Angle Modulation MCQ Quiz - Objective Question with Answer for Angle Modulation - Download Free PDF

Last updated on Jun 26, 2025

Latest Angle Modulation MCQ Objective Questions

Angle Modulation Question 1:

What is the bandwidth occupying 98% of power, as per Carson's rule for the following FM signal 10 sin?

  1. 20 KHz
  2. 60 kHz
  3. 100 kHz
  4. 1.01 MHz

Answer (Detailed Solution Below)

Option 3 : 100 kHz

Angle Modulation Question 1 Detailed Solution

Explanation:

 According to Carson's rule, the bandwidth of an FM signal that occupies 98% of its power can be expressed as:

Bandwidth (B) ≈ 2 × (Δf + fm)

Where:

  • Δf: The frequency deviation, which represents the maximum shift in the carrier frequency due to the modulation signal.
  • fm: The maximum frequency of the modulating signal.

In this case, the FM signal is given as:

10 sin(2π × 106t + 4 cos(2π × 104t))

From the given FM signal, we can extract the following parameters:

  • Carrier Frequency (fc): 106 Hz (1 MHz)
  • Frequency of the Modulating Signal (fm): 104 Hz (10 kHz)
  • Frequency Deviation (Δf): The frequency deviation is determined by the modulation index multiplied by the frequency of the modulating signal. In this case, the modulation index is 4, so:

Δf = Modulation Index × fm

Δf = 4 × 104

Δf = 40 kHz

Now, applying Carson's rule to calculate the bandwidth:

Bandwidth (B) ≈ 2 × (Δf + fm)

B ≈ 2 × (40 kHz + 10 kHz)

B ≈ 2 × 50 kHz

B ≈ 100 kHz

Angle Modulation Question 2:

In television broadcasting the total channel bandwidth required to transmit video and audio signals is:

  1. 2.4 MHz
  2. 5.785 MHz
  3. 5.578 MHz
  4. 4.2 MHz

Answer (Detailed Solution Below)

Option 2 : 5.785 MHz

Angle Modulation Question 2 Detailed Solution

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Explanation:

Television Broadcasting Channel Bandwidth

Definition: In television broadcasting, the total channel bandwidth refers to the range of frequencies required to transmit both video and audio signals simultaneously. This bandwidth ensures that all components of the television signal, including visual content, audio content, and any additional data, are transmitted without interference or loss of quality.

Details of Bandwidth Requirements:

Television broadcasting utilizes a combination of video and audio signals that must be transmitted within a specific frequency range. The video signal requires a larger portion of the bandwidth as it contains detailed visual information, while the audio signal occupies a smaller portion for sound transmission. To prevent interference between the two signals and to maintain clarity, sufficient separation (guard bands) is also included within the allocated bandwidth.

The total channel bandwidth for standard analog television broadcasting is determined by the following components:

  • Video Signal Bandwidth: The video signal typically requires a bandwidth of approximately 4.2 MHz. This bandwidth allows for the transmission of detailed visual information, such as brightness (luminance) and color (chrominance).
  • Audio Signal Bandwidth: The audio signal requires a much smaller bandwidth, generally around 0.2 MHz (200 kHz). This is sufficient for transmitting high-quality monaural or stereo audio.
  • Guard Band: To avoid interference between the video and audio signals, a guard band of approximately 0.15 MHz (150 kHz) is included. This ensures that the signals do not overlap and degrade each other’s quality.

Total Bandwidth Calculation:

The total channel bandwidth required for transmitting both video and audio signals can be calculated as:

Total Bandwidth = Video Signal Bandwidth + Audio Signal Bandwidth + Guard Band

Substituting the values:

Total Bandwidth = 4.2 MHz + 0.2 MHz + 0.15 MHz

Total Bandwidth = 4.575 MHz

However, in practice, the total channel bandwidth is slightly rounded up to accommodate additional factors, such as signal synchronization (horizontal and vertical sync pulses) and other engineering considerations. This results in a standard channel bandwidth of approximately 5.785 MHz for analog television broadcasting.

