Optical Communications Multiple Choice Questions on “Analog Systems”.
1. Determine dispersion equalization penalty if total RMS pulse broadening is 4.8ns, BT is 25 Mbits/s.
a) 0.03 dB
b) 0.08 dB
c) 7 dB
d) 0.01 dB
Answer: a
Explanation: Dispersion equalization penalty is denoted by DL. It is given by-
DL = 2 (2σTBT√2)4. Here σT=RMS pulse broadening.
2. Determine RMS pulse broadening with mode coupling if pulse broadening is 0.6 over 8km.
a) 1.6ns
b) 1.7ns
c) 1.5ns
d) 1.4ns
Answer: b
Explanation: Total RMS pulse broadening with mode coupling is given by-
σT = σ√L. Here σT = RMS pulse broadening, L = length of the fiber.
3. Determine dispersion equalization penalty with mode coupling of 1.7ns if BT is 25 Mbits/s.
a) 4.8 * 104dB
b) 4 * 104dB
c) 4.2 * 104dB
d) 3.8 * 104dB
Answer: c
Explanation: Dispersion equalization penalty is denoted by DL. With mode coupling, it is given by-
DL=2 (2σTBT√2)4. Here σT=RMS pulse broadening.
4. Determine dispersion equalization penalty without mode coupling if BT is 150 Mbits/s and total rms pulse broadening is 4.8ns.
a) 34 dB
b) 33 dB
c) 76.12 dB
d) 34.38 dB
Answer: d
Explanation: Dispersion equalization penalty is denoted by DL(WM). It is given by-
DL(WM) = 2 (2σTBT√2)4. Here σT = RMS pulse broadening, (WM) = without mode coupling.
5. Determine ratio of SNR of coaxial system to SNR of fiber system if peak output voltage is 5V, quantum efficiency of 70%, optical power is 1mW, wavelength of 0.85μm.
a) 1.04 * 104
b) 2.04 * 104
c) 3.04 * 104
d) 4.04 * 104
Answer: a
Explanation: Ratio of SNR of coaxial system to SNR of fiber system is given by-
Ratio = V2hc/2KTZ0ηPiλ. Here, η=quantum efficiency, Pi = 0ptical power in mW, V=optical output voltage.
6. Determine the peak output voltage if efficiency is 70%, wavelength is 0.85μm and output power is 1mW.
a) 7V
b) 8V
c) 5V
d) 6V
Answer: b
Explanation: Peak output voltage is given by-
V2 = (2KTZ0ηPiλ * Ratio)/hc. Here, η = quantum efficiency, Pi=0ptical power in mW, V=optical output voltage.
7. Determine the efficiency of a coaxial cable system at 17 degree Celsius with peak output voltage 5V, 0.85 μm wavelength and SNR ratio of 1.04 * 104.
a) 80%
b) 70%
c) 40%
d) 60%
Answer: b
Explanation: The efficiency of a coaxial cable system is η=V2hc/2KTZ0ηPiλ * Ratio. Hereη=quantum efficiency, Pi = 0ptical power in mW, V=optical output voltage.
8. Determine the wavelength of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 100Ω impedance, optical power of 1mW, 70% quantum efficiency.
a) 0.39μm
b) 0.60μm
c) 0.85μm
d) 0.98μm
Answer: c
Explanation: The wavelength can be determined by –
λ = V2hc/2KTZ0ηPi * Ratio. Hereη=quantum efficiency, Pi = 0ptical power in mW, V = optical output voltage.
9. Determine the impedance of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 0.85μmwavelength, optical power of 1mW, 70% quantum efficiency and SNR ratio of 1.04 * 104.
a) 80Ω
b) 50Ω
c) 90Ω
d) 100Ω
Answer: d
Explanation: The impedance is given by-Z0=V2hc/2KTPi * Ratio. Hereη=quantum efficiency, Pi = Optical power in mW, V=optical output voltage.
10. The 10-90% rise times for components used in D-IM analog optical link is given. (LED=10ns, Intermodal=9ns/km, Chromatic=2ns/km, APD = 3ns). Link is of 5km. Determine the total rise time.
a) 62ns
b) 53ns
c) 50ns
d) 52ns
Answer: d
Explanation: Total rise time is given by-
Tsyst=1.1[Ts2+Tn2+Tc2+TD2]1/2. Here Ts = rise time, Tn = intermodal time, Tc = Chromatic time.
11. The 10-90% rise times for components used in D-IM analog optical link is given. (LED=10ns, Intermodal=9ns/km, Chromatic=2ns/km, APD = 3ns). Link is of 5km. It has an optical bandwidth of 6MHz. Determine maximum permitted system rise time.
a) 58.3ns
b) 54ns
c) 75ns
d) 43.54ns
Answer: a
Explanation: The maximum permitted system rise time is given by-
Tsyst(Max) = 0.35/Bopt. Here, Bopt=Optical Bandwidth.
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