sinusoid has \(\mathrm{THD} = 0\); the larger the harmonic content, the larger the figure.
| Context | Typical or permitted THD |
|---|---|
| Laboratory signal generator | below 0.1 % |
| Quality audio amplifier | below 0.01 % |
| Public electricity supply (voltage) | limited to about 5 % |
| Switched-mode power supply (current) | often above 100 % if uncorrected |
| Square wave | 48.34 % |
Why harmonics matter in power systems. They cause additional \(I^{2}R\) heating without delivering useful power; they increase transformer core losses, which rise faster than linearly with frequency; and the triplen harmonics — the third, ninth and fifteenth — are particularly troublesome in three-phase systems, because they arrive in phase in all three lines and therefore add in the neutral instead of cancelling. A four-wire installation feeding many switched-mode loads can have a neutral current exceeding the line currents, a hazard entirely absent from the balanced analysis of Chapter 12.
The remedies were introduced earlier: a delta winding traps triplen currents in a circulating loop (Section 13.10), and a series LC branch tuned to a troublesome harmonic diverts it before it spreads (Section 14.11).
Problem. Compute the total harmonic distortion of a square wave.
Solution. From Example 17.3 the harmonic amplitudes are
The factor \(4/\pi\) is common to numerator and denominator and cancels, leaving
The sum inside is \(\sum_{n\text{ odd}} 1/n^{2}\) with the \(n=1\) term removed. That full sum is a standard result:
Nearly half the fundamental's RMS value is contained in the harmonics — which is why a square wave sounds so different from a sine of the same frequency, and why it is entirely unsuitable as a power waveform.
A cross-check. The square wave's RMS value is 1 by Chapter 11's table. Parseval's theorem gives \(F_{\mathrm{rms}}^{2} = \tfrac{1}{2}\sum A_{n}^{2} = \tfrac{1}{2}\left(\tfrac{4}{\pi}\right)^{2}\left(\tfrac{\pi^{2}}{8}\right) = 1\) \(\checkmark\) — the harmonic decomposition reproduces the correct RMS value exactly.
Practical Engineering Applications
A full-wave rectifier converts a sinusoid into \(\left|V_{m}\sin\omega t\right|\), whose period is half that of the input. Its Fourier series is worth knowing:
The DC term \(2V_{m}/\pi = 0.6366V_{m}\) is the useful output — the average value met in Section 11.4. Everything else is ripple, dominated by the first term at \(2\omega\) with amplitude \(4V_{m}/3\pi = 0.4244V_{m}\).
Two features of that series guide the filter design. The lowest ripple component is at twice the supply frequency, not at the supply frequency — a full-wave rectifier is much easier to smooth than a half-wave one for exactly this reason. And the amplitudes fall as \(1/n^{2}\), so a filter that adequately suppresses the first ripple component will handle the rest comfortably.
Problem. A full-wave rectifier produces \(\left|100\sin 314t\right|~\mathrm{V}\). Find the DC output and the amplitude of the first three ripple components.
Solution. Using the series above with \(V_{m} = 100~\mathrm{V}\):
The ripple amplitudes are \(4V_{m}/\left[\pi\left(4n^{2}-1\right)\right]\):
| \(n\) | Frequency | Amplitude | Relative to DC |
|---|---|---|---|
| 1 | \(2\omega = 628~\mathrm{rad/s}\) | 42.44 V | 66.7 % |
| 2 | \(4\omega = 1257~\mathrm{rad/s}\) | 8.488 V | 13.3 % |
| 3 | \(6\omega = 1885~\mathrm{rad/s}\) | 3.638 V | 5.71 % |
The unsmoothed output is 63.66 V of DC with 42.44 V of ripple riding on it — quite unusable as a supply. But the ripple falls off rapidly, and a low-pass filter attenuating the 628 rad/s component by a factor of 100 would reduce the dominant ripple to 0.42 V while affecting the DC not at all.
