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Communications MCQ Set 1

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

View Answer

Answer: c [Reason:] It was in the beginning of 20th 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: d [Reason:] 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

View Answer

Answer: b [Reason:] 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.

Communications MCQ Set 2

1. Which are the two main sources of noise in photodiodes without internal gain?
a) Gaussian noise and dark current noise
b) Internal noise and external noise
c) Dark current noise & Quantum noise
d) Gaussian noise and Quantum noise

View Answer

Answer: c [Reason:] The two main sources of noise in photodiodes without internal gain are dark current noise and quantum noise. They are regarded as shot noise on the photocurrent. These noise are together called as analog quantum noise.

2. The dominating effect of thermal noise over the shot noise in photodiodes without internal gain can be observed in wideband systems operating in the range of ________
a) 0.4 to 0.5 μm
b) 0.8 to 0.9 μm
c) 0.3 to 0.4 μm
d) 0.7 to 0.79 μm

View Answer

Answer: b [Reason:] When the photodiode is without internal avalanche gain, the detector load resistor and active elements’ thermal noise in the amplifier tends to dominate. It is seen in wideband systems operating in the 0.8 to 0.9 μmwavelength band. This is because the dark currents in the silicon diodes can be made very small.

3. A silicon p-i-n photodiode incorporated in an optical receiver has following parameters:
Quantum efficiency = 70%
Wavelength = 0.8 μm
Dark current = 3nA
Load resistance = 4 kΩ
Incident optical power = 150nW.
Bandwidth = 5 MHz
Compute the photocurrent in the device
a) 67.7nA
b) 81.2nA
c) 68.35nA
d) 46.1nA

View Answer

Answer: a [Reason:] The photocurrent is given by Ip= ηP0eλ/hc Where η= Quantum efficiency P0= Incident optical power e = electron charge λ = Wavelength h = Planck’s constant c= Velocity of light.

4. In a silicon p-i-n photodiode, if load resistance is 4 kΩ, temperature is 293 K, bandwidth is 4MHz, find the thermal noise in the load resistor
a) 1.8× 10-16A2
b) 1.23× 10-17A2
c) 1.65× 10-16A2
d) 1.61× 10-17A2

View Answer

Answer: d [Reason:] The thermal noise in the load resistor is given by – it2= 4KTB/RL Where T = Temperature B = Bandwidth RL= Load resistance.

5. ________________ is a combination of shunt capacitances and resistances
a) Attenuation
b) Shunt impedance
c) Shunt admittance
d) Thermal capacitance

View Answer

Answer: c [Reason:] Admittance is a measure of how easily a circuit will allow a current to flow. It is the inverse of impedance and is measured in Siemens. It is a combination of shunt capacitances and resistances.

6. ______________ is used in the specification of optical detectors
a) Noise equivalent power
b) Polarization
c) Sensitivity
d) Electron movement

View Answer

Answer: a [Reason:] Noise equivalent power is defined as the amount of incident optical power per unit bandwidth required to produce an output power equal to detector output noise power. Noise equivalent power is the value of incident power which gives an output SNR of unity.

7. A photodiode has a capacitance of 6 pF. Calculate the maximum load resistance which allows an 8MHz post detection bandwidth.
a) 3.9 kΩ
b) 3.46 kΩ
c) 3.12 kΩ
d) 3.32 kΩ

View Answer

Answer: d [Reason:] The load resistance is given by- RL= 1/2πCdB Where B = Post detection bandwidth Cd= Input capacitance RL= Load resistance.

8. The internal gain mechanism in an APD is directly related to SNR. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The internal gain mechanism in an APD increases the signal current into the amplifier. This improves the SNR because the load resistance and amplifier noise remains unaffected.

9. ____________ is dependent upon the detector material, the shape of the electric field profile within the device.
a) SNR
b) Excess avalanche noise factor
c) Noise gradient
d) Noise power

View Answer

Answer: b [Reason:] Excess avalanche noise factor is represented as F (M). Its value depends upon the detector material, shape of electric field profile and holes and electrons inclusion. It is a function of multiplication factor.

