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光纤通信技术课后部分习题答案

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光纤通信技术课后部分习题答案 Sep. 26 th P41 2.6 Calculate the energy of a single photon at 1300 nm and at 1500 nm. Solution: A single photon at 1300 nm: )(10529.1 101300 10626.6103 / 19 9 348 JchhfE p         A single photon at 1550 nm: )(10282.1 10...

光纤通信技术课后部分习题答案
Sep. 26 th P41 2.6 Calculate the energy of a single photon at 1300 nm and at 1500 nm. Solution: A single photon at 1300 nm: )(10529.1 101300 10626.6103 / 19 9 348 JchhfE p         A single photon at 1550 nm: )(10282.1 101550 10626.6103 / 19 9 348 JchhfE p         2.7 a. How many photons per second emanate from a laser diode radiating at 1300 nm if its power is 1 mW? b. How many photons per second emanate at 1550 nm? Solution: Using the result above: a: )(101101 33 JPtE   NEE p  15 19 3 10540.6 10529.1 10      pE E N b: NEE p  15 19 3 10798.7 10282.1 10      pE E N P81 3.12 For a specific fiber, NA = 0.2375 and n1 = 1.4860. Find n2 (n cladding). Solution: 2 2 2 1 nnNA  hence 2 2 22 4860.12375.0 n Hence n2 = 1.4669 3.22 An optical fiber with attenuation of 0.25 dB/km is used for 20-km transmission. The light power launched into the fiber is 2mW. What is the output power? Solution: Loss(dB) = -10 lg (Pout/Pin) A(dB/km) = loss (dB) / fiber length(km) Hence, 20 2 lg10 25.0 mW Pout  Hence, Pout = 0.6325 mW. 3.25 Find the maximum transmission distance for a fiber link with an attenuation of 0.3 dB/km if the power launched in is 3mW and the receiver sensitivity is 100 W. Solution: Loss(dB) = -10 lg (Pout/Pin) A(dB/km) = loss (dB) / fiber length(km) Hence, hfiberlengt 3 10100 lg10 3.0 3   Hence, fiber length = 49.24 km 3.7 The core refractive index is 1.4513 and the cladding index is 1.4468. What is (1) the critical propagation angle? (2) the acceptance angle? (3) the numerical aperture? Solution: (1) The critical angle is:               5131.4 4513.1 4468.1 1sin1sin 2 1 2 1 21 n n c (2): The acceptance angle is:       1148.135131.4sin4513.1sin2sinsin22 11 1 ca n  (3): The NA: NA = sin a = 0.1142 3.21 What does the term “transparent windows” mean? Specify three peak wavelengths for the transparent windows in modern optical fibers. Solution: The term “transparent windows” means that the regions where absorption reaches minimum value in the typical spectral attenuation diagram (Loss (dB/km) - (nm) diagram).The three peak wavelengths for them are: 850 nm, 1300 nm, and 1550 nm. 3.29 What is the number of modes  VN . for a graded-index fiber if d is 50 m, NA is 0.200, and the operating wavelength is 1300 nm? Solution: The V number is: 2.24200.0 101300 1050 9 6         NA d V the number of modes is: 146 4 2.24 4 22 VN 3.30 How many modes can support a step-index optical fiber whose d = 8.3 m, n- 1core = 1.4513, n2dad = 1.4468, and  = 1550 nm? Solution: 92.14468.14513.1 101550 103.8 22 9 6 2 2 2 1         nn d V V < 2.405, N = 1 P71 E3.47 A graded-index fiber has n1 = 1.486 and NA = 0.200. What is the bit rate for a 1-km link? Solution: t BR cn NAL cN NAL tGI       4 1 3232 31 4 3 1 4 where N1 core group index of refraction Hence, sGbitsbit NAL cn BRGI /92.4/1092.4 200.0101 486.110388 9 43 38 4 3 1       Oct. 17 th 4.30 A graded-index fiber has the following characteristics: NA = 0.200, dcore = 50m and  = 1300 nm. What power is carried by the fiber’s cladding? S: 17.24200.0 1300 1050 3    Z NA d V   %9.3 17.243 22 3 22    VP P total clad Hence, about 3.9% of the total power is carried by the fiber’s cladding. 