= Sinusoidal steady state
The \b[<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.<6.200 Notes: Sinusoidal Steady State by Karl K. Berggren>[p3, start-up transient assumption; pp10–11, complex exponential solution]{ref}
Using <peak phasors>, write the source as $v_s(t)=\operatorname{Re}(\widehat V_s e^{i\omega t})$. If $H(i\omega)$ is the relevant <transfer function>, its <particular solution> is
$$
v_{\mathrm{ss}}(t)=\operatorname{Re}\!\left[H(i\omega)\widehat V_s e^{i\omega t}\right].
$$
The 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 $t\ge0$, 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 equation>
$$
\tau\dot v_C+v_C=V_0\cos\omega t.
$$
Put $x=\omega\tau$. Substitution verifies the sinusoidal <particular solution> $V_0(\cos\omega t+x\sin\omega t)/(1+x^2)$. Adding $Ae^{-t/\tau}$ and imposing the <initial condition> fixes $A=-V_0/(1+x^2)$, giving
$$
\boxed{v_C(t)=\frac{V_0}{1+x^2}\left(\cos\omega t+x\sin\omega t-e^{-t/\tau}\right).}
$$
Its 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>, is
$$
\frac{|v_C(t)-v_{\mathrm{ss}}(t)|}{V_0/\sqrt{1+x^2}}
=\frac{e^{-t/\tau}}{\sqrt{1+x^2}}.
$$
With chosen $R=1\,\mathrm{k\Omega}$, $C=1\,\mathrm{\mu F}$, $V_0=1\,\mathrm V$ and $\omega=1000\,\mathrm{rad/s}$, 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 $4.259\tau$.
\Image[/rc-sinusoidal-turn-on.png]
{title=A phasor response omits the exponential switch-on transient}
{description=Original RC example with time constant 1 millisecond and driving angular frequency 1000 radians per second. The exact capacitor response starts at zero, approaches its sinusoidal particular solution and crosses a one-percent relative transient threshold after about 4.259 time constants.}
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.<Laboratory 11: Sinusoidal Steady State Response of RC and RL Circuits by MIT OpenCourseWare>[pp1–2, natural decay and sinusoidal amplitude/phase observations]{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 equation
$$
L\ddot q+\frac qC=V_0\cos\omega_0t
$$
has the <particular solution> $q_p(t)=V_0t\sin(\omega_0t)/(2L\omega_0)$. 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.
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