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The exponential settling bound for RC sinusoidal turn-on measures the decaying error relative to the persistent sinusoid's amplitude, rather than dividing by a sinusoid that repeatedly crosses zero. For the RC circuit in sinusoidal steady state, put $\tau =RC>0$ and let $v_{\mathrm{ss}}$ be the sinusoidal particular solution. Subtracting its equation from the full circuit equation givesFor source $V_0\cos \omega t$ and an initially uncharged capacitor, the parent article derives $v_{\mathrm{ss}}(0)=V_0/[1+(\omega \tau {)}^2]$ and steady amplitude $A=V_0/\sqrt{1+(\omega \tau {)}^2}$. Thus the normalized transient is exactlyBecause it decreases monotonically, an error threshold $0<ϵ<1$ is met at all subsequent times onceThis is a transient-to-amplitude specification, not a pointwise relative error at each zero crossing.

For an original comparison, choose $\tau =2\mathrm{ms}$, $\omega \tau =3$, and $ϵ=0.01$. The stated cosine turn-on needs $t_{ϵ}\approx 6.908\mathrm{ms}$. If the switching wave phase is unknown, the initial steady-state value can instead have magnitude as large as $A$. With zero initial capacitor electric potential difference, the worst-switching-phase settling bound for a first-order circuit is $e^{-t/\tau }$, requiring $\tau \log 100\approx 9.210\mathrm{ms}$. The shorter first result depends on the specified switching phase.

An unknown stored initial energy changes the leading mismatch again; the same exponential identity still applies. A bench test must therefore record the switching phase, initial capacitor electric potential difference, and the chosen normalization. MIT's RC laboratory measures the natural decay with square-wave excitation and then compares sinusoidal amplitudes and phases on an oscilloscope.^[ref]

A harmonic component is a sinusoid whose frequency is a positive integer multiple of a specified fundamental frequency: $f_n=n{f}_1$. The fundamental itself is the first harmonic component. In a periodic Fourier series with period $T=1/f_1$,the constant $a_0$ is a DC component, and the positive-frequency sinusoidal terms are the harmonic components. A component may have zero amplitude; a missing fundamental does not require the spacing of the remaining components to change. An overtone is a component above the lowest mode, but physical overtones need not be exact integer multiples.

For an original generation model, apply $v_{\mathrm{in}}(t)=1\mathrm{V}\cos (2\pi 1000t)$ to a memoryless nonlinear device with $v_{\mathrm{out}}=v_{\mathrm{in}}+\beta v_{\mathrm{in}}^2$ and $\beta =0.2{\mathrm{V}}^{-1}$. The identity ${\cos }^{2}x=(1+\cos 2x)/2$ givesThe nonlinearity produces both a DC offset and a second harmonic component; they are different parts of the frequency spectrum. The second harmonic component's peak amplitude is $0.1\mathrm{V}$, while the fundamental's is $1\mathrm{V}$.

A linear time-invariant device in sinusoidal steady state cannot create this extra frequency from a single sinusoidal input; a nonlinear response, a different input component, or an unexcluded transient needs to be considered. In an actual measurement, the source's own harmonic components and the receiver's frequency response must also be accounted for. Anritsu's primary measurement procedure separately acquires the fundamental and selected harmonic components, with measurement calibration covering their frequencies.^[ref] A discrete Fourier transform can instead separate a suitably sampled record; its bin amplitudes still depend on normalization and record length.^[ref]

James L. Kirtley Jr. is an electrical engineer whose work concerns electric machines and electric power systems. MIT's departmental profile identifies him as a professor emeritus of electrical engineering, with energy and robotics among his research areas.^[ref]

The engineering problem is to match a motor's behaviour to its application. MIT reported in 2016 that Kirtley worked on customized electric motors for the MIT Cheetah robot, where brief peaks of torque serve running and jumping. The same interview describes his interest in electric motors interacting with microgrids, particularly their dynamic behaviour during supply disturbances.^[ref] A short burst and sustained operation place different demands on the system; the largest instantaneous output alone does not describe its operating role.

