武汉理工 光纤通信复习材料(英文)

1.Make a choice （共十题 每题1分）10p

(1).Which of the following dispersion dose not exist in single-mode optical fiber? （ D ） A .material dispersion B.waveguide dispersion C.polarization-mode dispersion D.intermodal dispersion (2).The unit of the fiber attenuation coefficient is (C)

A. dB

B. dBm

C. dB/km

D. dBm/km (3).the bands of Optical fiber communication is (B)

A.0.01um-0.39um

B. 0.8um-2.0um

C.0.39um-0.79um

D.100um-1000um

(4).If the optical input power of a Optical amplifier is 10mW ，and the optical output power is 100mW as well ,then its output gain level is (A)

A.10dB

B.20dB

C.30dB

D.40dB

(5)In order to make sure of the system BER conditions , if the minimum optical input power of the receiver is 1 uW, the sensitivity of the receiver must be (A)

A.－30dBm

B.－40dBm

C.－20dBm

D.－43dBm

(6)The principal light sources used for fiber optical communications applications are ：（D ） A.OA 、LD B.PIN 、APD C.PD 、LED D.LD 、LED

(7)laser action is the result of three key process ，which one of the following is not be included ？（D ） A.photon absorption B.spontaneous emission C.stimulated emission https://www.360docs.net/doc/ed3670672.html,pound radiation (8) A single mode fiber usually has a core diameter of A.

A. 10m

B. 62.5nm

C. 125nm

D. 50mm

(9)To make sure that the APD photo-detector works properly, a sufficiently D is applied across the p-n junction. A. high forward-bias voltage B. low forward-bias voltage C. low reverse-bias voltage D. high reverse-bias voltage (10) When DFA fiber amplifier uses as light Repeaters, its main effect is B. A. amplifying and regenerating the signal B. regenerating the signal C. amplifying the signal D. reducing the signal noise (11) In graded-index optical fiber, the numerical aperture NA can be expressed as C. A.

21n n - B. ?2a C. ?2n 1 D. 21n n a -

(12) In practical SMFs, the core diameter is just below the cutoff of the first higher-order mode; that is, for V slightly A.

A. <2.4

B. > 2.4

C. =3

D. =3.5

(13) It is well known that the total dispersion in the single-mode regime is composed of two components: C. A. mode-partition noise, inter- symbol Interference B. frequency chirp, modal dispersion C. material dispersion, waveguide dispersion D. modal dispersion, waveguide dispersion

(14) At present, erbium doped fiber amplifier’s maximum small signal gain is around A. A. 40 dB B. 30 dB C. 20 dB D. 10 dB (15)Which of the following doesn’t belong to passive optical components B A. Directional coupler B. Semiconductor laser C. Optical fiber connector D. Optical attenuator (16) The A mode has no cut off and ceases to exist only when the core diameter is zero. A. HE11 B. TE01 C. TM01 D. EH11

(17)When the phase difference is an integral multiple of A, the two modes will beat and the input polarization state

will be reproduced.

A. 2π

B. π

C. 1800

D. π/2 (18)which one of the following model can transmit in the single-mode optical fiber ?(A)

A.HE11

B. TE01

C. TM01

D. EH11

2. Write the full name of the following acronym（共十题每题1分）10p

(1)DCF: dispersion compensating fiber

(2)CNR: carrier to noise ratio

(3)SRS: Stimulated Raman Scattering

(4)SOA: Semiconductor optical amplifier

(5)NA: numerical aperture

(6)PON: passive optical network

(7)SLM: single longitudinal mode

(8)NEP: noise-equivalent power

(9)DSF: dispersion shift fiber

(10)SONET: synchronous optical network

(11)A TM: asynchronous transfer mode

(12)ISDN: integrated services digital network

(13)WDM: wavelength-division multiplexing

(14)SDH: Synchronous digital hierarchy

(15)TLLM: transmission-line laser model

(16)ONSL: optical network simulation layer

(17)OVPO: outside vapor-phase oxidation

(18)V AD: vapor-phase axial deposition

(19)CDMA: code-division multiple access

(20)FDM: frequency-division multiplexing

(21)DSP: digital signal processing

(22)MCVD: modified chemical vapor deposition

(23)EDFA: erbium-doped fiber amplifier

(24)FDDI: fiber distributed data interface

(25)SIOF: step index optical fiber

(26)GIOF: graded index optical fiber

(27)SQW: single quantum-well

(28)ARQ: automatic reapt request

(29)FEC: forward error correction

3. Filling blanks(共20空每空1分) 20p

(1) According the mode which propagate in the fiber, the fiber can be divided into (single mode) fiber) and (multimode) fiber.

