250+ TOP MCQs on Control and Data Entry and Answers

Avionics Multiple Choice Questions on “Control and Data Entry”.

1. What type of beams are used for the tactical control panel?
a) Matrix array of infrared beams
b) Light beams
c) Lasers
d) Microwave beams
Answer: a
Clarification: A typical tactical control panel uses a matrix of infrared beams across the surface of the display containing various functional keys. Touching a specific function key on the display surface interrupts the x and y infrared beams which intersect over the displayed key function and hence signal the operation of that particular key function.

2. What is DVI control?
a) Direct voice input
b) Digitally vectored input
c) Digital voice input
d) Direct vectored input
Answer: a
Clarification: DVI control is direct voice input control through which the pilot can enter flight and navigational commands through voice.

3. In a DVI control, which system recognizes the pilot’s control?
a) Flight data computer
b) Air Data Computer
c) Speech recognition system
d) Central command center
Answer: c
Clarification: The commands given by the pilot is recognized by a speech recognition system which compares the spoken audio with preloaded vocabulary templates. If the command is recognized indication is given either aurally or visually to confirm the command.

4. The main reason for DVI control system in aircraft is _____________
a) less weight
b) latest technology
c) reduce pilot work load
d) space constraints
Answer: c
Clarification: DVI control system is used to control the aircraft using voice commands, thus the pilot need not have to operate push buttons, switches. This reduces the workload on the pilot during heavy workload operations such as combat.

5. What are isolated word recognizers?
a) Systems that can recognize anyone’s voice
b) System that is not fully voice commanded
c) Systems that require a pause between each word to recognize
d) Systems that only accept one-word command
Answer: c
Clarification: An isolated word recognizer requires pauses between each word said as it cannot recognize fully connected speech. The DVI control in an aircraft must not be an isolated voice recognizer to ensure smooth man-machine interface.

6. The DVI control system need not be super sensitive to the voice of the pilot.
a) True
b) False
Answer: b
Clarification: The DVI control system must be super sensitive to the voice of the pilot so that it can be operated in cockpit noise environment. In a fast jet combat aircraft, the background noise levels can be very high.

7. What is the reason for using speech synthesizers in aircrafts?
a) Communication with ATC
b) Communication with another aircraft
c) Used in flight data recorder
d) Warning messages to pilot
Answer: d
Clarification: Speech synthesizers are used to provide voiced warning messages to the pilot of system malfunctions and failures. They also aid DVI systems to provide feedback that a spoken command has been correctly recognized.

8. The audio/tactile management system combined with FBW and automated engine control is called ________
a) autopilot
b) flight management system
c) voice, throttle, stick control
d) flight control computer
Answer: c
Clarification: When the audio/tactile system, FBW, and the automated engine control is combined it is called the Voice, Throttle, Stick control. It is used in fighter aircraft such as Eurofighter Typhoon to reduce pilot workload.

9. What is the role of eye trackers in cockpits?
a) Improve concentration
b) Improve accuracy for targeting
c) Monitor pilot health
d) Assist in high g maneuvers
Answer: b
Clarification: Using an eye tracker the pilot’s gaze can be measured with accuracy which helps in better targeting. A fast and accurate data system can be achieved by this means without any discomfort to the pilot. They can track the eye by exploiting the principle of corneal reflection.

10. Why is the stereo sound warning system used?
a) Better sound clarity
b) Warning sound from the actual direction of a threat
c) To cancel the heavy background noise
d) To hear sound external to the aircraft
Answer: b
Clarification: A stereo sound warning system is used to produce a warning signal with the direction of threat. It consists of speakers mounted all around the cockpit, if the threat is approaching from the left side, the speakers on the left side produce signals to warn the pilot.

11. Only the pilot of a particular aircraft can use its DVI control system?
a) True
b) False
Answer: b
Clarification: The DVI control system is a universal system which can recognize the voice of anyone if spoken clearly. This is done for maintenance purposes and for an emergency.

here is complete set of 1000+ Multiple Choice Questions and Answers.

250+ TOP MCQs on Electromagnetic Spectrum and Answers

Avionics Multiple Choice Questions on “Electromagnetic Spectrum”.

