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Side losses need not destroy the method. Let $\theta$ be the small temperature disturbance about a steady background in surroundings at fixed temperature. For a rod of cross-sectional area $S$ and perimeter $\mathcal{P}$, a uniform real surface heat transfer coefficient $h_s$ produces a loss per unit volume of $H\theta$, with $H=h_s\mathcal{P}/S$. Assume constant $k,\rho ,c_p,H$, a cross-section nearly uniform in temperature, and no distributed heat source. The balance becomesHere $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, while $\lambda$ has units ${\mathrm{s}}^{-1}$. This is a distributed version of Newton's law of cooling; a constant linearized contribution from thermal radiation may be included over a small temperature range. It does not represent arbitrary time-dependent coupling to a surrounding material with its own thermal storage.

For the decaying exponential branch, the phasor equation givesso thatReal positive side loss increases $a$ and decreases $b$ at fixed angular frequency and thermal diffusivity. Their product eliminates $\lambda$, yielding the combined inferenceThis result requires that attenuation and lag belong to the same exponential complex wavenumber at the same frequency, and that the loss term is real. It does not generally survive a superposition caused by a nearby finite end, spatially varying coefficients, transverse temperature gradients, or unequal sensor delays. The inferred $\lambda =\alpha (a^2-b^2)$ supplies an additional model check; a negative value signals inconsistency with this passive-loss model rather than negative physical cooling.

The reciprocal of electrical impedance, relating electric current and electric potential difference phasors by $\tilde{I}=Y\tilde{V}$. Its unit is the siemens, $\mathrm{S}={\Omega }^{-1}$. Electrical admittances add for branches in a parallel circuit.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

An actual teaching experiment turns the phasor definition into readings. Leipzig University's laboratory handout Measurements with an oscilloscope, experiment E 3e, uses a function generator, a two-channel oscilloscope, a known resistor and a capacitor. Figure 5 places the capacitor above the resistor, with a common return below the resistor. Channel 1 measures the applied terminal electric potential difference $v_1$; channel 2 measures $v_R$. The handout asks for peak-to-peak values and relative timing, then comparison of measured and calculated electrical impedances with $R=3\mathrm{k\Omega }$, $C=0.47\mu \mathrm{F}$ at $100$ and $300\mathrm{Hz}$. It also asks students to check the oscilloscope time base against a digitally counted generator frequency. These are apparatus and procedural specifications, not reported measured results.^[ref]

For this arrangement, Ohm's law gives $i(t)=v_R(t)/R$ if probe loading and the sensing resistor's electrical reactance are negligible. Record both voltage magnitudes with the same convention: peak, peak-to-peak, or root mean square. After the transient settles, compare matched rising zero crossings $t_1$ of $v_1$ and $t_R$ of $v_R$, with period $T=1/f$. ThenThe recovered electrical impedance includes the known resistor. For a separate unknown component in series with that resistor, subtract $R$ from the inferred total electrical impedance. A current crossing earlier than the applied voltage crossing gives $\Delta t<0$, hence negative $\varphi$. For the stipulated series RC impedance example at $1000\mathrm{Hz}$, this predicts $\Delta t=-0.160717\mathrm{ms}$; it is a calculated timing, not a lab observation. Timing supplies the complex argument that an amplitude ratio alone cannot recover.

As additional measurement controls, first feed the same sinusoid to both channels to check relative delay and polarity; then replace the reactive chain with a known resistor, expecting nearly zero phase difference. Keep both probe returns at the common return shown in figure 5. A reversed voltage polarity adds half a turn to its inferred wave phase. Match crossings with the same slope and choose $-T/2<\Delta t\le T/2$; adjacent cycles otherwise change $\varphi$ by a full turn. A sweep continuous in frequency helps track that branch. These controls test the signed inference rather than assuming that the scope's delay readout has the desired sign.

