Category: Avionics Objective Questions

Avionics Assessment Questions and Answers on “Noise Suppression Effects of FM”.

1. Which of the following is not a source of noise in FM?
a) Weather
b) Electronic circuits
c) Lightning
d) Velocity of motion

Answer: d
Clarification: Lightning strikes, ignition systems, motors, electronic circuits and weather create interference signals called noise. They are generally of high frequency and spikes of voltages.

2. Which circuit in FM receivers cancels or filters out noise?
a) Filter circuits
b) Anti-noise circuits
c) Limiter circuits
d) delimiter circuits

Answer: c
Clarification: The receivers in FM contain limiter circuits which restrict the amplitude of the received signal. The noise which only affects the amplitude of the signal causes variations in amplitude of the signal. The limiter circuit clips off any variation in amplitude to filter out most of the noise.

3. Noise introduces a frequency variation into the signal.
a) True
b) False

Answer: a
Clarification: The change in amplitude by the noise introduces a phase shift which is reflected as a small frequency variation in the signal. These frequency variations changes or distorts the signal.

4. What is the frequency deviation produced by noise if the signal to noise ratio is 3:1 and modulating frequency is 800Hz?
a) 152.32Hz
b) 482.5Hz
c) 132.8Hz
d) 271.Hz

Answer: d
Clarification: ɸ=sin^-1(N/S)=sin^-1(1/3)=sin-1(0.3333)=19.47°=0.34rad
δ=ɸ(fm)=0.34(800)=271.8Hz.

5. What is the noise to signal ratio if the phase shift introduced by noise is 0.75°?
a) 0.5
b) 0.966
c) 0.25
d) 1.75

Answer: b
Clarification: ɸ=sin^-1(N/S)
(N/S)=sin(ɸ) =sin(75°)=0.966.

6. What is the frequency deviation produced by noise if the modulating frequency is 400Hz and the phase difference introduced by noise is 0.43rad?
a) 124Hz
b) 163Hz
c) 172Hz
d) 200Hz

Answer: c
Clarification: δ=ɸ(fm)=0.43(400)=172Hz.

7. Calculate the phase difference by noise if the Signal to noise ratio is 7:2.
a) 15.366°
b) 17.5°
c) 13.65°
d) 8.21°

Answer: a
Clarification: ɸ=sin^-1(N/S)=sin^-1(2/7)=sin^-1(0.265)=15.366°.

8. What is the ratio of the shift produced by the noise to the maximum allowed deviation if the modulating frequency is 300Hz and the phase shift by noise is 25° and the maximum allowable deviation is 5kHz?
a) 1.3
b) 0.530
c) 0.153
d) 0.0261

Answer: d
Clarification: δ=ɸ(f_m)= 0.436(300)=130.89Hz
Frequency deviation by noise / maximum deviation = 130.89/5000=0.0261.

9. What is the signal to noise ratio if the maximum allowable frequency deviation is 4kHz and the frequency deviation by noise is 156.42Hz?
a) 0.0364
b) 0.0391
c) 25.75
d) 20.45

Answer: c
Clarification: N/S = Frequency deviation by noise / maximum deviation = 156.42/4000 =0.0391
S/N = (N/S)^-1=25.57.

10. What is the technique in which the high frequency components are amplified more than the low frequency components in FM?
a) Garble
b) Preemphasis
c) Detoriation
d) Selective amplification

Answer: b
Clarification: Preemphasis is a technique in which the high frequency signals are amplified more than the lower frequency signals to have better resistance to noise. It is usually used to transmit sounds from musical instruments which have high frequency harmonics.

11. Which of the following pairs of resistors and capacitors can be used in high pass filters for preemphasis circuits?
a) 100Ω and 0.075μF
b) 100Ω and 0.75μF
c) 150Ω and 0.75μF
d) 16Ω and 0.75μF

Answer: b
Clarification: The time constant for a high pass filter to be used in the preemphasis circuit is 75μs. R x C = 75μs. Thus when R is 100Ω and C is 0.75μF, R x C = 75μs.

