Optical Communications Multiple Choice Questions on “Optical Amplifiers – Semiconductor Optical Amplifiers”.
1. For linear as well as in nonlinear mode _______________ are most important network elements.
a) Optical amplifier
b) Optical detector
c) A/D converter
d) D/A converters
Answer: a
Explanation: In single-mode fiber system, signal dispersion is very small, hence there is attenuation. These systems don’t require signal regeneration as optical amplification is sufficient so optical amplifier are most important.
2. The more advantages optical amplifier is ____________
a) Fiber amplifier
b) Semiconductor amplifier
c) Repeaters
d) Mode hooping amplifier
Answer: b
Explanation: Semiconductor optical amplifiers are having smaller size. They can be integrated to produce subsystems. Thus are more profitable than other optical amplifier.
3. ________________ cannot be used for wideband amplification.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber amplifiers provide a very narrow spectral bandwidth. These bandwidth can be around 50 MHz, hence cannot be employed for wideband amplification.
4. ____________ is used preferably for channel selection in a WDM system.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber provides amplification of a particular channel. This amplification can be done without boosting other channels besides that particular channel.
5. For used in single-mode fiber __________ are used preferably.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: Semiconductor optical amplifiers have low power consumption. There single mode structure makes them appropriate and suitable for used in single mode fiber.
6. Mostly ____________ are used in nonlinear applications.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) FPAs
Answer: d
Explanation: FPAs have a resonant nature. This can be combined with their high internal fields. They provide pulse shaping and bi-stable elements. Thus, are used widely in nonlinear application.
7. _______________ is superior as compared to _________________
a) TWA, FPA
b) FPA, TWA
c) EDFA, FPA
d) FPA, EDFA
Answer: a
Explanation: In TWA operating in single-pass amplification mode, the Fabry-Perot resonance is suppressed by facet reflectivity reduction. This affects in increasing of amplifier spectral bandwidth. This makes them less dependence of transmission characteristics on fluctuations in biased current, input signal polarization. Thus FPA are superior to TWA.
8. ______________ are operated at current beyond normal lasing threshold current, practically.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: The anti-reflection facet coatings affects in the form of increasing lasing current threshold. This causes SOAs to be operated at current beyond normal lasing threshold current.
9. An uncoated FPA has peak gain wavelength 1.8μm, mode spacing of 0.8nm, and long active region of 300 v. Determine RI of active medium.
a) 4.25×106
b) 3.75×107
c) 3.95×107
d) 4.25×109
Answer: b
Explanation: n=λ2/2δλL=1.8×10-6/2×0.8×10-9×300×10-6=3.75×107.
10. Determine the peak gain wavelength of uncoated FPA having mode spacing of 2nm,and 250μmlong active region and R.I of 3.78.
a)2.25×10-4
b)4.53×10-8
c)1.94×10-6
d)4.25×109
Answer: c
Explanation: The peak gain wavelength is given by
λ2=n2δλL=3.78×2×2×10-9×250×10-6=1.94×10-6m.
11. An SOA has net gain coefficient of 300, at a gain of 30dB. Determine length of SOA.
a) 0.32 m
b) 0.023 m
c) 0.245 m
d) 0.563 m
Answer: b
Explanation: The length of SOA is determined by
L = Gs(dB)/10×g×loge = 30/10×300×0.434`= 0.023 m.
12. An SOA has length of 35.43×10-3m, at 30 dB gain. Determine net gain coefficient.
a) 5.124×10-3
b) 1.12×10-4
c) 5.125×10-3
d) 2.15×10-5
Answer: c
Explanation: The net gain coefficient of SOA is given by
g = L×10×loge/Gs(dB) = 35.43×10-3×10×0.434/30
=5.125×10-3.
13. An SOA has mode number of 2.6, spontaneous emission factor of 4, optical bandwidth of 1 THz. Determine noise power spectral density.
a) 1.33×10-3
b) 5.13×1012
c) 3.29×10-6
d) 0.33×10-9
Answer: a
Explanation: The noise power spectral density Past is
Past = mnsp(Gs-1) hfb
= 2.6×4(1000-1)×6.63×10-34×1.94×1014×1×1012
= 1.33×10-3W.
14. An SOA has noise power spectral density of 1.18mW, spontaneous emission factor of 4, optical bandwidth of 1.5 THz. Determine mode number.
a) 1.53 × 1028
b) 6.14 × 1012
c) 1.78 × 1016
d) 4.12 × 10-3
Answer: a
Explanation: The mode number is determined by
m = Past/nsp(Gs-1) hfB
= 1.18×10-3/4(1000-1)×6.63×10-34×1.94×1014×1.3×1012
= 1.53 × 10-34.
.