4.51 Calculate the pulse spreading caused by chromatic dispersion for BF04431-02 multimode grade-index fiber from SpecTran, (See Figure 3.20) operating at 1300 nm. Assume  = 70 nm. S: Taking 0 = 1342.5 nm, S0 = 0.097 kmnm ps 2           3 4 00 4    S D kmnm ps         329.4 1300 5.1342 1300 4 097.0 3 4 tgmat = D() = 4.329 x 70 = 303 ps/km = 0.3 ns/km 4.55 What is meant by the term dispersion power penalty? Calculate the disperson power penalty for the pulse spread obtained in Problem 4.51. The BR is 2.5 Gbit/s. S: The amount of transmitting power required for compensating the increase bit error rate (BER) cause by attenuation is called dispersion power penalty. In Problem 4.51, ttotal = tmat = 18 ns (if fiber length is 100 km) PD(dB) = -10 log10 {exp[-(1/4 ) (ttotal) 2 (BR)2]} = -10 log10 {exp[-1/4 x (18 x 10 -9 ) 2 ( x 2.5 x 109)2]} = 2.17 x 10 4 dB, enough to burn any type of fiber, impossible. if the fiber length is 1 km: P 4.34  2 4  t 2  BR 2  PD=2.17 dB ( realizable!) 4.56 What is the bit-rate length limitation caused by chromatic disperson for the Spectran fiber referred to in Problem 4.51? S: (BR x L)max = 1 / [4 D()]    D)2(41   1210329.427041  kmsGb  /17.1 Chapter 5-6 5.13 Compute the pulse spread caused by chromatic dispersion if a fiber has a zero- dispersion wavelength at 1312 nm, a zero-dispersion slope of kmnm ps 2 090.0 , a length of 100 km, and operates at 1310 nm. The LD’s spectral width is 1 nm. S:     kmnm ps SD   18.0)13121310(090.000    psLDtchrom 18100118.0   5.17 Calculate pulse spread caused by polarization-mode dispersion if kmpsDPMD /2.0 and L = 120 km. S: psLDt PMDPMD 19.21202.0  6.2 What is the V-number of the Low-PMD SM from Plasma Optical Fibre BV, assuming that the core diameter is equal to 8.3 m and the relative index equals 0.36%? S: 889.1%36.0 101215 467.1103.8 2 9 6         c dn V 6.8 What length of DCF is needed to compensate for chromatic dispersion at a 120-km span of Low-PMD fiber from Plasma Optical Fibre BV (see Figure 5.17) if DDCF() = -127 ps/nm*km and  = 1550 nm? S: compenschrom tt      DCFDCF LDLD   4 DCFDCF LDLS          3 4 0 0    DCFL127120 1550 1310 1550 4 092.0 3 4        L = 16.50 km 6.10 a. What means do we have to cope with PMD? b. What bit rate can be achieved at a 120-km span with Low-PMD Fiber from Plasma Optical Fibre BV? S a: We use PM fiber or PM components together to make a PM Link. We also develop PMD compensation techniques. b: psLDt PMDPMD 19.21202.0  sGbit t BR PMD PMD /114 19.24 1 4 1      P246 7.2 A drying gas flows through a preform during the consolidation phase of the OVD process. Why is this important? S: It is the remove residual moisture because optical fiber’s absorption properties are caused by some molecules of the hydroxide anion OH - , which is contained in water. Without the process above, these molecules will be incorporated in silica during fabrication and be very hard to eliminate. OH - molecules have major peaks of absorption at some wavelengths. Moreover, every foreign particle found inside the core will slightly modify the refractive index and cause scattering, including water molecules. Thus it is important to dry the perform. 7.6 List four basic components of a fiber cable and explain their functions. S: The four basic components of a fiber cable are fibers, buffer tube, strength member, and outer jacket. The plastic buffer tube is the first shield protecting the fiber from environment damage. The strength member is to release the fiber from mechanical stress during installation and operation. Outer jacket assembles the entire construction and shields it from an adverse environment. 