McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 1. |
The major drawback of frequency multipliers is that they have: |
| A. | higher attenuation |
| B. | complex construction methods |
| C. | complex design |
| D. | none of the mentioned |
| Answer» C. complex design | |
| Explanation: designing a good quality frequency multiplier is more difficult since it non-linear analysis, matching at multiple frequencies, stability analysis and thermal considerations. considering all these issues for designing a multiplier makes it very complex. | |
| 2. |
Oscillators operating at millimeter wavelength are difficult to realize and are also less efficient. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: as frequency increases to the millimeter wave range, it becomes increasingly difficult to build fundamental frequency oscillators with good power, stability and noise characteristics. an alternative approach is to produce a harmonic of a low frequency oscillator through the use of frequency multiplier. | |
| 3. |
A major disadvantage of frequency multipliers is that they multiply the noise factor along with frequency. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: a disadvantage of frequency multipliers is that noise levels are also increased by the multiplication factor. | |
| 4. |
If a frequency multiplier has a multiplication factor of 10, then the increase in noise level due to frequency multiplication is: |
| A. | 10 db |
| B. | 20 db |
| C. | 25 db |
| D. | 15 db |
| Answer» B. 20 db | |
| Explanation: for a frequency multiplier, the increase in noise power is given by 20 log n, where n is the multiplication factor of the multiplier. substituting in the below equation, increase in noise level is 20 db. | |
| 5. |
In a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with at all frequencies other than required harmonic. |
| A. | real impedances |
| B. | reactive impedance |
| C. | complex impedance |
| D. | none of the mentioned |
| Answer» B. reactive impedance | |
| Explanation: in a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with reactive impedance at all frequencies other than required harmonic nfo. if the diode junction capacitance has a square –law i-v characteristic , it is necessary to terminate unwanted harmonics with short circuit. | |
| 6. |
Resistive multipliers are more efficient as compared to reactive multipliers. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: resistive multipliers generally use forward biased schottky-barrier diodes to provide non linear characteristic. resistive multipliers have low efficiency but have better bandwidth. | |
| 7. |
Reactive multipliers have a disadvantage that they cannot be used at very high frequencies and they become less efficient. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: at millimeter frequencies, varactor diode exhibits resistive property. hence, at high frequency the multiplier becomes lossy and also does not offer high bandwidth, which is a major disadvantage. | |
| 8. |
For a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is: |
| A. | 50 % |
| B. | 25 % |
| C. | 75 % |
| D. | 12.5 % |
| Answer» B. 25 % | |
| Explanation: for a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is given by 1/m2 where m is the multiplication factor. for a factor 2 multiplier, maximum theoretical conversion efficiency is 25 %. | |
| 9. |
is a device that converts electrons to photons or vice-versa. |
| A. | antenna |
| B. | electron gun |
| C. | photon amplifier |
| D. | microwave tube |
| Answer» A. antenna | |
| Explanation: antenna is a device that converts electrons into photons or vice versa. a transmitting antenna converts electrons into photons while a receiving antenna converts photons into electrons. | |
| 10. |
The basic equation of radiation that is applied to any antenna irrespective of the type of the antenna is: |
| A. | il= qv |
| B. | iq = lv |
| C. | i/l=q/v |
| D. | none of the mentioned |
| Answer» A. il= qv | |
| Explanation: basic equation of radiation is given by il=qv. i is the time change in current, l is the length of the current element, q is the charge v is the acceleration of the charge. | |
| 11. |
The number of patterns radiation pattern required to specify the characteristic are : |
| A. | three |
