Amplitude Modulation Questions and Answers

Q: Which best explains why AM survives in low-cost consumer equipment?

The whole case is detector simplicity: a diode, a capacitor and a resistor, with no local oscillator, no phase tracking and no discriminator. That is what put radios into every home in the twentieth century and what keeps AM in the cheapest equipment today. Its noise performance is the poorest of the analogue schemes. High-level modulated transmitters need linear modulators even if the radio chain is class C. And single sideband occupies less spectrum.

Q: An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 2 MHz and 1 kHz respectively. An appropriate value for the time constant of the envelope detector is :

An appropriate time constant is 100 µs.The RC time constant of an envelope detector must be long compared with the carrier period, so the output does not follow the individual carrier cycles, but short enough to follow the modulating envelope:\[ \begin{aligned} \frac{1}{f_{c}} &= \frac{1}{2\ \text{MHz}} = 0.5\ \mu s \\ \frac{1}{2\pi f_{m}} &= \frac{1}{2\pi \times 1\ \text{kHz}} \approx 159\ \mu s \end{aligned} \]100 µs lies between these limits. 200 µs exceeds \(1/(2\pi f_{m})\) and would cause diagonal clipping, 2000 µs is far too slow, and 0.050 µs is shorter than a carrier period, so the output would ripple at the carrier frequency.

Q: An AM has carrier frequency and modulating signal frequency 2 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector in demodulator is :

An appropriate time constant is 40 μs.The RC time constant of an envelope detector must be long compared with the carrier period, so that the capacitor does not discharge between carrier peaks, but short compared with the modulating period, so that it can follow the envelope:\[ \begin{aligned} \frac{1}{f_{c}} &= \frac{1}{2\ \text{MHz}} = 0.5\ \mu\text{s} \\ \frac{1}{f_{m}} &= \frac{1}{2\ \text{kHz}} = 500\ \mu\text{s} \end{aligned} \]40 μs lies comfortably between the two. 0.08 μs and 0.10 μs are shorter than a carrier period, so the output would ripple at the carrier frequency, and 120 μs is closer to the modulating period and risks diagonal clipping.

Q: When the modulating signal controls the amplitude of a carrier wave, it is known as :

It is amplitude modulation.In amplitude modulation the instantaneous amplitude of a high-frequency carrier is made to vary in step with the modulating (message) signal, while the carrier frequency stays fixed:\[ s(t) = A_{c}[1 + \mu m(t)]\cos\omega_{c}t \]In frequency modulation the message varies the carrier's frequency, and in phase modulation its phase, with the amplitude constant in both. Pulse amplitude modulation varies the heights of a train of pulses rather than the amplitude of a continuous carrier wave.

101.

Consider the following statements about the AM modulation index.

1. It sets the fraction of power in the sidebands

2. It sets the transmitted bandwidth

3. It can be measured from the envelope extremes

Which of the statements given above is or are correct?

A.1 and 3 only
B.1 and 2 only
C.2 and 3 only
D.1, 2 and 3

Answer: A

The first and third are correct. The second is the key contrast with angle modulation: in AM the bandwidth is twice the highest message frequency and is entirely independent of the index, whereas in frequency modulation the index is precisely what determines the bandwidth. Confusing the two is one of the commonest errors in the subject, and it leads to nonsense such as expecting a lightly modulated AM signal to occupy a narrower channel.

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102.

Which best explains why AM survives in low-cost consumer equipment?

A.It occupies the least spectrum per channel
B.A diode and two components recover the message
C.It offers the best signal-to-noise ratio available
D.Its transmitters need no linearity

Answer: B

The whole case is detector simplicity: a diode, a capacitor and a resistor, with no local oscillator, no phase tracking and no discriminator. That is what put radios into every home in the twentieth century and what keeps AM in the cheapest equipment today. Its noise performance is the poorest of the analogue schemes. High-level modulated transmitters need linear modulators even if the radio chain is class C. And single sideband occupies less spectrum.

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103.

An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 2 MHz and 1 kHz respectively. An appropriate value for the time constant of the envelope detector is :

A.2000 µs
B.0.050 µs
C.100 µs
D.200 µs

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

An appropriate time constant is 100 µs.

