Multiple choice Questions and Answers on Frequency Modulation

Q.1. In Frequency Modulation –

a. Amplitude of the carrier remains same
b. Frequency of the carrier varies in accordance with the modulating signal
c. The number of side bands are infinite
d. All of the above

ANSWER:  (d) All of the above

Q.2 Frequency deviation in FM is

a. Change in carrier frequency to the frequency above and below the centre frequency
b. Formation of side bands
c. The variation of the instantaneous carrier frequency in proportion to the modulating signal
d. All of the above

ANSWER:  (d) All of the above

 Q.3. Carrier swing is defined as

a. The total variation in frequency from the lowest to the highest point
b. Frequency deviation above or below the carrier frequency
c. Width of the side band
d. None of the above

ANSWER:  (a) The total variation in frequency from the lowest to the highest point

Q.4. The amount of frequency deviation in FM signal depends on

a. Amplitude of the modulating signal
b. Carrier frequency
c. Modulating frequency
d. Transmitter amplifier

ANSWER: ( a) Amplitude of the modulating signal

Q.5. Drawbacks of using direct method for generation of FM signal are

a. Does not give high stability to FM signal frequency
b. Distorted FM signal is generated due to harmonics of modulating signal
c. Cannot be used for high power FM generation
d. Both a and b

ANSWER: (d) Both a and b

Q.6. Advantage of using direct method for generation of FM signal is

a. It gives high stability to FM signal frequency
b. Distortion free FM signal is generated
c. High power FM generation is possible
d. None of the above

ANSWER:  (c) High power FM generation is possible

Q.7. What are the disadvantages of using balanced slope detector for demodulation of FM signal?

a. The detector operates only for small deviation in frequency
b. Low pass filter of the detector produces distortion in the detection
c. Both a and b
d. None of the above

ANSWER:  (c) Both a and b

Q.8. Drawbacks of Tuned Radio Receiver are

a. Oscillate at higher frequencies
b. Selectivity is poor
c. Bandwidth of the TRF receiver varies with incoming frequency
d. All of the above

ANSWER: (d) All of the above

Q.9. Sensitivity is defined as

a. Ability of receiver to amplify weak signals
b. Ability to reject unwanted signals
c. Ability to convert incoming signal into Image Frequency
d. Ability to reject noise

ANSWER: (a)  Ability of receiver to amplify weak signals

Q.10. In radio receivers, varactor diodes are used for

a. Tuning
b. Demodulation
c. Mixing
d. None of the above

ANSWER:  (a) Tuning

Q.11. The standard value for Intermediate frequency (IF) in double conversion FM receivers is

a. 455 KHz
b. 580 KHz
c. 10.7 MHz
d. 50 MHz

ANSWER:  (c) 10.7 MHz

Q.12. Amplitude limiter in FM receivers are used to

a. Remove amplitude variations due to noise
b. Filteration
c. Demodulation
d. Amplification

ANSWER:  (a) Remove amplitude variations due to noise

Q.13. Pre emphasis is done

a. For boosting of modulating signal voltage
b. For modulating signals at higher frequencies
c. In FM before modulation
d. All of the above

ANSWER:  (d) All of the above

Q.14. De-emphasis is

a. is restoring of original signal power
b. is done at the detector output of the receiver
c. is the inverse process of Pre emphasis
d. All of the above

ANSWER:  (d) All of the above

Q.15. Pre emphasis is done before

a. Before modulation
b. Before transmission
c. Before detection at receiver
d. After detection at receiver

ANSWER:  (a) Before modulation

Q.16. What is the effect on the deviation d of an FM signal when it is passed through a mixer?

a. Doubles
b. Reduces
c. Becomes half
d. Remains unchanged

ANSWER:  (d) Remains unchanged

Q.17. Armstrong method is used for the generation of

a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC

ANSWER:  (b) Indirect FM

Q.18. The modulation index of FM is given by

a. μ = frequency deviation/ modulating frequency
b. μ = modulating frequency /frequency deviation
c. μ = modulating frequency/ carrier frequency
d. μ = carrier frequency / modulating frequency

ANSWER:(a) μ = frequency deviation/ modulating frequency

Q.19. Disadvantages of FM over AM are

a. Prone to selective fading
b. Capture effect
c. Poorer signal to noise ratio at high audio frequencies
d. All of the above