| B. | four |
| C. | two |
| D. | five |
| Answer» A. three | |
| Explanation: the three patterns required are, θ component of the electric field as the function of the angles as θ and φ, the φ component of the electric field as the function of the angles θ and φ, the phase of these fields as a functions of the angle φ and θ . | |
| 12. |
The beam width of the antenna pattern measured at half power points is called: |
| A. | half power beam width |
| B. | full null beam width |
| C. | beam width |
| D. | none of the mentioned |
| Answer» A. half power beam width | |
| Explanation: the beam width of an antenna measure at half of the maximum power received by an antenna or the 3 db beam width of the antenna is termed as half null beam width. | |
| 13. |
An antenna has a field pattern E (θ) =cos θ. cos 2θ. The first null beam width of the antenna is: |
| A. | 450 |
| B. | 900 |
| C. | 1800 |
| D. | 1200 |
| Answer» B. 900 | |
| Explanation: half power beam width of the antenna is obtained by equating the field pattern of the antenna to 0.707 (half power point) and finding θ. 2θ gives the value of beam width. twice the half power beam width gives the first null beam width. with the same steps applied, the half power beam width of the antenna is 450. first null beam width is 900. | |
| 14. |
The solid area through which all the power radiated by the antenna is: |
| A. | beam area |
| B. | effective area |
| C. | aperture area |
| D. | beam efficiency |
| Answer» A. beam area | |
| Explanation: the beam area is the solid angle through which all of the power radiated by the antenna would stream if p (θ, φ) maintained its maximum value over beam area and zero elsewhere. this value is approximately equal to the angles subtended by the half power points of the main lobe in the two principal planes. | |
| 15. |
A is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space. |
| A. | transmitting antenna |
| B. | receiving antenna |
| C. | radar |
| D. | mixer |
| Answer» A. transmitting antenna | |
| Explanation: a transmitting antenna is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space. it appears as an electrical circuit on one side, provides an interface with a propagating plane wave. | |
| 16. |
Dipole antennas are an example for: |
| A. | wire antennas |
| B. | aperture antennas |
| C. | array antennas |
| D. | none of the mentioned |
| Answer» A. wire antennas | |
| Explanation: dipoles, monopoles, oops, yagi-uda arrays are all examples for wire antennas. these antennas have low gains, and are mostly used at lower frequencies. | |
| 17. |
antennas consist of a regular arrangement of antenna elements with a feed network |
| A. | aperture antennas |
| B. | array antennas |
| C. | printed antennas |
| D. | wire antennas |
| Answer» B. array antennas | |
| Explanation: array antennas consist of a regular arrangement of antenna elements with a feed network. pattern characteristics such as beam pointing angle and side lobe levels can be controlled by adjusting the amplitude and phase excitation of array elements. | |
| 18. |
A parabolic reflector used for reception with the direct broadcast system is 18 inches in diameter and operates at 12.4 GHz. The far-field distance for this antenna is: |
| A. | 18 m |
| B. | 13 m |
| C. | 16.4 m |
| D. | 17.3 m |
| Answer» D. 17.3 m | |
| Explanation: far field distance for a reflector antenna is given by 2d2/λ. d is the diameter and λ is the operating signal wavelength. | |
| 19. |
of an antenna is a plot of the magnitude of the far field strength versus position around the antenna. |
| A. | radiation pattern |
| B. | directivity |
| C. | beam width |
| D. | none of the mentioned |
| Answer» A. radiation pattern | |
| Explanation: radiation pattern of an antenna is a plot of the magnitude of the far field strength versus position around the antenna. this plot gives the detail regarding the region where most of the energy of antenna is radiated, side lobes and beam width of an antenna. | |
| 20. |
Antennas having a constant pattern in the azimuthal plane are called |
| A. | high gain antenna |
| B. | omni directional antenna |
| C. | unidirectional antenna |
| D. | low gain antenna |
| Answer» B. omni directional antenna | |