10. For silicon APDs, the value of excess noise factor is between _________
a) 0.001 and 0.002
b) 0.5 and 0.7
c) 0.02 and 0.10
d) 1 and 2

View Answer

Answer: c [Reason:] The excess noise factor (K) is same as that of the multiplication factor. In case of holes, the smaller values of K produce high performance and therefore the performance is achieved when k is small. For silicon APDs, k=0.02 to 0.10.

11. __________ determines a higher transmission rate related to the gain of the APD device.
a) Attenuation
b) Gain-bandwidth product
c) Dispersion mechanism
d) Ionization coefficient

View Answer

Answer: b [Reason:] Gain-bandwidth product is defined as Gain multiplied by the bandwidth. Gain is a dimensionless quantity but the gain-bandwidth product is therefore measured in the units of frequency.

12. _________________ APDs are recognized for their high gain-bandwidth products.
a) GaAs
b) Alloy-made
c) Germanium
d) Silicon

View Answer

Answer: d [Reason:] Silicon APDs possess large asymmetry of electron and hole ionization coefficient. Thus, they possess high gain-bandwidth products. These APDs do not operate at high transmission rates.

13. APDs do not operate at signal wavelengths between 1.3 and 1.6μm. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] APDs having high gain-bandwidth products do not operate at signal wavelengths between 1.3 and 1.6 μm.Hence, these APDs are not prefered for use in receivers operating at high transmission rates.

Communications MCQ Set 3

1. How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
a) Four
b) One
c) Three
d) Two

View Answer

Answer: a [Reason:] A full equivalent circuit for the digital optical fiber receiver includes four circuits. These are the detector circuit, noise sources, and amplifier and equalizer circuit.

2. __________ compensates for distortion of the signal due to the combined transmitter, medium and receiver characteristics.
a) Amplification
b) Distortion
c) Equalization
d) Dispersion

View Answer

Answer: c [Reason:] Equalization adjusts the balance between frequency components within an electronic signal. It compensates for distortion of the signal. The distortion may be due to the transmitter, receiver etc.

3. ____________ is also known as frequency-shaping filter.
a) Resonator
b) Amplifiers
c) Attenuator
d) Equalizer

View Answer

Answer: d [Reason:] Equalizer, often called as frequency-shaping filter has a frequency response inverse to that of the overall system frequency response. In wideband systems, it boosts the high frequency components to correct the overall amplitude of the frequency response.

4. The phase frequency response of the system should be ____________ in order to minimize inter-symbol interference.
a) Non-Linear
b) Linear
c) More
d) Less

View Answer

Answer: b [Reason:] An equalizer is used as frequency shaping filter. The phase frequency response of the system should be linear to acquire the desired spectral shape for digital systems. This, in turn, minimizes the inter-symbol interference.

5. Noise contributions from the sources should be minimized to maximize the receiver sensitivity. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Noise sources include transmitter section, medium and the receiver section. As the noise increases, the sensitivity at the receiver section decreases. Thus, noise contributions should be minimized to maximize the receiver sensitivity.

6. How many amplifier configurations are frequently used in optional fiber communication receivers?
a) One
b) Two
c) Three
d) Four

View Answer

Answer: c [Reason:] Three amplifier configurations are used in optical fiber communication receivers. These are voltage amplifiers, semiconductor optical amplifier and current amplifier. Voltage amplifier is the simplest and most common amplifier configuration.

7. How many receiver structures are used to obtain better receiver characteristics?
a) Two
b) One
c) Four
d) Three

View Answer

Answer: d [Reason:] The various receiver structures are low-impedance front end, high-impedance front end and trans-impedance front-end. The noise in the trans-impedance amplifier will always exceed than the front end structure.