(2) The most common misalignment occurring in practice, which also causes the greatest power loss, is (axial displacement).

(3) The electromagnetic energy of a guided mode is carried partly in the (core) and partly in the (cladding).

(4)The basic attenuation mechanisms in a fiber are (absorption), (scattering) and (radiative losses) of the optical waveguide.

(5) The two main optical amplifier types can be classified as (SOAs) and (DFAs).

(6) Networks are traditionally divided into: (LANs), (MANs),(WANs).

(7)The principal light sources used for fiber optic communications applications are (ILDs) and (LEDs) .

(8) The dominant noise generated in an optical amplifier is (ASE).

(9) The two most common samples of these spontaneous fluctuations are (shot noise) and (thermal noise).

(10)According to the refractive index of the core, the fiber can be divided into (step-index) fiber and (graded-index) fiber .

(11)The total dispersion in single-mode fibers consists mainly of (material) and (waveguide) dispersion.

(12) The most meaningful criterion for measuring the performance of a digital communication system is the

(average error probability). In an analog system the fidelity criterion is usually specified in terms of a (peak signal-to-noise ratio).

(13) The simplest transmission link is a (point-to-point line) that has a transmitter on one end an a receiver on the other.

(14)The commonly used materials for fiber lasers are (erbium) and (neodymium).

(15)Absorption is related to the fiber material, whereas scattering is associated both with the (fiber material) and with (structural imperfections) in the optical waveguide.

(16)The two basic LED configurations being used for fiber optics are (surface emitters) and (edge emitters).

(17) The basic schemes for improving the reliability are (ARQ) and (FEC).

4. Give a brief description of following terms and questions（共5题每题3分）15p

(1) Briefly describe the key system features of WDM.

1) capacity upgrade

2) transparency

3) wavelength routing

4) wavelength switching

5) the connectors used to join individual fiber cables to each other and to the source and detector

(2) Briefly describe there major goals of SDH.

1) Avoid the problems of PDH

2) Achieve higher bit rates

3) Better means for operation, administration and Maintenance

(3) List at least three advantages of SOA.

1) Small size, and easy to be integrated with semiconductor circuits.

2) Fabrication is simple and with low power consumption, long life-span and low cost.

3) Gain response is very quick and well suited for switching and signal processing in optical networks application.

4) Can amplify optical signal and process signal in the same time such as switch, so can be used in wavelength converting and optical switch.

(4) List more than three disadvantages of SOA.

1) The coupling loss with optical fiber is too large

2) Sensitive to polarization

3) Noise figure is high(~8 dB)

4) crosstalk

5) Easy to be affected by temperature, low stability

(5) Stimulated Emissions

If a photon of energy hv12 impinges on the system while the electron is still in its excited state, the electron is immediately stimulated to drop to the ground state and give off a photon of energy hv12.

(6)Dynamic range

System dynamic range is the maximum optical power range to which any detector must be able to respond. (7) What conditions should be met to achieve a high signal-to-noise ratio?

1) The photodetector must have a high quantum efficiency to generate a large signal power.

2) The photodetector and amplifier noises should be kept as low as possible.

(8) Please write the three basic categories of degradation of light sources

1) internal damage

2) ohmic contact degradation

3) damage to the facets of laser diodes

(9)List the three factors largely determining the frequency response of an LED

1) the doping level in the active region

2) the injected carrier lifetime Ti in the recombination region

3) the parasitic capacitance of the LED.

