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UPWs Normally Incident on Lossless Half Space

Uniform Plane Waves Normally Incident on a Lossless Half Space.


The current sheet solution in Lecture 25 provided us with much information on the properties of uniform plane waves:
. `E ×`H equals the power flow density, as we saw in the previous lecture, and also gives the direction of propagation.
.E ^ `H and both `E and `H are perpendicular to the direction of propagation.
. Ratios of perpendicular components of `E and `H = ± h
. l = 2p /b and u =1/ me Now imagine that a UPW is “incident” on a half space as shown in the figure below. Our quest now will be to determine `E and H everywhere.

In this lecture, we will consider the reflection and transmission of UPWs normally (i.e., perpendicularly) incident on this planar boundary.
Comments
• The incident UPW is produced by an infinite current sheet located somewhere in region 1. We’ll just specify an incident UPW with electric field Ei rather than solving the current
sheet problem again. • Because of the half space, we expect some “reflection” and “transmission” at the interface.
• The total field in region 1 is the sum of an incident and a reflected wave. The total field in region 2 is just the transmitted field.
Solution for Reflected and Transmitted Waves
Proceeding with the solution, we first draw the vector triplets (E , H , and direction) associated with each UPW (incident, reflected, and transmitted).

It is important to choose `Er and Et in the same direction as `Ei . The directions of the corresponding magnetic field vectors can be found using the RHR. Next, we write the solutions for `E and `H in each region making sure to use b1 ,h1 in region 1 and b2 ,h2 in region 2:
Region 1–

Region 2–

To determine the unknown coefficients `E1- and `E2+ (assuming `E1+ is known) we apply the boundary conditions:
• `Etan continuous at z = 0: `Ex1|z=0 =` Ex2|z=0
Using (1) and (3) in this boundary condition equation, then

Similar to our analysis of TLs, we will define the ratio

as the electric field reflection coefficient at z = 0 and

as the electric field transmission coefficient at z = 0. Substituting these definitions into (5) we find
1+ T = T
• Htan continuous at z = 0:
From (2) and (4)
Hy1|z=0 = Hy2|z=0
or

Solving (8) and (9) for and gives

You probably recognize the form of (10) from your previous work with transmission lines where

As we can see in (10) and (11), for UPWs the amount of reflection and transmission at an interface depends on the contrast in the intrinsic impedances of the two half spaces. We have now completed the solution for the UPW incident on a half space:
For z ³ 0


Example N28.1: A UPW is incident on the half space shown.

Determine (a) the reflection and transmission coefficients at the interface y = 0, and (b) the total magnetic field in each region.
.
The first step is to sketch the vector triplets for the incident, reflected, and transmitted UPWs:

(a) From (3) and (14)
E2+ = E1+
In this example, it is given that

Using

(b) From the given `Ei and the sketch above

To determine `Hr , first write `Er

From this `Er and the sketch above

Consequently, in region 1

while in region 2 using the given `Et and the sketch above