His AC Power Flow in Linear Networks by James L. Kirtley Jr. connects the phasor representation of electric potential difference and electric current to physical energy transfer. In its peak-amplitude convention, complex electrical power is $S=V_{\mathrm{pk}}{I}_{\mathrm{pk}}^{\ast }/2$. With RMS phasors $V=V_{\mathrm{pk}}/\sqrt{2}$ and $I=I_{\mathrm{pk}}/\sqrt{2}$, this becomes $S=V{I}^{\ast }$. The complex conjugate ensures that the relative phase enters the active power.^[ref]

For an original load example in sinusoidal steady state, take $V=20\mathrm{V}$ and electrical impedance $Z=(3+4i)\Omega$, using RMS phasors and the passive sign convention: electric current enters the positive-potential terminal. A series resistor and inductor realize this model at the specified angular frequency when $R=3\Omega$ and $\omega L=4\Omega$. ThenThus active power is 48 watts, reactive power is 64 vars, apparent power is 80 volt-amperes, and power factor is $48/80=0.6$. Write $\varphi =\arctan (4/3)$ and $x=\omega t$. The corresponding electric potential difference and electric current areMultiplying and applying a trigonometric identity gives instantaneous powerIts cycle average is 48 watts, while its range is $[-32,128]$ watts. The negative portions represent temporary return of stored energy from this ideal load; the electrical resistance still dissipates a positive average. A synchronized product of the two waveforms reveals this behaviour, whereas multiplying their root mean square magnitudes gives only the apparent power.

Reactive power overcompensation occurs when capacitive correction exceeds the inductive reactive power that it is intended to cancel. Under the passive sign convention and a positive-frequency sinusoidal supply, the source then sees negative net reactive power, producing a leading power factor. This can happen when a fixed capacitor bank remains connected as the inductive load falls. Eaton's application paper discusses excessive leading operation and the use of switched correction stages.^[ref]

Consider an ideal fixed RMS electric potential difference $V$ at angular frequency $\omega$, supplying a load that absorbs constant $P>0$ and $Q_L>0$. Let $Q_{\mathrm{cap}}=\omega C{V}^2\ge 0$ be the magnitude of the correction's reactive power; its signed contribution is $-Q_{\mathrm{cap}}$. Adding ideal shunt capacitance givesThe model holds the load and supply fixed while changing $C$. It omits losses and additional network dynamics. Optimal shunt capacitance for sinusoidal power factor correction follows by minimizing the squared root mean square electric current:For $C<C_{\ast }$ the remaining reactive power is positive and the power factor is lagging. For $C>C_{\ast }$ it is negative and the power factor is leading. The overcompensation threshold is $C_{\ast }$, but the current reduction interval for shunt reactive compensation is wider. Comparing the square of the corrected and uncorrected electric current givesThus source RMS current is reduced for $0<C<2C_{\ast }$, equals the uncorrected value at $C=2C_{\ast }$, and is larger for $C>2C_{\ast }$. An overcompensated operating point between $C_{\ast }$ and $2C_{\ast }$ still draws less source electric current than the uncorrected load.

For an original example, keep the parent article's chosen $P=1000\mathrm{W}$, $Q_L=750\mathrm{var}$, $V=200\mathrm{V}$ and 50 hertz. The uncorrected source draws 6.25 amperes at power factor0.8. Full cancellation needs $C_{\ast }\approx 59.68\mu \mathrm{F}$ and reduces source electric current to 5 amperes. A 100 microfarad correction gives $Q_{\mathrm{net}}\approx -506.64\mathrm{var}$ and 5.605 amperes, with a leading factor about0.8920. A 150 microfarad correction instead gives about $-1134.96\mathrm{var}$ and7.563 amperes, worse than no correction. The crossing is $2C_{\ast }\approx 119.37\mu \mathrm{F}$.

The optimum changes with the load. If the example load's $P$ and $Q_L$ both halve while its original optimal correction remains connected, then $P=500\mathrm{W}$ and $Q_{\mathrm{net}}=-375\mathrm{var}$. Source electric current is 3.125 amperes and the factor is 0.8 leading. This dependence motivates automatic power factor correction. The cited controller description measures system electric potential difference and electric current, using external current transformers, and switches discrete capacitor bank stages to follow a target.^[ref]

Retain the phase or sign of $Q_{\mathrm{net}}$ when reporting the result. A bare power factor magnitude does not distinguish leading from lagging; the manufacturer FAQ also illustrates how a meter's signed display depends on its convention.^[ref] The calculation assumes sinusoidal steady state; current distortion factor is a separate consideration for nonsinusoidal current.