1. Cps is the unit of what?
a) Frequency
b) Amplitude
c) Phase
d) Gain

Answer: a
Clarification: Cps or cycles per second is an alternative unit for the frequency of an EM wave. The other unit for frequency is Hertz which is more commonly used around the world.

2. The range of electromagnetic signals encompassing all frequencies is referred to as?
a) EM waves
b) EM frequency
c) EM spectrum
d) EM radiation

Answer: c
Clarification: The range of electromagnetic signals encompassing all frequencies is referred to as Electromagnetic spectrum. It includes radio waves, microwaves, optical waves, x-rays, gamma rays and cosmic waves.

3. The distance that the wave travels in one cycle is called as?
a) Displacement
b) Wavelength
c) Cycle length
d) Cycle amplitude

Answer: b
Clarification: The wavelength of a wave is the distance the wave travels in one cycle. The wavelength of a signal is represented by the Greek letter λ (lambda) and has the same units as length.

4. What is the wavelength of a signal with a frequency of 150Mhz?
a) 10m
b) 2m
c) 5m
d) 20m

Answer: b

5. What is the frequency of the EM wave with a wavelength of 12cm?
a) 2.4Ghz
b) 2.5Ghz
c) 250 Mhz
d) 0.25Mhz

Answer: b

6. A signal travels a distance of 75 ft in the time it takes to complete 1 cycle. What is its frequency?
1 m 5 3.28 ft
a) 13Mhz
b) 13.12Mhz
c) 14Mhz
d) Cannot be determined

Answer: b
Clarification: 1m=3.28ft
75/3.28=22.86m
Frequency= 300/22.86 =13.12Mhz.

7. The maximum peaks of an electromagnetic wave are separated by a distance of 8 in. What is the frequency?
a) 256.4Khz
b) 1477.8Khz
c) 1477.8Mhz
d) 256.7Mhz

Answer: c
Clarification: Wavelength =8 in.= 0.203 m
Frequency = 300/0.203 = 1477.8Mhz.

8. SHF Em waves are used in?
a) Ground to ground communication
b) Satellite communication
c) Aircraft to ATC communication
d) Underwater communication

Answer: b
Clarification: SHFs or Super High Frequency EM waves are those which lie between the frequency range of 3 to 30GHz range. These microwaves frequencies are widely used in satellite communication and radar.

9. Which one of the following is not a source of infrared radiation?
a) Human bodies
b) Light bulbs
c) Sun
d) Books

Answer: d
Clarification: Infrared radiation is generally associated with heat. Infrared is produced by light bulbs, human bodies, and basically any physical equipment that generates heat.

10. Which one of the following is not a reason for not using X-rays in communication?
a) Highly attenuated
b) High power is required
c) Stopped by ionizing effects
d) Ionizing radiation

Answer: c
Clarification: X-rays are not stopped by ionizing effects and that was the only reason they were ever thought they could be used in communication. In re-entry vehicles, a temporary communication cut off is experienced due to ionizing effects. X- rays were hard to modulate and focus and hence gradually let down.

11. The portion of the electromagnetic spectrum occupied by a signal is called ________
a) Signal spectrum
b) Bandwidth
c) Frequency width
d) Signal strength

Answer: b
Clarification: Bandwidth is that portion of the electromagnetic spectrum occupied by a signal. It is also the frequency range over which a receiver or other electronic circuit operates. More specifically, bandwidth is the difference between the upper and lower frequency limits of the signal or the equipment operation range.

12. What is the bandwidth of a signal having 928Mhz and 902Mhz as its upper and lower frequencies?
a) 26Mhz
b) 26Hz
c) 1830Hz
d) 1830Mhz

Answer: a
Clarification: Bandwidth = upper frequency – lower frequency = 928Mhz – 902Mhz =26Mhz.

13. What is the upper frequency of a signal with a bandwidth of 6Mhz, if the lower frequency limit is 54Mhz?
a) 60Mhz
b) 48Mhz
c) 60Hz
d) 48Hz

Answer: a
Clarification: Bandwidth = upper frequency – lower frequency
Upper frequency = bandwidth + lower frequency = 54Mhz + 6Mhz = 60Mhz.