Measure the loaded terminal electric potential difference on channel 1 at every frequency. The generator's output impedance can change the applied magnitude; its front-panel setting is not the measured terminal value. The generator's internal electrical impedance is outside the terminal ratio above, while the known sensing resistor is inside it. Probe loading, finite timing resolution, component tolerances and distorted waveforms limit the inference. A repeat at lower drive amplitude tests whether one linear electrical impedance still fits. Electrical reactance bench measurement develops these loading and reference-direction issues further.

Input impedance is the ratio of a small applied input-voltage phasor to the resulting input-current phasor, with the output termination and operating point specified. A finite input impedance loads a voltage source. For an ideal operational amplifier the input impedance is infinite, so neither input terminal draws current.

A non-ideal amplifier is not specified by infinite input impedance alone: its open-loop gain and output impedance must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit finite-gain operational-amplifier integrator model.

Writing Kirchhoff's current law at the input and output nodes givesThe second equation gives $y=(sC{r}_o-A)v/(1+sC{r}_o)$. Substitution in the first gives the general transfer functionFor sinusoidal phasors, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the amplitude ratio and their arguments give the relative phase.

The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It givesThus below ${\omega }_l=[(1+A_0)RC{]}^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that frequency it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq 1/(\omega RC)$ and phase $+90^{\circ }$. Constant finite gain by itself predicts no further high-frequency turnover.

For a more realistic sketch, use the standard dominant-pole open-loop gain $A(s)=A_0/(1+s/{\omega }_a)$, still with $r_o=0$. ThenThe exact frequency response isAssuming $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, the turnover frequencies are well separated: ${\omega }_l\simeq 1/(A_{0}RC)$ and ${\omega }_h\simeq A_0{\omega }_a$, the approximate unity-gain angular frequency of the amplifier. The regimes arewith magnitude slopes $0,-20,-40\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^{\circ }$ and $-180^{\circ }$ denote the same phase.

If finite $r_o$ is retained and $A(i\omega )\to 0$, the general expression instead approaches $r_o/(R+r_o)$ at very high frequency: the capacitor couples the input to the output through a passive voltage divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,{\omega }_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.

Field equations. In a homogeneous source-free nonmagnetic medium, $\mathbf{D}=ϵ{\epsilon }_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. Maxwell's equations becomeThe first equation follows from $\nabla \cdot \mathbf{D}=0$ for spatially constant nonzero $ϵ$. When the effective relative permittivity is negative below, its charge response is already included in $\mathbf{D}$; it must not also be counted as an extra free source.

Amplitude relations and bulk dispersion. Work with complex phasors proportional to $e^{ikx-Kz-i\omega t}$. For $\mathbf{H}=(0,H_y,0)$, the Ampere-Maxwell law givesso the amplitudes must obeyThey are not arbitrary independent constants: the dispersion relation alone would not make arbitrary amplitudes satisfy the equations. These relations give $ik{E}_x-K{E}_z=0$, so the electric divergence vanishes; the magnetic divergence vanishes because there is no $y$ dependence. The only nonzero component of Faraday's law of induction isSubstitution then givesThis proves that all four field equations hold for the specified transverse magnetic polarization when both the amplitude relations and the dispersion relation are imposed.

Interface matching. Let region 1 be vacuum above the interface, with ${ϵ}_1=1$, $K={\beta }_1>0$; region 2 has ${ϵ}_2=ϵ$, $K={\beta }_2<0$. These signs make the evanescent fields decay as $z\to \pm \infty$. In the absence of free surface current, tangential $H_y$ is continuous, giving a common interface amplitude. Tangential $E_x$ is also continuous, henceNormal $D_z$ is continuous as well: both sides have $D_z=-k{H}_y/\omega$. This does not require $E_z$ itself to be continuous; the material can have polarization surface charge. The normal magnetic field is zero on both sides.

Existence and surface dispersion. The opposite signs of ${\beta }_1$ and ${\beta }_2$ require $ϵ<0$. The two bulk equations areUse ${\beta }_2^2={ϵ}^2{\beta }_1^2$ to eliminate the decay constants. For the nontrivial interface branch,A positive ${\beta }_1^2$ together with real propagating $k$ gives $ϵ<-1$. A negative permittivity between $-1$ and zero is insufficient. This lossless planar surface-plasmon dispersion describes a surface plasmon polariton; $ϵ=1$ would instead be identical media and supplies no decaying bound mode.

