# Amplitude Modulation MCQ || Amplitude Modulation Questions and Answers

1. In TV transmission which modulation is employed for picture and sound.

1. AM for picture and FM for sound are employed
2. FM for picture and AM for sound are employed
3. FM is employed for both picture and sound
4. AM is employed for both picture and sound

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.

2. In an amplitude modulated signal, modulating frequency is 10 kHz and the modulation index is 0.6. What should be the best-suited RC time constant for the envelope detector?

1. 0.52 msec
2. 0.02 msec
3. 0.20 msec
4. 0.35 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:

fm = 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

3. In Amplitude Modulation, the __________ of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal.

1. Real Value
2. Instantaneous value
3. Both Real and Instantaneous value
4. None of the above

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.

4. A transmitter puts out a total power of 2.09 Watts of 30% AM signal. How much power is contained in each of the sidebands?

1. 2.09 watts
2. 2 Watts
3. 0.09 watts
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:

Ps = Pcµ2/2

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:

Ps1 = Pcµ2/4

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

5. What will be the total modulation index if a wave is an amplitude modulated by three sine waves with modulation indices of 25%, 50%, and 75%?

1. Mt = 1.5
2. Mt = 0.93
3. Mt = 1.22
4. Mt = 1

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:

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

μeff = 0.935

6. What is the line connecting the positive and negative peaks of the carrier waveform called?

1.  Peak line
2. Maximum amplitude ceiling
3. Modulation index
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.

7. If mf is modulation index of FM, What is the value of ratio (S/N)WBFM/(S/N)AM for 100% amplitude modulation with identical total transmitted power?

1. $\frac{9}{2}m_f^2$
2. $\frac{3}{2}m_f^2$
3. $\frac{3}{2}m_f^3$
4. $\frac{9}{2}m_f^3$

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

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

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

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

8. In a DSB-SC system with 100% modulation, the power saving is

1. 50 %
2. 66 %
3. 75 %
4. 100 %

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_cPower 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 %

9. What is the reference line for the modulating signal?

1. Zero line
2. Carrier peak line
3. Modulated peak line
4. Un-modulated 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.

10. A carrier wave of peak voltage 15V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.

1. 19 V
2. 18 V
3. 9 V
4. 3 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

Scroll to Top