Direct Detection Receiver Performance Considerations ( Optical Communication ) MCQs – Optical Communication MCQs
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Optical Communication MCQs – Direct Detection Receiver Performance Considerations ( Optical Communication ) MCQs
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Noise
1. _____________ refers to any spurious or undesired disturbances that mask the received signal in a communication system.
a) Attenuation
b) Noise
c) Dispersion
d) Bandwidth
Answer: b
Explanation: Noise is an unwanted and undesirable quantity. It affects the received signal in a communication system. In optical fiber communication systems, noise is due to the spontaneous fluctuations rather than erratic disturbances.
2. How many types of noise are observed because of the spontaneous fluctuations in optical fiber communication systems?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: There are three types of noise because of the spontaneous fluctuations in optical fiber communication systems. These are thermal noise, the dark current noise and quantum noise. These noise types are not caused by the electronic interference.
3. ______________ is caused due to thermal interaction between the free electrons and the vibrating ions in the conduction medium.
a) Thermal noise
b) Dark noise
c) Quantum noise
d) Gaussian noise
Answer: a
Explanation: Thermal noise is basically a spontaneous fluctuation caused due to thermal interaction of electrons and ions. It is especially prevalent in resistors at room temperature. Thermal noise is measured in the form of current and is called as thermal noise current.
4. A small leakage current still flows from the device terminals even if there is no optical power incident on the photo detector.
a) True
b) False
Answer: a
Explanation: A reverse leakage current that flows from the device terminals is called as dark current. This dark current contributes to the total system noise. This gives random fluctuations about the average particle flow of the photocurrent.
5. ___________ distribution provides the description the random statistics of light emitted in black body radiation.
a) Poisson
b) Cumulative
c) Probability
d) Bose-Einstein
Answer: d
Explanation: Incoherent light is emitted by independent atoms and therefore there is no phase relationship between the emitted photons. The property dictates an exponential intensity distribution which is identical to Bose-Einstein distribution.
6. The probability of zero pairs being generated when a light pulse is present is given by which of the following equation?
a) P(0/1) = exp(-Zm)
b) P(x) = exp (Zm)
c) P(y) = x (0) + x(1)
d) P(z) = P(-Zm)
Answer: a
Explanation: The probability of zero pairs being generated when a light pulse is present is given by equation –
P (0/1) = exp(-Zm)
Where, P (0/1) represents the system error probability p(e) and Zm is variance of the probability distribution.
7. The minimum pulse energy needed to maintain a given bit-error-rate (BER) which any practical receiver must satisfy is known as ___________
a) Minimal energy
b) Quantum limit
c) Point of reversed
d) Binary signaling
Answer: b
Explanation: A perfect photo detector emits no electron-hole pairs in the absence of illumination. The error probability determines a standardized fundamental limit in digital optical communications. This limit is termed as quantum limit.
8. A digital optical fiber communication system requires a maximum bit-error-rate of 10-9. Find the average number of photons detected in a time period for a given BER.
a) 19.7
b) 21.2
c) 20.7
d) 26.2
Answer: c
Explanation: The probability of error is given by-
P(e) = exp(-Zm)
Where, Zm = No. of photons
Here P(e) = 10-9, therefore Zm is calculated from above relation.
9. For a given optical fiber communication system, P(e) = 10-9, Zm = 20.7, f = 2.9×1014, η = 1. Find the minimum pulse energy or quantum limit.
a) 3.9×10-18
b) 4.2×10-18
c) 6.2×10-14
d) 7.2×10-14
Answer: a
Explanation: The minimum pulse energy or quantum limit is given by –
Emin = Zmhf/η
Where, Zm = Number of photons
h = Planck’s constant
f = frequency
η = Quantum efficiency.
10. An analog optical fiber system operating at wavelength 1μmhas a post-detection bandwidth of 5MHz. Assuming an ideal detector and incident power of 198 nW, calculate the SNR (f = 2.99×1014Hz).
a) 46
b) 40
c) 50
d) 52
Answer: c
Explanation: The SNR is given by –
S/N = ηP0/2hfB
Where, η = 1 (for ideal detector)
P0 = incident power
h = Planck’s constant
B = Bandwidth.
11. The incident optical power required to achieve a desirable SNR is 168.2nW. What is the value of incident power in dBm?
a) -37.7 dBm
b) -37 dBm
c) – 34 dBm
d) -38.2 dBm
Answer: a
Explanation: Incident power in denoted by P0. It is given by –
P0 = 10log10(P0(watts))
Where P0(watts) = incident power in Watts/milliWatt.
12. In the equation given below, what does τstands for?
Zm = ηP0τ/hf
a) Velocity
b) Time
c) Reflection
d) Refractive index
Answer: b
Explanation: In the given equation, Zm is the variance of the probability distribution. The number of electrons generated in time τis equal to the average of the number of photons detected over this time period Zm. Hence, τ is the time and P0 is the incident power, ηis the quantum efficiency and f is the frequency.
Receiver Noise
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
Answer: c
Explanation: 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
Answer: b
Explanation: 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
Answer: a
Explanation: 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
Answer: d
Explanation: 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
Answer: c
Explanation: 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
Answer: a
Explanation: 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
Answer: d
Explanation: 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
Answer: a
Explanation: 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
Answer: b
Explanation: 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
Answer: c
Explanation: 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
Answer: b
Explanation: 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
Answer: d
Explanation: Silicon APDs possess a 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.
a) True
b) False
Answer: a
Explanation: 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.
