Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Mid Infrared and Far Infrared Lasers”.

1. The parameters having a major role in determining threshold current of efficiency of injection laser are ___________
a) Angle recombination and optical losses
b) Frequency chirping
c) Relaxation oscillation
d) Mode hopping
Answer: a
Explanation: Optical losses due to free carrier absorption are more because of their dependence on square of the wavelength. Also irradiative recombination through Auger recombination contributes to it. Both these effects cause more problems in md-infrared wavelengths and so are of much importance art high temperature due to high concentration of free carriers. They also limit maximum operating temperatures.

2. Auger current is mostly ___________________ for material with band gap providing longer wavelength emission.
a) Unaffected
b) Lesser
c) Larger
d) Vanishes
Answer: c
Explanation: The total current required for injection laser threshold is more than that provided to radioactive recombination as Auger current is added. This current depends on electronic band structure of material and often consists of different Auger transitions. So it is larger for materials with band gaps providing longer wavelength emission.

3. Injection lasers operating in smaller wavelengths are subjected to increased carrier losses.
a) True
b) False
Answer: b
Explanation: Injection lasers operating in longer wavelengths (mid and far infrared) are subjected to increased carrier losses as compared to devices operating up to 1.6μm. This is from nonradiative recombination through Auger interaction. This recombination energy is dissipated as thermal energy to other free carriers. If band gap of semiconductor is increased, occurrence of these events gets increased.

4. Devices based on quaternary PbSnSeTe and their ternary compounds, emit at wavelength?
a) Between 3-4 μm
b) Longer than 4 μm
c) Between 3.5 to 4.2 μm
d) Between 2 to 3 μm
Answer: b
Explanation: Quaternary devices emit at wavelength longer than 4μm. Auger effects are less in these alloys which provide lower current thresholds and higher maximum operating temperature.

5. Replacing Sn with Eu, Cd or Ge in some _________________ the band gap.
a) Remove the band gap
b) Does not affect
c) Decreases
d) Increases
Answer: d
Explanation: When in a particular alloy laser for example PbSnSeTe, if Sn is replaced with Eu, Cd or Ge, there is an increase in band gap. This increase in band gap provides the laser to operate in shorter wavelength.

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6. Lasing obtained in __________ when 191 mW of pump light at a wavelength of 0.477 μm is launched into laser.
a) Ternary PbSnSeTe alloy laser
b) Quaternary PbSnSeTe alloy laser
c) Doped Fluoro-zirconate fiber
d) Ternary PbEuTe alloy laser
Answer: c
Explanation: When Fluoro-zirconate fiber lasers are doped with Erbium helium or thulium, there are emission at 2-3 μm wavelength range. But lasing was obtained in this doped Fluorozirconate fiber at a wavelength of 0.477μm.

7. The thulium doped fiber laser when pumped with alexandrite laser output at 0.786 μm, the laser emits at ___________
a) 0.6 μm
b) 0.8 μm
c) 2.3 μm
d) 1.2μm
Answer: c
Explanation: The thulium system emits at 2.3 μmwhen subjected to alexandrite laser at 0.786 μm. this system is four levels in which the pump band is upper lasing level at 2.3μm.

8. The diode-cladding-pumped Erbium praseodymium-doped fluoride device operates at wavelength.
a) Around 3 μm
b) 4 μm
c) 2.6 μm
d) 1.04 μm
Answer: a
Explanation: The diode-cladding-pumped Erbium praseodymium-doped fluoride device operates at a wavelength of 3 μm. This laser is capable of producing a very high output power of about 1W or more. It consists of double clad fluoride fiber.

9. A technique based on inter-sub band transition is known as ___________
a) Auger recombination
b) Frequency chirping
c) Inter-valence band absorption
d) Quantum cascading
Answer: d
Explanation: The quantum cascaded laser is a layered semiconductor device having a series of coupled quantum wells grown on GaAs or Imp substrate. This principle of QC lasers provides emission of an optical signal around full wavelength range. Quantum mechanical band structure determines the emitted wavelength.