Correct Option Analysis:

The correct option is:

Option 2: 5.785 MHz

This option correctly represents the total channel bandwidth required to transmit both video and audio signals in analog television broadcasting. It accounts for the combined bandwidth of the video signal, audio signal, and necessary guard bands, as explained above.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 2.4 MHz

This option is incorrect because 2.4 MHz is far too small to accommodate both video and audio signals in television broadcasting. The video signal alone requires a bandwidth of approximately 4.2 MHz, which exceeds this value. Therefore, this option does not meet the requirements for transmitting both signals.

Option 3: 5.578 MHz

This option is close to the correct value but is slightly lower than the standard bandwidth of 5.785 MHz. It does not account for the full guard band and other additional considerations necessary for high-quality signal transmission.

Option 4: 4.2 MHz

This option represents only the video signal bandwidth and does not include the bandwidth for the audio signal or the guard band. As a result, it is incomplete and incorrect for representing the total channel bandwidth in television broadcasting.

Conclusion:

In television broadcasting, the total channel bandwidth required to transmit both video and audio signals is approximately 5.785 MHz. This value includes the bandwidth for the video signal, the audio signal, and a guard band to prevent interference. Understanding the components and their contributions to the total bandwidth is essential for ensuring high-quality signal transmission in analog television systems.

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Angle Modulation Question 3:

Which of the following statements about the AM detection using PLL circuit is/are correct?
S1: It has higher noise immunity than the conventional peak detector type AM detector.
S2: The PLL is locked to the carrier frequency of the AM signal.

  1. Only S1
  2. Only S2
  3. Neither S1 nor S2
  4. Both S1 and S2

Answer (Detailed Solution Below)

Option 4 : Both S1 and S2

Angle Modulation Question 3 Detailed Solution

Explanation:

AM Detection Using PLL Circuit

Definition: Phase-Locked Loop (PLL) circuits are used for a variety of applications in communication systems, including frequency synthesis, demodulation, and signal synchronization. In the context of AM (Amplitude Modulation) detection, PLLs are used to demodulate the AM signal by locking onto the carrier frequency and extracting the modulating signal.

Working Principle: A PLL circuit consists of a phase detector, a low-pass filter, and a voltage-controlled oscillator (VCO). The phase detector compares the phase of the incoming AM signal with the phase of the VCO output. The resulting error signal is filtered and used to adjust the VCO frequency. When the PLL is locked, the VCO frequency matches the carrier frequency of the AM signal, and the error signal corresponds to the modulating signal (audio or data).

Advantages:

  • Higher noise immunity compared to conventional peak detector type AM detectors.
  • Precise demodulation of the AM signal even in the presence of noise and signal distortion.
  • Ability to lock onto the carrier frequency, providing stable and accurate demodulation.

Disadvantages:

  • Increased complexity and cost compared to simple peak detector circuits.
  • Requires careful design and tuning to ensure proper locking and demodulation.

Applications: PLL-based AM detectors are used in high-fidelity radio receivers, communication systems, and other applications where accurate and reliable demodulation of AM signals is critical.

Correct Option Analysis:

The correct option is:

Option 4: Both S1 and S2

This option correctly identifies both key aspects of AM detection using PLL circuits:

  • S1: It has higher noise immunity than the conventional peak detector type AM detector.
    This statement is true because PLL circuits can reject noise and maintain lock on the carrier frequency, leading to better demodulation performance in noisy environments.
  • S2: The PLL is locked to the carrier frequency of the AM signal.
    This statement is also true because the fundamental operation of a PLL involves locking onto the carrier frequency of the input signal, ensuring accurate extraction of the modulating signal.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Only S1

While S1 is correct, this option is incomplete because it does not acknowledge the importance of S2, which is also a true statement about PLL-based AM detectors.