14. Higher frequencies can have more channels for a particular bandwidth.
a) True
b) False

Answer: a
Clarification: The benefit of using the higher frequencies for communication carriers is that a signal of a given bandwidth represents a smaller percentage of the spectrum at the higher frequencies than at the lower frequencies. It means that there can be more number of channels with the same bandwidth in higher frequency than lower frequencies.

15. Which organization regulates international EM spectrum division?
a) ITU
b) FCC
c) NTIA
d) WCC

Answer: a
Clarification: The International Telecommunications Union (ITU), an agency of the United Nations that is headquartered in Geneva, Switzerland, comprises 189 member countries that meet at regular intervals to promote cooperation and negotiate national interests. The ITU brings together the various countries
to discuss how the frequency spectrum is to be divided up and shared. Because many of the signals generated in the spectrum do not carry for long distances, countries can use these frequencies simultaneously without interference.

250+ TOP MCQs on Digital Signal Processing and Answers

Avionics Multiple Choice Questions on “Digital Signal Processing”.

1. Any digital computer can be used for DSP.
a) True
b) False
Answer: a
Clarification: DSP is the use of a fast digital computer or digital circuitry to perform processing on digital signals. Any digital computer with sufficient speed and memory can be used for DSP.

2. What happens after the signal is passed through the analog to digital converter in a DSP?
a) Changed back to analog
b) Stored in a RAM
c) Amplified
d) Attenuated
Answer: b
Clarification: When the signal is converted from analog to digital it is a sequence of binary numbers which is stored in the RAM. A user defined code that is usually stored in the ROM performs mathematical and other manipulations after which it is converted back into analog signals.

3. Who is credited with creating the stored program concept?
a) John Von Neumann
b) Larry Page
c) Alan Turing
d) Ken Thompson
Answer: a
Clarification: Physicist John Von Neumann is generally credited with creating the stored program concept that is the basis of operation of all digital computers. Binary words representing computer instructions are stored sequentially in a memory to form a program. The instructions are fetched and executed one at a time at high speed.

4. What is the accessibility limitation of only one data or instruction set at a time from the memory called?
a) Von Neumann limitation
b) Von Neumann limit
c) Von Neumann speed
d) Von Neumann bottleneck
Answer: d
Clarification: The accessibility limitation of only one data or instruction set at a time from the memory is called as Von Neumann bottleneck. This has the effect of greatly limiting the execution speed.

5. Which type of architecture uses different storage space for program code and the data?
a) Von Neumann architecture
b) Harvard architecture
c) Fragmented architecture
d) Split cell architecture
Answer: b
Clarification: In a Harvard architecture microprocessor, there are two memories, a program or instruction memory, usually a ROM, and a data memory, which is a RAM. Also, there are two data paths into and out of the CPU between the memories. Because both instructions and data can be accessed simultaneously, very high-speed operation is possible.

6. What is the reason for the need of high speed DSP?
a) Less power consumption at higher speeds
b) Better processing capabilities
c) High sampling frequency
d) Easily programmable
Answer: c
Clarification: The time taken for input/output and the processing time together must be smaller than the sampling period to ensure the continuous flow of data. Since high sampling frequencies are needed for accurately converting the analog signal to digital, high speed DSP is a must.

7. Selectivity in a DSP is better than its analog equivalent.
a) True
b) False
Answer: a
Clarification: With DSP, the filters can have characteristics far superior to those of equivalent analog filters. Selectivity can be better because of the ease of controlling binary numbers, and the passband or reject band can be customized to the application.

8. Reduction in the number of binary words required to represent an analog signal is called ________
a) Undersampling
b) Oversampling
c) Data compression
d) Data minimization
Answer: c
Clarification: Data compression is a process that reduces the number of binary words needed to represent a given analog signal. Since analog to digital conversion produces a huge amount of data, for transmission it is a necessity that data is compressed.

9. What is the process of examining the frequency content of a signal?
a) Signal decoding
b) Spectrum analysis
c) Signal analysis
d) Data analysis
Answer: b
Clarification: Spectrum analysis is the process of examining a signal to determine its frequency content. Algorithms such as discrete Fourier transform (DFT) and FPGA is used to analyze the frequency content of an input signal.

