Electronics Guide

Period, Fundamental, and Harmonics

A signal f(t) is periodic when f(t + T) = f(t) for every t, and its period T is the smallest positive number for which this holds. Its fundamental frequency is f₀ = 1/T in hertz, or ω₀ = 2πf₀ = 2π/T in radians per second. The nth harmonic has the frequency nf₀, so a 1 kHz waveform has its second harmonic at 2 kHz and its third at 3 kHz. The fundamental is the first harmonic, and the constant term is the DC component, the average value over one period.

A sum of sinusoids is periodic only when every ratio of their frequencies is a rational number. Sinusoids at 50 Hz and 60 Hz together repeat every 0.1 s, and within that period they are the fifth and sixth harmonics of 10 Hz. Sinusoids at 1 Hz and √2 Hz together never repeat. Components of a power-system waveform at frequencies that are not integer multiples of the line frequency are called interharmonics; Harmonics and Power Factor describes how they are measured.

Why Sinusoids

A stable linear time-invariant circuit driven by a sinusoid settles to a sinusoid of the same frequency, changed only in amplitude and phase. No other periodic waveform keeps its shape in general: a square wave comes out of an RC filter with rounded corners. Once a periodic input is written as a sum of sinusoids, phasor analysis, as developed in AC Circuit Analysis, handles each one, and superposition adds the results.

Orthogonality

Harmonics of different orders are orthogonal. Integrated over any interval of length T, and for positive integers m and n,

∫ cos(mω₀t) cos(nω₀t) dt = ∫ sin(mω₀t) sin(nω₀t) dt = 0 when m and n differ, and T/2 when m = n

∫ sin(mω₀t) cos(nω₀t) dt = 0 for all m and n

The integral of each sinusoid alone over a period is also zero. Multiplying a waveform by one harmonic and averaging therefore isolates that harmonic's coefficient, which is how the coefficient formulas below work and how a lock-in amplifier extracts one frequency from a noisy signal.

Amplitude and Phase Form

Each pair of terms combines into a single cosine:

f(t) = A₀ + ∑_n=1^∞ A_n cos(nω₀t + φ_n)

with A₀ = a₀, A_n = √(a_n² + b_n²), a_n = A_n cos φ_n, and b_n = −A_n sin φ_n. The phase φ_n is the angle of the complex number a_n − jb_n. The arctangent of −b_n/a_n gives it directly only when a_n > 0; otherwise add or subtract 180°, a correction that the two-argument arctangent function of most programming languages makes automatically.

A_n is the peak amplitude of the nth harmonic, and its RMS value is A_n/√2. Plotted against frequency, the A_n form the amplitude spectrum and the φ_n form the phase spectrum. Delaying a waveform changes its phases but not its amplitudes, which is why published tables of harmonic amplitudes agree while tables of phases often do not. Some texts expand in sines, A_n sin(nω₀t + θ_n), where θ_n = φ_n + 90°.

One-Sided and Two-Sided Spectra

The exponential form places a line of height |c_n| at both +nf₀ and −nf₀, a two-sided spectrum; the negative frequencies are bookkeeping, since the two exponentials together make one real cosine. A one-sided spectrum folds each pair into a single line of height A_n = 2|c_n|. Before comparing a plot with a table, check whether it shows |c_n|, the peak amplitude A_n, or the RMS value A_n/√2; they differ by factors of 2 and √2, or 6.02 dB and 3.01 dB.

Properties of the Coefficients

In the table, f(t) and g(t) share the period T and have coefficients c_n and d_n.

Differentiation emphasizes high harmonics, and where f(t) jumps, the derivative contains impulses, which the rule includes. Integration needs a zero average, because a nonzero average integrates to a ramp, which is not periodic.

These rules save integration. The even triangle wave with peaks of ±A, equal to +A at t = 0, falls with slope −4A/T for 0 < t < T/2 and rises with slope +4A/T for the next half-period, so its derivative is the square wave of the next section's worked example with a height of B = −4A/T, whose coefficients for odd n are c_n = −j2B/(nπ). Dividing them by jnω₀ gives the triangle wave's coefficients, c_n = 4A/(n²π²) for odd n, so a_n = 8A/(n²π²), without a second integration.

