Optical Communication Set 5

Electrical Engineering

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This set of Optical Communication Multiple Choice Questions & Answers (MCQs) focuses on Optical Communication Set 5

Q1 | A semiconductor laser crystal of length 5 cm, refractive index 1.8 is used as an optical source. Determine the frequency separation of the modes.
  • 2.8 ghz
  • 1.2 ghz
  • 1.6 ghz
  • 2 ghz
Answer: 1.6 ghz
Explaination:
the modes of laser are separated by a frequency internal δf and this separation is given by-
Q2 | Doppler broadening is a homogeneous broadening mechanism.
  • true
  • false
Answer: false
Explaination:
doppler broadening is a inhomogeneous broadening mechanism. in this broadening, the individual groups of atoms have different apparent resonance frequencies. atomic collisions usually provide homogeneous broadening as each atom in collection has same resonant frequency and spectral spread.
Q3 | An injection laser has active cavity losses of 25 cm-1 and the reflectivity of each laser facet is 30%. Determine the laser gain coefficient for the cavity it has a length of 500μm.
  • 46 cm-1
  • 51 cm-1
  • 50 cm-1
  • 49.07 cm-1
Answer: 49.07 cm-1
Explaination:
the laser gain coefficient is equivalent to the threshold gain per unit length and is given by –
Q4 | Longitudinal modes contribute only a single spot of light to the laser output.
  • true
  • false
Answer: true
Explaination:
laser emission includes the longitudinal modes and transverse modes. transverse modes give rise to a pattern of spots at the output. longitudinal modes give only a spot of light to the output.
Q5 | Considering the values given below, calculate the mode separation in terms of free space wavelength for a laser. (Frequency separation = 2GHz, Wavelength = 0.5 μm)
  • 1.4×10-11
  • 1.6×10-12
  • 1×10-12
  • 6×10-11
Answer: 1.6×10-12
Explaination:
the mode separation in terms of free space wavelength is given by-
Q6 |                            is essentially a crude form of Amplitude shift keying.
  • analog modulation
  • digital intensity modulation
  • photodetector
  • receiver structure
Answer: digital intensity modulation
Explaination:
many techniques have been developed to amplitude modulate an optical signal. digital intensity modulation used in direct detection systems is essentially a crude form of ask in which the received signal is detected using square law detector.
Q7 | Almost                    of the transmitter power is wasted in the use of external modulators.
  • half
  • quarter
  • one-third
  • twice
Answer: half
Explaination:
all external modulators suffer the drawback that around half of the transmitted power is wasted. to avoid this, non-synchronous detection can be employed.
Q8 | The line width in the range                  of bit rate is specified for ASK heterodyne detection.
  • 8%
  • 2 to 8%
  • 10 t0 50%
  • 70%
Answer: 10 t0 50%
Explaination:
the ask modulation scheme can be used with laser sources exhibiting the line widths comparable with the bit transmission rate. for ask heterodyne detection, line width range of 10 to 50% is usually specified.
Q9 |                  does not require an external modulator.
  • fsk
  • dsk
  • psk
  • ask
Answer: fsk
Explaination:
fsk involves the frequency deviation property of the directly modulated semiconductor laser used in wideband systems. unlike ask, it does not require an external modulator, which in turn, avoids the wastage of transmitted power.
Q10 | The frequency deviation at frequencies above 1 MHz is typically                          
  • 10 to 20 ma-1
  • 100 to 500 ma-1
  • 1000 to 2000 ma-1
  • 30 to 40 ma-1
Answer: 100 to 500 ma-1
Explaination:
the carrier modulation effect occurs at the frequencies above 1 mhz. at the phase of carrier modulation, the frequency deviation is about 100 to 500 ma-1.
Q11 |                        offers the potential for improving the coherent optical receiver sensitivity by increasing the choice of signalling frequencies.
  • mfsk
  • mdsk
  • mpsk
  • mask
Answer: mfsk