10. What is the program that is used to speed the spectrum analysis process?
a) DDFT
b) FDFT
c) FGPA
d) Fast Fourier transforms
Answer: d
Clarification: The DFT is a complex program that is long and time-consuming to run. In general, computers are not fast enough to perform DFT in real time as the signal occurs. Therefore, a special version of the algorithm has been developed to speed up the calculation. Known as the fast Fourier transform (FFT), it permits real-time signal spectrum analysis.

11. Which of the following is not possible when the signal is analog?
a) Phase shifting
b) Equalization
c) Modulation
d) Data compression
Answer: d
Clarification: Data compression is done by checking redundancy in data. Data redundancy checking is only possible when the data is in digital form and hence data compression cannot be done in analog signals.

250+ TOP MCQs on Satellite Subsystems and Answers

Avionics Multiple Choice Questions on “Satellite Subsystems”.

1. Which of the following is not a satellite subsystem?
a) Ground station
b) Power system
c) Telemetry tracking
d) Communication subsystem
Answer: a
Clarification: The communication subsystem is the most important part of the satellite. It requires varies additional systems like the power system, propulsion system, telemetry system for its proper functioning. The ground system however is not one of the satellite subsystem and is independent of the satellite. It is just a transponder to monitor and command the satellite.

2. Which of the following is not a part of the propulsion subsystem of a satellite?
a) Gyroscope
b) Jet thruster
c) AKM
d) Fuel control system
Answer: a
Clarification: The propulsion subsystem consists of the AKM(Apogee kick motor), jet thruster and the fuel control system. Gyroscopes and other attitude systems fall under the attitude control subsystem.

3. Which of the following are common baseband signals transmitted from the earth ground station?
a) Navigational data, computer data, video
b) Computer data, navigational data, voice
c) Voice, video, computer data
d) Computer data
Answer: c
Clarification: An earth station takes the signals to be transmitted, known as baseband signals, and modulates a microwave carrier. The three most common baseband signals are voice, video, and computer data.

4. Which of the following components receives, translates the signal frequency and re-transmits the signal in a satellite?
a) Repeater
b) Relay
c) Transponder
d) Transducer
Answer: c
Clarification: The uplink signals from earth are amplified, translated in frequency, and re-transmitted on the downlink to one or more earth stations. The component that performs this function is known as a transponder.

5. Why is there a huge spectrum space between the transmitted and received signal in satellite communication?
a) Reduce interference
b) Maximum efficiency
c) Less attenuation
d) To reduce space occupied by filters
Answer: a
Clarification: Because of the close proximity of the transmitter and the receiver in the satellite, the high transmitter output power for the downlink is picked up by that satellite receiver. Naturally, the uplink signal is totally obliterated. Furthermore, the transmitter output fed back into the receiver input causes oscillation. To avoid this problem, the receiver and transmitter in the satellite transponder are designed to operate at separate frequencies. In this way, they will not interfere with each other.

6. Which of the following transponders convert the uplink signal to downlink signal using two mixers
a) Single conversion transponders
b) Dual conversion transponders
c) Regenerative transponders
d) Dual mixer transponder
Answer: b
Clarification: A dual-conversion transponder makes the frequency translation in two steps with two mixers. No demodulation occurs.

7. In a regenerative transponder, the signal is demodulated and modulated again before transmission.
a) True
b) False
Answer: a
Clarification: A regenerative repeater demodulates the uplink signal after the frequency is translated to some lower intermediate frequency. The recovered baseband signal is then used to modulate the downlink signal.

8. What is the number of transponders if the satellite uses 12 channels of frequency and frequency reuse is implemented?
a) 12
b) 6
c) 24
d) 3
Answer: c
Clarification: Since in frequency reuse each channel can be used twice the numbers of transponders are also doubled. 12 x 2 = 24 transponders, two for each frequency.

