光纤通信技术答案 2

4.6 Solution:
For a light wave，if the electrical field and/or magnetic field part can be described by sine (or cosine) function; than we call the light wave a time-harmonic EM field. For wave equations, the time harmonic EM field provides an effect way to significantly simplify the solution.
4.3 Solution:
Because the circuit and the electrical field we deal with is usually stable, and the relation ∂D ∂t << J is always true so that we can omit the displacement currents part.
Pin , in fact, the actual loss value is always negative, so the attenuation converts to positive.
3.31 Solution:
The physical reason is that an EM wave propagating inside an optical fiber has to meet the boundary-condition requirements. Its phase ( ω t − β z ) has to be equal to 2π × k ,where k is an integer. EM waves that meet this requirement will exist as a stable pattern, or mode. EM waves that do not meet this requirement will not appear. This is why optical fiber supports EM waves-modes-that make stable patterns and does not support others.
3.17 Solution:
a. Bending induced loss depends on the bending radius and the operating wavelength. b. The specifications describe the relationship between bending radius and loss induced. c. at a specific wavelength, the smaller the radius is, the more power the fiber lose.
2 n12 - n2 , we can find more than one.For example, n1 =1.4860, n2 =1.4669 & n1 =1.4800, n2 =1.4608.
This formula reveals that the average and relative difference of n1 and n2 are more crucial to N.A. than themselves. In other words, N.A. might be the same for various combinations of n1 and n2 with the same average and relative difference.
3.32 Solution:
Δtmod al = tc - t0 = L /(v cos α c ) - L / v = Ln1 n1 - n2 ( ) = ( Ln1 / c)Δ c n2
3.38 Solution:
For short distance networks, I’ll just choose normal MMF since it coats less. While for long distance networks, the transport quality is more important and the huge capacity of SMF will surely meet our needs. And for intermediate distance, we should take both the two aspects into account and then decide which the best one is.
−
0.25× 20 10
= 0.632(mW )
3.23 Solution:
It’s a positive number. According to the formula Loss ( dB ) = - 10 log10 ( Pout / P in ) , we get a positive loss value just because the negative sign is used manually. Since Pout is always less than
3.18 Solution:
The coating is used for protection of the core. For different coatings, the endurance ability is different. So, for the same test condition, the attenuation caused by coating are different.
3.46 Solution:
Manufactures specify fiber bandwidth as bandwidth-length product. They measure the bandwidth of a specific piece of fiber and multiply the result by the fiber length.
4.2 Solution:
The first equation, ∇ ⋅ D =
ρV ,express the idea that the flow source of an electric field is a
free charge, Q. The second equation, ∇ ⋅ B = 0 , states that there is no flow source for a magnetic field. The third equation, ∇ × E = ∂B time-varying magnetic field. The fourth equation ∇ × H = J + ∂D
3.22 Solution:
P − 10 log out P in , we can easily get the output power: From the formula: A( dB / km) = L
Pout = Pin × 10
− AL 10
= 0.002 × 10
Assignment 2: solutions
3.12 Solution:
The same type of calculation made for this problem:
n2 =
3.13 Solution:
n12 - NA2 = 1.48602 - 0.23752 = 1.4669
According to the formula NA =
3.14 Solution:
Attenuation is the decrease in light power during light propagation along an optical fiber over an unit distance. Loss is the difference of the power at the output end of a fiber, Pout, to the power launched into the fiber, Pin.
∂t
, represent that the vortex source of an electric field is a
∂t
, means that the vortex source of a time-varying
magnetic field is the time-changing density of conduction and displacement currents.