8.1 a. Calculate the intrinsic connection losses for two 6.25/125 graded-index multimode fibers manufactured by Plasma Optical Fibre (Figure 3.18) caused by (a) diameter mismatch and (b) NA mismatch. b. Calculate the intrinsic connection losses caused by the MFD mismatch of two singlemode fibers made by Plasma Optical Fibre (Figure 5.17). Solution: a: From Figure 3.18, it is found that the tolerance of core diameter to be 62.5  25 m, and NA 0.275  0.015. (a) dB a a Losscore 6952.0 5.25.62 5.25.62 log10log10 22 1 2                                  (b) 9485.0 015.0275.0 015.0275.0 log10log10 22 1 2                                  NA NA LossNA dB b: From Figure 5.17, it is found that MFD1 = 9.3 + 0.5 m, MFD2 = 9.3 – 0.5 m dB W W W W LossMFD 05.0 8.9 8.8 8.8 8.9 4 log104log10 2 2 2 1 1 2                                       8.3 Calculate the extrinsic connection losses for two SM fibers manufactured by Plasma Optical Fibre caused by (a) lateral misalignment for x = 0.5 m, (b) angular misalignment for  = 30”, and (c) end separation for Z = 0.5 m. (Figure 5.17). S: From Figure 5.17, I obtained that W0 = ½ MFD = ½ x 9.3 m,  = 1550 nm, n = 1.467 (a):             2 0 2 explog10 W xLoss lat   dB05.0 3.9 5.0 explog10 2 2 1 2                    (b): 3 3 2 1 0 103793.2 101550 "30sin3.9467.1sin       ZnZW T      dBTLossang 52 104587.2explog10  (c): dB S Lossend 5 2 105668.6 1 1 log10         8.22 A fiber link includes five splices at 0.02 dB/splice, four connectors at 0.2 dB/connector, transmitter power of –10 dBm, and receiver sensitivity of –25 dBm. What length of this link will be allowed if a singlemode fiber cable with attenuation of 0.3 dB/km is used and the required power margin is 3 dB? S: Transmitter power, Pin = -10 dBm Cable Loss 0.3 dB/km * L km = 0.3L dB Splicing Loss: 0.02 dB/splice * 5 splices = 0.1 dB Connector Loss: 0.2 dB / connector * 4 connectors = 0.8 dB Receiver Sensitivity, PRS = -25 dBm Power margin: 3 dB From the above, it follows that: Pout = Power margin + PRS = 3 + (-25) = -22 dBm Total Loss = Pin – Pout = -10 – (-22) = 12 dB Total Loss = Cable Loss + Splicing Loss + Connector Loss Hence, 0.3L + 0.1 + 0.8 = 12 L = 37 km 8.23 A local data link to be installed has the following characteristics: Maximum bit rate: 32 Mbit/s Line code: return-to-zero (RZ) format Fiber: 6.25/125 m Installation length: 2000 m Operating wavelength: 1300 nm Rise time of lightwave equipment: 4 ns ID spectral width: 2 nm Will this fiber support the required bit rate? Prove your answer. S: 222 riseltwrisesystrisefib    2 2 35.0 riseltw BW           nsnsns MHz 18.106289.1034 32 35.0 22 2        (Required fib-rise) (From Figure 3.18) BWmodal = 500 MHz*km/2 km = 250 MHz BWel-modal = 0.707 BWmodal = 0.707 x 250 = 176.75 MHz mod-rise = 0.35 / BWel-modal = 1.9802 ns chrom-rise = tchrom = D()L (From Example 3.35) = 4.38 ps/(nm*km) x 2 km x 2 nm = 17.52 ps nsrisechromriserisefib 9803.101752.09802.1 2222 mod    compared with the required rise time of 10.18ns, the chosen fiber will support this link. Nov 10 th P361 9.16 Evaluate power in mW and dBm (1) coupled into a 62.5/125 m fiber and (2) radiated by a 1.3-m (2) radiated by a 1.3-m SLED (Figure 9.8[a]) at 100 mA. Assume that the LED radiates as a Lambertian source. S: From Figure 3.18, it was found that NA = 0.275 for a 62.5/125 m fiber. Pin = P0(NA) 2 = 0.275 2 P0 (1): Looking at Figure 9.8[a], when at 100 mA a 1.3 m SLED coupling power