McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 101. |
Doppler broadening is a homogeneous broadening mechanism. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: 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. | |
| 102. |
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. |
| A. | 46 cm-1 |
| B. | 51 cm-1 |
| C. | 50 cm-1 |
| D. | 49.07 cm-1 |
| Answer» D. 49.07 cm-1 | |
| Explanation: the laser gain coefficient is equivalent to the threshold gain per unit length and is given by – | |
| 103. |
Longitudinal modes contribute only a single spot of light to the laser output. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: 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. | |
| 104. |
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) |
| A. | 1.4×10-11 |
| B. | 1.6×10-12 |
| C. | 1×10-12 |
| D. | 6×10-11 |
| Answer» B. 1.6×10-12 | |
| Explanation: the mode separation in terms of free space wavelength is given by- | |
| 105. |
is essentially a crude form of Amplitude shift keying. |
| A. | analog modulation |
| B. | digital intensity modulation |
| C. | photodetector |
| D. | receiver structure |
| Answer» B. digital intensity modulation | |
| Explanation: 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. | |
| 106. |
Almost of the transmitter power is wasted in the use of external modulators. |
| A. | half |
| B. | quarter |
| C. | one-third |
| D. | twice |
| Answer» A. half | |
| Explanation: all external modulators suffer the drawback that around half of the transmitted power is wasted. to avoid this, non-synchronous detection can be employed. | |
| 107. |
The line width in the range of bit rate is specified for ASK heterodyne detection. |
| A. | 8% |
| B. | 2 to 8% |
| C. | 10 t0 50% |
| D. | 70% |
| Answer» C. 10 t0 50% | |
| Explanation: 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. | |
| 108. |
does not require an external modulator. |
| A. | fsk |
| B. | dsk |
| C. | psk |
| D. | ask |
| Answer» A. fsk | |
| Explanation: 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. | |
| 109. |
The frequency deviation at frequencies above 1 MHz is typically |
| A. | 10 to 20 ma-1 |
| B. | 100 to 500 ma-1 |
| C. | 1000 to 2000 ma-1 |
| D. | 30 to 40 ma-1 |
| Answer» B. 100 to 500 ma-1 | |
| Explanation: 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. | |
| 110. |
offers the potential for improving the coherent optical receiver sensitivity by increasing the choice of signalling frequencies. |
| A. | mfsk |
| B. | mdsk |
| C. | mpsk |
| D. | mask |
| Answer» A. mfsk | |
| Explanation: 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. | |
| 111. |
Eight level FSK and binary PSK yields an equivalent sensitivity. |
| A. | false |
| B. | true |
| Answer» B. true | |
| Explanation: 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. | |
| 112. |
External modulation for modulation format allows the most sensitive coherent detection mechanism. |
| A. | fsk |
| B. | dsk |
| C. | psk |
| D. | ask |
| Answer» C. psk | |
| Explanation: external modulation for psk is usually straightforward. it is therefore utilized to provide the modulation format which allows the most sensitive coherent detection mechanism. | |
| 113. |
can potentially provide spectral conservation through the use of multilevel signalling. |
| A. | m-ary psk |
| B. | mfsk |
| C. | ask |
| D. | dfsk |
| Answer» A. m-ary psk | |
| Explanation: 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. | |
| 114. |
The digital transmission on implementation of polarization modulation which involves polarization characteristics of the transmitted optical signal is known as |
| A. | frequency shift keying |
| B. | amplitude shift keying |
| C. | phase shift keying |
| D. | polarization shift keying |
| Answer» D. polarization shift keying | |
| Explanation: polarization shift keying is abbreviated as polsk. polsk requires additional receiver complexity than other modulation formats. | |
| 115. |
is fully depleted by employing electric fields. |
| A. | avalanche photodiode |
| B. | p-i-n diode |
| C. | varactor diode |
| D. | p-n diode |
| Answer» A. avalanche photodiode | |
| Explanation: apd is fully depleted by electric fields more than 104v/m. this causes all the drifting of carriers at saturated limited velocities. | |
| 116. |
At low gain, the transit time and RC effects |
| A. | are negligible |
| B. | are very less |
| C. | dominate |
| D. | reduce gradually |
| Answer» C. dominate | |
| Explanation: low gain causes the dominance of transit time and rc effects. this gives a definitive response time and thus device obtains constant bandwidth. | |