12. Find the frequency at which the signal enhancement flattens out in preemphasis circuit if R1=50Ω ,R2=70Ω and C=0.45μF.
a) 12126.09Hz
b) 550Hz
c) 10036..52Hz
d) 9004.56Hz

Avionics Multiple Choice Questions on “Standing Waves”.

1. Which of the following does not cause standing waves?
a) Short circuit
b) Open circuit
c) High frequency high power signal
d) Impedance is not matched
Answer: c
Clarification: If the load at the end of a line is an open circuit or a short circuit or has an impedance other than the characteristic impedance of the line, the signal is not fully absorbed by the load. This causes the signal to get reflected back the line and cause a standing wave.

2. Which of the following is not true regarding standing wave?
a) In a standing wave the energy moves towards the power source
b) In a standing wave power loss occurs
c) Standing waves do not affect signal strength
d) Standing waves are not desirable
Answer: c
Clarification: Standing waves are not desirable. Some of the energy is reflected from the end of the line and actually moves back up the line. This reflection indicates that the power produced by the generator is not totally absorbed by the load. When there is a power loss naturally the strength of the signal goes down.

3. What is the load on a transmission line?
a) 0 Ω
b) Infinity
c) Some value between 0 and infinity
d) Some value between0 and 1
Answer: c
Clarification: In practice, however, the load on a transmission line is neither infinite nor 0Ω; rather, it is typically some value in between. The load may be resistive or may have a reactive component.

4. The transmission line in which the resistive impedance is equal to the characteristic impedance is called _____
a) Matched lines
b) Paralleled lines
c) Balanced lines
d) Unbalanced lines
Answer: a
Clarification: Ideally, a transmission line should be terminated in a load that has a resistive impedance equal to the characteristic impedance of the line. This is called a matched line.

5. How would the graph of wavelength vs voltage look like if the transmission line is matched (neglect resistive loss)?
a) Linearly increasing
b) Linearly decreasing
c) Exponentially decreasing
d) Constant
Answer: d
Clarification: If a voltmeter is moved down a matched line from the generator to load and the rms voltage values are plotted, the resulting wavelength versus voltage line will be flat.

6. In an unmatched line the actual signal on the line is ___________
a) The sum of forward and reflected signals
b) The difference of forward and reflected signals
c) The product of forward and reflected signals
d) The modulus of forward signal
Answer: a
Clarification: In an unmatched line, standing waves are formed which is a combination of forward and reflected signal. The signal actually on a line is simply the algebraic sum of the forward and reflected signals.

7. What is the Standing wave ratio if a 75Ω antenna load is connected to a 50Ω transmission line?
a) 1
b) 2
c) 1.5
d) 1.43
Answer: c
Clarification: Standing wave ratio = SWR = Load impedance/ Characteristic impedance = 75/50 = 1.5.

8. The ratio of the incident voltage wave V_i to the reflected voltage wave V_r is called the reflection coefficient.
a) True
b) False
Answer: b
Clarification: The ratio of the reflected voltage wave V_r to the incident voltage wave V_i is called the reflection coefficient. The reflection coefficient provides the current and voltage information on the line.

9. What is the reflection coefficient If a line is terminated in its characteristic impedance?
a) 0
b) Infinity
c) 2
d) 0.5
Answer: a
Clarification: The reflection coefficient of a line that is fully terminated in its characteristic impedance is 0. This is because there is no reflected voltage on the line.

10. What percentage of power is reflected if the reflection coefficient is 0.5?
a) 50%
b) 75%
c) 25%
d) 12%
Answer: c
Clarification: Reflected power= Γ²= 0.5²=0.25 = 25% of initial power.

