Angle Modulation MCQ Quiz - Objective Question with Answer for Angle Modulation - Download Free PDF
Last updated on May 30, 2025
Latest Angle Modulation MCQ Objective Questions
Angle Modulation Question 1:
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.
Answer (Detailed Solution Below)
Angle Modulation Question 1 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 2:
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.
Answer (Detailed Solution Below)
Angle Modulation Question 2 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 3:
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?
Answer (Detailed Solution Below)
Angle Modulation Question 3 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.
Angle Modulation Question 4:
Narrowband FM is used in which of the following applications?
Answer (Detailed Solution Below)
Angle Modulation Question 4 Detailed Solution
The correct answer is: 4) Two-way mobile radio communication
Explanation:
Narrowband FM (NBFM) is characterized by:
Small frequency deviation (typically ±5 kHz or less).
Limited bandwidth (just wide enough for voice communication).
Primary Applications:
- Two-way mobile radios (e.g., police, ambulance, taxi communications).
- Aviation and marine VHF radios.
- Amateur radio (ham) transmissions.
Why Not the Other Options?
- Television transmission → Uses wideband FM for audio (higher fidelity) or AM for video.
- High-fidelity audio systems → Requires wideband FM (e.g., ±75 kHz deviation for FM radio).
- FM radio broadcasting → Uses wideband FM (88–108 MHz band with large deviation).
Angle Modulation Question 5:
In a PLL-based FM demodulator, the output of the phase detector is proportional to _______.
Answer (Detailed Solution Below)
Angle Modulation Question 5 Detailed Solution
In a Phase-Locked Loop (PLL)-based FM demodulator, the phase detector is a key component that:
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Compares the input FM signal with the output of a Voltage-Controlled Oscillator (VCO).
-
Produces an output voltage proportional to the phase difference between the two signals.
How it Works in FM Demodulation:
-
FM signal → fed into the phase detector.
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Phase detector compares it with VCO output.
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The output of the phase detector reflects the instantaneous phase difference.
-
This varying voltage corresponds to the modulating signal (since frequency variations in FM lead to phase changes).
-
After low-pass filtering, it gives the original message signal.
Incorrect Options:
Option | Why It's Incorrect |
---|---|
1) Carrier frequency | The VCO tracks the input frequency; the detector works on phase, not frequency directly. |
2) Amplitude | PLL FM demodulation is amplitude-insensitive. |
3) Frequency deviation | While the output of the entire PLL relates to frequency deviation, the phase detector output is directly proportional to phase difference. |
Top Angle Modulation MCQ Objective Questions
The modulating frequency in frequency modulation is increased from 10 kHz to 20 kHz. The bandwidth is -
Answer (Detailed Solution Below)
Angle Modulation Question 6 Detailed Solution
Download Solution PDFFrequency 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 = \(2(\Delta f+f_m)\)
Calculation:
Given, (fm)1 = 10 kHz
(BW)1 = \(2\Delta f+20\)
(fm)2 = 20 kHz
(BW)2 = \(2(\Delta f+20)\)
(BW)2 = \(2\Delta f+40\)
(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
Answer (Detailed Solution Below)
Angle Modulation Question 7 Detailed Solution
Download Solution PDFConcept:
In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency.
Mathematically, this is defined as:
\(m_f=\frac{Δ f}{f_m}\)
mf = Modulation index
Δf = Frequency deviation
fm = Modulating frequency
Calculation:
Given Δf = 50 kHz
fm = 100 Hz
\(m_f=\frac{50~kHz}{100~Hz}\)
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 _______________.
Answer (Detailed Solution Below)
Angle Modulation Question 8 Detailed Solution
Download Solution PDFThe 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.
- They are Normal distribution curves.
Which is true in a frequency modulated system ?
Answer (Detailed Solution Below)
Angle Modulation Question 9 Detailed Solution
Download Solution PDFFrequency 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).
Answer (Detailed Solution Below)
Angle Modulation Question 10 Detailed Solution
Download Solution PDFAnalysis:
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:
\(\phi(t)=A_ccos[ω_ct+g(k_k,m(t))]\)
m(t) = the modulating signal
ωc = Carrier frequency
kk becomes kc for FM and kp for PM.
The average power (Pavg) is given by:
\(P_{avg}=\frac{A_c^2}{2}\) (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.
Answer (Detailed Solution Below)
Angle Modulation Question 11 Detailed Solution
Download Solution PDFConcept:
In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency.
Mathematically, this is defined as:
\(m_f=\frac{Δ f}{f_m}\)
mf = Modulation index
Δf = Frequency deviation
fm = Modulating frequency
Calculation:
Given Δf = 1000 Hz = 1 kHz
fm = 1 kHz
\(m_f=\frac{1~kHz}{1~kHz}=1\)
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?
Answer (Detailed Solution Below)
Angle Modulation Question 12 Detailed Solution
Download Solution PDFConcept:
WBFM is given as:
\(WBFM\;\left( t \right) = {A_c}\;\mathop \sum \limits_{n = - \infty }^\infty {J_n}\left( \beta \right)\cos 2\pi \left( {{f_c} + n{f_m}} \right)t\;\;;\;\;\beta > 1\)
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?
Answer (Detailed Solution Below)
Angle Modulation Question 13 Detailed Solution
Download Solution PDFFrequency 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 -
Answer (Detailed Solution Below)
Angle Modulation Question 14 Detailed Solution
Download Solution PDF- 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
Answer (Detailed Solution Below)
Angle Modulation Question 15 Detailed Solution
Download Solution PDFAGC (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.