| 117. |
The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as |
| A. | is negligible |
| B. | very less |
| C. | increases gradually |
| D. | dominates |
| Answer» D. dominates | |
| Explanation: high gain causes avalanche buildup time to dominate. thus the | |
| 118. |
Often pulse shape is obtained from APD. |
| A. | negligible |
| B. | distorted |
| C. | asymmetric |
| D. | symmetric |
| Answer» C. asymmetric | |
| Explanation: 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. | |
| 119. |
Fall times of 1 ns or more are common. |
| A. | false |
| B. | true |
| Answer» B. true | |
| Explanation: 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. | |
| 120. |
Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μm wavelength. |
| A. | 0.459 |
| B. | 0.7 |
| C. | 0.312 |
| D. | 0.42 |
| Answer» A. 0.459 | |
| Explanation: the responsivity of a rapd is given by- | |
| 121. |
Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w. |
| A. | 6μm |
| B. | 7.21μm |
| C. | 0.112μm |
| D. | 3μm |
| Answer» C. 0.112μm | |
| Explanation: the wavelength can be found from the responsivity formula given by- | |
| 122. |
49 A/w. |
| A. | 0.91 μw |
| B. | 0.32 μw |
| C. | 0.312 μw |
| D. | 0.80 μw |
| Answer» D. 0.80 μw | |
| Explanation: the photocurrent is given by ip | |
| 123. |
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 |
| Answer» A. 0.735 | |
| Explanation: the photocurrent is given by ip | |
| 124. |
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 |
| Answer» B. 27.100 | |
| Explanation: the multiplication factor of photodiode is given by- | |
| 125. |
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 |
| Answer» D. 18.29 | |
| Explanation: the multiplication factor of photodiode is given by- | |
| 126. |
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 |
| Answer» A. 0.01 μa | |
| Explanation: the multiplication factor of photodiode is given by- | |
| 127. |
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. four | |
| 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. | |
| 128. |
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. equalization | |
| 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. | |
| 129. |
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. linear | |
| 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. | |
| 130. |
Noise contributions from the sources should be minimized to maximize the receiver sensitivity. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 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. | |
| 131. |
How many amplifier configurations are frequently used in optional fiber communication receivers? |
| A. | one |
| B. | two |
| C. | three |
| D. | four |
| Answer» C. three | |
| 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. | |
| 132. |
How many receiver structures are used to obtain better receiver characteristics? |
| A. | two |
| B. | one |
| C. | four |
| D. | three |
| Answer» D. three | |
| 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. | |
| 133. |
The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 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. | |
| 134. |
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. 13.3 khz | |
| Explanation: the maximum bandwidth obtained without equalization is given by – b = 1/2Πrtlct | |
| 135. |
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. greater dynamic range | |
| 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. | |
| 136. |
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. current mode amplifier | |
| 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. | |
| 137. |
is also known as frequency- shaping filter. |
| A. | resonator |
| B. | amplifiers |
| C. | attenuator |
| D. | equalizer |
| Answer» D. equalizer | |
| 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. | |
| 138. |
affects both the fiber attenuation and dispersion. |
| A. | refractive index |
| B. | micro-bending |
| C. | connectors |
| D. | splices |
| Answer» B. micro-bending | |
| Explanation: effects such as micro-bending with a resultant mode coupling affect both the fiber attenuation and dispersion. it does not provide the overall characteristics of the transmission link. | |
| 139. |
Which of the following is not included in the optical fiber link measurement test? |
| A. | attenuation measurement |
| B. | dispersion measurement |
| C. | splice loss measurement |
| D. | receiver sensitivity |
| Answer» D. receiver sensitivity | |