11. What is the resistive load if SWR= 3.05 and Zo =75Ω?
a) 1.23Ω
b) 51.23Ω
c) 254.2Ω
d) 24.59Ω
Answer: d
Clarification: Z_l = Z_o/SWR = 75/3.05 = 24.59Ω.

Avionics Multiple Choice Questions & Answers on “Hyperbolic Radio Systems – 2”.

1. What is used to indicate that a baseline is not usable?
a) Blank
b) Blink
c) Pulse
d) Continuous wave
Answer: b
Clarification: The secondary stations blink to notify the user that a master secondary pair is unusable. Blink is repetitive on/off pattern of the first two pulses of the secondary signal.

2. Which mode of Loran-C requires a minimum of 3 transmitters with iterative computation to obtain a fix?
a) Hyperbolic mode
b) Rho-Rho-Rho
c) Rho-Rho-Theta
d) Rho-Theta-Zulu
Answer: b
Clarification: There are some Loran-C users who do not employ Loran-C in a hyperbolic mode but rather in the direct range rho-rho-rho mode. The rho-rho-rho process involves a minimum of three transmitter stations and the use of an iterative computation to obtain fix.

3. Which of the following is false about Direct ranging in Loran-C?
a) Uses Rho-Rho mode
b) Requires minimum 2 stations
c) Cost efficient
d) High stable user frequency standard
Answer: c
Clarification: Direct ranging Rho-Rho mode requires a minimum of two stations, a highly stable user frequency standard and precise knowledge of the time of transmission of the signal. The use of this mode is limited by the high cost of stable frequency standard.

4. Which of the following advantages does the differential Loran-C provide?
a) Stable user frequency
b) No seasonal errors
c) Cost efficient
d) Very high range of 500miles
Answer: b
Clarification: The corrections using differential Loran-C are generally valid for the co-relation distance of approximately 100 miles from the reference station. Real time co-relations remove both seasonal and diurnal errors can be broadcast.

5. Which of the following methods breaks the signal into finite segments to find the ASF?
a) Millington’s method
b) Bill’s method
c) Differential method
d) Table look up process
Answer: a
Clarification: In Millington’s method, the signal between the transmitter and the receiver is broken down into finite segments of different conductivity levels, based on conductivity maps. The incremental phase delay is then computed as a function of range and conductivity for each path segment summed and averaged to provide an estimate of ASF.

6. What is the number of systems used for redundancy in a transmitting station?
a) 2
b) 1
c) 5
d) 3
Answer: a
Clarification: Each transmitter station is physically divided into two groups of units to provide system redundancy. At the appropriate interfaces switching units are provided between them.

7. Which of the following derives all the signals needed by the transmitter from the timer?
a) PATCO
b) ECD
c) HCG
d) TOPCO
Answer: a
Clarification: Dual redundant pulse amplitude and timing controllers or PATCO accept timing signals from the timer and derive from this all the signals needed by the transmitter. Signals generated by the PATCO include start triggers, charging triggers, digital amplitude reference signals, amplitude compensation signals, and megatron reference trigger.

8. In Omega system, a given frequency is transmitted by only one station at any given time.
a) True
b) False
Answer: a
Clarification: In the Omega system, each station transmits continuous wave signals on four common frequencies and one station unique frequency. The signal frequencies are time shared among the stations so that a given frequency is transmitted by only one station at any given time.

9. HCGs are the power generators of the Loran-C transmitter stations.
a) True
b) False
Answer: a
Clarification: Each HCG or half-cycle generators contributes to the power contained in the Loran-C pulse. 32 HCGs comprises the standard set. The basic set can be expanded in multiples of eight HCGs.

10. What is the use of notch filters in Loran-C receivers?
a) Reduce interference from other frequencies
b) Reduce interference from other stations
c) Reduce interference from sky waves
d) Reduce atmospheric noise
Answer: a
Clarification: Loran-C signal reception can be impaired by interference from other signal broadcast on slightly different frequencies. To avoid degradation of S/N associated with these interfering sources, Loran -C sets are equipped with notch filters.