Frequency range and long wavelengths. With $ϵ=1-{\omega }_p^2/{\omega }^2$, the existence condition becomesThe upper value is a limiting frequency approached as $k\to \infty$, not an attained maximum at finite $k$. Substitution into the dispersion relation givesFor $\omega \ll {\omega }_p$, choosing propagation in the positive $x$ direction,It approaches the vacuum light line from larger $k$. The vacuum decay constant is then ${\beta }_1\simeq {\omega }^2/(c{\omega }_p)$, so confinement becomes weak at low frequency. This calculation assumes the stated real, lossless effective permittivity; dissipation would require complex propagation and decay constants.

Effective permittivity. Use peak phasors with time dependence $e^{-i\omega t}$. The Ampere-Maxwell law isThe conductive contribution to complex permittivity is thereforeHere ${ϵ}^{\prime }$ is the source's effective relative permittivity, not just the real part of a complex response. The positive imaginary term is the loss sign appropriate to this time convention.

The penetrating electric wave. Combine this equation with Faraday's law of induction. In a uniform source-free medium the transverse phasor satisfiesFor a good conductor of electrical conductivity $\sigma$, $\sigma \gg \omega ϵ{\epsilon }_0$, so $k^2\simeq i{\mu }_0\sigma \omega$. The root which decays into $z>0$ isChoose the surface phasor amplitude $E_0$ real. Taking the real part of ${\tilde{E}}_x=E_0{e}^{ikz}$ givesThe skin depth $\delta$ is an amplitude-decay distance in metres.

The magnetic wave and phase. The curl of an $x$-polarized field varying in $z$ is in the $y$ direction. Since $\nabla \times \tilde{\mathbf{E}}=i\omega {\mu }_0\tilde{\mathbf{H}}$,ThusThe phase sign follows the specified convention. Equivalently the surface impedance of a good conductor is $E_x/H_y=(1-i)/(\sigma \delta )$.

How the fields differ from those in vacuum. The two metal fields have the same exponential envelope and spatial period $2\pi \delta$, but their phase arguments differ by $\pi /4$.

An individual travelling wave in the lossless exterior has no attenuation, real wave impedance and in-phase electric and magnetic fields. The exterior total field also includes the reflected wave and may show interference, so it need not share that simple phase relation everywhere. At the interface the tangential fields match; $E_0$ is the field just inside the metal, not necessarily the incident amplitude. A good conductor reflects strongly and its wave impedance is much smaller than the vacuum impedance.

Average inward energy flux. For peak phasors, the Poynting vector averaged over a cycle is $⟨\mathbf{S}⟩=\mathrm{Re}(\tilde{\mathbf{E}}\times {\tilde{\mathbf{H}}}^{\ast })/2$. The fields giveConsequently the constant in the question isThe flux is in the $+z$ direction and has units $\mathrm{W}{\mathrm{m}}^{-2}$.

Local energy conservation. The Poynting flux and local heating in a skin layer satisfyThe decrease becomes Joule heating in each layer. The Poynting theorem includes this transfer from field energy to the metal; averaged electromagnetic energy storage has zero time derivative in the periodic state, so no energy is lost from the complete balance.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile Electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let Electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective Electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile Electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction Electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

In the $e^{i\omega t}$ phasor convention, the total series RL circuit impedance is $R+i\omega L$. The resistor voltage follows the current, so its transfer function relative to the applied voltage isAt the stated condition $R=\omega L$,Thus the resistor-voltage amplitude is $\boxed{V_{R,0}=V_{\mathrm{in},0}/\sqrt{2}}$ and it lags by $45^{\circ }$. For a cosine input it is $V_R(t)=V_{\mathrm{in},0}\cos (\omega t-\pi /4)/\sqrt{2}$.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.