Amplitude Modulation MCQ || Amplitude Modulation Questions and Answers

Answer.4. 8%

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

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 %

Answer.3. Amplitude

Explanation

Amplitude Modulation (AM) is preferred for picture transmission on 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.

Answer.2. Distortion

Explanation

Signal distortion results when the rate of change of phase shift with frequency over the necessary bandwidth of the signal is not constant. When the amplitude of the modulating signal is greater than the amplitude of the carrier, distortion will occur.

The modulation index is equal to

µ = A_m/A_c

Where

Ac = amplitude of the carrier wave

Am = amplitude of the modulating signal,

In practice, μ is kept ≤ q1 to avoid distortion.

If A_m > A_c

Then μ > 1

Answer.2. 2 kHz

Explanation

The frequency spectrum of an Amplitude Modulated wave is given by:

Bandwidth in AM = (f_c + f_m) – (fc – fm)

Bandwidth in AM = 2fm

Calculation:

fc = 10 kHz

Upper sideband = fc + fm = 11 kHz

∴ The message frequency is:

fm = 11 – 10 = 1 kHz

Hence, Bandwidth is:

= 2f_m = 2 × 1 kHz = 2 kHz

Answer.3. 1%

Explanation

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:

Given μ = 20% = 0.2

P_c = Am²/2

Total Sideband power is:

P_s1 = P_cµ²/4

Assume A_m = 1

Ps₁ = 0.2²/4

P _sb  = 1%

Answer.2. Error information

Explanation

• Amplitude distortion is the unintended modification of a signal, typically while undergoing amplification.
• Intermodulation distortion is a result of nonlinearities in the system such that one frequency component tends to modulate another frequency component—e.g., a high audio frequency modulating a low audio frequency.
• Overmodulation results in spurious emissions by the modulated carrier, and distortion of the recovered modulating signal. This means that the envelope of the output waveform is distorted.
• Distortion occurs when the modulating signal amplitude is greater than the amplitude of the carrier, causing incorrect information to be transmitted. In amplitude modulation, it is particularly important that the peak value of the modulating signal be less than the peak value of the carrier.

Answer.2. 5.56 kW

Explanation

Power relation in VSB wave is given as

${P_{VSB}} = \frac{{{K^2}}}{4}{P_c} + F\:\left( {\frac{{{K^2}}}{4}{P_c}} \right)$

Where

F = represents a fraction

K = modulation index

Given that:

P_VSB = 0.625, K = 60% = 0.6, F = 25% = 0.25

P_C = ?

Then:

${P_{VSB}} = \frac{{{K^2}}}{4}{P_c} + F\;\left( {\frac{{{K^2}}}{4}{P_c}} \right)$

${P_{VSB}} = \frac{{{K^2}}}{4}{P_c}\left( {1 + F} \right)$

$0.625 = \frac{{{{\left( {0.6} \right)}^2}}}{4}{P_c}\left( {1 + 0.25} \right)$

${P_c} = \frac{{0.625 \times 4}}{{1.25 \times 0.36}} \simeq 5.56\;kW$

Answer.1. Modulator

Explanation

• A modulator is an electronic circuit that superimposes a low-frequency (information) signal onto a high-frequency (carrier) signal for the purpose of wireless transmission.
• The circuit used for producing AM is called a modulator. It has two inputs, the carrier, and the modulating signal, and the resulting output is the modulated signal. Amplitude modulators compute the product of the carrier and modulate signals.
• AM generation involves the mixing of a carrier and an information signal. In low-level modulation, the message signal and carrier signal are modulated at low power levels and then amplified. The advantage of this technique is that a small audio amplifier is sufficient to amplify the message signal.

Answer.2. 535 kHz to 1605 kHz

Explanation

• The lowest frequency portion comes in radio waves generally, have wavelengths range between 1 mm to 100 km or frequencies between 300 GHz to 3 kHz.
• There are some subcategories in between these waves like AM and FM radio.
• Amplitude Modulation (AM): Range 535 to 1605 kHz in this type of wave Amplitude varies and frequency is constant.
• Frequency Modulation (FM): Range 88 to 108 MHz in this kind of wave Frequency varies while the amplitude is constant.

Answer.2. 33.33%

Explanation

The transmission efficiency of an AM wave is defined as the percentage of total power contributed by the sidebands.

For a sinusoidal AM signal, it is given by:

$\eta = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}} \times 100$

μ = Modulation index

The maximum efficiency is obtained for μ = 1, i.e.

${\eta _{max}} = \frac{1}{{2 + 1}} \times 100$

η_max = 33.33 %