= Solution
$Z$ is the terminating load impedance, $Z_0$ the line's <characteristic impedance>, and $k$ the propagation <wave number>; $l$ is its <length>. In the lossless convention printed, an open circuit gives $Z_{\rm in}=-iZ_0\cot kl$, a short circuit gives $iZ_0\tan kl$, and a matched load $Z=Z_0$ gives \b[$Z_{\rm in}=Z_0$ for every <length>]. The divergent or zero limits at isolated electrical <lengths> are understood continuously.
Put $\theta=kl$, $c_\theta=\cos\theta$, $s_\theta=\sin\theta$. The voltage/current <matrix> of one section is
$$
M_j=\begin{pmatrix}c_\theta&iZ_js_\theta\\is_\theta/Z_j&c_\theta\end{pmatrix}.
$$
For a pair beginning with $Z_1$, write $M_1M_2=\left(\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right)$. Its entries are $A=c_\theta^2-(Z_1/Z_2)s_\theta^2$, $D=c_\theta^2-(Z_2/Z_1)s_\theta^2$, $B=i(Z_1+Z_2)s_\theta c_\theta$, and $C=i(1/Z_1+1/Z_2)s_\theta c_\theta$. The <alternating-impedance transmission line> has unchanged input $X$, its <Bloch impedance>, obeying $X=(AX+B)/(CX+D)$. Hence, away from the removable dividing zeros,
$$
iX^2(Z_1+Z_2)+X(Z_1^2-Z_2^2)\tan\theta-iZ_1Z_2(Z_1+Z_2)=0,
$$
$$
\boxed{X=\frac{i(Z_1-Z_2)\tan\theta\ \pm\sqrt{4Z_1Z_2-(Z_1-Z_2)^2\tan^2\theta}}2.}
$$
For real positive $Z_1,Z_2$, select the branch carrying <energy> into the line or decaying into it. A positive real part selects the pass-band root; infinitesimal passive loss resolves the stop-band branch. At $\theta=m\pi$ the cell <matrix> is the identity, so the exact fixed-point condition alone does not determine $X$; at odd quarter-wave <lengths> the appropriate limiting impedance can be zero or infinite.
In a <transmission-line stop band>, a lossless semi-infinite ladder totally reflects because its physical input is reactive. Equivalently its cell has
$$
\tfrac12\operatorname{tr}M=1-\frac{(Z_1+Z_2)^2}{2Z_1Z_2}\sin^2\theta<-1.
$$
Thus, for nonzero contrast, the stop-band interiors satisfy
$$
\boxed{|\tan kl|>\frac{2\sqrt{Z_1Z_2}}{|Z_1-Z_2|},\quad\text{equivalently}\quad
\sin^2kl>\frac{4Z_1Z_2}{(Z_1+Z_2)^2}.}
$$
With $\theta_c=\arcsin[2\sqrt{Z_1Z_2}/(Z_1+Z_2)]$, these are $m\pi+\theta_c<kl<(m+1)\pi-\theta_c$; band-edge reflection follows by a limit. If $k=\omega/u$, multiply the endpoints by $u/l$ for <angular frequencies>. Identical impedances have no stop band. A finite ladder can transmit by tunnelling, unlike the semi-infinite idealization.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-1-periodic-line.png]
{title=Calculated Bloch trace and reflected power for a chosen alternating-impedance line}
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