| Explanation: omni directional antennas radiate em waves in all direction. if the radiation pattern for this type of antenna is plotted, the pattern is a constant signifying that the radiated power is constant measured at any point around the antenna. | |
| 21. |
Beamwidth and directivity are both measures of the focusing ability of an antenna. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: beamwidth and directivity are both measures of the focusing ability of an antenna. an antenna with a narrow main beam will have high directivity, while a pattern with low beam will have low directivity. | |
| 22. |
If the beam width of an antenna in two orthogonal planes are 300 and 600. Then the directivity of the antenna is: |
| A. | 24 |
| B. | 18 |
| C. | 36 |
| D. | 12 |
| Answer» B. 18 | |
| Explanation: given the beam width of the antenna in 2 planes, the directivity is given by 32400/θ*∅, where θ,∅ are the beam widths in the two orthogonal planes. substituting in the equation, directivity of the antenna is 18. | |
| 23. |
If the power input to an antenna is 100 mW and if the radiated power is measured to be 90 mW, then the efficiency of the antenna is: |
| A. | 75 % |
| B. | 80 % |
| C. | 90 % |
| D. | insufficient data |
| Answer» C. 90 % | |
| Explanation: antenna efficiency is defined as the ratio of radiated power to the input power to the antenna. substituting the given data in the efficiency equation, the efficiency of the antenna is 90%. | |
| 24. |
If an antenna has a directivity of 16 and radiation efficiency of 0.9, then the gain of the antenna is: |
| A. | 16.2 |
| B. | 14.8 |
| C. | 12.5 |
| D. | 19.3 |
| Answer» A. 16.2 | |
| Explanation: gain of an antenna is given by the product of radiation efficiency of the antenna and the directivity of the antenna. | |
| 25. |
Gain of an antenna is always greater than the directivity of the antenna. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: gain of an antenna is always smaller than the directivity of an antenna. gain is given by the product of directivity and radiation efficiency. radiation efficiency can never be greater than one. so gain is always less than or equal to directivity. | |
| 26. |
A rectangular horn antenna has an aperture area of 3λ × 2λ. Then the maximum directivity that can be achieved by this rectangular horn antenna is: |
| A. | 24 db |
| B. | 4 db |
| C. | 19 db |
| D. | insufficient data |
| Answer» C. 19 db | |
| Explanation: given the aperture dimensions of an antenna, the maximum directivity that can be achieved is 4π a/λ2, where a is the aperture area and λ is the operating wavelength. substituting the given values in the above equation, the maximum directivity achieved is 19 db. | |
| 27. |
A rectangular horn antenna has an aperture area of 3λ × 2λ. If the aperture efficiency of an antenna is 90%, then the directivity of the antenna is: |
| A. | 19 db |
| B. | 17.1 db |
| C. | 13 db |
| D. | 21.1 db |
| Answer» B. 17.1 db | |
| Explanation: given the aperture dimensions of an antenna, the directivity that can be achieved is ap4π a/λ2, where a is the aperture area and λ is the operating | |
| 28. |
A resistor is operated at a temperature of 300 K, with a system bandwidth of 1 MHz then the noise power produced by the resistor is: |
| A. | 3.13×10-23 watts |
| B. | 4.14×10-15 watts |
| C. | 6.14×10-15 watts |
| D. | none of the mentioned |
| Answer» B. 4.14×10-15 watts | |
| Explanation: for a resistor noise power produced is given by ktb, where t is the system temperature and b is the bandwidth. substituting in the above expression, the noise power produced is 4.14×10-15 watts. | |
| 29. |
With an increase in operating frequency, the background noise temperature: |
| A. | increases |
| B. | decreases |
| C. | remains constant |
| D. | remains unaffected |
| Answer» A. increases | |
| Explanation: the plot of frequency v/s background noise temperature shows that with the increase of the signal frequency, the | |
| 30. |
The noise temperature of an antenna is given by the expression: |
| A. | radtb + (1-rad) tp |
| B. | (1-rad) tp |
| C. | radtb |
| D. | none of the mentioned |
| Answer» A. radtb + (1-rad) tp | |
| Explanation: the noise temperature of an antenna is given by the expression radtb + (1-rad) tp. here, tb is the brightness temperature and tp is the physical | |
| 31. |
