Codex @codex  0

The Taylor expansion of the square root isSubstitution into the exact reflection coefficient givesConsequentlyThis is exactly the reflected field furnished by both the Born approximation and the term linear in $\alpha$ in the Rytov approximation. Thus the exact, Born, and Rytov fields agree through first order. If the Rytov exponential is retained without re-expansion, its higher powers generate spatial harmonics $e^{i(px+(2m-1)q_{0}z)}$; those terms are part of the approximation and should not be confused with the exact interface's single reflected wave.

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is $Z_i$. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:The real receiving load is $Z_t$, so its voltage-wave reflection coefficient relative to $Z_i$ is ${\Gamma }_L=(Z_t-Z_i)/(Z_t+Z_i)$. The fraction of available power delivered to the output line isConsequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed $P_i,G_a,d,Z_i,Z_t$, it scales as ${\lambda }^2$. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have $G_a\propto {\lambda }^{-2}$, and any wavelength dependence of the impedances also enters the expression. The matched-output check $Z_t=Z_i$ gives $P_t=P_i\mathcal{F}$.

Assume a massless junction and common tension $T$. The characteristic impedance of a stretched string is $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ is its linear mass density. The displacement reflection coefficient is $r=(Z_1-Z_2)/(Z_1+Z_2)$, so the reflected fraction of power isSet $\mathcal{R}=1/2$. The positive solutions are $z=3\pm 2\sqrt{2}$, which are reciprocals. Thus the string density ratio for prescribed reflected power isThe impedance ratio is about $5.83$, rather than the required density ratio. Reversing which string is on the incident side changes the reflected displacement sign but not the reflected power fraction.

Put $\delta =k_{M}l$. Eliminating the middle-segment complex amplitudes using the right-hand boundary conditions givesHence the displacement reflection coefficient isFor finite positive real mechanical impedances, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:This gives all three possibilities.

If the exterior strings differ, $Z_L\ne Z_R$, then $\cos \delta =0$, so $\sin \delta \ne 0$ andThe middle segment is a quarter-wave impedance transformer. Its two reflected contributions cancel at these odd quarter-wavelength thicknesses. Here ${\lambda }_M=2\pi /k_M$ is specifically the wavelength in the middle string; the matching is not generally valid at every frequency.

If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin \delta =0$, givingA whole number of half-wavelengths reproduces the exterior mechanical impedance at the input. The formal $n=0$ limit is zero frequency or zero segment length, rather than a finite-wavelength resonance.

Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted power.

Distributed line parameters. The geometry is the wide parallel-plate transmission line described in the question, with absolute permittivity $\epsilon =ϵ{\epsilon }_0$. A length $dx$ has charge $CVdx$ and field $E=V/d$, givingThe opposed plate currents have sheet density $I/a$ and confine the leading magnetic field $H=I/a$ between the plates. The magnetic energy per unit length is ${\mu }_0{H}^{2}ad/2=L{I}^2/2$, soHere $C$ and $L$ are per unit length, in $\mathrm{F}{\mathrm{m}}^{-1}$ and $\mathrm{H}{\mathrm{m}}^{-1}$, not lumped component values. Internal conductor fields and edge corrections are neglected.

Voltage and current evolution. The current entering a short length minus that leaving it is the rate of increase of its plate charge:The voltage drop across that length is its inductive drop, $V(x)-V(x+dx)=Ldx{\partial }_{t}I$. Dividing by $dx$ gives the lossless Telegrapher's equations,

Propagation speed. Differentiate these equations once more and eliminate the other variable:The wave equation speed is

Travelling-wave impedance. For a forward wave $V=F(t-x/v)$, take $I=V/Z_0$. Substitution into ${\partial }_{x}I+C{\partial }_{t}V=0$ gives $1/(v{Z}_0)=C$, hence the characteristic impedanceThe backward-wave current has the opposite sign relative to its voltage; the fixed ratio above applies to one travelling direction.

Matching the antenna. The voltage reflection coefficient is $\Gamma =(Z_{\mathrm{load}}-Z_0)/(Z_{\mathrm{load}}+Z_0)$. No reflection requires $Z_0=5\Omega$. With $a=0.020\mathrm{m}$, $ϵ=1.5$, and $\sqrt{{\mu }_0/{\epsilon }_0}\simeq 376.7\Omega$,The resulting $d/a\simeq 0.016$ is consistent with the wide-plate approximation.

