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Erwin Louis Hahn's 1950 apparatus enclosed liquid samples in the coil of a tuned LC circuit, coupled to a gated radio transmitter and radio receiver. One transmitter quadrupled a crystal oscillator's 7.5 MHz to 30 MHz. A separate proton resonance regulator stabilized the static magnetic field.^[ref]

Waiting for thermal equilibrium prepared nuclear polarization. Oscilloscope photographs compared glycerol proton echoes at different field homogeneity: echo width changed; echo time followed twice the pulse separation. Receiver saturation obscured early free induction decay; transmitter leakage required care. The delayed signal tested refocusing; diffusion could still attenuate it.^[ref]

Hahn analyzed equal pulses as well as noting improved recovery with a second pulse twice as long; the ideal sequence above isolates that latter case.^[ref]

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

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.

Guglielmo Marconi's 1909 lecture describes antennas with earth connections and separate coupled receiving circuits. His 1900 system coupled a closed capacitor–inductor–spark gap loop to a transmitting antenna. Coil spacing adjusted coupling; inductance and capacitance adjusted tuning. All four circuits were tuned together. By 1909 his company's stations mainly used magnetic receivers, rather than coherers.^[ref]

Those are a participant's retrospective descriptions, not a complete account of priority. An earth connection is part of this antenna arrangement; it is not a universal requirement for radio communication. Nor does tuning an isolated LC circuit ensure that its frequency remains unchanged when strongly coupled to another circuit. The coupled resonant circuit model quantifies that distinction.

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]