Page 437 - Engineering Electromagnetics, 8th Edition

CHAPTER 12 Plane Wave Reflection and Dispersion 419

We thus have
E − = 23 E + (30)
x20
x20
We then write the magnetic field amplitudes in terms of electric field amplitudes
through
1
H + = E + (31a)
y20
η 2 x20
and
1 1
H y20 =− E − =− 23 E + (31b)
−
x20
x20
η 2 η 2
We now define the wave impedance, η w ,asthe z-dependent ratio of the total elec-
tric field to the total magnetic field. In region 2, this becomes, using (28a) and (28b),
E xs2 E + e − jβ 2 z + E − e jβ 2 z
x20
x20
η w (z) = =
+
H ys2 H y20 e − jβ 2 z + H − e jβ 2 z
y20
Then, using (30), (31a), and (31b), we obtain
e + 23 e
− jβ 2 z jβ 2 z
η w (z) = η 2
e − jβ 2 z − 23 e jβ 2 z
Now, using (29) and Euler’s identity, we have
(η 3 + η 2 )(cos β 2 z − j sin β 2 z) + (η 3 − η 2 )(cos β 2 z + j sin β 2 z)
η w (z) = η 2 ×
(η 3 + η 2 )(cos β 2 z − j sin β 2 z) − (η 3 − η 2 )(cos β 2 z + j sin β 2 z)
This is easily simplified to yield
η 3 cos β 2 z − jη 2 sin β 2 z
η w (z) = η 2 (32)
η 2 cos β 2 z − jη 3 sin β 2 z
We now use the wave impedance in region 2 to solve our reflection problem. Of
interest to us is the net reflected wave amplitude at the first interface. Since tangential
E and H are continuous across the boundary, we have
E xs1 + E − = E xs2 (z =−l) (33a)
+
xs1
and
H ys1 + H ys1 = H ys2 (z =−l) (33b)
+
−
Then, in analogy to (7) and (8), we may write
E + + E − = E xs2 (z =−l) (34a)
x10
x10
and
E x10 E − E xs2 (z =−l)
+
x10
− = (34b)
η 1 η 1 η w (−l)
where E + and E − are the amplitudes of the incident and reflected fields. We call
x10
x10
η w (−l) the input impedance, η in ,to the two-interface combination. We now solve