Page 30 - Fiber Optic Communications Fund

Electromagnetics and Optics 11

Substituting Eqs. (1.52) and (1.53) into Eq. (1.50), we obtain
xy z
⎡ ⎤
E E H
⎢ ⎥ x x y
∇× E = = y − z =− y. (1.54)
z y t
⎢ x y z ⎥
⎢ ⎥
⎣E
x 00 ⎦
Equating y- and z-components separately, we find
E x H y
=− , (1.55)
z t
E x
= 0. (1.56)
y
Substituting Eqs. (1.52) and (1.53) into Eq. (1.51), we obtain
xy z
⎡ ⎤
H H E
⎢ ⎥ y y x
∇× H = =− x + z = x. (1.57)
⎢ x y z ⎥ z x t
⎢ ⎥
⎣ 0 H 0 ⎦
y
Therefore,
H y E x
=− , (1.58)
z t
H y
= 0. (1.59)
x
Eqs. (1.55) and (1.58) are coupled. To obtain an equation that does not contain H , we differentiate Eq. (1.55)
y
with respect to z and differentiate Eq. (1.58) with respect to t,
2
E H y
x =− , (1.60)
z 2 tz
2
2
H y E x
=− . (1.61)
zt t 2
Adding Eqs. (1.60) and (1.61), we obtain
2
E x E x
= . (1.62)
z 2 t 2
The above equation is called the wave equation and it forms the basis for the study of electromagnetic wave
propagation.

1.5.3 Free-Space Propagation
2
2
2
For free space, = = 8.854 × 10 −12 C ∕Nm , = = 4 × 10 −7 N/A , and
0 0
1 8
≃ 3 × 10 m/s, (1.63)
c = √

0 0