Low is the G/T ratio of an antenna, higher is its efficiency. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in the g/t ratio of an antenna, g is the gain of an antenna and t is the antenna noise temperature. higher the g/t ratio of an antenna better is the performance of the antenna. | |
| 32. |
has a constant power spectral density. |
| A. | white noise |
| B. | gaussian noise |
| C. | thermal noise |
| D. | shot noise |
| Answer» A. white noise | |
| Explanation: thermal noise has a power spectral density for a wide range of frequencies. its plot of frequency v/s noise power is a straight line parallel to y axis. | |
| 33. |
Slotted line is a transmission line configuration that allows the sampling of: |
| A. | electric field amplitude of a standing wave on a terminated line |
| B. | magnetic field amplitude of a standing wave on a terminated line |
| C. | voltage used for excitation |
| D. | current that is generated by the source |
| Answer» A. electric field amplitude of a standing wave on a terminated line | |
| Explanation: slotted line allows the sampling of the electric field amplitude of a standing wave on a terminated line. with this device, swr and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found. | |
| 34. |
If the standing wave ratio for a transmission line is 1.4, then the reflection coefficient for the line is: |
| A. | 0.16667 |
| B. | 1.6667 |
| C. | 0.01667 |
| D. | 0.96 |
| Answer» A. 0.16667 | |
| Explanation: ┌= (swr-1)/ (swr+1). substituting for swr in the above equation for reflection co-efficient, given swr is 1.4, reflection co-efficient is 0.16667. | |
| 35. |
If the reflection coefficient of a transmission line is 0.4, then the standing wave ratio is: |
| A. | 1.3333 |
| B. | 2.3333 |
| C. | 0.4 |
| D. | 0.6 |
| Answer» B. 2.3333 | |
| Explanation: swr= (1+┌)/ (1-┌). where ┌ is the reflection co-efficient. substituting for the reflection co-efficient in the equation, swr is 2.3333. | |
| 36. |
In the expression for phase of the reflection coefficient, Lmin stands for : |
| A. | distance between load and first voltage minimum |
| B. | distance between load and first voltage maximum |
| C. | distance between consecutive minimas |
| D. | distance between a minima and immediate maxima |
| Answer» A. distance between load and first voltage minimum | |
| Explanation: lmin is defined as the distance between the terminating load of a transmission line and the first voltage minimum that occurs in the transmission line due to reflection of waves from the load end due to mismatched termination. | |
| 37. |
If the normalized load impedance of a transmission line is 2, then the reflection co- efficient is: |
| A. | 0.33334 |
| B. | 1.33334 |
| C. | 0 |
| D. | 1 |
| Answer» A. 0.33334 | |
| Explanation: zl=z0 (1+┌)/ (1-┌), this is the expression for load impedance. normalized load impedance is the ratio of load impedance to the characteristic impedance, taking zll/z0 as 2, the reflection co-efficient is equal to 0.33334. | |
| 38. |
Link budget consists of calculation of |
| A. | useful signal power |
| B. | interfering noise power |
| C. | useful signal & interfering noise power |
| D. | none of the mentioned |
| Answer» C. useful signal & interfering noise power | |
| Explanation: the link analysis and its output, the link budget consists of calculations and tabulations of useful signal power and interfering noise power at the receiver. | |
| 39. |
Which is the primary cost for degradation of error performance? |
| A. | loss in signal to noise ratio |
| B. | signal distortion |
| C. | signal distortion & loss in signal to noise ratio |
| D. | none of the mentioned |
| Answer» C. signal distortion & loss in signal to noise ratio | |
| Explanation: there are two primary causes for the degradation of error performance. | |
| 40. |
Which factor adds phase noise to the signal? |
| A. | jitter |
| B. | phase fluctuations |
| C. | jitter & phase fluctuations |
| D. | none of the mentioned |
| Answer» C. jitter & phase fluctuations | |
| Explanation: when a local oscillator is used in signal mixing, phase fluctuations and jitter adds phase noise to the signal. | |
| 41. |
Antennas are used |
| A. | as transducer |
| B. | to focus |
| C. | as transducer & to focus |
| D. | none of the mentioned |
