Amplitude Modulation MCQ Quiz - Objective Question with Answer for Amplitude Modulation - Download Free PDF
Last updated on Jul 9, 2025
Latest Amplitude Modulation MCQ Objective Questions
Amplitude Modulation Question 1:
A carrier is amplitude modulated with modulation index of 60% yielding transmitted power of 472 W. How much power can be saved by suppressing carrier and one sideband of the modulated signal?
Answer (Detailed Solution Below)
Amplitude Modulation Question 1 Detailed Solution
Amplitude Modulation and Power Calculation
Problem Statement: A carrier signal is amplitude modulated with a modulation index of 60%, resulting in a transmitted power of 472 W. We are tasked with determining how much power can be saved by suppressing the carrier and one sideband of the modulated signal.
Solution:
To solve this problem, let us analyze the power distribution in an amplitude-modulated signal.
Step 1: Power Components in an Amplitude-Modulated Signal
In amplitude modulation (AM), the total transmitted power (Pt) consists of:
- Carrier power (Pc).
- Power in the two sidebands (PUSB and PLSB), which are equal in magnitude.
The total transmitted power is given by:
Pt = Pc + PUSB + PLSB
The power in each sideband is determined by the modulation index (m), and is given as:
PUSB = PLSB = (m2/4) × Pc
Thus, the total transmitted power can be expressed as:
Pt = Pc (1 + m2/2)
Step 2: Calculate the Carrier Power (Pc)
From the problem, the modulation index (m) is 60%, or m = 0.6. The total transmitted power (Pt) is 472 W. Substituting these values into the formula for total transmitted power:
472 = Pc (1 + (0.6)2/2)
First, calculate (0.6)2/2:
(0.6)2/2 = 0.36/2 = 0.18
Now substitute this value into the equation:
472 = Pc (1 + 0.18)
472 = Pc × 1.18
Pc = 472 / 1.18
Pc ≈ 400 W
Thus, the carrier power is 400 W.
Step 3: Calculate the Sideband Power
The power in the sidebands is given by:
Psidebands = PUSB + PLSB = (m2/2) × Pc
Substitute m = 0.6 and Pc = 400 W:
Psidebands = (0.6)2/2 × 400
Psidebands = 0.36/2 × 400
Psidebands = 0.18 × 400
Psidebands = 72 W
The total power in the sidebands is 72 W, with each sideband contributing 36 W.
Step 4: Power Saved by Suppressing the Carrier and One Sideband
When the carrier and one sideband are suppressed, the remaining power is the power in the other sideband. Thus, the power saved is:
Power saved = Pt - Premaining
The remaining power is the power in one sideband:
Premaining = PLSB or PUSB = 36 W
Substitute the values:
Power saved = 472 - 36
Power saved = 436 W
Final Answer: The power saved by suppressing the carrier and one sideband is 436 W.
Amplitude Modulation Question 2:
A DSB-SC signal is being detected synchronously. The phase error in the locally generated carrier will _________.
Answer (Detailed Solution Below)
Amplitude Modulation Question 2 Detailed Solution
Explanation:
5Double Sideband Suppressed Carrier (DSB-SC) is a type of amplitude modulation where the carrier is suppressed, and only the sidebands carry the information. In synchronous detection, a locally generated carrier is used at the receiver to demodulate the signal. The phase and frequency of this locally generated carrier must align precisely with the transmitted carrier for accurate demodulation.
Phase Error in Synchronous Detection: A phase error occurs when there is a phase difference between the locally generated carrier at the receiver and the carrier used during transmission. This phase mismatch can significantly affect the demodulated output.
Impact of Phase Error: In the case of a DSB-SC signal, the demodulated output is given by the product of the transmitted signal and the locally generated carrier. When there is a phase error, the output signal gets distorted. Mathematically, the transmitted DSB-SC signal can be expressed as:
Transmitted Signal: s(t) = A × cos(ωct) × m(t)
Here:
- A = Amplitude of the signal
- ωc = Angular frequency of the carrier
- m(t) = Modulating signal
The locally generated carrier at the receiver can be expressed as:
Local Carrier: cos(ωct + θ)
Where θ is the phase error. The demodulated output is obtained by multiplying the transmitted signal with the local carrier:
Demodulated Output:
y(t) = s(t) × cos(ωct + θ)
Substituting the expression for s(t):
y(t) = [A × cos(ωct) × m(t)] × cos(ωct + θ)
Using the trigonometric identity:
cos(x) × cos(y) = 0.5 × [cos(x-y) + cos(x+y)]
y(t) = 0.5 × A × m(t) × [cos(θ) + cos(2ωct + θ)]
The term cos(2ωct + θ) represents a high-frequency component that is filtered out during demodulation, leaving:
y(t) = 0.5 × A × m(t) × cos(θ)
This shows that the amplitude of the demodulated signal is scaled by cos(θ). If there is a phase error θ, the output amplitude is reduced by the factor cos(θ). Additionally, the phase error introduces phase distortion in the demodulated signal.