A Procedure

1. Sketch one period, read the period T, and write ω₀ = 2π/T.
2. Choose a time origin that makes the waveform simple, preferably one that gives it even or odd symmetry.
3. Write f(t) piecewise over one period.
4. Find a₀, the area under one period divided by T, often by inspection.
5. Use symmetry to identify the coefficients that vanish, and integrate the rest. Ramps need integration by parts: ∫ t cos(kt) dt = cos(kt)/k² + t sin(kt)/k and ∫ t sin(kt) dt = sin(kt)/k² − t cos(kt)/k.
6. Simplify with sin(nπ) = 0 and cos(nπ) = (−1)ⁿ, which usually splits the result into separate expressions for odd and even n.
7. Check the result: the DC term against the average, a partial sum against the sketch at one or two instants, and the sum of the squared coefficients against the RMS value by Parseval's theorem.

Worked Example: Square Wave

Take f(t) = +A for 0 < t < T/2 and f(t) = −A for T/2 < t < T.

1. The average is zero, so a₀ = 0. The waveform is odd, so every a_n = 0.
2. Integrate over 0 to T: b_n = (2/T)[∫₀^T/2 A sin(nω₀t) dt − ∫_T/2^T A sin(nω₀t) dt].
3. The first integral is A[1 − cos(nπ)]/(nω₀) and the second is A[cos(nπ) − 1]/(nω₀), because nω₀T/2 = nπ.
4. Since ω₀T = 2π, b_n = [2A/(nπ)][1 − (−1)ⁿ], which is 4A/(nπ) for odd n and 0 for even n.

The result is

f(t) = (4A/π)[sin(ω₀t) + (1/3) sin(3ω₀t) + (1/5) sin(5ω₀t) + …]

At t = T/4 every sine is ±1 and the bracket becomes 1 − 1/3 + 1/5 − 1/7 + …, which sums to π/4, so the series gives A, as it should. In exponential form, c_n = −j2A/(nπ) for odd n.

Worked Example: Sawtooth

Take f(t) = 2At/T for −T/2 < t < T/2, a ramp from −A to +A that drops back at the end of each period.

1. The waveform is odd, so a₀ = 0 and every a_n = 0.
2. By odd symmetry, b_n = (4/T) ∫₀^T/2 (2At/T) sin(nω₀t) dt = (8A/T²)[sin(nω₀t)/(nω₀)² − t cos(nω₀t)/(nω₀)], evaluated from 0 to T/2.
3. At t = T/2 the sine is zero and the cosine is (−1)ⁿ, and (T/2)/(nω₀) = T²/(4nπ), so the bracket equals −T²(−1)ⁿ/(4nπ). At t = 0 it is zero.
4. Therefore b_n = 2A(−1)ⁿ⁺¹/(nπ).

f(t) = (2A/π)[sin(ω₀t) − (1/2) sin(2ω₀t) + (1/3) sin(3ω₀t) − …]

Hidden Symmetry and the Time Origin

Even and odd symmetry depend on where t = 0 sits; half-wave symmetry does not, because it compares the waveform with itself half a period later. The odd square wave of the previous section, advanced by T/4, becomes even:

f(t) = (4A/π)[cos(ω₀t) − (1/3) cos(3ω₀t) + (1/5) cos(5ω₀t) − …]

A DC offset hides symmetry in the same way: a square wave between 0 and V is V/2 plus a square wave of ±V/2, so it has a DC term of V/2 and odd harmonics of 2V/(nπ). Subtract the average before looking for symmetry.

Symmetry in Circuits

A sinusoid has half-wave symmetry, and a time-invariant circuit that answers an inverted input with an exactly inverted output preserves it however nonlinear the circuit is, so its steady-state output contains only odd harmonics. Symmetric clipping by back-to-back diodes, a balanced push-pull stage, and the magnetizing current of a core with a symmetric B-H loop all distort this way. Any asymmetry, such as half-wave rectification, one-sided clipping, a mismatched push-pull pair, or DC bias in a core, adds even harmonics and usually a DC shift, so a second harmonic in a supposedly symmetric system points to an imbalance. In a four-wire three-phase system with balanced nonlinear loads, harmonics whose order is a multiple of three add in the neutral instead of canceling, as Three-Phase Circuits and Power explains.

