Sinusoidal steady state

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]

Noncommuting damping and observation-time limits at resonance means that the late-time approximation of a damped circuit is not uniform as its dissipation tends to zero. A precise comparison needs an observable: the unnormalized resonant sinusoid itself has no ordinary limit as time tends to infinity. Use the series RLC circuit with zero initial conditions, source $V_0\cos {\omega }_{0}t$, and ${\omega }_0=1/\sqrt{LC}$. For $R>0$, its steady-state electric charge and electric current areThe remaining electrical resistance sets the resonant amplitude, as in the DOE series-resonance model.^[ref]

Define a dimensionless state error relative to that amplitude:For each fixed $R>0$, the homogeneous solution decays and $D_R(t)\to 0$. Conversely, on a fixed finite observation interval the actual circuit solution converges to the finite undamped solution as $R\downarrow 0$. The two terms containing $R$ then vanish, giving $D_R(t)\to 1$. Thus the genuine iterated limits areThis states the failure without treating an oscillating quantity as though it had a time limit.

For weak damping, put $\gamma =R/(2L)$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The exact zero-state chargeboth verifies the decaying transient and recovers the undamped resonant response in undamped resonant forcing has no bounded sinusoidal steady state by a fixed-time expansion.

For an original timing comparison, $L=10\mathrm{mH}$ gives envelope decay times $2L/R=0.1\mathrm{s}$ at $R=0.2\Omega$ and $1\mathrm{s}$ at $R=0.02\Omega$. At a fixed $50\mathrm{ms}$ the respective exponential multipliers are about $0.607$ and $0.951$. A smaller electrical resistance can make the sinusoidal late-time approximation worse over a fixed acquisition window. MIT's steady-state notes explicitly require the startup transient to have disappeared.^[ref]

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]

The worst-switching-phase settling bound for a first-order circuit maximizes the initial transient over the source's unknown switching wave phase. For the stable RC circuit, let its sinusoidal particular solution be $v_{\mathrm{ss}}(t)=A\cos (\omega t+\psi )$, where $A>0$, and let $u_0$ be the initial capacitor electric potential difference. The error identity in exponential settling bound for RC sinusoidal turn-on givesSince $\cos \psi$ ranges from minus one to one,For a validated initial bound $|u_0|\le U$, the sharp uniform bound is therefore $(1+U/A)e^{-t/\tau }$. It is attained by choosing the extreme initial value and the opposite steady-state wave phase. A normalized error tolerance $0<ϵ<1$ is guaranteed thereafter ifThere is no division by the instantaneous sinusoid, so its zero crossings cause no singular relative-error specification.

For an original example, choose $\tau =4\mathrm{ms}$, $A=0.5\mathrm{V}$, $U=0.25\mathrm{V}$, and $ϵ=0.02$. The worst-case settling time is $4\log 75\approx 17.270\mathrm{ms}$. Assuming an uncharged capacitor instead would give $4\log 50\approx 15.648\mathrm{ms}$. That shorter guarantee requires the additional initial-state information.

A phase-locked test can measure a smaller mismatch, even zero for a specially matched initial condition, without contradicting the worst-case result. A valid comparison records both the switching wave phase and the capacitor's initial state. The MIT laboratory's square-wave natural-decay observation and sinusoidal amplitude/wave phase measurement provide the two underlying bench measurements.^[ref] The result assumes the same linear first-order model throughout switching; additional modes or a time-varying component require another transient bound.