This is the design problem a power-supply filter solves, and Chapter 14's low-pass filter is the tool. Note that the DC term passes untouched because a filter's response at \(\omega = 0\) is its DC gain — the harmonics are separated by frequency, exactly as the Fourier decomposition describes them.
A spectrum analyser displays the amplitude spectrum of a signal directly, and it reveals things an oscilloscope cannot. A distorted amplifier shows harmonics of its test tone; an intermittent connection shows broadband noise; a switching power supply shows a comb of lines at its switching frequency and multiples.
The touch-tone telephone of Section 14.11 is a case in point: two tones are transmitted simultaneously, and on an oscilloscope their sum is an unintelligible beating waveform. On a spectrum analyser it is two clean lines whose frequencies identify the key immediately.
Because harmonics injected by one consumer degrade the supply for everyone, they are regulated. Standards such as IEEE 519 and IEC 61000-3-2 set limits on both the individual harmonic amplitudes and the total distortion that equipment may inject, and large installations are required to demonstrate compliance.
The pressure comes from the proliferation of non-linear loads. A resistive load draws a sinusoidal current from a sinusoidal supply; a rectifier-fed switched-mode supply — which is what nearly every computer, LED lamp and phone charger contains — draws current only in short pulses near the voltage peak, giving a current waveform rich in odd harmonics. The load is linear in neither sense of the word, and the analysis of this chapter is what quantifies the consequence.
Summary and Key Formulas
Any periodic function satisfying the Dirichlet conditions equals a sum of sinusoids at multiples of its fundamental frequency.
The coefficients follow from orthogonality, with \(1/T\) for the DC term and \(2/T\) for the harmonics.
Even symmetry kills all \(b_{n}\); odd symmetry kills \(a_{0}\) and all \(a_{n}\); half-wave symmetry kills all even harmonics.
At a discontinuity the series converges to the mean of the two values, and truncation produces a 9 % Gibbs overshoot that never diminishes.
The amplitude and phase spectra are line spectra containing the same information as the waveform.
The exponential form uses a single coefficient formula and a two-sided spectrum with \(\left|c_{n}\right| = A_{n}/2\).
Circuits are analysed by harmonic superposition: solve at each frequency, recompute every impedance, and add in the time domain.
RMS values add in quadrature, and average power is the sum of harmonic powers — different harmonics transfer no power to each other.
Total harmonic distortion measures the non-fundamental content; a square wave's is 48.34 %.
Triplen harmonics add rather than cancel in a three-phase neutral, which is a real hazard with non-linear loads.
| Result | Formula | Notes |
|---|---|---|
| Trigonometric series | \(f = a_{0}+\sum\left(a_{n}\cos n\omega_{0}t+b_{n}\sin n\omega_{0}t\right)\) | \(\omega_{0} = 2\pi/T\) |
| DC coefficient | \(a_{0} = \dfrac{1}{T}\displaystyle\int_{0}^{T}f\,dt\) | factor \(1/T\) |
| Cosine coefficient | \(a_{n} = \dfrac{2}{T}\displaystyle\int_{0}^{T}f\cos n\omega_{0}t\,dt\) | factor \(2/T\) |
| Sine coefficient | \(b_{n} = \dfrac{2}{T}\displaystyle\int_{0}^{T}f\sin n\omega_{0}t\,dt\) | |