(10)Write the three basic types of two-level binary line codes that can be used for optical fiber transmission links. 1) non-return-to-zero (NRZ) format 2) return-to-zero (RZ) format 3) phase-encoded (PE) format

(11) Please write the three different mechanisms causing absorption briefly 1) Absorption by atomic defects in the glass composition. 2) Extrinsic absorption by impurity atoms in the glass material. 3) Intrinsic absorption by the basic constituent atoms of the fiber material. (12) The disadvantage of Raman amplifier

Need large output power pump laser. As Raman Scattering, the energy is transferred from high frequency to low frequency. Cross talk will affect signal. 5. Figure （共1题 每题5分）5p

（1）

Please draw the basic step for an automatic-repeat-request (ARQ) error-correction scheme. Solution:

(2)Please draw out the basic elements of the optical receiver.(5p)

Solution:

(3)Please draw out the basic elements of an analog link and the major noise contributions. Solution:

光发送机

电模拟输入

信号

谐波失真互调失真RIN

激光削波

光纤信道模式失真损耗GVD

光放大器ASE 噪声

光检测器散弹噪声热噪声

放大器噪声APD 倍增噪声

电模拟输出到RF 接收机

(3) consider the encoder shown in Fig.1that changes NRZ data into a PSK https://www.360docs.net/doc/ed3670672.html,ing this encoder,draw the NRZ and PSK waveforms for the data sequence 0001011101001101.

clock

/2

PSK data

NRZ data

frequency A

frequency B

Fig.1

Solution:

6. Calculation Problems(共3-4题，统计40分) 40p (1) A wave is specified by 8cos 2(20.8)y t z π=-,where

y is expressed in micrometers and the

propagation constant is given in 1m μ-.Find (a) the amplitude,(b) the wavelength,(c) the angular frequency, and (d) the displacement at time 0t = and 4z m μ=.

Solution:

The general form is:

y = (amplitude) cos()cos[2(/)]t kz A vt z ωπλ-=-.Therefore

(a) amplitude

8m μ=

(b) wavelength:

11/0.8m λμ-= so that 1.25m λμ=

(c) 22(2)4v ωπππ===

(d) At 0t = and 4z m μ= we have

18cos[2(0.8)(4)]8cos[2(3.2)] 2.472y m m πμμπ-=-=-=

(2) A certain optical fiber has an attenuation of 0.6dB/km at 1300nm and 0.3dB/km at 1550nm.Suppose the following two optical signals are launched simultaneously into the fiber: an optical power of 150W μ at 1300nm and an optical power of 100W μ at 1550nm. What are the Solution:

power levels in

W μof these two signals at (a) 8km and (b) 20km?

Since the attenuations are given in dB/km, first find the power levels in dBm for 100W μ and 150W μ. These are, respectively, P(100W μ) = 10 log (100 W μ/1.0 mW) = 10 log (0.10) = - 10.0 dBm P(150W μ) = 10 log (150

W μ/1.0 mW) = 10 log (0.15) = - 8.24 dBm

(a) At 8 km we have the following power levels:

P 1300(8 km) = - 8.2 dBm – (0.6 dB/km)(8 km) = - 13.0 dBm = 50W μ

P 1550(8 km) = - 10.0 dBm – (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5W μ (b) At 20 km we have the following power levels:

P 1300(20 km) = - 8.2 dBm – (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55W μ P 1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1W μ

(3) A double-heterojunction InGaAsP LED emitting at a peak wavelength of 1310nm has radiative and nonradiative recombination times of 25 and 90ns, respectively. The drive current is 35mA. (a) Find the internal quantum efficiency and the internal power level.

(b) If the refractive index of the light source material is n=3.5, find the power emitted from the device. Solution: (a) From Eq. int

11/r nr

r

τητττ=

=

+, the internal quantum efficiency is

int 1

0.783125/90

η=

=+，

and from Eq.int int int I hcI

p hv q q ηηλ== the internal power level is

int (35)

(0.783)26(1310)

hc mA p mW q nm ==

(b) From Eq.int

e int 2

p (1)

n t p P n n η==+, 2

1

260.373.5(3.51)

P mW mW ==+ (4) An LED with a circular emitting area of radius 20m μ has a lambertian emission pattern with a

1002

()W cm

sr ?axial radiance at a 100mA drive current. How much optical power can be coupled into a

step-index fiber having a 100m μ core diameter and NA=0.22? How much optical power can be coupled from this source into a 50m μ core-diameter graded-index fiber having 12.0, 1.48n α== and 0.01?=?

Solution:

The source radius is less than the fiber radius, so Eq.

222222,1()2LED step s o s o P r B NA r B n ππ==? holds:

22223222,()(210)(100/)(0.22)191LED step s o P r B NA cm W cm W

ππμ-==?=

From Eq.