Explaination:
multilevel fsk includes 4- level or 8-level fsk. it improves the receiver sensitivity by reducing the deviation and increasing the usage of signalling frequencies.
Q12 | Eight level FSK and binary PSK yields an equivalent sensitivity.
  • false
  • true
Answer: true
Explaination:
binary psk and 8-level fsk provides an equivalent sensitivity. the main drawback of 8-level fsk is that it yields an equivalent sensitivity to binary psk at the expense of a greater receiver bandwidth requirement.
Q13 | External modulation for                   modulation format allows the most sensitive coherent detection mechanism.
  • fsk
  • dsk
  • psk
  • ask
Answer: psk
Explaination:
external modulation for psk is usually straightforward. it is therefore utilized to provide the modulation format which allows the most sensitive coherent detection mechanism.
Q14 |                    can potentially provide spectral conservation through the use of multilevel signalling.
  • m-ary psk
  • mfsk
  • ask
  • dfsk
Answer: m-ary psk
Explaination:
in m-ary schemes, the spectral efficiency is increased by the factor log2 m.this is purely for m-level schemes which can provide multilevel signalling patterns.
Q15 | The digital transmission on implementation of polarization modulation which involves polarization characteristics of the transmitted optical signal is known as
  • frequency shift keying
  • amplitude shift keying
  • phase shift keying
  • polarization shift keying
Answer: polarization shift keying
Explaination:
polarization shift keying is abbreviated as polsk. polsk requires additional receiver complexity than other modulation formats.
Q16 |                is fully depleted by employing electric fields.
  • avalanche photodiode
  • p-i-n diode
  • varactor diode
  • p-n diode
Answer: avalanche photodiode
Explaination:
apd is fully depleted by electric fields more than 104v/m. this causes all the drifting of carriers at saturated limited velocities.
Q17 | At low gain, the transit time and RC effects
  • are negligible
  • are very less
  • dominate
  • reduce gradually
Answer: dominate
Explaination:
low gain causes the dominance of transit time and rc effects. this gives a definitive response time and thus device obtains constant bandwidth.
Q18 | The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as                
  • is negligible
  • very less
  • increases gradually
  • dominates
Answer: dominates
Explaination:
high gain causes avalanche buildup time to dominate. thus the
Q19 | Often                      pulse shape is obtained from APD.
  • negligible
  • distorted
  • asymmetric
  • symmetric
Answer: asymmetric
Explaination:
asymmetric pulse shape is acquired from apd. this is due to relatively fast rise time as electrons are collected and fall time dictated by transit time of holes.
Q20 | Fall times of 1 ns or more are common.
  • false
  • true
Answer: true
Explaination:
the use of suitable materials and structures give rise times between 150 and 200 ps. thus fall times of 1 ns or more are common which in turn limits the overall response of device.
Q21 | Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μm wavelength.
  • 0.459
  • 0.7
  • 0.312
  • 0.42
Answer: 0.459
Explaination:
the responsivity of a rapd is given by-
Q22 | Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w.
  • 6μm
  • 7.21μm c) 0.112μm
  • d) 3μm
Answer: d) 3μm
Explaination:
the wavelength can be found from the responsivity formula given by-
Q23 | 49 A/w.
  • 0.91 μw
  • 0.32 μw
  • 0.312 μw
  • 0.80 μw
Answer: 0.80 μw
Explaination:
the photocurrent is given by ip
Q24 | Determine the Responsivity of optical power of 0.4μW and photocurrent of 0.294 μA.
  • 0.735
  • 0.54
  • 0.56
  • 0.21
Answer: 0.735
Explaination:
the photocurrent is given by ip
Q25 | Compute multiplication factor of RAPD with output current of 10 μAand photocurrent of 0.369μA.
  • 25.32
  • 27.100
  • 43 d) 22.2
Answer: 27.100
Explaination:
the multiplication factor of photodiode is given by-
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