10. In a QC laser, a same electron can emit number of photons.
a) True
b) False
Answer: a
Explanation: The QC laser operates by pumping a energy level and then using the energy in a controlled manner. This gives some energy each time over several steps. And since a QC laser structure includes a series of energy levels the same electron emits a number of photons while cascading down through each energy level.

11. The phenomenon resulting in the electrons to jump from one state to another each time emitting of photon is known as ___________
a) Inter-valence band absorption
b) Mode hopping
c) Quantum cascading
d) Quantum confinement
Answer: d
Explanation: In Quantum confinement, charge carriers are trapped in a small area and this occurs in quantum wells at nanometer scale. When the quantum layer size raises to a size comparable to emission wavelength, the electron motion becomes perpendicular to plane of layer. Due to this, the electrons jump from one state to another each time from one state to another.

12. A QC laser is sometimes referred as ___________
a) Unipolar laser
b) Bipolar laser
c) Gain guided laser
d) Non semiconductor laser
Answer: a
Explanation: A QC laser utilizes only n-type of charge carriers. Their operation is entirely based on electrons and holes play no part in this, so they are known as unipolar lasers.

13. In QC lasers, it is possible to obtain different output signal wavelengths. This can be achieved by ___________
a) Inter-valence band absorption
b) Mode hopping
c) Quantum cascading
d) Selecting layers of different thickness
Answer: d
Explanation: In QC laser, electrons emit energy. This energy emitted at this stage determines wavelength of radiation and it depends only on thickness of the layer. Thus output signal wavelength is dependent on thickness of lasers.

14. QC lasers ______________ the performance characteristics.
a) Have negligible effects
b) Does not affects
c) Improves
d) Degrades
Answer: c
Explanation: QC lasers are based on inter sub band transition techniques. They have ability of carrying large amount of currents. A single electron is enough to generate number of photons. Thus, provides an increase in output signal power which is greater than thousands at same wavelength due to large number of cascaded stages.

15. An MQW cascaded laser is more advantageous because of ___________
a) Mode hopping
b) Auger recombination
c) Control over layers of material
d) Properties of material
Answer: c
Explanation: In MQW cascaded layers, cascading creates number of injector/collector and active region in single stage. Each region contains a single quantum wells. Such structures permit maximum injection/collection of current and thereby produce a large number of photons. This formation of any injector/collector and active regions is achieved through precise control of several hundreds of layers of the material, where each layer should only be few nanometers thick.

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Optical Communications Interview Questions and Answers for Experienced people on “Stability of the Fiber Transmission Characteristics”.

1. ____________ results from small lateral forces exerted on the fiber during the cabling process.
a) Attenuation
b) Micro-bending
c) Dispersion
d) Stimulated Emission
Answer: b
Explanation: Optical fibers must be designed so that the transmission characteristics of the fiber are maintained after the cabling process. The main problem which occurs in the cabling process is the meandering of the axis of the fiber core on a microscopic scale within the cable form. This phenomenon is called as micro-bending.

2. Microscopic meandering of the fiber core axis that is micro-bending is caused due to ___________
a) Environmental effects
b) Rough edges of the fiber
c) Large diameter of core
d) Polarization
Answer: a
Explanation: Micro-bending can be generated at any stage during manufacturing process, cable installation process or during service. This is mainly due to environmental effects, mainly varying temperatures causing differential expansion or contraction.

3. How many forms of modal power distribution are considered?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two forms of modal power distribution are considered. The first form is seen when a fiber is excited by a diffuse Lambertian source, and is called as fully filled mode distribution. The second form occurs when, due to mode coupling and attenuation, the distribution of optical power becomes invariant with the distance of propagation along the fiber, and is called as steady-state mode distribution.

4. What does micro-bending losses depend on _____________
a) Core material
b) Refractive index
c) Diameter
d) Mode and wavelength
Answer: d
Explanation: Micro-bending losses cause differential expansion or contraction. These losses are mode dependent. The number of modes is a function inverse to the wavelength of the transmitted light and thus micro-bending losses are wavelength dependent.

