Lossy propagation

Lossy propagation may be represented as with circular cross-section by working out the lossy $z$ direction wavenumber, (Bruneau et al [44]). Starting from the lossy boundary condition gives $k_n$ as

$$k_n = \pm \sqrt{A_n + I_n - iR_n}$$ (2.70)

where $A_\nu$ is the square of the non-lossy version of $k_n$ which in rectangular geometry is

$$A_\nu =k^2 - \left(\frac{n_x \pi}{a}\right)^2 - \left(\frac{n_y \pi}{b}\right)^2.$$ (2.71)

The real part of the correction to $k_n^2$ is [44]

$$I_\nu = 2k\left( (2-\delta_{n_x,0}) \frac{\mbox{Re}(\epsilon... ...} +(2-\delta_{n_y,0}) \frac{\mbox{Re}(\epsilon_y)}{b} \right)$$ (2.72)

where the boundary specific admittances are

$$\epsilon_x = \left( 1 - \left(\frac{n_x \pi}{k a}\right)^2 \right) \epsilon_v + \epsilon_t$$ (2.73)

$$\epsilon_y = \left( 1 - \left(\frac{n_y \pi}{k b}\right)^2 \right) \epsilon_v + \epsilon_t$$ (2.74)

with $\epsilon_v = (1+i) \mbox{ } 2.03 \times 10^{-5} f^{1/2}$ and $\epsilon_t = (1+i) \mbox{ } 0.95 \times 10^{-5} f^{1/2}$. The imaginary part of the correction to $k_n^2$ is [44]

$$R_\nu = 2k\left((2-\delta_{n_x,0}) \frac{\mbox{Im}(\epsilon_... ...} +(2-\delta_{n_y,0}) \frac{\mbox{Im}(\epsilon_y)}{b}\right).$$ (2.75)

Using the same method as for cylindrical geometry, $k_n$ is the sum of real and imaginary parts

$$k_n = \chi_n + i \kappa_n$$ (2.76)

where $\chi_n$ and $\kappa_n$ are given by

$$\chi_n= \frac{1}{\sqrt{2}} \sqrt{\left\{A_n + I_n + \sqrt{(A_n + I_n)^2 + R_n^2}\right\}}$$ (2.77)

and

$$\kappa_n=-\frac{1}{\sqrt{2}} \sqrt{\left\{-(A_n + I_n) + \sqrt{(A_n + I_n)^2 + R_n^2}\right\}}.$$ (2.78)