Square and Triangle Waves

Both waves have half-wave symmetry, so both contain only odd harmonics. The square wave's harmonics fall as 1/n, which puts the third and fifth 9.54 dB and 13.98 dB below the fundamental; the triangle wave's fall as 1/n², which puts them 19.08 dB and 27.96 dB below. The fundamental carries 81.1 percent of a square wave's mean-square value but 98.6 percent of a triangle wave's. The square wave's fundamental, 4A/π ≈ 1.273A, exceeds the wave's own peak, so a filter that keeps only the fundamental turns a ±5 V square wave into a sine wave of 6.37 V peak.

Sawtooth Waves

A sawtooth has no half-wave symmetry, so it contains every harmonic, with amplitudes falling as 1/n and, in the odd form of the table, alternating in sign. The ramp from 0 to A equals a DC term of A/2 plus a sawtooth of ±A/2 delayed by half a period, so its harmonics have amplitudes of A/(nπ). The ramp carriers of pulse-width modulators and the sweeps of timebase circuits have this spectrum.

Pulse Trains and Duty Cycle

For the pulse train in the table, the exponential coefficients are c_n = AD sinc(nD), where sinc(x) = sin(πx)/(πx); some texts define sinc without the factors of π. The lines are spaced f₀ = 1/T apart, but the shape of their envelope depends only on the pulse width: it is 3 dB down near 0.443/τ and zero at every multiple of 1/τ, so a narrower pulse spreads its energy over more harmonics. At D = 0.25, every fourth harmonic vanishes. Duty cycles of D and 1 − D give the same harmonic amplitudes, because a pulse train and its complement differ only in sign and DC level.

Real pulses have finite edges. For a trapezoidal pulse train with equal rise and fall times t_r, each measured from 0 to 100 percent, and a width τ measured at half height, c_n = AD sinc(nD) sinc(nt_r/T). The envelope of the one-sided amplitudes is flat up to 1/(πτ), falls at 20 dB per decade up to 1/(πt_r), and falls at 40 dB per decade beyond. Slower edges therefore cut the upper spectrum, which is the basis of the edge-rate control described in Radiated Emission Sources.

Rectified Sine Waves

The half-wave rectified sine has a DC value of 0.318A, a component of 0.5A at the source frequency, and even harmonics of 0.212A, 0.042A, and 0.018A at two, four, and six times the source frequency, but no odd harmonics above the fundamental. The full-wave rectified sine repeats twice per source cycle, so the table indexes its lines to the source frequency: 0.637A of DC, then 0.424A, 0.085A, and 0.036A at two, four, and six times the source frequency. Both spectra fall as 1/n², because both waveforms are continuous with abrupt changes of slope. The ripple factor, the ratio of the RMS ripple to the DC value, is 1.21 for half-wave and 0.483 for full-wave rectification.

For filter design, a full-wave rectifier on a 60 Hz line leaves its largest ripple component at 120 Hz, while a half-wave rectifier leaves a larger one at 60 Hz. These spectra assume ideal diodes and a resistive load with no smoothing capacitor. With a capacitor, the diodes conduct in short pulses, and the line current becomes a pulse train with a different spectrum.

RMS Value from Harmonics

With F₀ = a₀ for the DC value and F_n = A_n/√2 for the RMS value of the nth harmonic, the theorem becomes

F_rms = √(F₀² + F₁² + F₂² + F₃² + …)

The RMS values of components at different frequencies combine as the square root of the sum of their squares, never arithmetically, whatever their phases. Phase does change the peak value, so two waveforms with the same harmonic amplitudes share an RMS value but can have different crest factors. A current with a fundamental of 10 A RMS, a fifth harmonic of 2 A, and a seventh harmonic of 1.4 A has an RMS value of √(100 + 4 + 1.96) A = 10.29 A.

The square wave checks the theorem. Its mean-square value is A², and the series requires A² = ½ ∑ [4A/(nπ)]² over odd n, which holds because the sum of 1/n² over odd n is π²/8. The fundamental alone supplies 8/π², or 81.1 percent, of the mean-square value; harmonics through the ninth bring the total to 96.0 percent, and harmonics through the 99th to 99.6 percent.