5. The fiber should be________________ to avoid deterioration of the optical transmission characteristics resulting from mode-coupling-induced micro-bending.
a) Free from irregular external pressure
b) Coupled with plastic
c) Large in diameter
d) Smooth and in a steady state
Answer: a
Explanation: Micro-bending losses results from environmental effects such as temperature variation. The irregular external pressure deteriorates the quality of transmission through the fiber. Thus, controlled coating and cabling of the fiber is essential in order to reduce the cabled fiber attenuation.

6. The diffusion of hydrogen into optical fiber affects the ______________
a) Transmission of optical light in the fiber
b) Spectral attenuation characteristics of the fiber
c) Core of the fiber
d) Cladding of the fiber
Answer: b
Explanation: The hydrogen absorption by an optical fiber increases optical fiber losses. It forms absorption peaks where the hydrogen diffuses into interstitial spaces in the glass. At high temperatures, these losses can increase and reduced if the hydrogen source is removed.

7. __________ can induce a considerable amount of attenuation in optical fibers.
a) Micro-bending
b) Dispersion
c) Diffusion of hydrogen
d) Radiation Exposure
Answer: d
Explanation: The optical transmission characteristics of the fiber cables can be degraded by exposure to nuclear radiation. The nature of this attenuation depends upon fiber structures, optical intensity, wavelength, etc. The radiation-induced attenuation comprises both permanent and temporary components which makes the exposure irreversible and reversible respectively.

8. The radiation-induced attenuation can be reduced through photo-bleaching.
a) True
b) False
Answer: a
Explanation: Photo-bleaching can be exploited to study the diffusion of molecules. It is used to remove the radiation exposure by quenching auto-fluorescence. It helps to increase signal-to-noise ratio of the fiber and thus reduces attenuation.

9. The losses due to hydrogen absorption and reaction with fiber deposits can be temporary.
a) True
b) False
Answer: b
Explanation: Hydrogen absorption occurs in two mechanisms. First phenomenon affects silica-based glass fibers whereas the second one occurs when hydrogen reacts with the fiber deposits to give P-OH, Ge-OH absorption. These losses are permanent.

10. The losses caused due to hydrogen absorption mechanisms are in the range of ___________
a) 20 dB/km to 25 dB/km
b) 10 dB/km to 15 dB/km
c) 25 dB/km to 50 dB/km
d) 0 dB/km to 5 dB/km
Answer: c
Explanation: The diffusion of hydrogen into optical fiber leads to an increase in optical fiber losses, causing damage to spectral loss characteristics. This phenomenon gets vibrant at higher temperatures. The losses caused due to such absorption are greater than 25 dB/km.

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Optical Communications Multiple Choice Questions on “Ray Theory Transmission”.

1. Who proposed the idea of transmission of light via dielectric waveguide structure?
a) Christian Huygens
b) Karpon and Bockham
c) Hondros and debye
d) Albert Einstein
Answer: c
Explanation: It was in the beginning of 20^th century where Hondros and debye theoretical and experimental study demonstrated that information can be transferred as a form of light through a dielectric waveguide.

2. Who proposed the use of clad waveguide structure?
a) Edward Appleton
b) Schriever
c) Kao and Hockham
d) James Maxwell
Answer: c
Explanation: The invention of clad waveguide structure raised the eyebrows of the scientists. The proposals by Kao and Hockham proved beneficial leading in utilization of optical fibre as a communication medium.

3. Which law gives the relationship between refractive index of the dielectric?
a) Law of reflection
b) Law of refraction (Snell’s Law)
c) Millman’s Law
d) Huygen’s Law
Answer: b
Explanation: Snell’s Law of refraction states that the angle of incidence Ø1 and refraction Ø2 are related to each other and to refractive index of the dielectrics.
It is given by n1sinØ1 = n2sinØ2
where n1 and n2 are the refractive indices of two mediums. Ø1 and Ø2 are angles of incidence and refraction.

4. The light sources used in fibre optics communication are ____________
a) LED’s and Lasers
b) Phototransistors
c) Xenon lights
d) Incandescent
Answer: a
Explanation: LED’s and Lasers are the light sources used in optical communication. During the working process of optical signals they are both supposed to be switched on and of rapidly and accurately enough to transmit the signal. Also they transmit light further with fewer errors.