into a 62.5 m fiber is 75 W. Thus, dBm mW W PdBm 2.11 1 75 lg10   . (2): The power radiated by a 1.3-m SLED at 100 mA is     dBm mW mW dBmPmW NA P P in 044.0 1 99.0 lg10,99.0 275.0 075.0 20  9.19 The 1.3 m-SLED shown in Figure 9.8(a) has a spectral width of 155 nm at 25C. What will be the spectral width if the temperature changes to 90C? S: From Figure 9.8(a), /T = 0.38 nm/C Hence, 38.0 2590 155    x x = 179.7 nm 9.32 How many longitudinal modes can a Fabry-Perot laser diode generate if the length of its resonator is 0.3 mm and the operating wavelength is 1550 nm? The width of the gain curve is 9 mm. S: N - N+1 =  2 /2L = 1550 2 / (2 x 0.3 x 10 6 ) = 4 nm The width of the gain curve is 9mm, thus it can generate 2 longitudinal modes. 9.41 Calculate the slope efficiency of a DFB LD whose specifications are given in Figure 9.19 on page 356. S: S = (PR – Pth) / (Itot – Ith) =  mAmWmAmA mW 135.0 1587 07.9    CH10 P428 10.14 Calculate the external quantum efficiency of a laser diode if its slope efficiency is equal to 0.08 mW/mA and the operating wavelength is 1550 nm. S: Ep = hc/ = 6.63 x 10 -34 x 3 x 10 8 /(1550 x 10 -9 ) = 1.2832 x 10 -19 J S(W/A) = (Ep/e) ext ext = %98.9 102832.1 106.108.0 19 19        A W E eS p 10.15 Calculate the power efficiency of the DFB laser diode whose data sheet is given in Figure 9.19. S: From Figure 9.19, It was found the forward current to be IF = 65 mA, assuming that R = 1 ,  = 1550 nm, Pout = 2 mW. p = RI e Eg I Pout  2 = %5.3 4805.065 2   mWmW mW 10.32 Explain the meaning of the term “chirp” as applied to an LD. How does this phenomenon affect the LD’s transmission characteristics? S: Chirp is the deviation of laser frequency from its radiation-center frequency. Chirp results in a broadening of a laser’s linewidth. 10.34 Calculate the laser noise power detected by a receiver for the following data: RIN = -150 dB/Hz, Preceived= 80 W, and BW = 140 MHz. S: RIN(dB / Hz) = 10 lg [ RIN (1/Hz)] = -150 (dB/Hz) Hence, RIN = 10 -15 (1/Hz) RIN(1/Hz) = / [

2 BW] =        HzWBWPHzRINPN 62261522 101401080101   = 3 x 10 -8 W = 0.03 W CH11 P493 11.5 The responsivity of a PD is 0.9 A/W and its saturation power is 2 mW. What is the photocurrent if the received power is (1) 1mW (2) 2mW (3) 3mW? S: p = RP, R = 0.9 A/W, PC = 2 mW (1): P = 1 mW, P = RP = 0.9 x 1 mW = 0.9 mA (2): P = 2 mW, P = RP = 0.9 x 2 mW = 1.8 mA (3): P = 3 mW > PC = 2mW; and as saturation means no more increase, ~ 1.8 mA. 11.23 What is the bandwidth of a germanium photodiode with w = 20 m and vsat ~ 10 5 m/s? S: pss m V W s msat tr 200102 10 1020 10 5 6         sGbit ps BW RCtr PD /796.0 2002 1 2 1       11.48 Calculate the RMS and bandwidth-normalized values of total noise current for a PD whose data are given in Figure 11.24 if the average input power is 0.1 uW,  = 1550 nm, RL = 50 k, and the diode operates at room temperature. S: Shot-noise current Ip * = RPin * = 0.86 A/W x 0.1 W = 0.086 A   nAGHzAcBWiei PDps 42.70.2086.0106.122 19*    HznaAcei BW i i p PD s SN /1066.1086.0106.122 419*    Thermal-noise current nAGHz k kBW Rc Tk i kJPD B t 7.250.250 3001038.14 4 23          Hz nA BW i i PD t tN 41075.5  Dark-noise current id * = 1.6 nA nAGHznAcBWei PDdd 98.00.215106.122 19*   52.19 10ddN PD i nAi BW Hz    1/f noise current Since a PD is used in high-speed application, this current is negligible. Total noise current The RMS value: nAiiii dtsnoise 77.2698.07.2542.7 222222  The RMS value of bandwidth-normalized total current noise:       Hz naiiii dNtNsNnoiseN 4252424222 1099.51019.21075.51066.1   