8. The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The noise in the trans-impedance amplifier exceeds that incurred by the high-impedance amplifier. Hence, the trans-impedance front-end provides a greater bandwidth without equalization than the high-impedance front end.

9. A high-impedance amplifier has an effective input resistance of 4MΩ. Find the maximum bandwidth that may be obtained without equalization if the total capacitance is 6 pF and total effective load resistance is 2MΩ.
a) 13.3 kHz
b) 14.2 kHz
c) 15.8 kHz
d) 13.9 kHz

View Answer

Answer: a [Reason:] The maximum bandwidth obtained without equalization is given by – B = 1/2ΠRTLCT Where, RTL= Total load resistance CT= Total capacitance.

10. A high-input-impedance amplifier has following parameters:
Total effective load resistance = 2MΩ.
Temperature = 300 K
Find the mean square thermal noise current per unit bandwidth for the high-impedance configuration.
a) 8.9×10-27A2/Hz
b) 8.12×10-27A2/Hz
c) 8.29×10-27A2/Hz
d) 8.4×10-27A2/Hz

View Answer

Answer: c [Reason:] the mean square thermal noise current per unit bandwidth for the high-impedance configuration is given by – iT2= 4KT/RTL Where, K = constant T = Temperature (Kelvin) RTL= total effective load resistance.

11. The mean square thermal noise current in the trans-impedance configuration is _________ greater than that obtained with the high-input-impedance configuration.
a) 30
b) 20
c) 15
d) 10

View Answer

Answer: b [Reason:] 13 dB noise penalties are incurred with the trans-impedance amplifier over that of the high-input-impedance configuration. It is the logarithmic function of the noise current value. However, the trans-impedance amplifiers can be optimized for noise performance.

12. The major advantage of the trans-impedance configuration over the high-impedance front end is ______________
a) Greater bandwidth
b) Less bandwidth
c) Greater dynamic range
d) Less dynamic range

View Answer

Answer: c [Reason:] Greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. This attenuation is obtained in the trans-impedance amplifier through the negative feedback and therefore the low frequency components are amplified by the closed loop. This increases the dynamic range.

13. The trans-impedance front end configuration operates as a __________ with negative feedback.
a) Current mode amplifier
b) Voltage amplifier
c) Attenuator
d) Resonator

View Answer

Answer: a [Reason:] The trans-impedance configuration overcomes the drawbacks of the high-impedance front end. It utilizes a low-noise, high-input-impedance amplifier with negative feedback. It operates as a current mode amplifier where high impedance is reduced by negative feedback.

Communications MCQ Set 4

1. Stimulated emission by recombination of injected carriers is encouraged in
a) Semiconductor injection laser
b) Gas laser
c) Chemist laser
d) Dye laser

View Answer

Answer: a [Reason:] Stimulated emission by use of optical cavity in crystal structure is used in semiconductor injection diodes. This provides the feedback of protons which gives injection laser many important advantages over other sources like LED’s.

2. In semiconductor injection laser, narrow line bandwidth is of the order
a) 1 nm or less
b) 4 nm
c) 5 nm
d) 3 nm

View Answer

Answer: a [Reason:] A narrow line bandwidth of order 1 nm or less is used. This narrow bandwidth is useful in minimizing the effects of material dispersion.

3. Injection laser have a high threshold current density of
a) 104Acm-2and more
b) 102Acm-2
c) 10-2Acm-2
d) 10-3Acm-2

View Answer

Answer: a [Reason:] Injection laser have a high threshold current density of 104Acm-2due to lack of matter and in-efficient light sources. These high current densities are largely utilized in pulse mode in order to minimize junction and thus avert damage.

4. ηT Is known as slope quantum efficiency. State true or false
a) True
b) False

View Answer

Answer: b [Reason:] ηD is known as slope quantum efficiency. It gives a measure rate of optical output power change with current and thus it determines slope of output characteristics in the region. So, ηT is referred to as slope quantum efficiency.

