250+ TOP MCQs on Noise Suppression Effects of FM and Answers

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