5. Attenuation in rectangular waveguides

So far, we've assumed that the walls of the guide consist of perfect conductors, which is obviously an ideal case of lossless propagation. In fact, the existence of currents in walls of finite conductivity implies the dissipation of energy by the Joule effect and, consequently, lossy propagation. The attenuation coefficient is given by : $\alpha =\frac{1}{2}\frac{\text{ Puissance perdue dans les parois }}{\text{ Puissance transmise dans le diélectrique }}$

The power lost at a point on the guide walls is : $\frac{1}{2}{R}_S{|\stackrel{}{}}^{}$