Correct Option Analysis:
The correct option is:
Option 1: Cause phase distortion and reduce the output also.
This option is correct because a phase error in the locally generated carrier introduces distortion in the output and reduces its amplitude by the factor cos(θ). The demodulated signal's quality depends on the alignment of the local carrier with the transmitted carrier, and any phase mismatch degrades the output.
Amplitude Modulation Question 3:
In Vestigial Sideband (VSB) demodulation, which of the following is used to recover the original modulating signal?
Answer (Detailed Solution Below)
Amplitude Modulation Question 3 Detailed Solution
Explanation:
Vestigial Sideband (VSB) Demodulation
Definition: Vestigial Sideband (VSB) modulation is a type of amplitude modulation (AM) where one sideband is partially transmitted along with the carrier signal. This technique is widely used in television broadcasting because it reduces bandwidth while retaining the necessary information for signal recovery. The demodulation process involves recovering the original modulating signal from the modulated signal.
Correct Option:
The correct answer is Option 3: Product modulator followed by a low-pass filter.
Working Principle:
In VSB demodulation, the original modulating signal is recovered using a product modulator followed by a low-pass filter. The steps involved are:
- Product Modulator: The received VSB signal is multiplied with a locally generated carrier signal. This process effectively demodulates the signal by shifting it back to its original frequency range.
- Low-Pass Filter: After the multiplication process, the resultant signal contains the original baseband signal along with higher-frequency components. A low-pass filter is used to remove these unwanted high-frequency components, leaving only the desired baseband signal.
This method ensures that the original modulating signal is accurately recovered, making it the most effective approach for VSB demodulation.
Advantages:
- Efficient recovery of the original signal.
- Reduced bandwidth compared to conventional AM systems.
- Widely used in applications such as television broadcasting due to its simplicity and effectiveness.
Important Information:
To further understand the analysis, let’s evaluate the other options:
Option 1: Phase detector followed by a notch filter
This option is incorrect because a phase detector is typically used in phase demodulation systems, not VSB demodulation. A notch filter is used to suppress specific frequency components but does not play a role in recovering the original modulating signal in VSB demodulation.
Option 2: Frequency discriminator followed by a band-pass filter
This option is incorrect because a frequency discriminator is primarily used in frequency modulation (FM) demodulation, not amplitude modulation (AM) or VSB modulation. A band-pass filter is used to isolate specific frequency ranges but does not contribute to the recovery of the original modulating signal in VSB systems.
Option 4: Signal detector followed by a high-pass filter
This option is incorrect as well. A signal detector is used in envelope detection, which is suitable for basic AM demodulation but not for VSB demodulation. A high-pass filter removes low-frequency components, which would result in loss of the baseband signal in VSB systems.
Conclusion:
In VSB demodulation, the use of a product modulator followed by a low-pass filter is the most effective method for recovering the original modulating signal. The other options involve techniques or components that are either unrelated to VSB demodulation or unsuitable for accurately recovering the baseband signal. Understanding the principles of modulation and demodulation is crucial for selecting the correct method in various communication systems.
Amplitude Modulation Question 4:
When a carrier is AM modulated by three single tone signals of modulation percentages 50%, 50% and 20%, then the effective modulation index of the resulting signal is:
Answer (Detailed Solution Below)
Amplitude Modulation Question 4 Detailed Solution
Concept:
When a carrier is amplitude modulated by multiple independent modulating signals, the effective modulation index is calculated using the square root of the sum of the squares of individual modulation indices.
Effective modulation index, \( m_{eff} = \sqrt{m_1^2 + m_2^2 + m_3^2} \)
Given:
Modulation percentages are 50%, 50%, and 20%.
So, \( m_1 = 0.5, \; m_2 = 0.5, \; m_3 = 0.2 \)
Calculation:
\( m_{eff} = \sqrt{(0.5)^2 + (0.5)^2 + (0.2)^2} = \sqrt{0.25 + 0.25 + 0.04} = \sqrt{0.54} \)
Amplitude Modulation Question 5:
The purpose of carrier modulation is to:
Answer (Detailed Solution Below)
Amplitude Modulation Question 5 Detailed Solution
Explanation:
The Purpose of Carrier Modulation:
Definition: Carrier modulation is the process of varying a high-frequency signal (called the carrier) in accordance with a lower frequency message signal (information signal). This technique is widely used in communication systems to transmit information over long distances effectively.