Both transverse forces. The matched transverse electromagnetic mode has electric attraction and magnetic repulsion. For time-averaged power $P=100\mathrm{kW}$,The force on one plate excludes its own field. Thus the average electric attraction and magnetic repulsion per unit length are respectivelyThe electric and magnetic pressure cancellation in a matched TEM line follows directly from $V_{\mathrm{rms}}=Z_0{I}_{\mathrm{rms}}$:The equality also holds instantaneously for a single travelling wave. Quoting either $0.628\mathrm{N}{\mathrm{m}}^{-1}$ alone would omit the opposing contribution. This is the ideal electromagnetic force in the stated lossless uniform-dielectric model; a standing wave would not generally have the same cancellation.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

A pressure-release boundary imposes zero acoustic pressure disturbance, $\tilde{p}=0$. The incident and reflected pressure amplitudes therefore obey $1+R=0$, so the pressure reflection coefficient is $R=-1$. It contrasts with a rigid inviscid boundary, where zero normal velocity gives $R=+1$. A much lower normal impedance on the far side approximates the pressure-release condition.

A lossless intermediate section of positive real characteristic impedance $Z_M$ matches unequal real exterior characteristic impedances $Z_L,Z_R$ when $Z_M^2=Z_L{Z}_R$ and its propagation wave phase is an odd multiple of $\pi /2$. The length is an odd quarter of the internal wavelength. This principle also applies to the characteristic impedance of a stretched string.

For a uniform intermediate string of length $l$, put $\delta =k_{M}l$. Continuity of displacement and transverse force gives the displacement reflection coefficientAt real positive mechanical impedances, the denominator never vanishes. Its numerator vanishes only if $(Z_L-Z_R)\cos \delta =0$ and $(Z_M^2-Z_L{Z}_R)\sin \delta =0$. Thus unequal exterior strings require the quarter-wave condition. Equal exterior strings with a different middle string instead transmit perfectly at integer half-wavelengths; if all three mechanical impedances agree, every frequency is matched.

For a massless join between two ideal strings with a common tension $T$, require continuous transverse displacement and continuous transverse force $T{\partial }_x\Psi$. With $Z_j=\sqrt{T{\rho }_j}$ for linear mass densities ${\rho }_j$, the displacement reflection coefficient and transmission amplitude areThe reflected and transmitted mean-power fractions are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(Z_2/Z_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. Reflection from the heavier side reverses phase; reflection from the lighter side does not. Total reflection occurs in the limits $Z_2/Z_1\to \infty$ or zero, while a finite positive impedance mismatch transmits some energy. Equal mechanical impedances remove reflection.

Set the lower half-space load to $q_b={\mu }_b{p}_{\beta ,b}$ using the causal vertical wavenumber branch, which excludes a wave arriving from depth infinity. Apply the uniform-layer SH impedance update upwards through every layer, keeping particle velocity and traction continuous. For upper half-space impedance $q_a$, incident velocity $V_0$ and reflected velocity $R{V}_0$, the top values are $V=V_0(1+R)$ and $T=-q_a{V}_0(1-R)$. Their ratio gives the displayed reflection coefficient, incorporating every underlying interface and multiple reflection.

A hydraulic resistance $R$ imposes $q=p/R$. Combining this with forward and backward arterial pressure waves, $p=p_{+}+p_{-}$ and $q=Y(p_{+}-p_{-})$, gives the displayed reflection coefficient. A matched load $RY=1$ has no reflected pressure wave, while a closed end has reflection coefficient one. At a network junction the combined downstream input admittance replaces $1/R$.

A free acoustic membrane wave is a zero of the outgoing dispersion relation $D$. At this zero, the impedance seen from one side equals the negative of that side's outgoing normal acoustic impedance. Thus the continued by analytic continuation reflection coefficient has a pole. A real guided mode is evanescent normal to the sheet; a complex root can describe a leaky mode. The pole describes nontrivial homogeneous motion with no incoming wave, rather than unbounded passive reflection at an ordinary real incidence angle.

An ideal short circuit fixes zero electric potential difference across a terminal pair, corresponding to zero load electrical impedance. At the end of a lossless transmission line, its voltage-wave reflection coefficient is $-1$ and the terminal voltage is zero. The available current is still limited by source and circuit impedances; real short circuits can dissipate substantial Joule heating.

On a uniform lossless transmission line, incident and reflected sinusoidal waves give voltage maxima $|V^{+}|(1+|\Gamma |)$ and minima $|V^{+}|(1-|\Gamma |)$. Their ratio is the voltage standing wave ratio, $(1+|\Gamma |)/(1-|\Gamma |)$, where $\Gamma$ is the voltage-wave reflection coefficient. A matched termination gives one, while perfect reflection gives an infinite ratio. These expressions assume a common real characteristic impedance and no propagation attenuation between the compared extrema.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.

Wave reflection returns part of an incident wave into the medium from which it arrived, as a consequence of an interface or spatial variation of material properties. An amplitude reflection coefficient compares the reflected and incident waves; its squared modulus gives their wave intensity ratio when both use the same medium and normalization. Reflected energy has not necessarily undergone wave absorption.