An average-responding meter scales the rectified average by 1.111, the ratio of RMS value to rectified average that holds for a sine wave, so it reads a square wave 11 percent high and a triangle wave 3.8 percent low. A true-RMS meter reads both correctly.

Power Carried by Harmonics

When a voltage and a current share the same fundamental period, the average power is

P = V₀I₀ + ∑_n=1^∞ V_nI_n cos θ_n

where V_n and I_n are the RMS values of the nth harmonics and θ_n is the phase angle between them. By orthogonality, a voltage harmonic and a current harmonic of different orders deliver no average power. In a resistor every harmonic pair is in phase and P = F_rms²/R: the ±5 V square wave across 10 Ω dissipates 2.5 W, of which the fundamental, 4.50 V RMS, supplies 2.03 W. When the supply voltage is a pure sinusoid, only the fundamental current delivers average power, and harmonic currents add RMS current and losses without adding power. That is the origin of distortion power factor, which Harmonics and Power Factor treats.

Dirichlet Conditions

The coefficients exist whenever |f(t)| has a finite integral over a period, but whether the partial sums approach the waveform is a separate question. The Dirichlet conditions answer it for practical waveforms. Suppose that over one period the waveform

• is absolutely integrable, so that ∫_T |f(t)| dt is finite;
• has a finite number of maxima and minima; and
• has a finite number of discontinuities, each a finite jump.

Then the series converges to f(t) wherever the waveform is continuous, and to the midpoint of the jump, [f(t⁻) + f(t⁺)]/2, at each discontinuity. The square wave's series gives exactly 0 at t = 0 and t = T/2, halfway between −A and +A. The conditions are sufficient but not necessary, and the waveforms met in circuit work satisfy them.

Kinds of Convergence

For any waveform with a finite RMS value, the series converges in mean square: the RMS value of the error between the waveform and its partial sum through harmonic N tends to zero as N grows. By Parseval's theorem, the mean square of that error is the sum of the omitted |c_n|². No other sum of sinusoids through harmonic N has a smaller RMS error, so the truncated series is the best least-squares fit of its order.

If the waveform is continuous and its derivative is piecewise continuous, as for a triangle wave or a rectified sine, the convergence is uniform: the largest error anywhere in the period shrinks to zero. Continuity alone does not guarantee convergence at every instant, since continuous functions exist whose series diverge at a point, but such functions do not arise in circuit analysis.

Smoothness and the Rate of Decay

The smoother the waveform, the faster its coefficients fall. Jumps give coefficients that fall as 1/n, as in the square wave, sawtooth, and pulse train; a continuous waveform whose slope jumps gives 1/n², as in the triangle wave and the rectified sines. In general, when f(t) and its first k − 1 derivatives are continuous and the kth derivative jumps, the coefficients fall as 1/n^k+1. On a logarithmic plot these envelopes fall at 20 dB and 40 dB per decade, the two slopes of the trapezoidal pulse envelope.

The Gibbs Phenomenon

Near a jump, every partial sum overshoots, and the overshoot does not shrink as harmonics are added. It narrows and moves toward the jump, but its height approaches a fixed fraction of the jump,

(1/π) ∫₀^π (sin x)/x dx − ½ ≈ 0.0895

or 8.95 percent, above the upper level on one side of the jump and below the lower level on the other. A ±A square wave jumps by 2A, so its partial sums peak near 1.179A. For odd N, the first peak of the square-wave sum through harmonic N lies at t = T/[2(N + 1)] after the jump, and an undershoot to about 0.903A follows at twice that time. The table shows the peak holding its height as it moves in, while the RMS error keeps falling.

Henry Wilbraham analyzed the overshoot in 1848, in The Cambridge and Dublin Mathematical Journal. J. Willard Gibbs missed it in a note of 1898 and described it in a correction that Nature published on 27 April 1899, and Maxime Bôcher named it the Gibbs phenomenon in a detailed analysis in 1906.

Tapering the coefficients trades overshoot for a slower edge: with the Lanczos sigma factors, sinc[n/(N + 1)], the square wave's overshoot falls to about 1.2 percent of the jump, and with Fejér's weights, 1 − n/(N + 1), it disappears.