Option 2: Only S2

Similar to Option 1, this option is incomplete as it only considers S2 and ignores the validity of S1, which is an important aspect of the advantages of PLL-based AM detectors.

Option 3: Neither S1 nor S2

This option is incorrect because both S1 and S2 are true statements about PLL-based AM detectors. Ignoring both statements would lead to a misunderstanding of the advantages and operation of PLL circuits in AM detection.

Conclusion:

Understanding the role of PLL circuits in AM detection is crucial for recognizing their advantages over conventional peak detector circuits. PLLs offer higher noise immunity and precise locking to the carrier frequency, ensuring accurate demodulation of AM signals. Both S1 and S2 are correct statements that highlight these key benefits, making Option 4 the accurate choice.

Angle Modulation Question 4:

Which of the following statements about FM detection using PLL circuit are correct?
S1: At the error amplifier output, we get demodulated FM output.
S2: FM signal is applied to the input of the PLL.

  1. Only S1
  2. Neither S1 nor S2
  3. Only S2
  4. Both S1 and S2

Answer (Detailed Solution Below)

Option 4 : Both S1 and S2

Angle Modulation Question 4 Detailed Solution

The correct answer is: 4) Both S1 and S2

Explanation:
S1: At the error amplifier output, we get a demodulated FM output.
True. In a Phase-Locked Loop (PLL) used for FM detection:

The error amplifier (or phase detector output) produces a voltage proportional to the phase difference between the input FM signal and the VCO signal.

Since frequency modulation (FM) is the derivative of phase modulation, this voltage directly represents the demodulated message signal.

S2: The FM signal is applied to the input of the PLL.
True. The FM signal is fed into the PLL's input (phase detector). The PLL locks onto the carrier frequency and tracks the frequency variations (modulation), allowing demodulation.

Angle Modulation Question 5:

According to Carson’s Rule, what is the approximate bandwidth required for a VHF/UHF two-way radio signal using FM with a 5 kHz maximum frequency deviation and a maximum audio frequency of 3 kHz?

  1. 30 kHz
  2. 20 kHz
  3. 10 kHz
  4. 16 kHz

Answer (Detailed Solution Below)

Option 4 : 16 kHz

Angle Modulation Question 5 Detailed Solution

Explanation:

Carson's Rule:

Carson's Rule is a mathematical formula used to estimate the bandwidth requirements for a frequency-modulated (FM) signal. According to Carson's Rule, the total bandwidth (BT) required for an FM signal can be approximated using the following formula:

BT = 2(Δf + fm)

Where:

  • Δf = Peak frequency deviation of the carrier signal (in this case, 5 kHz)
  • fm = Maximum modulating frequency (in this case, 3 kHz)

Using the given values:

  • Δf = 5 kHz
  • fm = 3 kHz

Applying these values to Carson's Rule:

BT = 2(5 kHz + 3 kHz) = 2(8 kHz) = 16 kHz

Therefore, the approximate bandwidth required for a VHF/UHF two-way radio signal using FM with a 5 kHz maximum frequency deviation and a maximum audio frequency of 3 kHz is 16 kHz.

Top Angle Modulation MCQ Objective Questions

The modulating frequency in frequency modulation is increased from 10 kHz to 20 kHz. The bandwidth is -

  1. Doubled
  2. Halved
  3. Increased by 20 kHz
  4. Increased tremendously

Answer (Detailed Solution Below)

Option 3 : Increased by 20 kHz

Angle Modulation Question 6 Detailed Solution

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Frequency Modulation:

In frequency modulation, the carrier amplitude remains constant, but its frequency is changed in accordance with the modulating signal. 

The bandwidth of a frequency modulated signal is:

BW = 

Calculation:

Given, (fm)1 = 10 kHz

(BW)1 = 

(fm)2 = 20 kHz

(BW)2

(BW)2 = 

(BW)2 = (BW)1 + 20

The bandwidth is increased by 20 kHz.