11. What type of transmission is used in Omega system?
a) Amplitude modulated wave
b) Continuous wave
c) Phase modulated wave
d) Frequency modulated wave
Answer: b
Clarification: At each Omega station, continuous wave signals on four common frequencies and one station unique frequency. The signal frequencies are time shared among the stations so that a given frequency is transmitted by only one station at any given time.

12. Which of the following type is not used in an Omega system?
a) Grounded tower
b) Insulated tower
c) Hanging tower
d) Valley span
Answer: c
Clarification: Since the Omega stations transmit in very low frequencies, the antennas are the largest physical structures in the stations. Three types of antennas are employed in the Omega system: grounded tower, insulated tower, and the valley span.

13. _____ acts as a coarse tuning device for the antenna?
a) Loop
b) Helix
c) Amplifiers
d) Filters
Answer: b
Clarification: The RF signal that is to be transmitted is transferred to the ‘Helix,’ a large helical coil that acts as a coarse tuning device for the antenna. The helix is equipped with separate taps for each signal frequency transmitted.

14. What type of transmitter is used in a Decca system?
a) Frequency synthesizer
b) Local oscillator
c) Crystal controlled
d) Digital
Answer: c
Clarification: A typical Decca chain consists of a master station and three slave stations. A station has a 2kW crystal controlled transmitter feeding a 300ft antenna.

15. What is the approximate range of Chayka system?
a) 200mi
b) 100mi
c) 300mi
d) 1000mi
Answer: d
Clarification: Chayka is a pulse phase radio navigation system similar to the Loran-C system. By using ground waves at low frequencies, the operating range is 1000mi and by using pulse techniques, sky wave contamination can be avoided.

Avionics Multiple Choice Questions on “Radar Altimeter”.

1. Why is the radio altimeter placed in the tail of aircraft?
a) Can be placed anywhere
b) When landing tail is closer to the ground
c) When landing nose is closer to the ground
d) Weight balance
Answer: b
Clarification: When an aircraft lands it has to do the flare maneuver where the pilot pitches the aircraft’s nose up to slow down the plane. Radio altimeter measures precise distance from the ground and since the tail is closer to the ground than the nose, to measure the minimum distance it is placed in the tail.

2. What is the maximum altitude that can be measured using a radio altimeter?
a) 100ft
b) 500ft
c) 50 ft
d) 5000ft
Answer: d
Clarification: Radio altimeters are generally used for measuring altitude during landing or take off. A radio altimeter gives the distance between the aircraft and the ground and not the pressure altitude. Maximum range of a radio altimeter is 5000ft but is usually switched to at 2500ft.

3. Altimeters that are specifically designed to provide mark signals at specific altitudes for automatic operations are called as _________
a) Radar altimeter
b) Pressure altimeter
c) Altitude marking radar
d) Mark altimeter
Answer: c
Clarification: Altitude marking radar are generally low altitude altimeter designed specifically to provide mark signals at specific altitudes for initiation of an automatic operation such as fuze triggering on submunitions or chute opening on lunar landing systems.

4. What is the function leading edge range tracker servo loop?
a) Provide range to the nearest return
b) Speed measurements
c) Vertical speed measurement
d) Automatic ground navigation
Answer: a
Clarification: To provide a range to the nearest return within the bounds of the antenna beam, many modern radar altimeters incorporate the gate in a pulse modulated radar or a filter in a frequency modulated radar over the leading edge of the return.

5. What is the frequency designated to radar altimeters?
a) 4.2 to 4.4 GHz
b) 2.4 to 4.2 GHz
c) 500 to 1 GHz
d) 13 to 14 GHz
Answer: a
Clarification: The frequency band of 4.2 to 4.4 GHz is assigned to radar altimeters. This frequency band is high enough to result in a reasonable small sized antenna to produce 40° to 50° beam but significantly low so that attenuation by rain is minimum.