| Answer» C. as transducer & to focus | |
| Explanation: antennas are used as transducer that converts electronic signals to electromagnetic fields and vice versa. they are also used to focus the electromagnetic energy in the desired direction. | |
| 42. |
Mechanism contributing to a reduction in efficiency is called as |
| A. | amplitude tapering |
| B. | blockage |
| C. | edge diffraction |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: mechanism contributing to a reduction in efficiency is called as amplitude tapering, spillover, edge diffraction, blockage, scattering, re-radiation and dissipative loss. | |
| 43. |
Space loss occurs due to a decrease in |
| A. | electric field strength |
| B. | efficiency |
| C. | phase |
| D. | signal power |
| Answer» A. electric field strength | |
| Explanation: due to the decrease in electric field strength there will be a decrease in signal strength as a function of distance. this is called as space loss. | |
| 44. |
Antenna’s efficiency is given by the ratio of |
| A. | effective aperture to physical aperture |
| B. | physical aperture to effective aperture |
| C. | signal power to noise power |
| D. | losses |
| Answer» A. effective aperture to physical aperture | |
| Explanation: the larger the antenna aperture the larger is the resulting signal power density in the desired direction. the ratio of effective aperture to physical aperture is the antenna’s efficiency. | |
| 45. |
Effective radiated power of an isotropic radiator can be given as a product of |
| A. | radiated power and received power |
| B. | effective area and physical area |
| C. | transmitted power and transmitting gain |
| D. | receiving power and receiving gain |
| Answer» C. transmitted power and transmitting gain | |
| Explanation: an effective radiated power | |
| 46. |
A dipole antenna is also called as? |
| A. | marconi antenna |
| B. | yagi antenna |
| C. | bidirectional antenna |
| D. | hertz antenna |
| Answer» D. hertz antenna | |
| Explanation: one of the most widely used antenna types is the half-wave dipole antenna. this antenna is also formally known as the hertz antenna after heinrich hertz, who first demonstrated the existence of electromagnetic waves. | |
| 47. |
The impedance at the center of the antenna is known as? |
| A. | characteristic impedance |
| B. | radiation resistance |
| C. | transmission impedance |
| D. | recovery resistance |
| Answer» B. radiation resistance | |
| Explanation: the transmission line is connected at the center. the dipole has an impedance of 73 v at its center, which is the radiation resistance. at the resonant frequency, the antenna appears to be a pure resistance of 73 v. | |
| 48. |
What happens when the radiation resistance of the antenna matches the characteristic impedance of the transmission line? |
| A. | no transmission occurs |
| B. | no reception occurs |
| C. | swr is maximum |
| D. | swr is minimum |
| Answer» D. swr is minimum | |
| Explanation: when the radiation resistance of the antenna matches the characteristic impedance of the transmission line, the swr is minimum and maximum power reaches the antenna. this allows maximum power to be transmitted. | |
| 49. |
The type of dipole antenna that has a higher band width is called as? |
| A. | conical antenna |
| B. | yagi antenna |
| C. | helical antenna |
| D. | marconi antenna |
| Answer» A. conical antenna | |
| Explanation: a common way to increase bandwidth in the antenna is to use a version of the dipole antenna known as the conical antenna. the overall length of the antenna is 0.73λ or 0.73(984)/f = 718.32/f. this is longer than the traditional one-half wavelength of a dipole antenna, but the physical shape changes the necessary dimensions for resonance. | |
| 50. |
The radiation pattern of a half-wave dipole has the shape of a |
| A. | doughnut |
| B. | sphere |
| C. | hemisphere |
| D. | circular |
| Answer» A. doughnut | |
| Explanation: the radiation pattern of any antenna is the shape of the electromagnetic energy radiated from or received by that antenna. typically that radiation is concentrated in a pattern that has a recognizable geometric shape. the radiation pattern of a half-wave dipole has the shape of a doughnut. | |
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