Optical Communications Multiple Choice Questions on “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.
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