ANSWER:  (d) All of the above

Q.20. What is the required bandwidth according to the Carson’s rule, when a 100 MHz carrier is modulated with a sinusoidal signal at 1KHz, the maximum frequency deviation being 50 KHz.

a. 1 KHz
b. 50 KHz
c. 102 KHz
d. 150 KHz

ANSWER: (c) 102 KHz

Explanation: 
According to Carson’s rule, bandwidth of FM is given by 2(Δf+ fm) where Δf is the deviation in frequency and fm is the frequency of sinusoidal signal. The required bandwidth is therefore calculated as
2 * (50KHz + 1KHz)
= 2 * 51 KHz
= 102 KHz

Q.21. The audio signal having frequency 500Hz and voltage 2.6V, shows a deviation of 5.2KHz in a Frequency Modulation system. If the audio signal voltage changes to 8.6V, calculate the new deviation obtained.

a. 17.2 KHz
b. 19.6 KHz
c. 25.6 KHz
d. 14.6 KHz

ANSWER:  (a) 17.2 KHz

Explanation: 
Deviation in FM is given by Δf = k_f * A_m
Therefore, k_f = Δf/ A_m
= 5.2/2.6
= 2
When voltage changes to 8.6V = A_m
New frequency deviation Δf = k_f * A_m
= 2* 8.6
= 17.2 KHz

Q.22. According to Carson’s rule, Bandwidth B and modulating frequency f_m are related as

a. B = 2(Δf + f_m) Hz
b. B = f_m Hz
c. B < 2f_m Hz
d. B > 2f_m Hz

ANSWER:  (a) B = 2(Δf + f_m) Hz

Q.23. What is the change in the bandwidth of the signal in FM when the modulating frequency increases from 12 KHz to 24KHz?

a. 40 Hz
b. 58 Hz
c. 24 Hz
d. Bandwidth remains unaffected

ANSWER:  (c) 24 Hz

Explanation: 
According to Carson’s rule, the bandwidth required is twice the sum of the maximum frequency deviation and the maximum modulating signal frequency. Or,
B=2(Δf +f_m) Hz
B= 2(Δf +12) Hz = 2 Δf + 24 Hz (1)
Assuming Δf to be constant,
B = 2 Δf + 48 Hz (2)
(2)-(1),
= 24Hz
Therefore the bandwidth changes by 24Hz.

Q.24. What is the maximum frequency deviation allowed in commercial FM broadcasting?

a. 100 KHz
b. 75 KHz
c. 15 KHz
d. 120 KHz

ANSWER:  (b) 75 KHz

Q.25. What is the maximum modulating frequency allowed in commercial FM broadcastings?

a. 40 KHz
b. 75 KHz
c. 15 KHz
d. 120 KHz

ANSWER: (c) 15 KHz

 Q.26. The ratio of actual frequency deviation to the maximum allowable frequency deviation is called:

a. Multi tone modulation
b. Percentage modulation
c. Phase deviation
d. Modulation index

ANSWER:  (b) Percentage modulation

Q.27. The range of modulating frequency for Narrow Band FM is

a. 30 Hz to 15 KHz
b. 30 Hz to 30 KHz
c. 30 Hz to 3 KHz
d. 3 KHz to 30 KHz

ANSWER:  (c) 30 Hz to 3 KHz

Q.28. FM is advantageous over AM as

a. The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM
b. The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted
c. There is less possibility of adjacent channel interference due to presence of guard bands
d. All of the above

ANSWER: All of the above

Q.29. VCO is used to generate

a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC

ANSWER:  (a) Direct FM

Q.30. Change in instantaneous phase of the carrier with change in amplitude of the modulating signal generates

a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC

ANSWER: (b) Indirect FM

Q.31. Phase-locked loop can be used as

a. FM demodulator
b. AM demodulator
c. FM receiver
d. AM receiver

ANSWER:  (a) FM demodulator

Q.32. The increase or decrease in the frequency around the carrier frequency is termed as

a. Figure factor
b. Frequency deviation
c. Modulation index
d. Frequency pectrum

ANSWER:  (b) Frequency deviation

Q.33. Carson’s rule is used to calculate

a. Bandwidth of FM signal
b. Signal to noise ratio
c. Modulation index
d. Noise figure

ANSWER:  (a) Bandwidth of FM signal

Q.34. The ratio of maximum peak frequency deviation and the maximum modulating signal frequency is termed as

a. Frequency deviation
b. Deviation ratio
c. Signal to noise ratio
d. Frequency spectrum