5. The ________ ray passes through the axis of the fiber core.
a) Reflected
b) Refracted
c) Meridional
d) Shew
Answer: c
Explanation: When a light ray is passed through a perfect optical fiber, any discontinuities at the core cladding interface would result in refraction rather than total internal reflection. Such light ray passes through the axis of fiber core and is called as meridional ray. This principle is used while stating the fundamental transmission properties of optical fiber.

6. Light incident on fibers of angles________the acceptance angle do not propagate into the fiber.
a) Less than
b) Greater than
c) Equal to
d) Less than and equal to
Answer: b
Explanation: Acceptance angle is the maximum angle at which light may enter into the fiber in order to be propagated. Hence the light incident on the fiber is less than the acceptance angle, the light will propagate in the fiber and will be lost by radiation.

7. What is the numerical aperture of the fiber if the angle of acceptance is 16 degree?
a) 0.50
b) 0.36
c) 0.20
d) 0.27
Answer: d
Explanation: The numerical aperture of a fiber is related to the angle of acceptance as follows:
NA = sin Ѳa
Where NA = numerical aperture
Ѳ = acceptance angle.

8. The ratio of speed of light in air to the speed of light in another medium is called as _________
a) Speed factor
b) Dielectric constant
c) Reflection index
d) Refraction index
Answer: d
Explanation: When a ray travels from one medium to another, the ray incident from a light source is called as incident ray. In passing through, the speed varies. The ratio of the speed of incident and the refracted ray in different medium is called refractive index.

9. When a ray of light enters one medium from another medium, which quality will not change?
a) Direction
b) Frequency
c) Speed
d) Wavelength
Answer: b
Explanation: The electric and the magnetic field have to remain continuous at the refractive index boundary. If the frequency is changed, the light at the boundary would change its phase and the fields won’t match. In order to match the field, frequency won’t change.

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Optical Communications Multiple Choice Questions on “Optical Ethernet”.

1. What is the exception in the similarities between the optical Ethernet and the Ethernet LAN?
a) Physical layer
b) Data-link layer
c) Refractive index
d) Attenuation mechanism
Answer: a
Explanation: Optical Ethernet is similar to the conventional Ethernet LAN with the exception of the physical layer. Physical layer includes the flow of data in the form of binary digits. This transmission takes place on the bit level.

2. Which technology is used by optical Ethernet?
a) GP-technology
b) HJ-technology
c) IP-technology
d) GB-technology
Answer: c
Explanation: Optical Ethernet is the fourth generation of the Ethernet family. The earlier generations include X.25, Frame Relay and ATM. Unlike these technologies, optical Ethernet uses IP-based technology.

3. When was the Gigabit Ethernet network developed?
a) 1977
b) 1988
c) 1990
d) 2002
Answer: b
Explanation: The Gigabit Ethernet (Gbe) network was developed in 1988. It was developed by merging two technologies namely 802.3 Ethernet and ANSI X3T11 fiber channel.

4. Optical Ethernet can operate at the transmission rates as low as ______________
a) 10 M bits per second
b) 40 M bits per second
c) 100 M bits per second
d) 1000 M bits per second
Answer: a
Explanation: Usually, high transmission rates define optical Ethernet. The ITU-T Recommendation specifies the physical layer for optical Ethernet. It can operate at transmission rates as low as 10 M bits per second.

5. How many types of optical Ethernet connections are developed?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: There are three different types of optical Ethernet connections. They are point-to-point, point-to-multipoint, and multipoint-to-multipoint. Multipoint refers to more number of connections on either side.

6. Which type of connection can be used as an Ethernet switch?
a) Point-to-point
b) Multipoint-to-multipoint
c) Multipoint-to-point
d) Point-to-multipoint
Answer: b
Explanation: The multipoint-to-multipoint configuration refers to the bus, tree or mesh topology. Such a mesh can be made to work as a switching hub with non-blocking switching features. It facilitates switching between different optical Ethernet users.

7. How many aspects are included in the standard Ethernet protocol?
a) One
b) Two
c) Four
d) Three
Answer: c
Explanation: Optical Ethernet follows standard Ethernet protocol. This protocol includes four different aspects: Frame, MAC, signaling components and the physical medium.

