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Karl K. Berggren is an electrical engineer whose research includes superconductivity-based electrical circuits, electronic devices and single-photon detectors for quantum applications. MIT's Research Laboratory of Electronics identifies him as a professor of electrical engineering and physics and as head of its quantum nanostructures and fabrication group.^[ref] These research areas connect the behavior of very small devices to the circuits needed to read and control them.

MIT's identified faculty profile also links an externally hosted portrait of Karl K. Berggren.

His 6.200 Notes: Sinusoidal Steady State by Karl K. Berggren explain why complex numbers simplify linear differential equations driven by sinusoidal inputs. For a series RC circuit observed across its capacitor, the equation $RCd{v}_C/dt+v_C=v_{\mathrm{in}}$ gives the late-time frequency responseThe notes obtain this result by solving for a complex exponential and taking its real part.^[ref]

For an original illustration, $R=1\mathrm{k\Omega }$ and $C=1\mu \mathrm{F}$ give a time constant of $1\mathrm{ms}$. At $\omega =1000\mathrm{rad}{\mathrm{s}}^{-1}$, $H=(1-i)/2$: a one-volt peak sinusoidal input produces a $1/\sqrt{2}$-volt peak output lagging by $\pi /4$. A separate term proportional to $e^{-t/(RC)}$ is the transient and depends on the starting state. Distinguishing that term from the forced sinusoid is what makes the steady-state calculation experimentally useful. This numerical illustration applies his teaching derivation; it is not a reported device measurement.

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]

A peak amplitude phasor is the complex amplitude that reconstructs a real single-frequency sinusoid without an extra amplitude multiplier:Its modulus is the sinusoid's peak amplitude, and its complex argument is the wave phase relative to the chosen cosine reference. This is the default convention in phasor. MIT's sinusoidal notes explicitly describe the cosine reference and phase/amplitude parameters.^[ref]

For an original example, take $v(t)=3\cos \omega t-4\sin \omega t$ volts. Expanding the real part shows that $\hat{V}=3+4i$ volts, with peak amplitude $5$ volts and wave phase $\arctan (4/3)\approx 53.13^{\circ }$. The corresponding root mean square phasor is $(3+4i)/\sqrt{2}$ volts, whose magnitude is about $3.536$ volts. The physical signal is the same under both conventions.

The peak phasor versus positive-frequency Fourier coefficient distinction follows fromThus the positive-frequency coefficient in this two-exponential expansion is $\hat{V}/2$, not $\hat{V}$. A real sinusoid requires its conjugate negative-frequency term as well. This algebra does not assign ordinary Fourier-transform function values to an infinite-duration sinusoid; those require the corresponding distributional interpretation.

A common time-origin change multiplies both voltage and current peak phasors by the same unit-modulus factor, leaving their relative phase and impedance ratio unchanged. Peak-versus-RMS normalization in complex electrical power is still essential when an amplitude convention changes rather than just the time origin.

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]