ANSWER:  (b) Deviation ratio

Q.35. Calculate the maximum frequency deviation for the FM signal
v(t) = 10 cos (6000t+ 5sin2200t)

a. 2200 Hz
b. 6000 Hz
c. 1750 Hz
d. 11000 Hz

ANSWER:  (c) 1750 Hz

Explanation: 
A standard FM signal is represented by
v(t) = A_c cos(2πf_ct + k_fsin2πf_mt)
A_c = carrier amplitude
f_c = carrier frequency
k_f = modulation index
f_m = modulating frequency = 2200/2π = 350 Hz
k_f = frequency deviation/modulating frequency
5 = freq deviation/ 350
Therefore, deviation = 5 * 350
= 1750Hz

Q.36. Calculate the dissipation in power across 20Ω resistor for the FM signal
v(t)= 20 cos(6600t+ 10sin2100t)

a. 5W
b. 20W
c. 10W
d. 400W

ANSWER:  (a) 5W

Explanation: 
A standard FM signal is represented by
v(t) = A_c cos(2πf_ct + k_fsin2πf_mt)
A_c = carrier amplitude
f_c = carrier frequency
k_f = modulation index
f_m = modulating frequency
k_f = frequency deviation/modulating frequency
the power dissipated across 20Ω resistor is given by
V_rms²/R
=(20/√2)²/R
= 5W

Q.37. What is the value of carrier frequency in the following equation for the FM signal?
v(t)= 5 cos(6600t+ 12sin2500t)

a. 1150 Hz
b. 6600 Hz
c. 2500 Hz
d. 1050 Hz

ANSWER:  (d) 1050 Hz

Explanation: 
A standard FM signal is represented by
v(t) = A_c cos(2πf_ct + k_fsin2πf_mt)
A_c = carrier amplitude
f_c = carrier frequency
k_f = modulation index
f_m = modulating frequency
k_f = frequency deviation/modulating frequency
therefore, f_c = 6600/2π
= 1050Hz

Q.38.  Calculate the modulation index in an FM signal when f_m (modulating frequency) is 250Hz and Δf (frequency deviation) is 5KHz.

a. 20
b. 35
c. 50
d. 75

ANSWER:  (a) 20

Explanation: 
Modulation index is the measure of how much the modulation parameter changes from its un modulated value. The modulation index of FM is given by
μ = frequency deviation/ modulating frequency
= Δf/ f_m
Where Δf is the peak frequency deviation i.e. the deviation in the instantaneous value of the frequency with modulating signal.
f_m is the value of modulating frequency
μ = 5000/250
= 20

Q.39. After passing the FM signal through mixer, what is the change in the frequency deviation Δ when the modulating frequency is doubled?

a. Becomes 2 Δ
b. Becomes Δ /2
c. Becomes Δ²
d. Remains unchanged

ANSWER:  (d) Remains unchanged

Q.40. In frequency modulation,

a. Armstrong method is used for generation
b. Multiple side bands are generated
c. The FM signal has infinite bandwidth
d. All of the above

ANSWER:  (d) All of the above

Q.41. Maximum frequency deviation and the maximum bandwidth allowed for commercial FM broadcast is

a. 80KHz, 160Khz
b. 75KHz, 200Khz
c. 60KHz, 170Khz
d. 75KHz, 250Khz

ANSWER: 7 (b) 5KHz, 200Khz

Q.42. Guard bands are provided in FM signal to

a. Prevent interference from adjacent channels
b. To increase the noise
c. To increase bandwidth
d. None of the above

ANSWER: (a)  Prevent interference from adjacent channels

Q.43. For a FM signal v(t) = 15 cos ( 10 * 10⁸t + 10 sin 1220t), calculate
1. Carrier frequency
2. Modulating frequency

a. 159.1MHz, 194.1Hz
b. 185.5MHz, 200.15Hz
c. 350.1MHz, 200.1Hz
d. 159.1Hz, 194.1Hz

ANSWER: 1 (a) 59.1MHz, 194.1Hz

Q.44. For a FM signal v(t) = 25 cos (15 * 10⁸t + 10 sin 1550t), calculate:

1. Modulation index
2. Maximum frequency deviation

a. 10, 3000.1Hz
b. 20, 1550.9Hz
c. 10, 2465.9Hz
d. 10, 2000.0Hz

ANSWER: (c)  10, 2465.9Hz

Explanation: 
Standard expression for FM signal is given by
v(t) = A cos ( ω_ct + mf sin ω_mt)
Comparing with the given equation,
Modulation index m_f = 10
Maximum frequency deviation is given by
m_f = Δf/f_m
Δf = m_f * f_m
Here f_m = 1550/2Π = 246.59 Hz
Δf = 10 * 246.59
= 2465.9Hz

Q.45.  For a FM signal v(t) = 20 cos ( 10 * 10⁸t + 30 sin 3000t), calculate the power dissipated by the FM wave in a 20Ω resistor.

a. 100 Watts
b. 10 Watts
c. 200 Watts
d. 20 Watts

ANSWER:  (b) 10 Watts

Explanation: 
Standard expression for FM signal is given by
v(t) = A cos ( Ω_ct + m_f sin Ω_mt)
Comparing with the given equation,
A = 20
The dissipated power is given by P = V²_rms/R
= (20/√2)²/ 20
= 10Watts

Q.46.  A 100MHz carrier is frequency modulated by 5 KHz wave. For a frequency deviation of 100 KHz, calculate the carrier swing of the FM signal.

a. 2000 KHz
b. 100 KHz
c. 105 KHz
d. 200 KHz

ANSWER: (d) 200 KHz

Explanation: 
Carrier frequency f_c = 100MHz
Modulating frequency f_m = 5 KHz
Frequency deviation Δf = 100 KHz
Carrier swing of the FM signal = 2 * Δf
= 2 * 100
= 200 KHz

Q.47.   A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.

a. 100
b. 50
c. 70
d. 90

ANSWER:  (b) 50

Explanation: 
Carrier frequency f_c = 100MHz
Modulating frequency f_m = 10 KHz
Frequency deviation Δf = 500 KHz
Modulation index of FM signal is given by
m_f = Δf/f_m
= 500 * 10³/ 10 * 10³
= 50

Q.48.  Narrow band FM has the characteristics:

a. The frequency sensitivity k_f is small
b. Bandwidth is narrow
c. Both a and b
d. None of the above

ANSWER: (c)  Both a and b

Q.49.  Wide band FM has the characteristics:

a. The frequency sensitivity k_f is large
b. Bandwidth is wide
c. Both a and b
d. None of the above

ANSWER:  (c) Both a and b

Q.50. Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz.

a. 170 KHz
b. 200 KHz
c. 100 KHz
d. 1000 KHz

ANSWER:  (a) 170 KHz

Explanation: 
Modulating frequency f_m = 10 KHz
Frequency deviation Δf = 75 KHz
According to Carson s rule, BW = 2(Δf + f_m)
= 2 (75 + 10)
= 170 KHz

Q.51. FM signal is better than AM signal because

a. Less immune to noise
b. Less adjacent channel interference
c. Amplitude limiters are used to avoid amplitude variations
d. All of the above

ANSWER:  (d) All of the above

Q.52.  FM is disadvantageous over AM signal because

a. much wider channel bandwidth is required
b. FM systems are more complex and costlier
c. Adjacent channel interference is more
d. Both a and b

ANSWER: (d) Both a and b

Q.53. The equation of the FM signal is  10 sin [2 π ×  10⁶ t +5 sin ( 2 π × 10³ t )]. The modulating frequency is  _____.

a. 10⁶ Hz

b. 5 Hz

c. 10³ Hz

ANSWER: (c) 10³ Hz

Q.54. If the deviation is 75 kHz and maximum modulating frequency is 5 kHz, what is the bandwidth of an FM wave ?

a. 80 kHz

b. 160 kHz

c. 40 kHz

d. 320 kHz

ANSWER: (b) 160 kHz

Q.55. With increase in the modulation index of an FM wave, the number of sidebands having significant amplitude will _______.

a. increase

b. decrease

c. remain constant

ANSWER: (a)  increase

Q.56. The transmitted power in an FM system is ______.

a. Dependent on the number of sidebands

b. Dependent on the carrier power and sidebands

c. Always constant

ANSWER: (c) Always constant

Q.57. The modulation index of a wideband FM system is _____.

a. > 1

b. < 1

c. = 1

d. none of these

ANSWER: (a) > 1

Q.58. To obtain the same S/N ratio at the same distance from the transmitter, FM transmitter has to transmit ______ power as compared to AM .

a. more

b. less

ANSWER: (b) Less

Q.59. Many FM transmitters can use the same carrier frequency .

a. True

b. False

ANSWER: (a) True