The sinusoidal steady state of a stable linear time-invariant system driven by a sinusoid is its persistent response at the driving frequency, after the relevant transient response has decayed. A phasor calculation finds that response; it does not generally satisfy the circuit's switch-on initial conditions. The late-time assumption is explicit in Berggren's treatment of sinusoidal circuit analysis.^[ref]

Using peak phasors, write the source as $v_s(t)=\mathrm{Re}({\hat{V}}_s{e}^{i\omega t})$. If $H(i\omega )$ is the relevant transfer function, its particular solution isThe complete response adds a homogeneous solution determined by the initial conditions. Approaching the displayed expression requires the modes affecting the measured output to decay; an unchanged amplitude and a fixed phase shift then characterize its frequency response.

For an original turn-on example, drive a series RC circuit with $v_s(t)=V_0\cos \omega t$ for $t\ge 0$, measure the capacitor's electric potential difference, and take $v_C(0)=0$. With time constant $\tau =RC>0$, Kirchhoff's voltage law gives the first-order linear differential equationPut $x=\omega \tau$. Substitution verifies the sinusoidal particular solution $V_0(\cos \omega t+x\sin \omega t)/(1+x^2)$. Adding $A{e}^{-t/\tau }$ and imposing the initial condition fixes $A=-V_0/(1+x^2)$, givingIts sinusoidal part has amplitude $V_0/\sqrt{1+x^2}$ and lags by $\arctan x$. The exponential settling bound for RC sinusoidal turn-on, relative to that amplitude, isWith chosen $R=1\mathrm{k\Omega }$, $C=1\mu \mathrm{F}$, $V_0=1\mathrm{V}$ and $\omega =1000rad/s$, we have $\tau =1\mathrm{ms}$ and $x=1$. At turn-on the phasor approximation predicts 0.5 volt while the capacitor is initially at zero. After 5 milliseconds, their difference is about 3.37 millivolts, or 0.476 percent of the sinusoidal amplitude. A 1 percent settling time threshold for this specific switch-on is about $4.259\tau$.

A useful measurement compares a long, low-frequency square-wave response with sinusoidal responses at several frequencies. An oscilloscope can identify the exponential decay and then measure amplitude ratios and phase shifts. MIT's laboratory uses that sequence to connect the time constant to the sinusoidal response; probe loading belongs in the measured circuit model.^[ref]

Undamped resonant forcing has no bounded sinusoidal steady state. For an ideal series LC circuit, let $q$ be electric charge and ${\omega }_0=1/\sqrt{LC}$. At resonance, the equationhas the particular solution $q_p(t)=V_{0}t\sin ({\omega }_{0}t)/(2L{\omega }_0)$. Its growing envelope prevents a bounded attracting sinusoidal response. This ideal counterexample explains why decay of the relevant homogeneous solutions is part of the physical steady-state assumption.

Undamped resonant forcing has no bounded sinusoidal steady state because the forcing continually adds energy to a mode that has no dissipation. For a series LC circuit driven from rest by $V_0\cos \omega t$, let $q$ be its electric charge and ${\omega }_0=1/\sqrt{LC}$. When $\omega \ne {\omega }_0$, imposing $q(0)=\dot{q}(0)=0$ on the linear ordinary differential equation givesBoth frequencies persist: the undamped homogeneous solution never disappears. Taking the limit $\omega \to {\omega }_0$ at fixed time produces the growing resonant response described in sinusoidal steady state. Its factor proportional to time cannot be removed by adding a bounded homogeneous solution.

For an original numerical example, choose $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$ and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4\mathrm{rad}{\mathrm{s}}^{-1}$. At exactly ten resonant periods, $t=20\pi /{\omega }_0\approx 6.283\mathrm{ms}$, the resonant solution has $q=0$ butAt twenty periods the electric current doubles and the stored magnetic energy quadruples. These are ideal-model values; a real source and winding have finite electrical resistance and ratings.

With positive series resistance $R$, the resonant phasor calculation instead gives current amplitude $V_0/R$. The DOE handbook explicitly identifies resistance as the remaining series electrical impedance at resonance.^[ref] For weak damping, decay takes a time of order $2L/R$, which grows as resistance decreases. These noncommuting damping and observation-time limits at resonance explain why an infinite steady-state amplitude is a failed limiting calculation, not the actual finite-time response. The late-time transient assumption underlying sinusoidal analysis must be checked.^[ref]