7.1.7 Wave impedances ZTM and ZTE

For a plane wave the ratio of the transverse fields is the intrinsic impedance $\eta$. In a guide the corresponding ratio is the wave impedance of the mode.

TM ($H_z=0$). From Eq. (7.1.11), $E_x=-\frac{\mathrm{j}\tau }{h^2}\frac{\partial E_z}{\partial x}$ and $H_y=-\frac{\mathrm{j}\omega \epsilon }{h^2}\frac{\partial E_z}{\partial x}$, while $E_y=-\frac{\mathrm{j}\tau }{h^2}\frac{\partial E_z}{\partial y}$ and $H_x=\frac{\mathrm{j}\omega \epsilon }{h^2}\frac{\partial E_z}{\partial y}$. The derivatives cancel in the ratios:

since $k/(\omega \epsilon )=\sqrt{\mu /\epsilon }=\eta$.

TE ($E_z=0$). Now $E_x=-\frac{\mathrm{j}\omega \mu }{h^2}\frac{\partial H_z}{\partial y}$, $H_y=-\frac{\mathrm{j}\tau }{h^2}\frac{\partial H_z}{\partial y}$, and similarly for $E_y$, $H_x$:

since $\omega \mu /k=\eta$.

Below cutoff $\tau =-\mathrm{j}\alpha$, and the impedances become imaginary:

A purely reactive $Z$ puts $E_T$ and $H_T$ in quadrature, so that $\frac{1}{2}\mathrm{Re}\{E\times H^{\ast }\}\cdot a_z=\frac{1}{2}\mathrm{Re}\{|E_T{|}^2/Z^{\ast }\}=0$: an evanescent mode stores energy near its source but carries no power.