Exercise Let s consider an optical fiber line operating in the 2 nd single-mode window with LASER emitting diode. An OOK modulation is adopted with a data rate of 00 Mb/s. The length of the fiber line is L=30Km. Fiber distortions are equalized in standard modality. Under such hypothesis it is required to:. Verify if the optimal working condition for an avalanche photodiode are respected. In case they are not respected, please assume to use a PIN diode as photodetector (parameters of photodetectors are set to the usual values). 2. ompute the bit-error-probability at the output of the demodulator (the noise figure of the electric receiver front-end equals to 6dB). From the Table in Slide no. 70 of Part 5: a window 2 a window 2 a window (single-mode) 3 a window 3 a window (single-mode) Attenuation 3 0.4 0.35 0.25 0.2 (db/km) F m (GHz km) 0.9 0.9 0.9 Fλ (GHz km) 0.0.2.2 0. 0.5 LED Fλ (GHz km) 0.6 4.0 4.0 0.2 0.25 ELED Fλ (GHz km) 2.0 6.5 375 6.3 6.3 LASER Transmitted -5-7 -34 - -35 (dbm) LED Transmitted -9 - -22-2 -23 (dbm) ELED Transmitted (dbm) LASER 5 3 4 2 We have: And a transmitted power ON: =0.35 / =375 =3 To solve the first question we have to evaluate the received energy-per-bit E and compare it to the threshold value of -2 db W /Hz, to see whether it is convenient or not to use an APD. E is the received optical energy when a light pulse is transmitted -> ON -> symbol. We know that: = "!!
=3 0.35 30= 7.5 = 37.5 # In the OOK modulation: E = $ =00 '()* + = 7.5 # $ We may verify that E < 2. /0 our photodetector. so it means that in this case it is preferable to use an APD as For the second question in the problem, we have that: By using the logarithmic form, we have: =2345 6 7 9: 67; 9, =203+.7, #.7>?@ AB $ =203+.7 D 37.5E.7 =65.52, From this value we have to subtract 6 db due to electric receiver noise figure and another possible contribution related to the frequency at which the LASER is working (standard equalization residual attenuation), i.e.: where we have: 6.02 5 $ I H H = /" H = " =375 =2.5 30 We get: 6.02 5 $ I=6.02 J 0K I 30 H 375 0 LM 3.5 0 OP That is absolutely negligible. So we may assume: 5 6 7 9 =59.52 From which it is clear that 0: the optical link does not introduce any perceivable bit error probability (quasi error free transmission). 2
Exercise 2 A ATV distribution system consists of an Hybrid Fiber oax (HF) transmission link. The digital TV signal coming from a distribution TV center multiplexes in TDM 20 digital high-quality MPEG-2 TV channels (bit-rate per channel: 2Mb/s). The distribution center is connected to the ATV hub by means of a fiber connection along a distance of 20 Km. The fiber is working in 2 nd single-mode window with LASER emitting diode. The choice of the photodetector depends on the received light energy per bit at the output of the fiber, as usual (parameters of photodetectors are set to the usual values). The ATV hub embarks a regenerative repeater demodulating the optical signal and retransmitting it through the cable line. The ATV broadcast is performed by transmitting the multiplexed TV streams on the cable with 64-QAM modulation to residential users placed at a distance of.6 Km from the hub (the kilometric attenuation of the cable is 2.5dB/Km@MHz). The waveform adopted for ATV transmission is a raised cosine with roll-off factor equal to 0.35. The transmission signal-to-noise ratio at the input of the coaxial cable is 30dB. Moreover, cable and fiber linear distortions are equalized in standard modality. Under these hypotheses, it is required to compute the bit-error-rate measured at the output of the receiver of the residential user (let s suppose that errors in different and heterogeneous network segments are independent and that the noise figure of fiber electric front-end equals to 7dB and the noise figure of the cable receiver is db). The regenerative repeater ensures that the error probability at the output of the two heterogeneous media are independent, so we may say that: S T =S T UVTWX +S T YZT[W Let us evaluate what happens along the fiber link. From the same table used for the previous exercise, we have: And a transmitted power ON: =0.35 / =375 =3 = "! =3 0.35 20= 4 = 34 #! $ =40 ]^ (TDM multiplexing is assumed to be ideal in absence of further indications) E = We may verify that E > 2. /0 our photodetector. = 3.9 0OP $ 4 0 _ 0 OAA = 0 # so it means that in this case it is preferable to use a PIN as 3
By using the logarithmic form for PIN diode, we have: 67; 9, =5+, # 0>?@ AB $, 6.02 5 $ I H The last term is negligible, as we may verify that: H = " =375 =.75 20 5 $ I 4.5 0 Oa H The fiber electric frontend causes an additive attenuation of 7 db, so: This means that: 67; 9, =5+, # 0>?@ AB $, =5 34 76 7=6 S T UVTWX =Q345 6 7 9: 0 The fiber link may be considered error free. The global error probability will be due to the cable link. To compute the bit error probability in 64-QAM we adopt an approximation: c apod 6 6c apod 6c apod 26c KOe 6c KOe = 7 4 234c fg 6c apod 26c KOe = 7 2 234c fg c fg is the average energy per transmitted bit, and it equals the average energy per transmitted symbol / k. 6c apod 7 2 234c fg In the standard equalization case: c apod 6 6c apod= 7 2 234c fg 4
5 c fg h 9, =, +4 0>?@ AB $, ]^ +0>?@ AB ; i + I jfkl mlj But the terms:, +4 0>?@ AB $, ]^ +0>?@ AB ; i orrespond to the input signal to noise ratio, as assigned by the exercise, equal to 30 db (note that 0>?@ AB ; i =0 as no channel coding has been applied). As a matter of fact, we have, by definition: # no,npq rs t m u =v w x y ]z{^ and s is used here to represent the Boltzmann constant. So, we have to compute I : Where: I =0>?@ AB } 4.34} 3 0>?@ AB D+~Ev ~ 4 y Given the raised cosine filter used, we have: And: We get: }= B 4 H H " i " i =.6 4.34 H= $ 2 D~+E= $ 2 ; i >?@ I ' D~+E=40.35 =4.5 ' 2 6 }= B 4 H H " i 4.34 =.95 I =0>?@ AB } 4.34} 3 0>?@ AB D+~Ev ~ 4 y= 9.46 5 c fg h 9, =, +4 0>?@ AB $ ]^, +0>?@ AB ; i + I jfkl mlj =2.54 And: c apod A a 6c apod= _ AI 25 uƒ _ = _ AI 2D.6E The value found for c apod represents the system bit error probability. 5