222,122[1()]2s LED graded s o r P r B n α

παα

=?-

+

232222,12

2(210)(100/)(1.48)(0.01)[1()]15925

LED graded P cm W cm W πμ-=?-=

(5)Suppose

an

avalanche

photodiode

has the

following

parameters:

1/231,1,0.85,,10L D L I nA I nA F M R η=====Ω

, and

1B kHz

=.Consider a sinusoidally

varying 850nm signal, which has a modulation index m=0.85 and an average power level 050P dBm =-, to

fall on the detector at room temperature. At what value of M does the maximum signal-to-noise ratio occur?

Solution:

Using Eq.

2222()()24/p P D L B L

i M S

N q I I M F M B qI B k TB R <>=+++ we have

22

005/2001

()22()24/D L B L

R P m M S N qB R P I M qI B k TB R =

+++

16

2

235/2191.215102.17610 1.65610

M M ---?=?+? The value of M for maximum S/N is found from Eq.2

24/()

x L B L

opt P D qI k T R M xq I I ++=

+, with x = 0.5:

Moptimum = 62.1.

(6)An LED operating at 1300 nm injects 25W μ of optical power into a fiber. If the attenuation between the LED and the photodetector is 40 dB and the photodetector quantum efficiency is 0.65, what is the probability that fewer than 5 electronhole pairs will be generated at the detector in a 1-ns interval ? Solution: From

?=

=τ

ηη

0)(hv

N hv

E

dt t P ,the average number of electron-hole pairs generated in a time t is

6.10)

/103)(106256.6()103.1)(101)(1025(65.0/8

346910=?????===----s m Js m s W hc Pt h E N ληνη Then,from Eq.(7-2)

%505.0120

133822!5)6.10(!)(6.106

.105=====---e e n e N n P N n

(7) An engineer has the following components available:

(a) GaAlAs laser diode operating at 850 nm and capable of coupling 1 mW (0 dBm) into a fiber.

(b) Ten sections of cable each of which is 500 m long, as a 4-dB/km attenuation, and has connectors on both ends. (c) Connector loss of 2dB/connector. (d) A pin photodiode receiver. (e) An avalanche photodiode receiver.

Using these components, the engineer wishes to construct a 5-km link operating at 20 Mb/s. If the sensitivities of the pin and APD receivers are -45 and -56 dBm, respectively, which receiver should be used if a 6-dB system operating margin is required? Solution: (a)Use

margin 2system L l P P P f C R S T ++=-=α，to analyze the link power budget.(a) For the pin

photodiode,with 11 joints

dB

L km dB dB dBm dBm in systemm L l P P P f C R S T 6)/4()2(11)45(0arg )(11++=--=++=-=α

Which gives L=4.25km. the teansmission distance cannot be met with these components.

(b)For the APD

dB L km dB dB dBm dBm 6)/4()2(11)56(0++=--

Which gives L=7.0km. the transmission distance can be met with these components.

(8) Suppose we want to frequency-division multiplex 60 FM signals. If 30 of these signals have a per-channel modulation index i m =3 percent and the other 30 signals have i m =4 percent, find the optical modulation index

of the laser. Solution:

The total optical modulation index is

%4.27])04(.30)03(.30[][2/1222/12=+==∑i

i m m

(9) An optical transmission system is constrained to have 500-GHz channel spacings. How many wavelength channels can be utilized in the 1536-to-1556-nm spectral band? Solution:

In terms of wavelength,at a central wavelength of 1546nm a 500-GHz channel spacing is

nm s s

m nm f c 410500/103)1546(1

98

22

=??=?=?-λλ The number of wavelength channels fitting into the 1536-to-1556 spectral band then is

54/)15361556(=-=nm nm N (10) The output saturation power

sat out P , is defined as the amplifier output power for which the amplifier gain G

is reduced by 3 dB (a factor of 2) from its unsaturated value 0G . Assuming 0G >>1, show that in terms of the

amplifier saturation power

sat amp P ,, the output saturation power is

sat amp sat out P G G P ,00,)

1(2

ln -=

Solution: Let

2/0G G = and 0,/22/G P P P sat out out in ==.then Eq.(11-15) yields

2ln 212,,00

sat

out sat amp P P G G += Solving for

s a t out P , and with 10>>G ,we have

sat amp sat amp sat amp sat out P P P G G P ,..00,693.0)2(ln 2

2

ln =≈-=