Universidade do Algarve Faculdade de Ciências e Tecnologia Departamento de Física Ano lectivo
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1 Universidade do Algarve Faculdade de Ciências e Tecnologia Departamento de Física Ano lectivo Unidade Curricular Sistemas de Comunicação Ótica Optical Communication Systems Mestrado Integrado em Engenharia Electrónica e Telecomunicações Mestrado em Engenharia Informática
2 Example 1: Consider a typical optical transmission link with good receiver design that can tolerate a power loss of 20 db at 1 Gb/s. Estimate the maximum transmission distance if: a) A MMF used for transmission at 850 nm has an attenuation of 5 db/km, and a dispersion of 0.1 ns/km. (Ans. Maximum transmission distance 4 km; dispersion of 0.4 ns; maximum capacity-distance product of 4 Gb/s-km.) b) A SMF with 0.5 db/km at 1300 nm is used. (Ans. Maximum transmission distance 40 km; little dispersion at 1300 nm; maximum capacity-distance product of 40 Gb/s-km; if the bit rate increases lets say to 10 Gb/s, the maximum power loss allowed would be 10 db, which leads to a reduction of the maximum transmission distance to 20 km. However, the capacity-distance product increases to 200 Gb/s.) c) A SMF with 0.2 db/km at 1550 nm is used. (Ans. Maximum transmission distance 100 km; little dispersion at 1550 nm when using a single frequency laser; maximum capacity-distance product of 100 Gb/s-km; if the bit rate increases lets say to 10 Gb/s, the capacity-distance product can be increased to 1000 Gb/s-km.)
3 Example 2: Consider an optical transmission link with good receiver design that can achieve a detection performance of 50 photons per bit, operating at 10 Gb/s. What would be the required optical power? Ans. Assuming operation at 1550 nm. The energy per bit is given by Eb=50xEf= 6.4E-18 J per bit. At 20 Gb/s, the required received optical power is: Prx=Ebx1E10 = 64 nw = -42 dbm. Example 3: Assuming in the example above the transmission power is 0 dbm, what would be the tolerable loss? Ans. The tolerable power loss is 42 db. Neglecting the dispersion eefct from using single mode source, and assuming operation at 1550 nm (0.2 db/km, the distance limit is 210 km; the capacity-distance product is 2100 Gb/s-km)
4 Example 4: Optical fiber soliton transmission with optical amplification, has been demonstrated at a distance more than km with a bit rate of 10 Gb/s. What is the capacity-distance product? Ans. Along the transmission line, power loss due to fibre attenuation is compensated by the use of EDFA; the pulse shape is maintained unchanged by the balance between fibre dispersion that broadens the pulse and fibre nonlinearity that compresses the pulse. The capacity-distance product can be as high as Gb/skm. Example 5: Consider a DWDM system of 100 channels with separation of 10 GHz. If the first channel is operated at 1550 nm, what is the total spectrum width used? Ans. The frequency corresponding to 1550 nm is 2E14 Hz. Because the channel separation is 10 GHz, the carrier frequency of the next channel is 2E4 Hz + 10E9 Hz, corresponding to a wavelength of nm. For 100 channels, this means the total spectrum width is 7 nm (channel separation 0.07 nm). (For a practical system this channel separation can be too small because the output wavelength can change with temperature or bias current. Typical wavelength shift due to temperature is 0.1 nm/k (compare with 0.07 nm channel separation.)
5 Example 6: A fiber of 50 km length has Pin=10 mw and Pout=1 mw. Find the loss in db/km. Ans. Loss(dB)=10log[1 mw/10mw]=-10 db. The loss per unit length of the fiber is: Loss(dB/km)=(-10 db/50 km)=-0.2 db/km. Example 7: A 10 km fiber optic communication system link has a fiber loss of 0.3 db/km. Find out the output power if the input power is 20 mw. Ans. Loss(dB)=0.3 db/km x 10 km = 3 db. Pout=20 mwx10^(-3/10)=10 mw
6 Link Power Budget Typical exercises: Example 8: A system has the following characteristics: Find the loss margin Lm. Ans. Lm(dB)=3 dbm -3 db -(40 km x 0.5 db/km) -1 db -3 db - (-36 dbm)=12 db
7 Example 8 (cont.): A system has the following characteristics: Find the loss margin Lm. Ans. Lm(dB)=3 dbm -3 db -(40 km x 0.5 db/km) -1 db -3 db - (-36 dbm)=12 db. Example 9: A receiber has a sensitivity Ps of -45 dbm for BER of E-12. What is the minimum powerthat must be incident on the detector? Ans. -45 dbm=10log(p/1 mw). P=1 mw x 10^-4.5=31.6 uw
8 Dispersion and fiber bandwidth (BW) Fiber bandwidth (BW):
9 Dispersion shifted fibers Fiber bandwidth (BW):
10 Example 10: a 2 km length multimode fiber has a modal dispersion of 1 ns/km and a chromatic dispersion of 100 ps/km-nm. It is used an LED of linewidth 40 nm. a) What is the total dispersion? B) calculate the bandwidth (BE) of the fiber. Ans. a) Dt modal = 2 km x 1 ns/km= 2 ns; Dchromatic = 2 km x 100 ps/km x 40 nm= 8 ns. Dtotal = 8.25 ns. b) BW=0.35/Dtotal=42.42 MHz. BWx distance= 85 MHz.km. Example 11: 50 km SMF, material dispersion of 10 ps/km and waveguide dispersion of -5 ps/km. Laser diode with a linewidth of 0.1 nm. a) What is the chromatic dispersion? b) What is Dtotal? c) Calculate the BW of the fiber. Ans. (D=dispersion) a)dchromatic =10 ps/km nm 5 ps/km nm=5 ps/km nm. b)dtotal = 50 km x 5 ps/km nm x 0.1 nm= 25 ps. c)bw=0.35/dtotal=14 GHz. BWx distance= 700 GHz.km
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