5. The total efficiency of an injection laser with GaAs active region is 12%. The applied voltage is 3.6 V and band gap energy for GaAs is 2.34 eV. Determine external power efficiency.
a) 7.8 %
b) 10 %
c) 12 %
d) 6 %

View Answer

Answer: a [Reason:] The total external power efficiency is defined as η= ηT(Eq/V)*100 = 0.12 (2.34/3.6) *100 = 7.8 % .

6. In a DH laser, the sides of cavity are formed by _______________
a) Cutting the edges of device
b) Roughening the edges of device
c) Softening the edges of device
d) Covering the sides with ceramics

View Answer

Answer: b [Reason:] In a DH laser, the sides of cavity are formed by roughening edges of the device. This is done so as to reduce the unwanted emission in these directions and limit the number of horizontal transversal modes.

7. A particular laser structure is designed so that the active region extends the edges of devices. State whether the following statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Laser structures are particularly designed so that the active region does not extend beyond the edges. This is done to reduce problems like difficult heat sinking, lasing from multiple filament in wide active areas, unsuitable light output geometry for efficient coupling and also to reduce required threshold current.

8. Gain guided laser structure are
a) Chemical laser
b) Gas laser
c) DH injection laser
d) Quantum well laser

View Answer

Answer: c [Reason:] DH injection lasers are known as gain guided laser structure. This is because the optical mode distribution along the junction plane is decided by optical gain.

9. Laser modes are generally separated by few
a) Tenths of micrometer
b) Tenths of nanometer
c) Tenths of Pico-meter
d) Tenths of millimeter

View Answer

Answer: b [Reason:] The spacing in between modes is about a few tenths of nanometer. The spacing of the modes depends on optical cavity length where each one corresponds to an integral number of lengths.

10. The spectral width of emission from the single mode device is
a) Smaller than broadened transition line-width
b) Larger than broadened transition line-width
c) Equal the broadened transition line-width
d) Cannot be determined

View Answer

Answer: a [Reason:] Single mode device has a smaller spectral width as compared to that of broadening transition line-width. This is because for a single-mode operation, the laser optical output must have only a single longitudinal and single transverse mode.

11. Single longitudinal mode operation is obtained by
a) Eliminating all transverse mode
b) Eliminating all longitudinal modes
c) Increasing the length of cavity
d) Reducing the length of cavity

View Answer

Answer: d [Reason:] Single longitudinal mode operation is obtained by reducing the length L of cavity. Length must be reduced until the frequency separation of adjacent modes is given in the equation δf= c/2nL is larger than gain curve. Then only single mode falling in transition line width will oscillate in laser cavity.

12 A correct DH structure will restrict the vertical width of waveguide region
a) 0.5μm.
b) 0.69 μm
c) 0.65 μm
d) Less than 0.4 μm

View Answer

Answer: d [Reason:] The vertical width DH structure should be less than 0.4μm. This allows only fundamental transverse mode and removes any interference of higher order transverse modes on emitted longitudinal waves.

13. The external power efficiency of an injection laser with a GaAs is 13% having band gap energy of 1.64 eV. Determine external power efficiency
a) 0.198
b) 0.283
c) 0.366
d) 0.467

View Answer

Answer: a [Reason:] The external power efficiency of an injection laser is given by ηep = ηT (Eq/V)*100 ηT= ηep /100 (v/Eg) = 13/100 (2.5/1.64) = 0.198.

Communications MCQ Set 5

1. ___________ has more sophisticated structure than p-i-n photodiode.
a) Avalanche photodiode
b) p-n junction diode
c) Zener diode
d) Varactor diode

View Answer

Answer: a [Reason:] Avalanche photodiode is second major type of detector in optical communications. This diode is more sophisticated so as to create a much higher electric field region.

2. The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as _______
a) Auger recombination
b) Mode hopping
c) Impact ionization
d) Extract ionization

View Answer

Answer: c [Reason:] In depletion region, almost all photons are absorbed and carrier pairs are generated. So there comes a high field region where carriers acquire energy to excite new carrier pairs. This is impact ionization.