11.52 Calculate SNR for a PD whose data are given in Figure 11.24 if the average input power is 0.1 uW,  = 1550 nm, and the diode is operating at room temperature with the following parameters: a. RL = 50 , RL = 500 k and RL = 50 M b. BWPD = 50 kHz, BWPD = 50 MHz, BWPD = 50 GHz Draw conclusions as to how load resistance and bandwidth affect SNR. S: R = 0.86 A/W, Pin * = 0.1 W, hence * * 22 2 in R p Pi  = 7396 nA 2 From 11.48, I obtain Hz nAi Hz nAi dNsN 54 1019.2,1066.1   For RL = 50 ,  L B tN R Tk i 4 Hz nA21082.1  For RL = 500 k,  L B tN R Tk i 4 Hz nA41082.1  For RL = 50 M,  L B tN R Tk i 4 Hz nA51082.1  Hence, for BWPD = 50 kHz, inoise = is 2 + it 2 + id 2 = (isN 2 + itN 2 + idN 2 ) BWPD            )50(1021.5 )500(1042.2 )50(53.446 )( ** 6 6 222 2 2 2 MR kR R BWiii p i p SNR L L L PDdNtNsNnoise ii For BWPD = 50 MHz,          )50(1021.5 )500(1042.2 )50(45.0 3 3 MR kR R SNR L L L For BWPD = 5 GHz,           )50(1.52 )500(2.24 )50(105.4 3 MR kR R SNR L L L From the results above, we can draw the conclusion that SNR increases with RL and decreases with BWPD, with BWPD being the dominant factor affecting SNR. 11.54 Calculate SNRs, SNRt, and overall SNR for Si and ZnGaAs avalanche PDs if M = 10, P = 0.1 uW, R = 0.9, RL = 50 k, BWPD = 20 GHz, and T = 300k. S: For an Si APD, Fs = 2.49   48.56249.2106.12 1.09.0 2 19      GHzc W BWeF RPSNR PDs s         83.12222 50 3001038.14 1.09.010 4 23 222222                   GHz k W BW R Tk PRM SNR PD L B t  (Overall SNR = 54) For an InGaAs, Fs = 12.78 SNRs = 0.11 278.12106.12 1.09.0 2 19      GHzc W BWeF RP PDs  SNRt = 1222.83 (Overall SNR = 11) 11.59 a. What is BER? b. How can you improve BER? S a: BER stands for bit-error rate, which is the number of the erroneous bits per total number of bits transmitted. b: The only way to minimize BER is to make th at thopt, which introduces digital SNR, Q. The larger the Q, the less the BER will be. 11.63 Why does BER depend on the Q parameter? S: 01 01 ii ii Q    When i0 = 0, Q = 01 1 ii i  , which is simply the ratio of signal current to noise current. BER depends on SNR, so it depends on Q. CH12 P582 12.12 What do we mean by the term add/drop procedure? Why is this a problem in TDM networks? S: When TDM or WDM is used, an add/drop node is required to extract the desired signal from the main data stream. The problem of an add/drop procedure is that one pushes the main data stream at a high bit rate while the individual channel operates at a much lower bit rate. Thus, the main data stream needs to be demultiplexed to the bit rate of the individual channel. 12.29 What is the channel spacing for a WDM with 40 channels? S: 0.8 nm (100 GHz) 12.58 Calculate the noise figure of an optical amplifier if the input-signal power is 250 W, the input-noise power is 25 nW in a 1-nm bandwidth, the output- signal is 50 mW, and the output-noise power is 15 W in a 1-nm bandwidth. S: Fn =     poweroutnoisepoweroutsignal powerinnoisepowerinsignal SNR SNR out in    / /         3 10151050 102510250 15 50 25 250 63 96       W mW nW W   12.70 Calculate the gain of an EDFA if the input power is 150 W, the output power is 55 mW, and the noise power is 20 W. S: Gain (dB) = 10 lg dB W WmW P PP in ASEout 42.23 250 2055 lg10       12.77 Calculate the noise figure of an EDFA if the input-signal power is 300 W, the input-noise power is 30 nW in a 1-nm bandwidth, the output-signal power is 50 mW, and the output-noise power is 20 W in a 1-nm bandwidth. S: Fn = 

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