Amplitude Modulation MCQ || Amplitude Modulation Questions and Answers

Answer.1. AM for picture and FM for sound are employed

Explanation

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 sidebands.
• This is only 1.4 % of the total channel bandwidth of 7 MHz. This results in efficient utilization of the channel.

Answer.2. 0.02 msec

Explanation

For envelope detector, the time constant must satisfy the following relation:

$RC \le \frac{1}{{{\omega _m}}}\left[ {\frac{{\sqrt {1 – {\mu ^2}} }}{\mu }\:} \right]$ —– (1)

For the good performance of the envelope detector:

$\frac{1}{{{\omega _c}}}$

Given:

f_m = 10 kHz

modulation index (μ) = 0.6

From equation 1

$RC \le \frac{1}{{2\pi \times 10}}\left[ {\frac{{\sqrt {1 – {{0.6}^2}} }}{{0.6}}\:} \right]$

= 0.02 msec

Answer.2. Instantaneous value

Explanation

In Amplitude Modulation, the amplitude of the carrier sine wave is varied by the value of the information signal.

The instantaneous value of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal.

The carrier frequency remains constant during the modulation process, But its amplitude varies in accordance with the modulating signal.

Answer.4. 0.045 watts

Explanation

Concept:

Power of a transmitted AM wave is given as:

$P_t = {P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)$

$P_t = {P_c} + P_c\frac{μ ^2}{2}$

Power in the carrier = Pc

Power in both the sidebands is given by:

P_s = P_cµ²/2

Since the power is distributed equally to the left and to the right side of the sideband, the power in one of the sidebands is given by:

P_s1 = P_cµ²/4

Analysis:

We know that:

${P_t} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)$

$2.09 = {P_c}\left( {1 + \frac{{0.09}}{2}} \right)$

$\therefore {P_c} = 2\;Watts$

Power in each sideband is:

${P_{SB}} = {P_c}~\frac{{{\mu ^2}}}{4} = \frac{{2\left( {0.09} \right)}}{4}$

= 0.045 Watts

Answer.2. M_t = 0.93 

Explanation

When a carrier is modulated by different waves having different modulation indexes, the effective (total) modulation index is given by:

${{\rm{μ }}_{{\rm{eff}}}} = \sqrt {{\rm{μ }}_1^2 + {\rm{μ }}_2^2+μ_3^2+…}$

Calculation:

With μ₁ = 0.25, μ2 = 0.50, μ3 = 0.75, the effective modulation index will be:

${{\rm{μ }}_{{\rm{eff}}}} = \sqrt {(0.25)^2 + (0.50)^2+(0.75)^2}$

μ_eff = 0.935

Answer.4. Envelope

Explanation

• The envelopes of a signal are the boundary within which the signal is contained. Envelopes contain some information of signals, though it is an imaginary curve, for example, demodulating amplitude modulated (AM) signals by them.
• The envelope is an imaginary line connecting the positive peaks and negative peaks of the carrier waveform give the exact shape of the modulating information signal.

Answer.1. $\frac{9}{2}m_f^2$

Explanation

Concept:

Signal to noise ratio in case of WBFM is given by

${\left( {\frac{S}{N}} \right)_{WBFM}} = \frac{3}{2}m_f^2$

Signal to noise ratio in case of AM is given by

${\left( {\frac{S}{N}} \right)_{AM}} = \frac{{{\mu ^2}}}{{{\mu^2} + 2}}$

Calculation:

Given:

100% amplitude modulation i.e.

μ = 1

${\left( {\frac{S}{N}} \right)_{WBFM}} = \frac{3}{2}m_f^2$

${\left( {\frac{S}{N}} \right)_{AM}} = \frac{{{1^2}}}{{{1^2} + 2}} = \frac{1}{3}$

$\frac{{{{\left( {\frac{S}{N}} \right)}_{WBFM}}}}{{{{\left( {\frac{S}{N}} \right)}_{AM}}}} = \frac{{\frac{3}{2}m_f^2}}{{\frac{1}{3}}} = \frac{9}{2}\;m_f^2$

Answer.2. 66 %

Explanation

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

&

Power Saved=\frac{P_c}{P_{total}}$ —(1)

Power Saved = Pc in DSB – SC

Power Saved =  2/(2 + µ²)

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 %

Answer.2. Carrier peak line

Explanation

The modulating signal uses the peak value of the carrier rather than zero as its reference point. The envelope varies above and below the peak carrier amplitude. The zero reference line of the modulating signal coincides with the peak value of the unmodulated carrier.

The formula of the modulation index is given by

m = PM/PC

Where

m is the modulation index

PM is the peak value of the modulating signal

PC is the peak value of the carrier signal.

Answer.3. 9 V

Explanation

The formula of the modulation index is given by

m = PM/PC

Where

m is the modulation index

PM is the peak value of the modulating signal

PC is the peak value of the carrier signal.

Given

Peak voltage of the carrier wave, PC =1 5V

Modulation index, m = 60%

PM = 60% × 15

PM = 9 V