| Amplitude–phase form | \(A_{n} = \sqrt{a_{n}^{2}+b_{n}^{2}}\), \(\phi_{n} = -\tan^{-1}\dfrac{b_{n}}{a_{n}}\) | ready for phasors |
| Even symmetry | \(b_{n} = 0\) | cosines only |
| Odd symmetry | \(a_{0} = a_{n} = 0\) | sines only |
| Half-wave symmetry | even harmonics vanish | odd harmonics only |
| Exponential series | \(f = \displaystyle\sum_{n=-\infty}^{\infty}c_{n}e^{jn\omega_{0}t}\) | two-sided |
| Exponential coefficient | \(c_{n} = \dfrac{1}{T}\displaystyle\int_{0}^{T}f\,e^{-jn\omega_{0}t}dt\) | one formula for all \(n\) |
| Form conversion | \(c_{n} = \dfrac{a_{n}-jb_{n}}{2}\), \(c_{-n} = c_{n}^{*}\) | \(\left|c_{n}\right| = A_{n}/2\) |
| RMS value | \(F_{\mathrm{rms}} = \sqrt{a_{0}^{2}+\tfrac{1}{2}\sum A_{n}^{2}}\) | Parseval; quadrature |
| Average power | \(P = V_{0}I_{0}+\tfrac{1}{2}\sum V_{n}I_{n}\cos\left(\theta_{n}-\phi_{n}\right)\) | same order only |
| Total harmonic distortion | \(\mathrm{THD} = \dfrac{\sqrt{\sum_{n\ge2}A_{n}^{2}}}{A_{1}}\) | square wave: 48.34 % |
| Square wave | \(\dfrac{4}{\pi}\sum_{n\text{ odd}}\dfrac{\sin n\omega_{0}t}{n}\) | amplitude 1 |
| Full-wave rectified | \(\dfrac{2V_{m}}{\pi}-\dfrac{4V_{m}}{\pi}\sum\dfrac{\cos 2n\omega t}{4n^{2}-1}\) | ripple at \(2\omega\) |
| Gibbs overshoot | 8.95 % of the jump | independent of terms |
Common Mistakes
Using \(2/T\) for the DC coefficient. \(a_{0}\) carries \(1/T\); only \(a_{n}\) and \(b_{n}\) carry \(2/T\). This doubles the DC term silently.
Integrating before checking symmetry. A minute spent inspecting the waveform can eliminate two thirds of the work.
Adding harmonic responses as phasors. Different frequencies cannot be combined in the phasor domain — convert each to the time domain first, exactly as in Section 10.8.
Forgetting to recompute impedances at each harmonic. \(jn\omega_{0}L\) changes with \(n\); using the fundamental's impedance throughout is a serious error.
Adding RMS values arithmetically. They add in quadrature: \(\sqrt{\sum F_{n}^{2}}\), not \(\sum F_{n}\).
Computing power between harmonics of different order. Orthogonality means they contribute exactly nothing.
Expecting the fundamental's amplitude to be less than the waveform's. A square wave of unit amplitude has a fundamental of \(4/\pi = 1.273\).
Believing more terms remove the Gibbs overshoot. They narrow it but never lower it below 8.95 %.
Assuming triplen harmonics cancel in a three-phase neutral. They arrive in phase and add — the opposite of the fundamental's behaviour.
Mixing amplitude and RMS forms in the power formula. With amplitudes the harmonic terms carry a factor of one half; with RMS values they do not.
Chapter Review
Work these before opening the answers. Check for symmetry before integrating.
P17.1 A waveform has \(f(-t) = -f(t)\). Which coefficients vanish, and what does the series contain?
Show answer
Odd symmetry: \(a_{0} = 0\) and every \(a_{n} = 0\). The series contains sine terms only, computed from \(b_{n} = \tfrac{4}{T}\int_{0}^{T/2}f\sin n\omega_{0}t\,dt\).P17.2 Find the fundamental frequency and the first three harmonic frequencies of a waveform of period 5 ms.
Show answer
Harmonics at 400 Hz, 600 Hz and 800 Hz.\[f_{0} = \frac{1}{T} = \frac{1}{0.005} = 200~\mathrm{Hz}, \qquad \omega_{0} = 2\pi\left(200\right) = 1257~\mathrm{rad/s}\]P17.3 A signal is \(f(t) = 4+6\cos 10t+3\cos 20t+2\cos 30t\). Find its RMS value.
