Input impedances of terminated sections of transmission line When dealing with optics and dielectric waveguides, a way to obtain the guided modes is to impose the "transverse resonance condition". 
Let the following be a transmission line with characteristic impedance $Z_0$ which is terminated at its both ends with loads $Z_1$ and $Z_2$. Its total length is $l$.

Let's now split the line at an arbitrary position $x$ (of course, $0 \leq x \leq l$) and let's compute the input impedances $Z_{in1}$ and $Z_{in2}$ in both the directions:

$$Z_{in1} = Z_0 \frac{Z_1 + j Z_0 \tan (\beta x)}{Z_0 + j Z_1 \tan (\beta x)}$$
$$Z_{in2} = Z_0 \frac{Z_2 + j Z_0 \tan [ \beta (l - x) ]}{Z_0 + j Z_2 \tan [ \beta (l - x) ]}$$
The above mentioned transverse resonance condition imposes that $Z_{in1} + Z_{in2} = 0$ regardless of the chosen position $x$. But:


*

*does the sum $Z_{in1} + Z_{in2}$ depend on $x$?

*And if not, why?



My attempt: the only useful relation that I found is
$$\tan [ \beta (l - x) ] = \frac{\tan(\beta l) - \tan (\beta x)}{1 + \tan(\beta l) \tan (\beta x)}$$
This leads to
$$Z_{in1} + Z_{in2} = Z_0 \frac{Z_1 + j Z_0 \tan (\beta x)}{Z_0 + j Z_1 \tan (\beta x)} + Z_0 \frac{Z_2 + j Z_0 \displaystyle \frac{\tan(\beta l) - \tan (\beta x)}{1 + \tan(\beta l) \tan (\beta x)}}{Z_0 + j Z_2 \displaystyle \frac{\tan(\beta l) - \tan (\beta x)}{1 + \tan(\beta l) \tan (\beta x)}}$$
Even trying to simplify the $1 + \tan(\beta l) \tan (\beta x)$ denominator in the second addend, the sum $Z_{in1} + Z_{in2}$ still has not a common denominator and a lot of terms depending on $x$. Is there another way to proceed? The only reasonable hypothesis that can be done is that the two loads $Z_1$ and $Z_2$ are purely reactive, that is: they are of the form $Z_1 = jX_1$ and $Z_2 = jX_2$, with $X_1, X_2 \in \mathbb{R}$. This, anyway, doesn't seem to simplify the above expression.
 A: Think from first principles in terms of the waves travelling in both directions. At any point $x$, looking rightwards towards impedance $Z_2$, the relationship between leftwards running $E^-$ and rightwards running $E^+$ electric fields is:
$$E^- = E^+ \,\Gamma_2\,\exp(-2\,i\,\beta\,(\ell-x))\tag{1}$$
where the reflexion co-efficient at the termination is $\Gamma_2 = (Z_2-Z_0)/(Z_1-Z_0)$. Do the same looking leftwards at the relationship between the two waves established by impedance $Z_1$:
$$E^+ = E^- \,\Gamma_1\,\exp(-2\,i\,\beta\,x)\tag{2}$$
Dividing (2) into (1) will eliminate both the electric fields. It will also eliminate $x$, leaving you with the resonance condition on the propagation constant $\beta$. Such a system will only support a discrete set of frequencies.
It is highly analogous to a pitched pipe wind musical instrument like a flute.

Now, as for the independence of the impedance resonance equation from $x$, this simply follows from the resonance condition from (1) and (2), which is:
$$\frac{1}{\Gamma_1}=\Gamma_2\,\exp(2\,i\,\beta\,\ell)\;\Leftrightarrow\;\tilde{\Gamma}_1(x)=\tilde{\Gamma}_2(x)^{-1}\tag{3}$$
where $\tilde{\Gamma}_j(x)$ have their obvious meanings as the reflexion co-efficients at the point $x$. The impedance equation is implied by, and implies a condition of the form $f\left(\tilde{\Gamma_1}\right) = f\left(\tilde{\Gamma}_2^{-1}\right)$, where $f$ is any bijective function  between the set of extended complex numbers (the Riemann sphere) and itself. In this case, $f$ is the Möbius transformation given by $f^{-1}(z) = \frac{z-Z_0}{z+Z_0}$ (indeed, any holomorphic bijection between the Riemann sphere and itself has to be some Möbius transformation, but that is an aside). You can easily verify that $f\left(\frac{1}{z}\right) = -f(z)$, which gives you the condition you seek, for all $x$. The fact that all such conditions can be returned to (3) shows that the resonance is independent of $x$.
The Möbius transformation $f$ is of course the grounding for the Smith chart. Changing $x$ simply rotates your Smith Chart about its origin.
