= Solution
Choose depth $z$ positive downwards and harmonic time dependence $e^{-i\omega t}$. At normal incidence, <P waves> and the two polarizations of <S waves> are independent scalar problems: a <P wave> uses the longitudinal modulus $\rho\alpha^2$, and an <S wave> uses the <shear modulus> $\rho\beta^2$. In a uniform layer with <wave speed> $c$ and <seismic impedance> $q=\rho c$, write <velocity> and compressive <traction> as
$$
V=Ae^{i\omega z/c}+Be^{-i\omega z/c},\qquad
P=-T=q\left(Ae^{i\omega z/c}-Be^{-i\omega z/c}\right).
$$
At a perfectly bonded interface the <velocity> and <traction> are continuous. For incident <velocity> <wave amplitude> one from medium 1 into medium 2, these two conditions are $1+r_v=t_v$ and $q_1(1-r_v)=q_2t_v$. Thus
$$
\boxed{r_v=\frac{q_1-q_2}{q_1+q_2},\qquad t_v=\frac{2q_1}{q_1+q_2}.}
$$
Reflection is determined by <seismic impedance>, not by speed alone. A step to larger <seismic impedance> reverses the reflected <velocity>/<displacement field> waveform; the pressure <reflection coefficient> has the opposite sign. Its transmitted pressure coefficient is $t_p=2q_2/(q_1+q_2)$. Equal <seismic impedance> eliminates the reflection even when speed and <wavelength> change. At a free surface <traction> vanishes and <velocity> reflects with coefficient $+1$; at a rigid surface <velocity> reflects with coefficient $-1$.
The mean <elastic-wave energy flux> is $\tfrac12\operatorname{Re}(PV^*)=q(|A|^2-|B|^2)/2$ for positive real $q$. Therefore the reflected and transmitted <energy> fractions are
$$
\mathcal R=|r_v|^2,\qquad \mathcal T=\frac{q_2}{q_1}|t_v|^2,\qquad
\boxed{\mathcal R+\mathcal T=1.}
$$
A <displacement field> <transmission coefficient> greater than one is compatible with <conservation of energy>; its flux must be weighted by the receiving <seismic impedance>.
For many layers, use the <seismic layer transfer matrix>. A layer of thickness $h_j$ has <wave phase> thickness $\delta_j=\omega h_j/c_j$ and
$$
\begin{pmatrix}V\\P\end{pmatrix}_{z+h_j}
=M_j\begin{pmatrix}V\\P\end{pmatrix}_{z},\qquad
M_j=\begin{pmatrix}\cos\delta_j&i\sin\delta_j/q_j\\iq_j\sin\delta_j&\cos\delta_j\end{pmatrix},\qquad\det M_j=1.
$$
This follows either by eliminating $A,B$ between the two faces or by integrating $V'=i\omega P/(\rho c^2)$ and $P'=i\omega\rho V$. For the complete stack put $M=M_N\cdots M_1=\begin{pmatrix}A&B\\C&D\end{pmatrix}$. With upper and lower <half-space> impedances $q_a,q_b$, an outgoing lower wave satisfies $P_b=q_bV_b$. Solving this condition gives the input <seismic impedance>
$$
Z_{\rm in}=\frac{q_bA-C}{D-q_bB},\qquad
\boxed{R=\frac{q_a-Z_{\rm in}}{q_a+Z_{\rm in}},\quad
T=\frac{2q_a}{q_aD+q_bA-C-q_aq_bB}.}
$$
Here $T$ is lower-face <velocity> divided by upper-face incident <velocity>, including the propagation <wave phase>. For lossless layers and real <frequency>, $|R|^2+(q_b/q_a)|T|^2=1$. This supplies a sensitive check on any spreadsheet or numerical implementation; it is not correct simply to multiply the individual interface transmissions and discard the <elastic waves>.