Where the Overshoot Appears

An ideal low-pass filter with a brick-wall cutoff passes a finite set of harmonics unchanged and removes the rest, so its response to a square wave is exactly a partial sum, with the full 9 percent overshoot however high the cutoff. Filters with gradual roll-offs taper the harmonics and overshoot less; a Gaussian response does not overshoot at all, and the Bessel filters that approximate it overshoot only slightly. Designing a finite impulse response filter by truncating an ideal impulse response is a Fourier series in frequency, so the ripple beside each band edge stays near 9 percent of the step in gain however many coefficients are used. That limits a rectangular-window design to about 21 dB of stopband attenuation, since 20 log₁₀ 0.0895 = −21.0 dB, which is why the window-based designs described in Digital Filter Implementation favor tapered windows. Overshoot seen on an oscilloscope usually has another cause, such as probe ground-lead resonance or an underdamped amplifier; Gibbs ringing requires a sharp truncation of the spectrum.

Worked Example: Square Wave into an RC Low-Pass Filter

A ±5 V, 1 kHz square wave drives an unloaded RC low-pass filter whose corner frequency, f_c = 1/(2πRC), is also 1 kHz. Values of R = 1.59 kΩ and C = 100 nF give 1.001 kHz; the calculation uses exactly 1 kHz. The response is H(jω) = 1/(1 + jω/ω_c), so at harmonic n the gain magnitude is 1/√(1 + n²) and the phase shift is −arctan n.

The input harmonics fall as 1/n, and above its corner the filter contributes nearly another factor of 1/n, so the output harmonics fall almost as 1/n². The output fundamental has an RMS value of 3.183 V, and summing every harmonic by Parseval's theorem gives an output RMS value of 3.225 V, so the output THD is 16.4 percent, down from 48.3 percent at the input.

A time-domain solution checks the result. In steady state each half-cycle charges the capacitor exponentially from one peak toward the opposite supply level, and symmetry gives a peak of V_p = 5 tanh[T/(4RC)] V = 5 tanh(π/2) V = 4.586 V at the end of each half-cycle. The Fourier sum approaches that peak slowly: 3.18 V from the fundamental alone, 4.27 V through the ninth harmonic, and 4.55 V through the 99th. The exact waveform's RMS value is 3.225 V, as the harmonic sum predicted. RMS values converge quickly, but sharp corners need many terms, so the question should decide the method.

Tuned Circuits Select Harmonics

A resonant circuit does the opposite: it passes one harmonic and rejects the rest. A series RLC band-pass filter with its output across the resistor, tuned to 3 kHz with a quality factor Q of 10, has the gain |H| = 1/√[1 + Q²(f/f_r − f_r/f)²], which is 0.0375 at 1 kHz and 0.0933 at 5 kHz. Driven by the same square wave, whose fundamental is three times its third harmonic, it delivers a 1 kHz component of 11.2 percent and a 5 kHz component of 5.6 percent of the 3 kHz output, and all the unwanted components together total 12.9 percent. Frequency multipliers work this way, and so does the tank circuit that restores a sinusoid from the current pulses of a class C stage, described in Power Amplifiers.

When Superposition Fails

Harmonic-by-harmonic analysis requires linearity. A diode, a saturating transistor, or a magnetic core creates frequencies the input did not contain. Simulators find the periodic steady state of such circuits by harmonic balance, which represents voltages and currents as Fourier series with unknown coefficients, as RF and Microwave Design Software describes.

From Series to Transform

A periodic waveform has a line spectrum, with energy only at DC and at multiples of f₀. A single pulse, which never repeats, has a continuous spectrum given by the Fourier transform,

F(jω) = ∫_−∞^∞ f(t) e^−jωt dt

If f₁(t) equals a periodic waveform over one period and is zero elsewhere, comparing this integral with the coefficient formula gives

c_n = (1/T) F₁(jnω₀)

The lines are samples of the transform of one period, taken every ω₀ and scaled by 1/T. A rectangular pulse of height A and width τ centered on t = 0 has F₁(jω) = Aτ sin(ωτ/2)/(ωτ/2), and sampling it at ω = nω₀ gives back c_n = AD sinc(nD). Lengthening T with τ fixed crowds the lines together and shrinks them in proportion to 1/T while their envelope keeps its shape; in the limit, Tc_n becomes the continuous transform. Conversely, the Fourier transform of a periodic waveform is a set of impulses, F(jω) = 2π ∑ c_nδ(ω − nω₀). The Laplace Transform article relates the transform to s = jω, and the Z-Transform article treats the discrete-time Fourier transform of sampled signals.