9. Why is it not possible to provide transmit function by wideband amplifier and mixer circuits?
a) Heavy attenuation
b) High power output over wideband is not possible
c) Economically not profitable
d) Weight of the system increases five fold
Answer: b
Clarification: it is generally not possible to generate very high output power over such wide bandwidth. The fact is that no components and circuits can do this well. The high-power amplifiers in most transponders are traveling-wave tubes that inherently have limited bandwidth. They operate well over a small range but cannot deal with the entire 500-MHz bandwidth allocated to a satellite.

10. Which of the following is not true?
a) Battery is only used as a back up
b) When in orbit, solar power is always available
c) Battery is used for initial satellite orientation and stabilization
d) The batteries are charged using solar power
Answer: b
Clarification: When a satellite goes into an eclipse or when the solar panels are not properly positioned, there is a temporary cut in solar power supply. In situations like this the batteries take over temporarily and keep the satellite operating. The batteries are not large enough to power the satellite for a long time; they are used as a backup system for eclipses, initial satellite orientation and stabilization, or emergency conditions.

11. Telemetry, command, and control (TC&C) subsystem allow a ground station to monitor and control conditions in the satellite.
a) True
b) False
Answer: a
Clarification: The telemetry system is used to report the status of the onboard subsystems to the ground station. The telemetry system typically consists of various electronic sensors whose data are selected by a multiplexer and then converted to a digital signal, which then modulates an internal transmitter. This transmitter sends the telemetry information back to the earth station, where it is recorded and monitored.

250+ TOP MCQs on Gyroscopes – 2 and Answers

Avionics online test on “Gyroscopes – 2”.

1. Which of the following systems do not use MEMS gyro?
a) Space launch vehicles
b) Flight control systems
c) Missile guidance
d) Car stability systems
Answer: a
Clarification: MEMS gyros are being exploited not only in avionic applications such as flight control, standby attitude and heading reference systems, and missile mid-course guidance but also in the automobile industry in car stability enhancement systems. Space launch vehicles use ring laser gyros which are more accurate.

2. Why is a tuning fork gyro better than the vibrating cylinder type gyro?
a) Reduce base motion
b) Low power consumption
c) Easy to manufacture
d) Low run up time
Answer: a
Clarification: The vibrating mass rate gyro is a simple instrument suffers from the unacceptable characteristic that the smallest linear motions applied to its base cause unacceptably large errors. To overcome the effect of base motion, it is necessary to use balanced oscillations in which the oscillations of one mass are counter-balanced by equal and opposite motion of a second equal mass as in the case of tuning fork gyro.

3. Which type of gyros works by calculating the difference in time taken by light to travel in a closed circle?
a) Ring laser gyro
b) MEMS gyro
c) Floated gyro
d) Dry tuned gyro
Answer: a
Clarification: Optical gyroscopes such as the ring laser gyro and the fibre optic gyro measure angular rate of rotation by sensing the resulting difference in the transit times for laser light waves travelling around a closed path in opposite directions.

4. What is the basic principle by which laser gyros work?
a) Coriolis effect
b) Conservation of momentum
c) Sagnac effect
d) Mass conservation
Answer: c
Clarification: The Sagnac effect is a relativistic phenomena of light in a rotating reference system. When laser beams circulate in a closed path that is rotating in an inertial space, the optical length seen by the co-rotating beam appears longer than that seen by counter rotating beam.

5. Which of the following type of gyro works by counting the nodes of standing waves of laser beams?
a) Ring laser gyro
b) Optical fiber gyro
c) Dry tuned gyro
d) Floated gyro
Answer: a
Clarification: In a Ring laser gyro, the counterpropagating beams form resonant nodes within the gyro cavity and create an electromagnetic standing wave that remains fixed with inertial space. When the housing of the gyro rotates, a detector can count nodes of the standing waves, each of which represents an angle.

6. A fiber optic gyro measures rotation using the _______ of the two beams?
a) Intensity difference
b) Resonance
c) Interference intensity
d) Wavelength difference
Answer: c
Clarification: In a fiber optic gyro, The counterpropagating beams are launched into an optical path and are recombined as they exit. The interference generated by the recombination depends on the optical phase difference between the two beams and therefore provides a measure of rotation.

7. In two mode RLGs, only linearly polarized modes can be resonant in the cavity.
a) True
b) False
Answer: a
Clarification: Two mode RLGs are planar by design so that only linearly polarized modes can be resonant in the cavity. Suppression of one of the two modes of polarization ensures stable operation.

