Parallel-Plate Waveguide

The fig shows two parallel plates of width  and of infinite length, separated by a distance d. one of the wave solution between these two plates is

Where  and  . At  and , the boundary conditions  and  are satisfied. The other boundary conditions  and  imply that  and  at  ,  and  at . The current on the bottom plate, I, and the voltage across the two plates, V, can be defined as

                                  , 

Note that the current on the top plate is of opposite sign with that on the bottom plate, implying that the currents on both plates flow in opposite directions.

Substituting the field expression into the Faraday’s law and Ampere’s law, respectively, we have

By using the definition in (1.15), (1.16) can be transformed into

Where  (henry/m),  (farad/m) are the per-unit-length inductance and capacitance, respectively, of the parallel-plate waveguide. Equations (1.17) and (1.18) are called telegrapher’s equations or transmission line equations, we have

                                                         (1.19)

Since . The solution to (1.19) are

                                                    ,

Where  is the characteristic impedance of the transmission line.

Electric Field Distribution and Surface Charge Distribution

                                 Magnetic Field Distribution and Surface Current Distribution