Page 389 - Engineering Electromagnetics, 8th Edition

CHAPTER 11 The Uniform Plane Wave 371

increases, z must now decrease to keep the argument constant. For simplicity, we will
restrict our attention in this chapter to only the positive z traveling wave.
As was done for transmission line waves, we express the real instantaneous fields
of Eq. (15) in terms of their phasor forms. Using the forward-propagating field in (15),
we write:

1 jφ 1 − jk 0 z jωt 1 jωt jωt
E x (z, t) = |E x0 |e e e + c.c. = E xs e + c.c. = Re[E xs e ] (19)
2
2
E x0
where c.c. denotes the complex conjugate, and where we identify the phasor electric
field as E xs = E x0 e − jk 0 z .As indicated in (19), E x0 is the complex amplitude (which
includes the phase, φ 1 ).

EXAMPLE 11.1
8
Let us express E y (z, t) = 100 cos(10 t − 0.5z + 30 ) V/m as a phasor.
◦
Solution. We first go to exponential notation,
8
E y (z, t) = Re 100e j(10 t−0.5z+30 )

◦
8
and then drop Re and suppress e j10 t , obtaining the phasor
E ys (z) = 100e − j0.5z+ j30 ◦

Note that a mixed nomenclature is used for the angle in this case; that is, 0.5z is in
radians, while 30 is in degrees. Given a scalar component or a vector expressed as a
◦
phasor, we may easily recover the time-domain expression.

EXAMPLE 11.2
Given the complex amplitude of the electric field of a uniform plane wave, E 0 =
100a x +20 30 a y V/m, construct the phasor and real instantaneous fields if the wave
◦
is known to propagate in the forward z direction in free space and has frequency of
10 MHz.
Solution. We begin by constructing the general phasor expression:

E s (z) = 100a x + 20e j30 ◦ a y e − jk 0 z

8
7
where k 0 = ω/c = 2π × 10 /3 × 10 = 0.21 rad/m. The real instantaneous form is
then found through the rule expressed in Eq. (19):
7
7
e
E(z, t) = Re 100e − j0.21z j2π×10 t a x + 20e j30 ◦ e − j0.21z j2π×10 t a y

e
7
7
◦
= Re 100e j(2π×10 t−0.21z) a x + 20e j(2π×10 t−0.21z+30 ) a y

7 7 ◦
= 100 cos (2π × 10 t − 0.21z)a x + 20 cos (2π × 10 t − 0.21z + 30 ) a y