8. The emission of photons by the gas medium that fills the RLG cavity is called as?
a) Laser discharge
b) Spontaneous discharge
c) Spontaneous emission
d) Laser emission
Answer: c
Clarification: The gas medium that supplies the gain for the laser also occasionally emits photons which are unrelated to the laser signal. This is known as spontaneous emissions are leads to noise and random walk in the RLG angle output.

9. What is the cause of lock-in phenomena in RLG?
a) Backscatter
b) Low intensity
c) High intensity
d) Low standing wave ratio
Answer: a
Clarification: Lock-in phenomenon leads makes the RLG insensitive to low angular rates. The cause of the lock-in phenomenon is backscatter within the cavity, usually resulting from imperfections in or particulates in the mirror surfaces.

10. Which method uses a turntable in RLG to reduce lock-in errors?
a) Intensity control
b) Rate biasing
c) Using mechanical dither
d) Random rate fixing
Answer: b
Clarification: The method of rate biasing employs a turntable that applies a constant rotation to the gyro. An angular encoder measures the relative angle between the instrument and its base.

To practice all areas of Avionics for online tests,

250+ TOP MCQs on Microwave Landing System and Answers

Avionics Multiple Choice Questions on “Microwave Landing System”.

1. What is the main weakness of the ILS system?
a) Less power
b) Less range
c) More noise
d) Environment sensitivity
Answer: d
Clarification: The main weakness of the ILS system was its sensitivity towards the environmental factors. Since the frequency used by the system is in MHz, it was more susceptible to atmospheric and weather interference.

2. What does the MLS use to overcome the weakness of ILS system?
a) Narrow beam width antennas
b) More power
c) Filers for separating noise
d) Array of antennas
Answer: a
Clarification: The main weakness of the ILS system is eliminated by using narrow beam width antennas which are physically small since the frequency is high. As frequency increases antenna size decreases.

3. Which of the following is not one of the basic components of the MLS system?
a) Azimuth ground station
b) Elevation ground station
c) DME
d) Radar
Answer: d
Clarification: A basic MLS consists of azimuth and elevation ground stations and a conventional DME for 3D positioning on approach course to 40° on either side of center line and to 15° elevation above the runway.

4. Why is a back azimuth station used in some MLS?
a) Better range
b) Less noise
c) Missed approach
d) Back up for main azimuth station
Answer: c
Clarification: An expanded MLS system uses back azimuth stations. It is generally used for departure and for missed approaches. It provides lateral guidance to 40° on either side of the center line.

5. What is the number of channels in the MLS system?
a) 10
b) 50
c) 4000
d) 200
Answer: d
Clarification: The MLS station transmit both angle and data functions on one of 200 frequencies between 5031.0 and 5190.7 MHz. The relatively high number of channels allows the use of multiple MLS in metropolitan cities.

6. Which of the following determines the required transmission power in MLS?
a) Noise
b) Bit error rate
c) Runway length
d) Aircraft size
Answer: b
Clarification: The acceptable bit error rate of the differential phase shift keying (DPSK) transmissions at the 20-nmi limit determines the transmitter power needed in the ground stations.

7. Using an audio tone to encode angles is an efficient method.
a) True
b) False
Answer: b
Clarification: Early MLS with mechanical scanned array used a varying audio tone to encode the pointing angle on the scanning beam pattern. The FAA adopted the time interval between successive passages of the unmodulated beam as an efficient means of angle encoding.

8. What is the typical scanning rate of the MLS antenna?
a) 20,000°/sec
b) 10000°/sec
c) 500°/sec
d) 37°/sec
Answer: a
Clarification: The very high scanning rate of 20,000°/sec provides about 40 samples per second of the angle data, a rate ten times higher than that needed to control the aircraft.

9. MLS ESA works on the principle of phase shifting.
a) True
b) False
Answer: a
Clarification: An Electronically scanned array antenna is an array of radiating elements with a feed network incorporating variable propagation delays. These arrays cause the antenna pattern to rotate by “phase shifting” the RF signal.