Receiver Structures
1. How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
a) Four
b) One
c) Three
d) Two
Answer: a
Explanation: 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
Answer: c
Explanation: 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
Answer: d
Explanation: 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
Answer: b
Explanation: 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.
a) True
b) False
Answer: a
Explanation: 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
Answer: c
Explanation: 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
Answer: d
Explanation: 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.
a) True
b) False
Answer: a
Explanation: 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
Answer: a
Explanation: 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
Answer: c
Explanation: 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
Answer: b
Explanation: 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
Answer: c
Explanation: 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
Answer: a
Explanation: 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.
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 -35dBm.
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 Ct = Cd + Ca. The figure of merit is optimized when Cd=Ca. 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.
High Performance Receivers
1. How many design considerations are considered while determining the receiver performance?
a) Three
b) Two
c) One
d) Four
Answer: a
Explanation: Three main considerations are utilized for determining the receiver performance. Noise performance is a major design consideration providing a limitation to the sensitivity. Other two considerations are bandwidth and dynamic range.
2. FET preamplifiers provide higher sensitivity than the Si-bipolar device.
a) True
b) False
Answer: a
Explanation: At low speeds, the FET preamplifiers provide higher sensitivity than the Si-bipolar device. It is apparent that below 10Mbits-1the Si MOSFET preamplifier provides a lower noise performance than GaAs MESFET.
3. What is the abbreviation of HBT?
a) Homo-junction unipolar transistor
b) Homo-junction bipolar transistor
c) Hetero-junction bipolar transistor
d) Hetero-Bandwidth transcendence
Answer: c
Explanation: HBT is abbreviated as Hetero-junction bipolar transistor. It comprises a selectively doped hetero-junction FET. It is a high-speed, low-noise transistor device.
4. What type of receivers are used to provide wideband operation, low-noise operation?
a) APD optical receivers
b) Optoelectronic integrated circuits (OEICs)
c) MESFET receivers
d) Trans-impedance front-end receivers
Answer: b
Explanation: A strategy for the provision of wideband, low-noise receivers involves the use of p-i-n photodiode detector along with the monolithic integration of the device with semiconductor alloy FETs. It has an operating wavelength of 1.1 to 1.6 μmranges.
5. ___________ circuits extends the dynamic range of the receiver.
a) Monolithic
b) Trans-impedance
c) Automatic Error Control (AEC)
d) Automatic Gain Control (AGC)
Answer: d
Explanation: AGC circuit extends the dynamic range by diverting excess photocurrent away from the input of the receiver. The receiver dynamic range is an important performance parameter as it provides a measure of the difference between the sensitivity and its overload level.
6. The sensitivity of the low-impedance configuration is ____________
a) Good
b) Poor
c) Great
d) Same as that of high-impedance configuration
Answer: b
Explanation: A receiver saturation level is determined by the value of the photodiode bias resistor. The photodiode bias resistor valve is indirectly proportional to the sensitivity but is directly proportional in low impedance configuration. The low resistor value provides less sensitivity in the low-impedance configuration.
7. What is generally used to determine the receiver performance characteristics?
a) Noise
b) Resistor
c) Dynamic range & sensitivity characteristics
d) Impedance
Answer: c
Explanation: Dynamic range and sensitivity characteristics involve a graph of received power level and the value of feedback resistor. The high value of photodiode bias resistor in the high impedance front end causes high sensitivity and a narrow dynamic range. These factors prove useful for determining the performance characteristics of receiver.
8. The __________ technique eliminates the thermal noise associated with the feedback resistor in the trans-impedance front end design.
a) Compensation
b) Resonating impedance
c) Electromagnetic
d) Optical feedback
Answer: d
Explanation: The optical feedback strategy proves most useful at low transmission rate. The use of optically coupled feedback has demonstrated dynamic ranges of around 40 dB for p-i-n receivers operating at modest bit rates. It removes thermal noise associated with the feedback resistor.
9. The removal of the feedback resistor in the optical feedback technique allows reciever sensitivity of the order of _______________
a) -54 dBm at 2Mbit/sec
b) -12 dBm at 2Mbit/sec
c) -64 dBm at 2Mbit/sec
d) -72 dBm at 2Mbit/sec
Answer: c
Explanation: The removal of feedback resistor in the optical feedback technique allows low noise performance. Low noise performance, in turn, affects sensitivity. The receiver sensitivity gets high of the order of -64 dBm at 2Mbit/sec transmission rates.
10. The optical feedback technique is useful at low transmission rates.
a) True
b) False
Answer: a
Explanation: The optical feedback technique is useful at low transmission rates because in this case the feedback resistors employed are smaller than the optimum value for low-noise performance. This is done to maintain the resistor at a practical size of 1MΩ. Large values of feedback resistor limits the dynamic range.
11. How many types of optical amplifier technologies are available.
a) One
b) Three
c) Four
d) Two
Answer: d
Explanation: There are two basic optical amplifier technologies available. They are semiconductor optical amplifiers and fiber amplifiers. Both these devices are utilized in the pre-amplification role.
12. The optimum filter bandwidth is typically in the range ________________
a) 0.1 to 0.3 nm
b) 0.5 to 3 nm
c) 0.1 to 0.3 μm
d) 0.5 to 3 μm
Answer: b
Explanation: The optimum fiber bandwidth is determined by detector noise, transmission rate and the transmitter chirp characteristics. It is typically in the range of 0.5 to 3 nmas it depends upon the filter insertion loss.