= Solution
For distributed <inductance> $L$ and <capacitance> $C$, a small segment obeys $V_z=-L I_t$ and $I_z=-C V_t$. Differentiation gives $V_{zz}=LCV_{tt}$ and $I_{zz}=LCI_{tt}$. For a forward wave,
$$
\boxed{v=\frac1{\sqrt{LC}},\qquad Z=\frac{V^+}{I^+}=\sqrt{L/C}.}
$$
A backward <voltage> wave has current $I^-=-V^-/Z$. This sign matters at an impedance discontinuity.
The source figure is the stacked-strip <parallel-plate transmission line> geometry, despite its use of the word coplanar. <Gauss's law> gives $C=\epsilon_0\epsilon_r a/d$. Opposite surface currents give <magnetic field> $B=\mu_0I/a$ between the strips by <Ampère's law>; the <magnetic flux> per unit <length> is $Bd$, hence $L=Bd/I=\mu_0d/a$. Therefore
$$
\boxed{v=\frac{c}{\sqrt{2.3}}\simeq1.98\times10^8\,\mathrm{m\,s^{-1}},\quad
Z=\frac da\sqrt{\frac{\mu_0}{2.3\epsilon_0}}
\simeq248.4\frac da\,\Omega.}
$$
No numerical $a/d$ is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.
Increasing the gap by $\alpha$ changes $L$ to $\alpha L$ and $C$ to $C/\alpha$, so $Z_2=\alpha Z_1$ while $v$ is unchanged. <voltage> and current continuity at the step give
$$
V_i+V_r=V_t,\qquad \frac{V_i-V_r}{Z_1}=\frac{V_t}{Z_2}.
$$
Consequently the <voltage> and energy-flux <transmission coefficients> are
$$
\boxed{t_V=\frac{2\alpha}{1+\alpha},\qquad
\mathcal T=\frac{Z_1}{Z_2}|t_V|^2=\frac{4\alpha}{(1+\alpha)^2},}
\qquad r_V=\frac{\alpha-1}{\alpha+1}.
$$
Here amplitudes are peak <voltage> amplitudes, with mean <power> $|V|^2/(2Z)$. The reciprocal gaps have $r_V(1/\alpha)=-r_V(\alpha)$: equal reflected <energy> but opposite <voltage> phase. A <voltage> <transmission coefficient> above one does not imply <energy> gain.
For a <finite impedance section of a transmission line>, put $\ell=l_2-l_1$ and $\delta=\omega\ell/v$. Multiple reflections inside the altered section interfere. Summing their <geometric series>, relative to the first boundary,
$$
t_{\rm whole}=\frac{(1-r_V^2)e^{i\delta}}{1-r_V^2e^{2i\delta}},\qquad
\boxed{\mathcal T_{\rm whole}=\left[1+\frac{(\alpha^2-1)^2}{4\alpha^2}\sin^2\delta\right]^{-1}.}
$$
Transmission is unity at $\delta=n\pi$ and minimal at $\delta=(n+1/2)\pi$, with minima $4\alpha^2/(1+\alpha^2)^2$. Resonances are separated by $\Delta f=v/(2\ell)$; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-line-resonances.png]
{title=Original calculated power transmission through a finite changed-impedance section between matched lines}
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