Page 32 - Fiber Optic Communications Fund

Electromagnetics and Optics 13

2
2
E E
x x 2
= , (1.74)
z 2 u 2
2
2
E E
x = x . (1.75)
t 2 u 2
Using Eqs. (1.74) and (1.75) in Eq. (1.68), we obtain
2
2
E x 1 E x
2
= . (1.76)
2
u 2 u 2
Therefore,
1
=± , (1.77)

( z ) ( z )
E = f t + or E = f t − . (1.78)
x x

The negative sign implies a forward-propagating wave and the positive sign indicates a backward-propagating
wave. Note that f is an arbitrary function and it is determined by the initial conditions as illustrated by the
following examples.
Example 1.1
Turn on a flash light for 1 ns then turn it off. You will generate a pulse as shown in Fig. 1.9 at the flash light
(z = 0) (see Fig. 1.10). The electric field intensity oscillates at light frequencies and the rectangular shape
shown in Fig. 1.9 is actually the absolute field envelope. Let us ignore the fast oscillations in this example
1
and write the field (which is actually the field envelope )at z = 0as
( )
t
E (t, 0)= f(t)= A rect , (1.79)
x 0
T
0
f(t) = A 0 rect (t/T 0 )
A 0

t
T 0 = 1 ns
Figure 1.9 Electrical field E (t, 0) at the flash light.
x

Screen

Flash light
1 m
z = 0 z = 1 m

Figure 1.10 The propagation of the light pulse generated at the flash light.

1 It can be shown that the field envelope also satisfies the wave equation.