= Solution
\b[Distributed circuit model.] A <transmission line> must be treated as a distributed system when its propagation <time> is appreciable compared with the <signal> timescale. Divide a uniform two-conductor line into elements of <length> $dx$. Each has series resistance $R'dx$ and <inductance> $L'dx$, and shunt <electrical conductance> $G'dx$ and <capacitance> $C'dx$. The primes denote quantities per unit <length>, with units $\mathrm{\Omega\,m^{-1}}$, $\mathrm{H\,m^{-1}}$, $\mathrm{S\,m^{-1}}$ and $\mathrm{F\,m^{-1}}$, respectively.
Kirchhoff's <voltage> and current laws give the <Telegrapher's equations>:
$$
\boxed{\frac{\partial V}{\partial x}=-R'I-L'\frac{\partial I}{\partial t},
\qquad \frac{\partial I}{\partial x}=-G'V-C'\frac{\partial V}{\partial t}.}
$$
For a lossless line, $R'=G'=0$, differentiation gives $V_{xx}=L'C'V_{tt}$ and the same <wave equation> for $I$. The propagation <speed> and <characteristic impedance> are
$$
\boxed{u=\frac1{\sqrt{L'C'}},\qquad Z_0=\sqrt{\frac{L'}{C'}}.}
$$
A general lossless <signal> has $V=V^+(t-x/u)+V^-(t+x/u)$ and $I=[V^+(t-x/u)-V^-(t+x/u)]/Z_0$. Thus <characteristic impedance> is the <voltage>/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no <energy>, because <energy> is carried away by propagation.
For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas are
$$
\gamma_{\rm prop}=\sqrt{(R'+i\omega L')(G'+i\omega C')},
\qquad Z_0=\sqrt{\frac{R'+i\omega L'}{G'+i\omega C'}}.
$$
Choose the square-root branches for forward attenuation and positive forward <power>. The <real part> of $\gamma_{\rm prop}$ is attenuation per unit <length>; its <imaginary part> is the phase <wavenumber>. For small losses, attenuation is approximately $R'/(2Z_0)+G'Z_0/2$, using the lossless real $Z_0$ in this estimate.
\b[Fields and geometry.] Two separated conductors support a <transverse electromagnetic mode>, with electric and <magnetic fields> transverse to the propagation axis. For a homogeneous <dielectric> and negligible conductor penetration, both examples below have $L'C'=\mu\epsilon$ and hence $u=1/\sqrt{\mu\epsilon}$. The distributed circuit model follows from the field <energy> and charges in each slice.
For a <coaxial cable> with inner radius $a$ and inner surface of the outer conductor at $b$, <Gauss's law> and <Ampère's law> give
$$
E_\rho=\frac{V}{\rho\ln(b/a)},\qquad H_\phi=\frac{I}{2\pi\rho},\quad a<\rho<b.
$$
The <electric field> is radial and the <magnetic field> circles the axis. Integration of <voltage> and field <energy> gives
$$
C'=\frac{2\pi\epsilon}{\ln(b/a)},\qquad
L'=\frac{\mu}{2\pi}\ln(b/a),\qquad
\boxed{Z_0=\frac1{2\pi}\sqrt{\frac\mu\epsilon}\ln\frac ba.}
$$
The equal and opposite return current on the outer conductor confines the ideal external field to the annulus.
For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The <electric field> crosses the gap, $E\simeq V/d$, while the <magnetic field> lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. Therefore
$$
C'=\frac{\epsilon w}{d},\qquad L'=\frac{\mu d}{w},\qquad
\boxed{Z_0=\frac d w\sqrt{\frac\mu\epsilon}.}
$$
For a forward wave, $\mathbf E\times\mathbf H$ points along the line. Integrating this <Poynting vector> over either cross-section gives instantaneous <power> $VI$. For complex peak amplitudes, average <power> is $\operatorname{Re}(VI^*)/2$, or $|V^+|^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an <energy flux> predominantly in the <dielectric> between them.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-2-2017-line-fields.png]
{title=Electric and magnetic fields in coaxial and parallel-plate transmission lines}
{description=Original transverse cross-sections for a forward TEM wave. Blue arrows show the electric field and orange arrows the magnetic field strength. Dots denote current out of the page and crosses current into it. Parallel-plate edge fields and coaxial conductor penetration are neglected.}
\b[Terminations and reflection.] At a load $Z_L$, require $V_L=Z_LI_L$. If $V^+$ and $V^-$ are the incident and reflected <voltage> amplitudes there, the total current is $(V^+-V^-)/Z_0$. Thus
$$
\boxed{\Gamma_V=\frac{V^-}{V^+}=\frac{Z_L-Z_0}{Z_L+Z_0},\qquad \Gamma_I=-\Gamma_V.}
$$
A matched load, $Z_L=Z_0$, absorbs the incident <power> with no reflection. An <open circuit> has $\Gamma_V=+1$, a current node and a doubled <voltage> at the end. A <short circuit> has $\Gamma_V=-1$, a <voltage> node and a doubled current there. A purely reactive load on a lossless line has $|\Gamma_V|=1$ and accepts no average <power>. For real $Z_0$, the reflected <power> fraction is $|\Gamma_V|^2$ and the accepted fraction is $1-|\Gamma_V|^2$.
Incident and reflected waves produce a standing-wave pattern with <voltage> <standing wave ratio> $(1+|\Gamma_V|)/(1-|\Gamma_V|)$. The <input impedance> of a lossless <length> $\ell$ is
$$
Z_{\rm in}=Z_0\frac{Z_L+iZ_0\tan(k\ell)}{Z_0+iZ_L\tan(k\ell)}.
$$
This gives $iZ_0\tan(k\ell)$ for a short and $-iZ_0\cot(k\ell)$ for an open. A quarter-wave section gives $Z_{\rm in}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return <time> is $2\ell/u$. Real conductor resistance, <dielectric loss> and departures from a TEM field profile limit the ideal model at high <frequency>.
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