The frequency deviation produced in a VHF carrier by a signal of 100 Hz is 50 kHZ. The frequency modulation index is

  1. 100 radians
  2. 250 radians
  3. 500 radians
  4. 750 radians

Answer (Detailed Solution Below)

Option 3 : 500 radians

Angle Modulation Question 7 Detailed Solution

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Concept:

In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency.

Mathematically, this is defined as:

mf = Modulation index

Δf = Frequency deviation

fm = Modulating frequency

Calculation:

Given Δf = 50 kHz

fm = 100 Hz

mf = 500 radians

A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.

Amplitude Modulation: The amplitude of the carrier is varied according to the amplitude of the message signal.

Frequency Modulation: The frequency of the carrier is varied according to the amplitude of the message signal.

Phase Modulation: The Phase of the carrier is varied according to the amplitude of the message signal.

 

Frequency curve is _______________.

  1. Asymptotic to y-axis
  2. Non-asymptotic to x-axis
  3. Asymptotic to x-axis
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Asymptotic to x-axis

Angle Modulation Question 8 Detailed Solution

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The correct answer is Asymptotic to the x-axis.

Key Points

  • A Frequency curve is a graph of frequency distribution where the line is smooth.
  • It is just like a frequency polygon.
  • In the polygon is line is straight, but in the curve the line is smooth.
  • It is an area diagram.
  • It is the graphical representation of frequency distribution.
  • The X-axis is marked with class intervals.
  • The Y-axis is marked with frequencies.
  • The beginning and end of the curve should touch the last class interval at the mad posts of the first and last interval.
  • The area of the curve is equal to that of a histogram.
  • The frequency curve is divided into 3 types based on the shape of the curve.
    • They are Normal distribution curves.
      • Positively skewed distribution curve.
      • Negatively skewed distribution curve.

Which is true in a frequency modulated system ?

  1. Variation of carrier amplitude does not affect quality of reception
  2. Amplitude of carrier is varied according to modulating signal
  3. The transmission does not require line of sight between transmitter and reciever
  4. Receiver uses diode detector

Answer (Detailed Solution Below)

Option 1 : Variation of carrier amplitude does not affect quality of reception

Angle Modulation Question 9 Detailed Solution

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Frequency Modulation:

  • Frequency Modulation is a modulation in which the frequency of the carrier wave is altered according to the instantaneous amplitude of the modulating signal, keeping phase and amplitude constant.
  • So, the variation in carrier amplitude and carrier phase does not affect the signal in the receiving end.
  • Line of sight (LoS) is a type of propagation that can transmit and receive data only where transmit and receive stations are in view of each other without any sort of an obstacle between them. Eg: FM radio, microwave, and satellite transmission.
  • Frequency Modulation works on the Line of sight propagation.


Types of FM detection:

  • Slope detection
  • Phase-locked loop detection
  • Foster Seeley detection
  • Ratio detector
  • Quadrature detectors.

Compute the average transmitted power of frequency modulated wave with carrier signal cos(2π1000t).

  1. 1.5
  2. 0.5
  3. 1
  4. 0.25

Answer (Detailed Solution Below)

Option 2 : 0.5

Angle Modulation Question 10 Detailed Solution

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Analysis:

The amplitude Ac is constant in a phase-modulated or a frequency modulated signal. RF power does not depend upon the modulation index.

A general expression for a phase or frequency modulated signal is:

m(t) = the modulating signal

ωc = Carrier frequency

kk becomes kc for FM and kp for PM.

The average power (Pavg) is given by:

 (Always)

We observe that the transmitted power is independent of the modulation index in the of FM.

Calculation:

A = 1

Hence, P = 1/2 = 0.5 W. 

In a frequency modulation system, maximum frequency deviation allowed is 1000 Hz and modulating frequency is 1 kHz. Determine modulation index β.