8. Which of the following is not included in the Ethernet frame format?
a) MAC
b) Preamble
c) Destination address
d) Source address
Answer: a
Explanation: The Ethernet frame format includes preamble, destination and source addresses, length, data and the check sequence. MAC is the protocol which is used for sharing the network nodes.

9. The _______________ provides point-to-point access to a bidirectional single-mode optical fiber.
a) Optical regenerator
b) Optical session
c) Optical distribution node
d) Optical buffer
Answer: c
Explanation: The ODN is abbreviated as Optical distribution node. It can access an optical fiber on a point-to-point basis. Thus, a single mode bidirectional optical fiber can be accessed by an Optical distribution node.

10. _______________ is the de-multiplexing technique used to split SONET bandwidth into logical groups.
a) SDH
b) Virtual concatenation
c) STS-1
d) Optical breakdown
Answer: b
Explanation: Virtual concatenation (VC) is basically a splitting technique. It can split the SONET bandwidth into groups. These groups may be transported or routed independently.

11. Ethernet switches support multiprotocol label switching.
a) True
b) False
Answer: a
Explanation: Ethernet switches support multiprotocol label switching. This feature is desired mainly in MAN. The use of such switches in LAN exceeds the network capabilities.

12. Length field in MAC frame ensures that the frame signals stay on the network in order to detect the frame within the correct time limit.
a) True
b) False
Answer: b
Explanation: MAC frame includes Length field to identify the type or length of the network protocols. The data field is used to ensure that the frame signals stay on the network long enough to detect the frame within the desired limit.

13. The ___________ protocol is not used when the Ethernet connections are configured for a full duplex operation.
a) TCP/IP
b) MAC
c) CSMA/CD
d) DTH
Answer: c
Explanation: In Ethernet connections, the full-duplex operation situation may lead to an increased frame dropping rate. The dropped frame cannot be detected without collision. Thus, CSMA protocol is not used in full duplex mode.

14. Optical Ethernet provides switching capabilities in layers ________
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 4
Answer: b
Explanation: Layer 2 and 3 are data link and network layers. IP routing is usually considered to be ⅔ switched network. Thus, unlike conventional LAN, optical Ethernet provides switching capabilities between layers 2 and 3.

15. The ______________ approach can provide interconnection among multiple site locations within 40 km range.
a) 3 Gbe
b) 5 Gbe
c) 1 Gbe
d) 10 Gbe
Answer: d
Explanation: The 10Gbe approach uses a 40 km window range. It can be used in LAN’s and MAN. It provides switch-to-switch network within data centers.

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Optical Communications Problems on “Dispersion Management and Soliton Systems”.

1. Calculate second order dispersion coefficient for path length L₂ 20km and L₁ 160km. Dispersion coefficient for L₂ is 17.
a) -2.125ps/nm km
b) -3.25ps/nm km
c) -3.69ps/nm km
d) -1.28ps/nm km
Answer: a
Explanation: The second order dispersion coefficient for path length is given by-
β₂₁ = -β₂₂L₂/L₁. Here, β₂₂ = Dispersion coefficient forL₂, L₂ and L₁ are path lengths.

2. Calculate the path length L₂ if L₁is 160, dispersion coefficient of L₂ is 17, dispersion coefficient of L₁ is -2.25 ps/nmkm.
a) 40 km
b) 20 km
c) 30 km
d) 10 km
Answer: b
Explanation: The path length L₂ is given by-
L₂ = β₂₁L₁/-β₂₂. Here, β₂₂ = Dispersion coefficient forL₂, β₂₁ = Dispersion coefficient for L₁, L₂ and L₁ are path lengths.

3. Calculate path length L₁ if L₂ is 20, dispersion coefficient of L₂ is 17, dispersion coefficient of L₁ is -2.25 ps/nmkm.
a) 180 km
b) 30 km
c) 160 km
d) 44 km
Answer: c
Explanation: The path length L1is given by-
L₁ = β₂₁L₂/-β₂₂. Here, β₂₂ = Dispersion coefficient forL₂, β₂₁ = Dispersion coefficient for L₁, L₂ and L₁ are path lengths.