Show answer
\[F_{\mathrm{rms}} = \sqrt{4^{2}+\tfrac{6^{2}}{2}+\tfrac{3^{2}}{2}+\tfrac{2^{2}}{2}} = \sqrt{16+18+4.5+2} = \sqrt{40.5} = 6.364\]P17.4 Find the THD of the signal in P17.3.
Show answer
The DC term is excluded — THD measures harmonic content relative to the fundamental only.\[\mathrm{THD} = \frac{\sqrt{3^{2}+2^{2}}}{6} = \frac{\sqrt{13}}{6} = \frac{3.606}{6} = 0.6009 = 60.09\,\%\]P17.5 A source \(v_{s} = 12+8\cos 50t~\mathrm{V}\) drives a series circuit of 20 \(\Omega\) and 0.2 H. Find \(i(t)\).
Show answer
DC: \(I_{0} = 12/20 = 0.6~\mathrm{A}\).
Fundamental: \(\mathbf{Z} = 20+j10 = 22.36\angle 26.57^{\circ}~\Omega\), so\[\mathbf{I}_{1} = \frac{8}{22.36}\angle-26.57^{\circ} = 0.3578\angle-26.57^{\circ}~\mathrm{A}\]\[i(t) = 0.6+0.3578\cos\left(50t-26.57^{\circ}\right)~\mathrm{A}\]P17.6 Find the average power delivered in P17.5, by two routes.
Show answer
By harmonics:By \(I_{\mathrm{rms}}^{2}R\): \(I_{\mathrm{rms}}^{2} = 0.36+\tfrac{1}{2}\left(0.3578\right)^{2} = 0.424\), so \(P = \left(0.424\right)\left(20\right) = 8.480~\mathrm{W}\) \(\checkmark\)\[P = \left(12\right)\left(0.6\right)+\tfrac{1}{2}\left(8\right)\left(0.3578\right)\cos 26.57^{\circ} = 7.200+1.280 = 8.480~\mathrm{W}\]P17.7 The exponential coefficients of a signal are \(c_{n} = 3/\left(n^{2}+1\right)\). Find \(A_{2}\), the amplitude of the second harmonic in the trigonometric form.
Show answer
The one-sided amplitude is twice the two-sided coefficient magnitude.\[A_{n} = 2\left|c_{n}\right| \;\Longrightarrow\; A_{2} = 2\left(\frac{3}{5}\right) = 1.200\]P17.8 A full-wave rectifier gives \(\left|50\sin\omega t\right|~\mathrm{V}\). Find the DC output and the dominant ripple amplitude.
Show answer
The ripple is 66.7 % of the DC — the same fraction as in Example 17.8, since it is independent of \(V_{m}\).\[V_{\mathrm{DC}} = \frac{2\left(50\right)}{\pi} = 31.83~\mathrm{V}, \qquad V_{\text{ripple}} = \frac{4\left(50\right)}{3\pi} = 21.22~\mathrm{V}\]P17.9 Why does a waveform with sharper edges require more harmonics?
Show answer
The rate at which coefficients decay reflects smoothness. A waveform with a jump discontinuity has coefficients falling as \(1/n\); one that is continuous but has a discontinuous slope falls as \(1/n^{2}\); each further degree of smoothness adds another power. A sharp edge is a rapid change, and only high-frequency components can change rapidly — so reproducing it needs many harmonics.P17.10 Three single-phase non-linear loads, each drawing a third-harmonic current of 5 A, are connected to the three phases of a four-wire system. Find the third-harmonic neutral current.