Correct Option Analysis:
The correct option is:
Option 2: Shift the message to higher frequency band for better radiation.
Carrier modulation plays a crucial role in communication systems by shifting the message signal to a higher frequency band. This is done for several essential reasons:
- Efficient Radiation: Low-frequency signals are not efficiently radiated by antennas due to their long wavelengths. Antennas need to be comparable in size to the wavelength of the signal for effective radiation, and low-frequency signals would require impractically large antennas. By modulating the signal to a higher frequency band, antennas of reasonable size can be used for efficient radiation.
- Reduced Interference: Shifting the message signal to a higher frequency band allows multiple signals to coexist in different frequency ranges without overlapping. This reduces interference and enables the simultaneous transmission of multiple signals through multiplexing techniques.
- Improved Signal Quality: High-frequency signals are less prone to attenuation and distortion during transmission, ensuring better signal quality over long distances.
- Ease of Signal Processing: High-frequency signals are easier to process and amplify using available electronic components, making the design of communication systems more practical and cost-effective.
Detailed Working:
In carrier modulation, the message signal modulates certain characteristics of the carrier signal (such as amplitude, frequency, or phase). Depending on the method of modulation used, the message signal is encoded into the carrier signal, which is then transmitted through the medium (e.g., air, cable). At the receiver end, the original message signal is extracted from the modulated carrier signal using demodulation techniques.
Types of Modulation:
- Amplitude Modulation (AM): The amplitude of the carrier signal is varied in proportion to the message signal.
- Frequency Modulation (FM): The frequency of the carrier signal is varied according to the message signal.
- Phase Modulation (PM): The phase of the carrier signal is varied based on the message signal.
Applications:
- Radio and television broadcasting.
- Satellite communication.
- Mobile telephony and wireless communication systems.
- Data transmission over networks.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Reduce the amplitude of the message for better radiation.
This option is incorrect. Reducing the amplitude of the message signal does not improve radiation. In fact, reducing the amplitude could lead to weaker transmission and poorer signal reception. The primary purpose of carrier modulation is not to alter the amplitude of the message signal for radiation but rather to shift it to a higher frequency band, which facilitates efficient radiation.
Option 3: Result in reduced performance in noise in some of the systems.
This option is misleading. Carrier modulation often improves the system's performance against noise, especially in techniques like frequency modulation (FM), which is known for its noise immunity. While certain modulation methods may be more susceptible to noise, the primary purpose of modulation is not related to reducing performance in noise but rather ensuring efficient transmission and reception of signals.
Option 4: Shift the message to lower frequency band for better radiation.
This option is incorrect. Shifting the message to a lower frequency band would result in poorer radiation efficiency due to the requirement for large antennas and increased susceptibility to attenuation and distortion. Carrier modulation is specifically used to shift the message signal to a higher frequency band to overcome these issues.
Conclusion:
Carrier modulation is a fundamental technique in communication systems, enabling efficient transmission and reception of signals over long distances. By shifting the message signal to a higher frequency band, it facilitates better radiation, reduces interference, and improves signal quality. Understanding the purpose and benefits of carrier modulation is essential for designing and analyzing modern communication systems.
Top Amplitude Modulation MCQ Objective Questions
Consider sinusoidal modulation in an AM system. Assuming no over-modulation, the modulation index (𝜇) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V is ________.
Answer (Detailed Solution Below)
Amplitude Modulation Question 6 Detailed Solution
Download Solution PDFConcept:
A single tone modulated signal is given as:
Maximum envelope Amax : Ac (1+μ)
Minimum envelope Amin : Ac (1-μ)
\(\frac{{{A_{max}}}}{{{A_{min}}}} = \frac{{1 + μ}}{{1 - μ}}\)
The above can be written as:
\(μ = \frac{{{A_{max}} - {A_{min}}}}{{{A_{max}} + {A_{min}}}}\)
Calculation:
Given: Amax = 3 V and Amin = 1 V
The modulation index will be:
\(μ = \frac{{{3} - 1}}{{3 + 1}}\)
μ = 0.5
A signal with frequency fm modulates a carrier fc (where fc ≫ fm). Then the output of the AM - DSB-SC signal will contain frequencies:
Answer (Detailed Solution Below)
Amplitude Modulation Question 7 Detailed Solution
Download Solution PDFThe general expression of a DSB-SC signal is given as:
sDSB-SC(T) = x(t) cosωct
For a single-tone message signal Am cosωmt, the DSB-SC signal will be:
sDSB-SC(T) = Amcosωct cosωmt
ωm = Frequency of the message signal
ωc = Carrier signal frequency
The spectrum as represented:
We observe the carrier is suppressed in DSB-SC modulation.