3. _______ is fully depleted by employing electric fields.
a) Avalanche photodiode
b) P-I-N diode
c) Varactor diode
d) P-n diode

View Answer

Answer: a [Reason:] APD is fully depleted by electric fields more than 104V/m. This causes all the drifting of carriers at saturated limited velocities.

4. At low gain, the transit time and RC effects ________
a) Are negligible
b) Are very less
c) Dominate
d) Reduce gradually

View Answer

Answer: c [Reason:] Low gain causes the dominance of transit time and RC effects. This gives a definitive response time and thus device obtains constant bandwidth.

5. At high gain, avalanche buildup time ________
a) Is negligible
b) Very less
c) Increases gradually
d) Dominates

View Answer

Answer: d [Reason:] High gain causes avalanche buildup time to dominate. Thus the bandwidth of device decreases as increase in gain.

6. Often __________ pulse shape is obtained from APD.
a) Negligible
b) Distorted
c) Asymmetric
d) Symmetric

View Answer

Answer: c [Reason:] 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.

7. Fall times of 1 ns or more are common. State whether the given statement is true or false.
a) False
b) True

View Answer

Answer: b [Reason:] 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.

8. Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μmwavelength.
a) 0.459
b) 0.7
c) 0.312
d) 0.42

View Answer

Answer: a [Reason:] The Responsivity of a RAPD is given by- R = ηeλ/hc A/w whereη=efficiency, λ=wavelength, h =Planck’s constant.

9. Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w.
a) 6μm
b) 7.21μm
c) 0.112μm
d) 3μm

View Answer

Answer: c [Reason:] The wavelength can be found from the Responsivity formula given by- R = ηeλ/hc. The unit of wavelength isμm.

10. Compute photocurrent of RAPD having optical power of 0.7 μw and responsivity of 0.689 A/W.
a) 0.23 μA
b) 0.489 μA
c) 0.123 μA
d) 9 μA

View Answer

Answer: b [Reason:] The photocurrent is given byIP=P0R. Here IP=photocurrent, P0=Power, R= responsivity.

11. Determine optical power of RAPD with photocurrent of 0.396 μAand responsivity of 0.49 A/w.
a) 0.91 μW
b) 0.32 μW
c) 0.312 μW
d) 0.80 μW

View Answer

Answer: d [Reason:] The photocurrent is given byIP=P0R. Here IP=photocurrent, P0=Power, R= responsivity. P0=IP/R gives the optical power.

12. Determine the Responsivity of optical power of 0.4μW and photocurrent of 0.294 μA.
a) 0.735
b) 0.54
c) 0.56
d) 0.21

View Answer

Answer: a [Reason:] The photocurrent is given by IP=P0R. Here IP=photocurrent, P0=Power, R= responsivity. R=IP/P0gives the responsivity.

13. Compute multiplication factor of RAPD with output current of 10 μAand photocurrent of 0.369μA.
a) 25.32
b) 27.100
c) 43
d) 22.2

View Answer

Answer: b [Reason:] The multiplication factor of photodiode is given by- M=I/IP where I= output current, IP=photocurrent.

14. Determine the output current of RAPD having multiplication factor of 39 and photocurrent of 0.469μA.
a) 17.21
b) 10.32
c) 12.21
d) 18.29

View Answer

Answer: d [Reason:] The multiplication factor of photodiode is given by- M=I/IP where I= output current, IP=photocurrent. I=M*IP gives the output current inμA.

15. Compute the photocurrent of RAPD having multiplication factor of 36.7 and output current of 7μA.
a) 0.01 μA
b) 0.07 μA
c) 0.54 μA
d) 0.9 μA

View Answer

Answer: a [Reason:] The multiplication factor of photodiode is given by- M=I/IP where I= output current, IP=photocurrent. IP=I/M Gives the output current inμA.