Show answer
The fundamentals are 120° apart, so the third harmonics are \(3\times120^{\circ} = 360^{\circ}\) apart — that is, in phase. They therefore add arithmetically:Three times each line's third-harmonic current, in a conductor often sized smaller than the lines. This is why modern four-wire installations feeding electronic loads require an oversized neutral.\[I_{n} = 3\left(5\right) = 15~\mathrm{A}\]
MCQ 1. The coefficient \(a_{0}\) represents:
(a) the fundamental (b) the average value (c) the RMS value (d) the peak valueShow answer
(b) the average value over one period — the DC component.MCQ 2. An odd function has a Fourier series containing:
(a) cosines only (b) sines only (c) both (d) a DC term onlyShow answer
(b) sines only, since \(a_{0}\) and every \(a_{n}\) vanish.MCQ 3. Half-wave symmetry means the series contains:
(a) even harmonics only (b) odd harmonics only (c) all harmonics (d) no harmonicsShow answer
(b) odd harmonics only. All even coefficients vanish, as does \(a_{0}\).MCQ 4. At a jump discontinuity the Fourier series converges to:
(a) the upper value (b) the lower value (c) their mean (d) zeroShow answer
(c) their mean, by the Dirichlet conditions.MCQ 5. The Gibbs overshoot near a discontinuity:
(a) vanishes with more terms (b) stays at about 9 % (c) grows with more terms (d) equals 50 %Show answer
(b). More terms narrow it but never reduce its height below 8.95 %.MCQ 6. The fundamental amplitude of a unit square wave is:
(a) 1 (b) \(2/\pi\) (c) \(4/\pi\) (d) \(\pi/4\)Show answer
(c) \(4/\pi = 1.273\) — larger than the wave itself, since the harmonics subtract near the peak.MCQ 7. RMS values of harmonics combine by:
(a) simple addition (b) quadrature addition (c) multiplication (d) averagingShow answer
(b) quadrature — \(\sqrt{\sum F_{n}^{2}}\), by Parseval's theorem.MCQ 8. A third-harmonic voltage and a fifth-harmonic current together deliver:
(a) full power (b) half power (c) zero average power (d) reactive power onlyShow answer
(c) zero, by orthogonality. Only harmonics of the same order transfer power.MCQ 9. The magnitude of an exponential coefficient \(\left|c_{n}\right|\) equals:
(a) \(A_{n}\) (b) \(A_{n}/2\) (c) \(2A_{n}\) (d) \(A_{n}^{2}\)Show answer
(b) \(A_{n}/2\), since the two-sided spectrum splits each line between \(\pm n\omega_{0}\).MCQ 10. In a three-phase four-wire system, third-harmonic currents in the neutral:
(a) cancel (b) add (c) are zero (d) lag by 120°Show answer
(b) add, because tripling a 120° displacement gives 360° — they arrive in phase.
Fourier's contemporaries doubted that a discontinuous function could equal a sum of smooth ones. Explain what the Gibbs phenomenon reveals about the sense in which the equality holds.
Why must a waveform with sharp edges contain high harmonics? Relate your answer to the bandwidth needed to transmit a digital signal.
The fundamental of a square wave is larger in amplitude than the square wave itself. Explain how this is possible without violating anything.
Harmonics of different order transfer no average power to one another. Explain why, and say what this implies for a load that draws current at frequencies the supply does not provide.
Triplen harmonics add in a three-phase neutral while fundamentals cancel. Explain the mechanism, and say why this problem has grown worse over the past thirty years.
The exponential Fourier series contains negative frequencies, which have no physical meaning. Justify their use, and explain what guarantees the result is nonetheless real.
The Fourier series decomposes a periodic signal into discrete harmonics. But most real signals are not periodic — a speech waveform, a single pulse, a switching transient — and for these the series has nothing to say.
Chapter 18 removes the restriction by a limiting argument of great elegance: let the period grow without bound. As \(T \to \infty\) the fundamental frequency \(\omega_{0} = 2\pi/T\) shrinks towards zero, the spectral lines crowd closer together, and in the limit the discrete line spectrum becomes a continuous one. The sum becomes an integral, and the Fourier series becomes the Fourier transform.
The transform will look familiar: it is the Laplace transform of Chapter 15 evaluated on the \(j\omega\) axis, and it connects directly to the frequency response of Chapter 14. Part 3 closes by showing that all three transforms are aspects of one idea.