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

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 (P_t) consists of:

• Carrier power (P_c).
• Power in the two sidebands (P_USB and P_LSB), which are equal in magnitude.

The total transmitted power is given by:

P_t = P_c + P_USB + P_LSB

The power in each sideband is determined by the modulation index (m), and is given as:

P_USB = P_LSB = (m²/4) × P_c

Thus, the total transmitted power can be expressed as:

P_t = P_c (1 + m²/2)

Step 2: Calculate the Carrier Power (P_c)

From the problem, the modulation index (m) is 60%, or m = 0.6. The total transmitted power (P_t) is 472 W. Substituting these values into the formula for total transmitted power:

472 = P_c (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 = P_c (1 + 0.18)

472 = P_c × 1.18

P_c = 472 / 1.18

P_c ≈ 400 W

Thus, the carrier power is 400 W.

Step 3: Calculate the Sideband Power

The power in the sidebands is given by:

P_sidebands = P_USB + P_LSB = (m²/2) × P_c

Substitute m = 0.6 and P_c = 400 W:

P_sidebands = (0.6)²/2 × 400

P_sidebands = 0.36/2 × 400

P_sidebands = 0.18 × 400

P_sidebands = 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 = P_t - P_remaining

The remaining power is the power in one sideband:

P_remaining = P_LSB or P_USB = 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.

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:

\( m_{eff} = \sqrt{(0.5)^2 + (0.5)^2 + (0.2)^2} = \sqrt{0.25 + 0.25 + 0.04} = \sqrt{0.54} \)

Concept:

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.

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

Calculation:

Given: Pc = 100 W and μ = 1

We can write:

\({P_s} = P_c\frac{{{μ^2}}}{2}\)

\(= 100 \times \frac{1^2}{2}\)

\(P_s=50~W\)

The total sideband power = 50 W

power in upper sideband + power in lower side band = 50 W

power in upper sideband = power in lower sideband = 25 W

The correct answer is: 3) Pulse Duration Modulation (PDM)

Explanation:

In communication engineering, Pulse Width Modulation (PWM) is also known as:

• Pulse Duration Modulation (PDM)

• Pulse Length Modulation (PLM)

Key Characteristics of PWM/PDM:

• Definition: The width (duration) of pulses varies based on the modulating signal, while the amplitude remains constant.

• Applications:

  • Motor speed control

  • LED dimming

  • Power delivery (DC-DC converters)

Concept:

VSB (vestigial sideband) transmission transmits one sideband fully and the other sideband partially thus, reducing the bandwidth requirement.

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:

Concept:

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

The general expression of a DSB-SC signal is given as:

sDSB-SC(T) = x(t) cosωct

For a single-tone message signal A_m cosω_mt, 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 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:

In 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.

Concept:

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.

P_t = P_C

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 %

In 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.

Amplitude 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.

Concept:

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 %

Concept:

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 %

Concept:

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.

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