| Explanation: it is necessary to perform some tests on the optical fiber link to enhance productivity. apart from receiver sensitivity, other measurement methods are required to test the fiber link. | |
| 140. |
In case of field measurements, the equipment must have power consumption keeping in mind the battery operation. |
| A. | low |
| B. | high |
| C. | negligible |
| D. | maximum |
| Answer» A. low | |
| Explanation: the design criteria allows you to distinguish in parameters required for adaptation of battery operation and equipment handling. the power consumption must be low for an equipment to handle. | |
| 141. |
Which of the following are not considered as environmental conditions required for field measurements? |
| A. | temperature |
| B. | humidity |
| C. | mechanical load |
| D. | power |
| Answer» D. power | |
| Explanation: the equipment must be reliable and provide accurate measurements under extreme environmental conditions such as humidity, temperature and mechanical load. power is an internal factor. | |
| 142. |
Which of the following cannot be used in equipment for field measurements? |
| A. | fiber |
| B. | connector |
| C. | external triggering |
| D. | environmental factor |
| Answer» C. external triggering | |
| Explanation: the equipment cannot usually make use of external triggering and regulating circuits between the transmitter and receiver. this is because of their wide spacing on the majority of the optical links. | |
| 143. |
Which sensors are used for alteration of spectral range in equipment? |
| A. | wide-area photodiodes |
| B. | circulators |
| C. | gyrators |
| D. | photogenic sensors |
| Answer» A. wide-area photodiodes | |
| Explanation: wide area photodiodes such as silicon, germanium diodes are used for alteration of spectral range. it is generally preferred to have a measurement range from | |
| 144. |
The handheld optical power meter has a measurement accuracy of |
| A. | 0.01 db |
| B. | 0.25 db |
| C. | 0.8 db |
| D. | 1 db |
| Answer» B. 0.25 db | |
| Explanation: the optical power meter detects the fiber type and switches to optical power measurements. it provides an accuracy of about +(or -) 25 db. | |
| 145. |
may be used for measurement of the absolute optical attenuation on a fiber link. |
| A. | silicon photodiodes |
| B. | ingaasp photodiodes |
| C. | optical power meters |
| D. | gyrators |
| Answer» C. optical power meters | |
| Explanation: optical power meter employs cut-back technique. it is used for the measurement of the optical attenuation. | |
| 146. |
A large-area photodiode is utilized in the receiver to eliminate any effects from differing fiber and faces. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the modulating voltage maintains the equilibrium between the transmitter and the receiver side. a large area photodiode is required to eliminate differing fiber and faces to maintain the equilibrium. | |
| 147. |
Optical time domain reflectometry is also called a backscatter measurement method. |
| A. | false |
| B. | true |
| Answer» B. true | |
| Explanation: otdr technique is used in both on-field and laboratory applications. it is also called a backscatter measurement method as it provides the measurement of attenuation on an optical link down its entire length. | |
| 148. |
When considering source-to-fiber coupling efficiencies, the is an important parameter than total output power. |
| A. | numerical aperture |
| B. | radiance of an optical source |
| C. | coupling efficiency |
| D. | angular power distribution |
| Answer» B. radiance of an optical source | |
| Explanation: radiance is the optical power radiated into a unit solid angle per unit emitting surface area. since this optical power is dependent on radiance, radiance is much important factor than optical power. | |
| 149. |
It is a device that distributes light from a main fiber into one or more branch fibers. |
| A. | optical fiber coupler |
| B. | optical fiber splice |
| C. | optical fiber connector |
| D. | optical isolator |
| Answer» A. optical fiber coupler | |
| Explanation: nowadays, requirements to divide combined optical signals for applications are increasing. optical fiber coupler is one such device that is used for dividing and combining optical signals. it is generally used in lans, computer networks etc. | |
| 150. |
Optical fiber couplers are also called as |
| A. | isolators |
| B. | circulators |
| C. | directional couplers |
| D. | attenuators |
| Answer» C. directional couplers | |
| Explanation: optical fiber couplers are | |
Done Studing? Take A Test.
Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.