Page 37 - Fiber Optic Communications Fund
P. 37
18 Fiber Optic Communications
Similarly, multiplying Eq. (1.104) by H ,wehave
y
H 2 y E x
=−H y . (1.107)
2 t z
Adding Eqs. (1.107 ) and (1.106) and integrating over the volume V, we obtain
[ 2 ] L [ ]
E 2 x H y H y E x
t ∫ 2 + 2 dV =−A ∫ E x z + H y z dz. (1.108)
V 0
On the right-hand side of Eq. (1.108), integration over the transverse plane yields the area A since E and H y
x
are functions of z only. Eq. (1.108) can be rewritten as
[ 2 ] L | L
E x 2 H y [ ] |
x
x
t ∫ 2 + 2 dV =−A ∫ z E H dz =−AE H y| | . (1.109)
y
V 0 |
|0
2
2
The terms E ∕2 and H ∕2 represent the energy densities of the electric field and the magnetic field, respec-
y
x
tively. The left-hand side of Eq. (1.109) can be interpreted as the power crossing the area A and therefore,
2
E H is the power per unit area or the power density measured in watts per square meter (W/m ). We define
x
y
a Poynting vector as
= E × H. (1.110)
The z-component of the Poynting vector is
= E H . (1.111)
z
y
x
The direction of the Poynting vector is normal to both E and H, and is in fact the direction of power flow.
In Eq. (1.109), integrating the energy density over volume leads to energy and, therefore, it can be
rewritten as
1 d
= (0)− (L). (1.112)
z
z
A dt
The left-hand side of (1.112) represents the rate of change of energy per unit area and therefore, has the
z
dimension of power per unit area or power density. For light waves, the power density is also known as the
optical intensity. Eq. (1.112) states that the difference in the power entering the cross-section A and the power
leaving the cross-section A is equal to the rate of change of energy in the volume V. The plane-wave solutions
for E and H are given by Eqs. (1.87) and (1.90),
x
y
E = E cos (t − kz), (1.113)
x x0
H = H cos (t − kz), (1.114)
y y0
E 2
2
= x0 cos (t − kz). (1.115)
z
The average power density may be found by integrating it over one cycle and dividing by the period T = 1∕f,
1 E 2 x0 T
2
z av = T ∫ 0 cos (t − kz)dt, (1.116)
1 E 2 x0 T 1 + cos [2(t − kz)]
= dt (1.117)
T ∫ 0 2
E 2 x0
= . (1.118)
2