6. Which of the following is true with respect with solid state transmitters in pulsed radar altimeters?
a) Input power is around 100W
b) Use receiver pre-amplifiers
c) High reliability
d) Small size
Answer: a
Clarification: The early pulsed radars were evolved to 5W solid state transmitters incorporating receiver pre-amplifiers. They also provided a high degree of reliability, low probability of intercept, small size, and high accuracy.

7. Which of the following is responsible for the reduced use of transmitter power in radar altimeter?
a) Low range
b) Interference with other bands
c) Receiver low noise amplifier
d) Radome
Answer: c
Clarification: The transmitter power depends on the sensitivity of the receiver. The receiver low noise amplifier typically has a 2 to 3 dB noise figure, resulting in a sensitivity level that allows relatively low transmitter power.

8. What type of waveform must be transmitted to reduce Doppler shit errors?
a) Square
b) Triangular
c) Sine
d) Cosine
Answer: b
Clarification: To reduce the errors due to the Doppler shift of the return, a triangular waveform is usually used to modulate the transmitter. Thus a positive Doppler shift will produce a negative frequency error on the rising modulation slope. By averaging the frequency count the error can be minimized.

Avionics Multiple Choice Questions on “Flight Management Systems – 1”.

1. Which one of the following is not a function of the FMS?
a) Flight guidance and control of flight path
b) Monitor and regulate speed of the aircraft
c) Automatically switch between different types of communication
d) Automatic control of engine thrust

Answer: c
Clarification: The tasks of the FMS include Flight guidance and lateral and vertical control of the aircraft flight path, Monitoring the aircraft flight envelope and computing the optimum speed for each phase of the flight and ensuring safe margins are maintained with respect to the minimum and maximum speeds over the flight envelope, Automatic control of the engine thrust to control the aircraft speed. However, the communication systems are taken care manually by the pilot.

2. What does the following figure represent?

a) Flight Management System
b) Flight Data System
c) Total autopilot system
d) Flight Management Computer

Answer: a
Clarification: The above figure represents flight management system. The throttle, Flight Management Computer, Data storage, Navigation and display systems, and the autopilot together is called the FMS.

3. The number of independent FMS in a typical commercial aircraft is?
a) 1
b) 2
c) 5
d) 7

Answer: b
Clarification: There are two independent FMS in a typical commercial aircraft’s cockpit. FMS-1 is on the Captain’s side and FMS-2 on the First Officer’s side to carry ou the flight management functions. The reason why both are independent is that when one fails the other can take over.

4. Which of the following is not a way cockpit-flight crew interfaces in the FMS?
a) Primary Flight Display
b) Multi Function Display
c) Keyboard and Cursor Control Unit
d) Target Detection and Locking System

Answer: d
Clarification: The cockpit interfaces to the flight crew provided by each FMS comprise a Navigation Display (ND), a Primary Flight Display (PFD), a Multi-Function Display(MFD), a Keyboard and Cursor Control Unit (KCCU) and an Electronic Flight Instrument System (EFIS) Control Panel (EFIS CP).

5. What type of cockpit flight crew interface is used to enter or modify the data on the MFD?
a) Keyboard and Cursor Control Unit
b) Control stick
c) Control Switches
d) Control levers

Answer: a
Clarification: The flight crew can navigate through the pages of FMS and can consult, enter or modify the data via the Keyboard and Cursor Control Unit (KCCU). The Keyboard and Cursor Control Unit (KCCU) enables the flight crew to navigate through the FMS pages on the MFD and enter and modify data on the MFD and can also perform some flight plan revisions on the lateral Navigation Display (ND).

6. What control does the EFIS Control Panel provide?
a) Control over graphical and textual FMS data
b) Control over flight plan
c) Control over flight performance
d) Navigation through FMS pages

Answer: a
Clarification: The EFIS Control Panel (EFIS CP) provides the means for the flight crew to control the graphical and textual FMS data that appear on the ND and PFD. Control over flight plan and performance is given by MFD. Navigation through the pages is by KCCU.

