Example of a Normally Incident UPW on a Lossless Half Space.
Example N29.1. A UPW is incident from free space onto a glass half space with er = 4 and mr = 1, as shown in the figurebelow. It is specified that
`E = ax1. e-jbz V/m
and f = 200 MHz.
(a) Determine the time domain `E and `H for the incident, reflected, and transmitted fields.
Therefore, for the incident fields:
where h0 = 376.73037 beta. We now have the complete time domain expressions for the incident `E and `H .
For the reflected fields:
Note that in turns out `Er points in the ax- direction, which is opposite our initial assumption in the figure above. But with `Hr pointing in the ay direction, the cross product `E ×`H points in the proper direction of propagation, which is the ?az direction for the reflected wave. For the transmitted fields in region 2 we have:
For the transmitted magnetic field:
(b) Compute the time average power transmitted through a 5- `M2 surface of the glass. The time average Poynting vector in region 2 is given as.
which has only a z component. Since `E2 and `H2 are not functions of x or y, the total power transmitted though this surface area is simply
(c) Compute the standing wave ratio (SWR).
From the VisualEM “Example 6.8” Worksheet
• In the first plot of the worksheet (“Total Ex in regions 1 and 2”) it is apparent that there is more oscillation in x per unit z in region 2 (z > 0) than in region 1 (z < 0). Why ? Because the wavelength is smaller in region 2 than region 1:
A smaller wavelength indicates a more electromagnetically “dense” material.
• The next plot in the worksheet shows that `E1 is the sum of incident and reflected waves.
• The final plot is an animation of the total electric field in both regions 1 and 2. Study this animation carefully and notice:
. The wave “pulsates” in region 1. This is what partial wave interference looks like in the time domain. There is no pulsation in region 2 since there is no interference there.