  1. 2
  2. 2000
  3. 1
  4. 1000

Answer (Detailed Solution Below)

Option 3 : 1

Angle Modulation Question 11 Detailed Solution

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Concept:

In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency.

Mathematically, this is defined as:

mf = Modulation index

Δf = Frequency deviation

fm = Modulating frequency

Calculation:

Given Δf = 1000 Hz = 1 kHz

fm = 1 kHz

A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.

Amplitude Modulation: The amplitude of the carrier is varied according to the amplitude of the message signal.

Frequency Modulation: The frequency of the carrier is varied according to the amplitude of the message signal.

Phase Modulation: The Phase of the carrier is varied according to the amplitude of the message signal.

What is the maximum frequency of modulating signal in FM system, if it has 8 significant sidebands, and the maximum bandwidth is 32 kHz?

  1. 4 kHz
  2. 8 kHz
  3. 16 kHz
  4. 2 kHz

Answer (Detailed Solution Below)

Option 4 : 2 kHz

Angle Modulation Question 12 Detailed Solution

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Concept:

WBFM is given as:

1\)

Actual BW of WBFM is so to Bandlimit WBFM signal all of its significant lower-order sidebands retained and higher-order insignificant sidebands should be eliminated.

Calculation:

BW = 32 KHz

Given FM system has 8 sidebands (significant) i.e. 8 sidebands on the positive side of frequency & 8 sidebands on the negative side of the frequency

So, It has sidebands up to order ‘8’.

According to carson rule:

Significant sideband upto order of (β + 1) has B.W

BW = (β + 1) 2f­m

So,

32 = 8 × 2 fm

fm = 2 kHz

Which method from the following is used in demodulating FM wave?

  1. Distribution discrimination method
  2. Amplitude discrimination method
  3. Delay discrimination method 
  4. Phase discrimination method

Answer (Detailed Solution Below)

Option 4 : Phase discrimination method

Angle Modulation Question 13 Detailed Solution

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Frequency modulated (FM) signal can generally be demodulated by the following methods:

Frequency Discrimination:

Phase Discrimination:

Phase-locked loop (PLL):

Choose the incorrect statement regarding FM -

  1. It has large number of sidebands.
  2. Its modulation index is always less than 1.
  3. Its BW is greater than that of AM.
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Its modulation index is always less than 1.

Angle Modulation Question 14 Detailed Solution

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  • The FM sidebands are dependent on both the level of frequency deviation and the frequency of the modulation.
  • The total FM spectrum consists of the carrier, plus an infinite number of sidebands spreading out on either side of the carrier at integral multiples of the modulating frequency as shown:
  •          

    • The parameters for the FM sidebands are determined by a formula using Bessel functions of the first kind.
    • In FM, modulation index for wide Band FM is greater than 1 and for Narrow Band FM, it is less than 1. (option 2 is incorrect)
    • B.W. required in AM is = 2fm. Hence, less bandwidth is required in case of AM and the bandwidth required in case of FM is 2(β + 1)fm. Therefore, the bandwidth required is more in FM as compared to AM.

Automatic gain control is used

  1. To maintain the tuning correct
  2. To reduce the volume of loud passages of music
  3. To increase the amplification at high frequencies
  4. To maintain the same volume of the output when stations of different strength are received

Answer (Detailed Solution Below)

Option 4 : To maintain the same volume of the output when stations of different strength are received

Angle Modulation Question 15 Detailed Solution

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AGC (Automatic Gain Control):

  •  Automatic gain control (AGC) works in FM radio transmitter/receiver that maintains Automatic controlling of weak and strong signals which is received by the radio receiver.
  • The automatic frequency control voltage of the FM transmitter VCO is DC voltage.
  • AGC maintains a constant level of the output signal based on the received signal nature, i.e. it maintains the same volume of the output when stations of different strengths are received.
  • AGC adjusts the gain of RF and IF amplifiers according to need.
  • AGC can handle problems like overloading and fading in the receiver.

 

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