The physical reason is repeated <wave interference>. In one layer between media 0 and 2, let $r_{ij},t_{ij}$ denote the <velocity> coefficients above and $\delta=\omega h/c_1$. The first transmitted wave has <wave amplitude> $t_{01}t_{12}e^{i\delta}$. Every extra round trip supplies $r_{10}r_{12}e^{2i\delta}$. Summing the <geometric series> gives
$$
T=\frac{t_{01}t_{12}e^{i\delta}}{1-r_{10}r_{12}e^{2i\delta}},\qquad
R=r_{01}+\frac{t_{01}t_{10}r_{12}e^{2i\delta}}{1-r_{10}r_{12}e^{2i\delta}}.
$$
In the time domain the successive arrivals are delayed by $2h/c_1$, producing reverberation. In the <frequency> domain the same delayed terms produce resonant maxima and minima. For a layer between identical <half-spaces> of impedance $q_0$,
$$
\boxed{\mathcal T=\left[1+\frac14\left(\frac{q_1}{q_0}-\frac{q_0}{q_1}\right)^2\sin^2\delta\right]^{-1}.}
$$
At $\delta=n\pi$ the layer transmits all the incident <energy>: the internally <elastic waves> cancel the direct reflection. Halfway between these resonances, a large impedance contrast gives strong reflection. For unequal exterior impedances, a quarter-wave layer with $q_1^2=q_aq_b$ transforms its lower load to $Z_{\rm in}=q_1^2/q_b=q_a$, again giving zero reflection at its design <frequency>. This matching is narrow-band.
Periodic alternation produces a <periodic seismic multilayer stop band>. For a cell of two layers with total thickness $d$, the cell <matrix> has reciprocal <eigenvalues> $e^{\pm iKd}$, so
$$
\cos Kd=\frac12\operatorname{tr}(M_2M_1)
=\cos\delta_1\cos\delta_2-\frac12\left(\frac{q_1}{q_2}+\frac{q_2}{q_1}\right)\sin\delta_1\sin\delta_2.
$$
When the absolute value of this expression exceeds one, the infinite periodic medium has no propagating cell mode; a long finite stack has exponentially small transmission. The <energy> is reflected, not absorbed. Pass bands instead contain oscillatory transmission peaks, becoming more numerous as the stack length increases. Disorder removes the exact cell periodicity but not multiple scattering or frequency-selective interference.
For a concrete reproducible investigation, take $q_H/q_L=4$, equal one-way travel times $\tau$ in both layer types, exterior impedance $q_L$, and <frequency> coordinate $\nu=2\omega\tau/\pi$. At $\nu=1$ each layer is a quarter <wavelength>; one cell has diagonal <matrix> entries $-4,-1/4$. For $N$ cells,
$$
\mathcal T(1)=\frac4{(4^N+4^{-N})^2}.
$$
It is about $0.22145$ for one cell, $6.10\times10^{-5}$ for four cells and $9.31\times10^{-10}$ for eight cells. The infinite-cell stop band is $2\arcsin(0.8)/\pi<\nu<2-2\arcsin(0.8)/\pi$. The accompanying calculation also shows the first transmitted arrivals for a single high-impedance slab: their <wave amplitudes> are $0.64(0.36)^n$ at times $(2n+1)\tau$. Their sum is one, consistent with unit zero-frequency transmission. These calculations illustrate both the spectrum and the pulse train without assuming access to the course's spreadsheet.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-79-layer-transmission.png]
{title=Original transfer-matrix calculations showing a periodic stack stop band and the transmitted reverberation train of one slab}
Finally, as $\omega\to0$ for a fixed finite stack, every $M_j\to I$, so the response approaches that of a direct interface between the exterior <half-spaces>. Equal exterior impedances give $R\to0,T\to1$. For <wavelengths> much longer than a repeating cell, expansion of $M_j$ to first order instead gives an effective medium with averaged <mass density> $\rho_{\rm eff}=\sum h_j\rho_j/d$ and harmonic-mean modulus $M_{\rm eff}^{-1}=\sum h_j/M_j^{\rm mod}/d$, where $M_j^{\rm mod}=\rho_jc_j^2$. Its speed is $\sqrt{M_{\rm eff}/\rho_{\rm eff}}$, not generally the arithmetic mean of the layer speeds. Attenuation makes the <energy> balance an inequality, while for a fixed oblique horizontal slowness the scalar SH calculation uses $q_j=\mu_jp_{\beta,j}$ and $\delta_j=\omega p_{\beta,j}h_j$. Oblique P-SV waves require coupled <displacement field> and <traction> <matrices> because of <mode conversion at a planar elastic interface>. Thick evanescent layers require scattering or impedance recursion rather than exponentially ill-conditioned transfer-matrix products.
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