4. Calculate second order dispersion coefficient for path length L₁ 20 km and L₁ 160 km. Dispersion coefficient for L₁ is -2.125*10^-12s/nmkm.
a) 20
b) 19
c) 18
d) 17
Answer: d
Explanation: The second order dispersion coefficient for path length is given by-
β₂₂=-β₂₁L₂/L₁. Here, β21 = Dispersion coefficient forL1, L₂ and L1 are path lengths.

5. Calculate dispersion slope for second path fiber if L₁ is 150, L₂ is 10 and s₁ is 0.075.
a) 1.125
b) 2.125
c) 3.125
d) 1.9
Answer: a
Explanation: Dispersion slope for second path fiber is s₂ = -s₁(L₁/L₂). Here s1 and s2 are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

6. Calculate dispersion slope for first path fiber if L₁ is 160, L₂ is 20 and s₂ is 0.6ps/nm km.
a) 0.1
b) 0.432
c) 0.236
d) 0.075
Answer: d
Explanation: Dispersion slope for first path fiber is s₁ = -s₂(L₁/L₂). Here s1 and s2 are dispersion slopes for L₁, L₂. L₂ And L₁ are path lengths.

7. Calculate L₂ if dispersion slope for first path fiber is 0.075 and L₁ is 160 km and s₂ is -0.6ps/nm km.
a) 20 km
b) 30 km
c) 40 km
d) 50 km
Answer: a
Explanation: L₂ is determined by –
L2 = (-s₁/s₂)*L₁. Here s₁ and s₂ are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

8. Calculate L₁ if dispersion slope for first path fiber is 0.075 and L₂ is 20 km and s₂ is -0.6ps/nm km.
a) 170 km
b) 160 km
c) 180 km
d) 175 km
Answer: b
Explanation: L₁ is determined by –
L₂ = (-s₁/s₂)* L₂. Here s₁ and s₂ are dispersion slopes for L₁, L₂. L₂ and L₁ are path lengths.

9. Calculate separation of soliton pulses over a bit period length if R₂ pulse width is 6 ps for bit period of 70 ps.
a) 5.9
b) 5.7
c) 5.8
d) 5.4
Answer: c
Explanation: The separation of soliton pulses over a bit period length is calculated by –
q₀ = T₀/2ς. Here ς = pulse width and T₀ = bit period.

10. Calculate RZ pulse width if bit period is 60ps and separation of soliton pulses is 5.4.
a) 5.5ps
b) 8.1ps
c) 4.3ps
d) 2.3ps
Answer: a
Explanation: RZ pulse width can be calculated by –
ς = T₀/q₀. Here ς = pulse width and T₀ = bit period.

11. Calculate bit period if RZ pulse width is 50ps and separation of soliton pulses is 5.6.
a) 570ps
b) 540ps
c) 430ps
d) 560ps
Answer: d
Explanation: Bit period can be calculated by –
T₀ = 2T₂q₀. Here T₂=pulse width and T₀=bit period.

12. Calculate value of dimensionless parameter if bit period is 45ps and RZ pulse width is 4 ps.
a) 5.625
b) 5.0
c) 4
d) 6.543
Answer: a
Explanation: Dimensionless parameter is given by –
q₀ = T₀/2ς. Here ς=pulse width and T₀=bit period.

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Optical Communications Multiple Choice Questions on “FET Pre – Amplifiers”.

1. ____________ is the lowest noise amplifier device.
a) Silicon FET
b) Amplifier-A
c) Attenuator
d) Resonator-B
Answer: a
Explanation: FET operates by controlling the current flow with an electric field produced by an applied voltage on the gate of the device. Silicon FET is fabricated for low noise devices. It is the lowest noise amplifier device available.

2. FET device has extremely high input impedance greater than _________
a) 10⁷ Ohms and less than 10⁸
b) 10⁶ Ohms and less than 10⁷
c) 10¹⁴ Ohms
d) 10²³ Ohms
Answer: c
Explanation: FET operation involves the applied voltage on the gate of the device. The gate draws virtually no current, except for leakage, giving the device extremely high input impedance.