In television transmission, the type of modulation used for video is
Answer (Detailed Solution Below)
Amplitude Modulation Question 8 Detailed Solution
Download Solution PDFIn TV Transmission the use of FM is made for Audio transmission and AM for Video transmission.
Vestigial Sideband Modulation (VSB) is used for video modulation in TV transmission due to the following reasons :
- Video signal exhibits a large bandwidth and significant low-frequency content which suggests the use of VSB.
- VSB (vestigial sideband) transmission transmits one sideband fully and the other sideband partially thus, reducing the bandwidth requirement.
- The circuitry for demodulation in the receiver should be simple and therefore cheap. VSB demodulation uses simple envelope detection.
The spectrum of a vestigial sideband is as shown:
Modulation system used for video modulation in TV transmission is
Answer (Detailed Solution Below)
Amplitude Modulation Question 9 Detailed Solution
Download Solution PDFIn TV Transmission the use of FM is made for Audio transmission and AM for Video transmission.
Vestigial Sideband modulation (VSB) is used for video modulation in TV transmission due to the following reasons :
- Video signal exhibits a large bandwidth and significant low-frequency content which suggests the use of VSB.
- VSB (vestigial side band) transmission transmits one side band fully and the other side band partially thus, reducing the bandwidth requirement.
- The circuitry for demodulation in the receiver should be simple and therefore cheap. VSB demodulation uses a simple envelope detection.
A carrier is modulated to a depth of 40%. The percentage increase in the transmitted power is
Answer (Detailed Solution Below)
Amplitude Modulation Question 10 Detailed Solution
Download Solution PDFConcept:
The total transmitted power for an AM system is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)
Pc = Carrier Power
μ = Modulation Index
Analysis:
When the carrier is not modulated, i.e. modulation index = 0, the transmitted power is the carrier power only, i.e.
Pt = PC
When modulated with a modulation index of 40%, the total power is calculated as:
\({P_t} = {P_c}\left( {1 + \frac{{{0.4^2}}}{2}} \right)\)
The percentage increase in power will be:
\(=\frac{{P_c}\left( {1 + \frac{{{0.4^2}}}{2}} \right)-P_c}{P_c}\times 100\)
\(=\frac{0.4^2}{2}\times 100~\%\)
= 8 %
Which modulator is used for the generation of the DSB-SC signal?
Answer (Detailed Solution Below)
Amplitude Modulation Question 11 Detailed Solution
Download Solution PDFIn a balanced modulator, 2 AM-modulators are connected in a way that the resultant signal does not contain the carrier spectrum, i.e. to generate a Double Side-band suppressed carrier (DSB-SC).
The circuit diagram of a balanced modulator is as shown:
\( {s_{AM'}}\left( t \right) = {A_c}\left[ {1 + {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)
\(\\ {s_{AM''}}\left( t \right) = {A_c}\left[ {1 - {k_a}m\left( t \right)} \right]\cos 2\pi {f_c}t\)
The resultant DSB signal is the difference between the two, i.e.
\({s_{DSB}}\left( t \right) = {s_{AM'}}\left( t \right) - {s_{AM''}}\left( t \right) \)
Hence,
\({s_{DSB}}\left( t \right) = \;2{A_c}{k_a}m\left( t \right)\cos 2\pi {f_c}t\)
\( {s_{DSB}}\left( t \right) = \;{A_c}'m\left( t \right)\cos 2\pi {f_c}t\)
1) Square law modulator is used for the generation of conventional AM signal which includes the carrier frequency component as well.
2) An Armstrong modulator is used to generate an FM signal.
3) Envelop detector is used for AM demodulation.
In TV transmission
Answer (Detailed Solution Below)
Amplitude Modulation Question 12 Detailed Solution
Download Solution PDFAmplitude Modulation (AM) is preferred for picture transmission in TV because of the following reasons:
- The distortion which arises due to interference between multiple signals is more in FM than AM because the frequency of the FM signal continuously changes.
- Steady production of the picture is affected because of this.
- If AM were used, the ghost image, if produced is steady.
- Also, the circuit complexity and bandwidth requirements are much less in AM than in FM.
