Ryerson University
Department of Electrical and Computer Engineering
Optical Communication Systems
Problem Set - Digital Transmission Systems
1. Make a graphical comparison and a spreadsheet calculation of the maximum attenuation
-limited transmission distance of the following two systems operating at 100 Mb/s:
System one operating at 850 nm
(a) GaAlAs laser diode: fiber coupled power 0 dBm
(b) Silicon avalanche photodiode: -50 dBm sensitivity
(c) Graded index fiber: 3.5 dB/km attenuation at 850 nm
(d) 1 dB/connector connector loss
System two operating at 1300 nm
(a) InGaAsP LED: fiber coupled power -13 dBm
(b) InGaAs PIN photodiode: -38 dBm sensitivity
(c) Graded index fiber: 1.5 dB/km attenuation at 1300 nm
(d) 1 dB/connector connector loss
Allow 6 dB system margin in each case
2. An engineer has the following components available:
(a) GaAlAs laser diode, operating at 850 nm, fiber coupled power 0 dBm
(b) Ten sections of cable each of which is 500 m long, has 4 dB/k m attenuation, has
connectors at both ends
(c) 2 dB/connector connector loss
(d) A PIN photodiode receiver, -45 dBm sensitivity
1
(e) An avalanche photodiode receiver, -56 dBm sensitivity
The engineer wishes to construct a 5-km link operating at 20 Mb/s. Which receiver
should be used -f a 6-dB operating margin is required.
3. A 90-Mb/s NRZ data transmission system that sends two DS3 channels uses a GaAlAs
laser diode with 1 nm spectral width. The rise time of the laser transmitter is 2 ns. The
transmission distance is 7 km over a graded index fiber that has 800 MHz-km bandwidth
distance product.
(a) If the receiver bandwidth is 90 MHz and q = 0.7, what is the system rise time? Does
this meet the required bit rate being less than 70% of a pulse width?
(b) If there is no mode mixing (q = 1) , what is the system rise time?
4. Plot the transmission distance versus the data rate of the following system. The transmit-
ter is GaAlAs laser diode operating at 850 nm with 1 nm line width, coupling 1 mW into
the fiber. The fiber has 3.5 dB/km attenuation at 850 nm and 800 MHz-km bandwidth.
The receiver sensitivity can be approximated by the equation PR = 9LogB − 68.5 where,
B is the data rate is Mb/s. For 1 ≤ B ≤ 1000 Mb/s, plot the attenuation limited trans-
mission distance, the modal dispersion limits with q = 0.5 and q = 1 and the material
dispersion limit. Consider 1 dB connector loss at each end and 6-dB system margin.
5. Repeat the above problem with the following parameters. Source is and LED operating Λ
= 1300 nm; Ps is 50 µW; 40 nm line width. The fiber has 1.5 dB/km attenuation at 1300
nm and 800 MHz-km bandwidth. The receiver sensitivity can be approximated by the
equation PR = 11.5LogB − 60.5 where, B is the data rate is Mb/s. For 10 ≤ B ≤ 1000
Mb/s, plot the attenuation limited transmission distance, the modal dispersion limits
with q = 0.5 and q = 1 and the material dispersion limit. Consider 1 dB connector loss
at each end and 6-dB system margin.
6. A 1550-nm single mode digital fiber optic link needs to operate at OC-12 bit rate over
80 km without amplifiers. A single mode laser launches 13 dBm into the fiber. The
2
fiber has 3.5 dB/km attenuation and there is a splice with 0.1 dB loss at every km. The
coupling loss at the receiver is 0.5 dB,receive is an avalanche photodiode with -39 dBm
sensitivity. Excess noise penalties are predicted to be 1.5 dB. Do the power budget and
find the system margin. What is the system margin at OC-48 bit rate with an APD of
-31 dBm sensitivity
1 Answers:
1. System 1: Ps − Pr = 0 dBm - (-50 dBm) = 50 dB
50 = 2(lc ) + αL + SM ⇒ L = 12 km
System 2: Similarly, L = 11.3 km
2. With PIN, Ps − Pr = 45dB = 11(lc ) + αL + SM ⇒ L = 4.25 km, less than 5 km, the
distance requirement is not met With APD, Ps − Pr = 56dB = 11(lc ) + αL + SM , L =
7 km, the distance requirement is met
3. Dmat = 0.07 ns/nm-km
(a) tsys = 4.9 ns, 0.7/B = 7.77 ns, tsys < 0.7/B, the rise time meets the system
requirement.
(b) with q =1, tsys = 5.85 ns, 0.7/B = 7.77 ns, tsys < 0.7/B, the rise time meets the
system requirement.
4. (a) L = (Ps − 9logB + 62.5 − 2lc )/α
(b) L = 10000/B
(c) L = (1273/B)1/q
5. (a) L = (39.5 − 11.5log(B))/1.5
(b) L = (1273/B)1/q
6. Ps − Pr = 13 − (−31) = 44dB = 0.5 + 79(ls p) + (0.35)80 + 1.5 + SM
System Margin = 6.1 dB