Line Spectra in Measurement

Instruments compute spectra from samples, usually with the fast Fourier transform described in Transform Processing. Suppose N equally spaced samples x[i] span exactly M periods of a waveform that has no harmonic at or above half the sampling rate. The discrete Fourier transform

X[k] = ∑_i=0^N−1 x[i] e^−j2πki/N

then gives the coefficients exactly: X[Mn]/N = c_n, so the amplitude of harmonic n is A_n = 2|X[Mn]|/N. Two conditions often fail in practice.

• When the record does not hold a whole number of periods, each line leaks into neighboring bins. A sine wave that starts at zero phase and completes 2.5 cycles in a 64-sample record shows its two largest bins at 0.71 and 0.58 of its true amplitude and spreads smaller amounts into every other bin. Window functions reduce the leakage, and power-quality analyzers avoid it by synchronizing their measurement windows to whole line cycles.
• A waveform with jumps has harmonics without end, so some always lie above half the sampling rate and alias onto lower lines. Sampling one period of a square wave as 32 samples at +A and 32 at −A reads its fundamental 0.04 percent high and its third harmonic 0.4 percent high.

Square-Wave and Quasi-Square-Wave Inverters

A full bridge that switches its output between +V_dc and −V_dc as a square wave produces a fundamental of 4V_dc/π peak, 1.27 times the DC input, with every odd harmonic at 1/n of the fundamental and a THD of 48.3 percent. Delaying the switching of one leg relative to the other inserts a zero-voltage interval of 2β centered on each zero crossing. The resulting quasi-square wave has odd quarter-wave symmetry, and the quarter-period integral from the symmetry table gives

b_n = [4V_dc/(nπ)] cos(nβ), for odd n

Choosing β = 30° makes cos 3β zero, which removes the third harmonic and every other harmonic whose order is a multiple of three. The fundamental falls only to cos 30°, or 86.6 percent of the square wave's, and the THD falls to 31.1 percent. Selective harmonic elimination extends the idea: each additional switching angle per quarter-cycle can cancel one more harmonic order. DC-AC Conversion (Inversion) describes the inverter circuits.

Pulse-Width Modulation

In continuous conduction, the switch node of a buck converter is a pulse train between 0 and V_in at the switching frequency f_sw. Its DC term, DV_in, is the output voltage of an ideal buck converter, and its harmonics lie at multiples of f_sw with amplitudes of [2V_in/(nπ)]|sin(nπD)|. The output LC filter keeps the DC term and, well above its corner frequency f_c, attenuates each line by a factor of about (nf_sw/f_c)². With V_in = 12 V, D = 0.25, f_sw = 500 kHz, and f_c = 10 kHz, the first switching harmonic, 5.40 V peak, falls to 2.16 mV, and the second contributes 0.38 mV. The complete harmonic sum gives a peak-to-peak ripple of 4.44 mV, matching the time-domain formula (1 − D)V_out/(8LCf_sw²). Both calculations assume an ideal filter with no load damping and a capacitor with no equivalent series resistance, whose contribution can dominate real ripple.

An inverter leg that switches between ±V_dc/2 by comparing a sinusoidal reference at f₁ with a triangular carrier at m_ff₁ produces a fundamental of m_aV_dc/2 peak for m_a ≤ 1, where the modulation index m_a is the ratio of the reference peak to the carrier peak. The switching harmonics cluster at and around multiples of the carrier frequency rather than near the fundamental. A numerical Fourier analysis of this waveform with m_a = 0.8 and m_f = 21 gives, in units of V_dc/2, 0.818 at the 21st harmonic, 0.220 at the 19th and 23rd, and 0.314 at the 41st and 43rd, while every harmonic from the 2nd through the 17th stays below 0.01. An output filter, or the inductance of a motor winding, can therefore attenuate the switching components while passing the fundamental. Randomizing the switching frequency spreads the lines into a continuous spectrum, as PWM Generation describes.