3. The properties of a bipolar transistor are superior to the FET.
a) True
b) False
Answer: b
Explanation: bipolar transistor operates by controlling the current flow with an electric field produced with a base current. The properties of a bipolar transistor are limited by its high trans-conductance than the FET.

4. Bipolar transistor is more useful amplifying device than FET at frequencies _____________
a) Above 1000 MHz
b) Equal to 1 MHz
c) Below 25 MHz
d) Above 25 MHz
Answer: d
Explanation: In FETs, the current gain drops to values near unity at frequencies above 25MHz. The trans-conductance is fixed with decreasing input impedance. Therefore, bipolar transistor is more useful amplifying device at frequencies above 25MHz.

5. High-performance microwave FETs are fabricated from ___________
a) Silicon
b) Germanium
c) Gallium arsenide
d) Zinc
Answer: c
Explanation: Since the mid- 1970s, the development of high-performance microwave FETs found its way. These FETs are fabricated from gallium arsenide and are called as GaAs metal Schottky field effect transistors (MESFETs).

6. Gallium arsenide MESFETs are advantageous than Silicon FETs.
a) True
b) False
Answer: a
Explanation: Gallium arsenide MESFETs are Schottky barrier devices. They operate with both low noise and high gain at microwave frequencies (GHz). Silicon FETs cannot operate with wide bands.

7. The PIN-FET hybrid receivers are a combination of ______________
a) Hybrid resistances and capacitances
b) Pin photodiode and low noise amplifier (GaAs MESFETs)
c) P-N photodiode and low noise amplifier (GaAs MESFETs)
d) Attenuator and low noise amplifier (GaAs MESFETs)
Answer: b
Explanation: The PIN-FET or p-i-n/FET receiver utilizes a p-i-n photodiode along with a low noise preamplifier (GaAs MESFETs). It is fabricated using thick-film integrated circuit technology. This hybrid integration reduces the stray capacitance to negligible levels.

8. PIN-FET hybrid receiver is designed for use at a transmission rate of _____________
a) 130 Mbits^-1
b) 110 Mbits^-1
c) 120 Mbits^-1
d) 140 Mbits^-1
Answer: d
Explanation: At 140 Mbits^-1, the performance of PIN-FET hybrid receiver is found to be comparable to germanium and alloy APD receivers. A digital equalizer is necessary as the high-impedance front end effectively integrates the signal at 140 Mbits^-1.

9. It is difficult to achieve higher transmission rates using conventional __________
a) Voltage amplifier
b) Waveguide Structures
c) PIN-FET or APD receivers
d) MESFET
Answer: c
Explanation: It is difficult to achieve higher transmission rates due to limitations in their gain bandwidth products. Also, the trade-off between the multiplication factor requirement and the bandwidth limits the performance of conventional receivers.

10. Which receiver can be fabricated using PIN-FET hybrid approach?
a) Trans-impedance front end receiver
b) Gallium arsenide receiver
c) High-impedance front-end
d) Low-impedance front-end
Answer: a
Explanation: Trans-impedance front-end receivers are fabricated using the PIN-FET hybrid approach. An example of such receivers consists of a GaAs MESFET and two complementary bipolar microwave transistors.

11. A silicon p-i-n photodiode utilized with the amplifier and the receiver is designed to accept data at a rate of ___________
a) 276Mbits^-1
b) 274 Mbits^-1
c) 278Mbits^-1
d) 302Mbits^-1
Answer: b
Explanation: A silicon p-i-n photodiode is used with the low-noise preamplifier. This preamplifier is based on a GaAs MESFET. Thus, a receiver using p-i-n photodiode accepts a data rate of 274 Mbits^-1 giving a sensitivity around -35dB_m.

12. What is usually required by FETs to optimize the figure of merit?
a) Attenuation of barrier
b) Matching with the depletion region
c) Dispersion of the gate region
d) Matching with the detector
Answer: d
Explanation: Total capacitance is given by C_t = C_d + C_a. The figure of merit is optimized when C_d=C_a. This requires FETs to be matched with the detectors. This requires FETs to be matched with the detectors. This procedure is usually not welcomed by the device and is not permitted in current optical receiver design.

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