On the other hand, FM is preferred for sound because of the following reasons:
- The bandwidth assigned to the FM sound signal is about 200 kHz, of which not more than 100 kHz is occupied by significant side bands.
- This is only 1.4 % of the total channel bandwidth of 7 MHz. This results in efficient utilization of the channel.
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 total power carried by sidebands of the AM wave (DSB) for tone modulation of μ = 0.4 ?
Answer (Detailed Solution Below)
Amplitude Modulation Question 13 Detailed Solution
Download Solution PDFConcept:
The generalized AM expression is represented as:
s(t) = Ac [1 + μa mn (t)] cos ωc t
The total transmitted power for an AM system is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)
Pc = Carrier Power
μ = Modulation Index
The above expression can be expanded to get:
\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)
The total power is the sum of the carrier power and the sideband power, i.e.
\({P_s} = P_c\frac{{{μ^2}}}{2}\)
Analysis:
Total sideband power is calculated as:
\({P_s} = P_c\frac{{{μ^2}}}{2}\)
% of sideband power is given as:
⇒ \( \frac{P_c \frac{μ^2}{2}}{{P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)}\)
⇒ \( \frac{μ^2}{μ^2+2}\)
putting μ = 0.4, we get
⇒ \( \frac{0.4^2}{0.4^2\;+\;2} \times 100 = 0.074 \times 100\)
= 7.4 %
In a DSB-SC system with 100% modulation, the power saving is
Answer (Detailed Solution Below)
Amplitude Modulation Question 14 Detailed Solution
Download Solution PDFConcept:
The generalized AM expression is represented as:
s(t) = Ac [1 + μa mn (t)] cos ωc t
The total transmitted power for an AM system is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)
Pc = Carrier Power
μ = Modulation Index
The above expression can be expanded to get:
\({P_t} = {P_c} + P_c\frac{{{μ^2}}}{2}\)
The total power is the sum of the carrier power and the sideband power, i.e.
\({P_s} = P_c\frac{{{μ^2}}}{2}\)
The power in a single sideband will be:
\({P_s} = \frac{1}{2}\times P_c\frac{{{μ^2}}}{2}\)
With \(P_c=\frac{A^2}{2}\), the above can be written as:
\({P_s} = \frac{1}{2}\times \frac{A_c^2}{2}\frac{{{μ^2}}}{2}\)
\({P_s} = \frac{{{A_c^2μ^2}}}{8}\)
\(Power Saved=\frac{P_c}{P_{total}}\) ---(1)
Power Saved = Pc in DSB - SC
\(Power \ Saved=\frac{2}{2 \ + \ μ^2}\)
Analysis:
When μ = 1, the transmitted power will be:
\({P_t} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)=\frac{3}{2}P_c\)
\(Power Saved=\frac{P_c}{P_c(1 \ + \ \frac{μ^2}{2})}\)
As μ = 1
\(Power \ Saved=\frac{2}{2 \ + \ 1^2}\times 100\)
Power Saved = 66 %
Transmission bandwidth (B) of AM signal is -
Answer (Detailed Solution Below)
Amplitude Modulation Question 15 Detailed Solution
Download Solution PDFConcept:
AM: In amplitude modulation, the amplitude of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal.
The time-domain representation of an amplitude-modulated signal is given as:
s(t) = Ac [1 + ka m(t)] cos 2πfct
s(t) = Ac cos 2πfct + Ac ka m(t) cos 2πfct
Where the carrier Ac cos ωct is modulated in amplitude and ka is the amplitude sensitivity of the modulator.
The frequency-domain representation of an amplitude-modulated signal is given as:
\(S(f) = \frac{A_c}{2}[\delta (f-f_c)+\delta (f+f_c)]+\frac{A_c k_a}{2}[M(f-f_c)+M (f+f_c)]\)
∴ We can see that the bandwidth of an AM signal is twice that of the maximum frequency present in the message signal.
Features: |
AM |
FM |
Noise immunity |
In AM, the message is stored in the form of variation in amplitude. Noise affects the amplitude of signal most so AM is less noise immune. |
In FM, the message is stored in the form of variation in frequency so it has better noise immunity. |
Bandwidth |
B.W. required in AM is = 2fm. Hence, less bandwidth is required in case of AM. |
B.W. required in FM is = 2(β+1)fm. Hence, more bandwidth is required in case of FM. |
Transmitted power |
Power transmitted in AM is given by: \({P_T} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\) As the modulation index ‘μ’ increases power in AM increases. |
In FM, power transmitted is always equal to the total power of the carrier before modulation. Hence, FM requires less power than AM. |