7. What is the number of Flight Management Computers used in a typical commercial aircraft?
a) 2
b) 1
c) 4
d) 3

Answer: d
Clarification: There are a total of three Flight Management Computers FMC -A, FMC -B, FMC -C. They are necessary to carry out the necessary functional computations. They can be reconfigured to maintain the system operation in the event of failures.

8. What operating mode of the FMS does the figure show?

a) Double mode
b) Independent mode
c) Single mode
d) Redundancy mode

Answer: b
Clarification: In the Independent Mode, FMS-1 and FMS-2 are both operative, but there is no data exchange between them because they disagree on one or more items such as aircraft position, gross weight, etc.

Avionics Multiple Choice Questions on “Digital Transmission of Data”.

1. Which of the following is false with respect to digital data transmission?
a) LAN is a digital data transmission
b) Can transmit binary data
c) Only restricted to communication between computers
d) Can transmit analog data
Answer: c
Clarification: Digital data communication was only restricted to communication between computers. Since analog signals can now be easily converted into digital signals using analog to digital converter, digital transmission can now transmit voice, video and other analog signals in digital form.

2. Why is digital data not easily affected by noise?
a) High power transmission
b) Cannot easily change binary 1 to 0
c) High frequency transmission
d) Low frequency transmission
Answer: b
Clarification: Noise is inevitable in a signal transmission. In a digital transmission the amplitude of the noise must be higher than the amplitude of the signal at binary ’1’, which is generally not the case, and vice versa. The receiver can differentiate between binary ‘1’ and ‘0’ with a significant amount of noise.

3. What is the process of digital communication where a threshold value is set at the receiver to reduce noise?
a) Signal threshold
b) Data regeneration
c) Signal regeneration
d) Signal cut off
Answer: c
Clarification: In the process of signal regeneration a cut off value or threshold value is set in the receiver so that the noise can be clipped off. When the values are set properly only the binary data would pass through the circuit. This produces a clean output pulse.

4. What is the number of error bits that occur for a given number of bits transmitted called?
a) Bit error
b) Error rate
c) Bit error rate
d) Error bit rate
Answer: c
Clarification: Bit rate error(BER) is the number of error bits that are present for a given number of bits transmitted. It depends on several factors like the environment, equipment and other considerations. BER is calculated for a particular set of conditions.

5. Data is transmitted in 512 bytes. What is the average number of errors expected if the system BER is 2:10,000?
a) 1
b) 0
c) 0.542
d) 0.819
Answer: d
Clarification: Total number of bits = 512 x 8 = 4096 bits
Average number of errors = (2/10,000) x 4096 = 0.819.

6. Why can Digital data use time division multiplexing?
a) Less power
b) High power
c) Discrete data
d) Continuous data
Answer: c
Clarification: Digital data is discrete and hence can be transmitted via time division multiplexing, whereas analog data is continuous and can only be transmitted as a whole. However, analog data can be converted into digital by analog to digital converters and then be transmitted via time division multiplexing. Time division multiplexing is one of the biggest advantages of digital communication.

7. Speed of transmission is increased by channel encoding.
a) True
b) False
Answer: b
Clarification: Speed of transmission is reduced when data is channel encoded. For example, to transmit 1 byte of data, the encoding process may add 3 extra bits for a total of 11 bits. If the bit time is 100 ns, then it takes 800 ns to send the data bits but 1100 ns to send the encoded data.

8. Which of the following bit has an error if odd parity is used?

	data	Parity bit
A	1000	0
B	1001	0
C	1100	1
D	1111	1

a) A
b) B
c) C
d) D
Answer: b
Clarification: When odd parity is used, the parity bit is 1 when the number of binary ‘1’ is even and the parity bit is ‘0’ when the number of binary 1 is odd. Here, option ‘b’ has even number of 1’s but the parity bit is 0. Thus option b has error.