The RC time constant of an envelope detector must be long compared with the carrier period, so the output does not follow the individual carrier cycles, but short enough to follow the modulating envelope:

\[ \begin{aligned} \frac{1}{f_{c}} &= \frac{1}{2\ \text{MHz}} = 0.5\ \mu s \\ \frac{1}{2\pi f_{m}} &= \frac{1}{2\pi \times 1\ \text{kHz}} \approx 159\ \mu s \end{aligned} \]

100 µs lies between these limits. 200 µs exceeds \(1/(2\pi f_{m})\) and would cause diagonal clipping, 2000 µs is far too slow, and 0.050 µs is shorter than a carrier period, so the output would ripple at the carrier frequency.

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104.

An AM has carrier frequency and modulating signal frequency 2 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector in demodulator is :

A.0.08 μs
B.40 μs
C.120 μs
D.0.10 μs

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

An appropriate time constant is 40 μs.

The RC time constant of an envelope detector must be long compared with the carrier period, so that the capacitor does not discharge between carrier peaks, but short compared with the modulating period, so that it can follow the envelope:

\[ \begin{aligned} \frac{1}{f_{c}} &= \frac{1}{2\ \text{MHz}} = 0.5\ \mu\text{s} \\ \frac{1}{f_{m}} &= \frac{1}{2\ \text{kHz}} = 500\ \mu\text{s} \end{aligned} \]

40 μs lies comfortably between the two. 0.08 μs and 0.10 μs are shorter than a carrier period, so the output would ripple at the carrier frequency, and 120 μs is closer to the modulating period and risks diagonal clipping.

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105.

Suppose the modulating signal is \(m(t) = 2\cos(2\pi f_m t)\) and the carrier signal is \(y_c(t) = A_c \cos(2\pi f_c t)\), which one of the following is a conventional AM signal without over modulation ?

A.\(x(t) = A_c m(t) \cos(2\pi f_c t)\)
B.\(x(t) = A_c \cos(2\pi f_c t) + \dfrac{A_c}{4} m(t) \cos(2\pi f_c t)\)
C.\(x(t) = A_c [1 + m(t) \cos(2\pi f_c t)\)
D.\(x(t) = A_c \cos(2\pi f_m t) \cos(2\pi f_c t)\) \(+\ A_c \sin(2\pi f_m t) \cos(2\pi f_c t)\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The conventional AM signal without overmodulation is \(x(t) = A_c\cos(2\pi f_c t) + \dfrac{A_c}{4} m(t) \cos(2\pi f_c t)\).

Substituting \(m(t) = 2\cos(2\pi f_m t)\):

\[ x(t) = A_c[1 + 0.5\cos(2\pi f_m t)]\cos(2\pi f_c t) \]

This is a carrier plus sidebands with modulation index 0.5, less than 1, so there is no overmodulation.

\(A_c m(t)\cos(2\pi f_c t)\) has no carrier term, so it is DSB-SC, not conventional AM. \(A_c[1 + m(t)]\cos(2\pi f_c t)\) has modulation index 2, which is overmodulation. The last option contains no separate carrier term either, so it is not conventional AM.

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106.

When the modulating signal controls the amplitude of a carrier wave, it is known as :

A.frequency modulation
B.amplitude modulation
C.phase modulation
D.pulse amplitude modulation

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

It is amplitude modulation.

In amplitude modulation the instantaneous amplitude of a high-frequency carrier is made to vary in step with the modulating (message) signal, while the carrier frequency stays fixed:

\[ s(t) = A_{c}[1 + \mu m(t)]\cos\omega_{c}t \]

In frequency modulation the message varies the carrier's frequency, and in phase modulation its phase, with the amplitude constant in both. Pulse amplitude modulation varies the heights of a train of pulses rather than the amplitude of a continuous carrier wave.

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107.

The modulation index of an AM wave is :

A.\(m = (V_{max} - V_{min})/(V_{max} + V_{min})\)
B.\(m = (V_{max} + V_{min})/(V_{max} - V_{min})\)
C.\(m = (V_{max} \times V_{min})/(V_{max} + V_{min})\)
D.\(m = (V_{max} \times V_{min})/(V_{max} - V_{min})\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The modulation index is \(m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}\).

An AM wave has envelope \(V_{c}(1 + m\cos\omega_{m}t)\), so its maximum and minimum amplitudes are \(V_{max} = V_{c}(1 + m)\) and \(V_{min} = V_{c}(1 - m)\). Subtracting and adding:

\[ \begin{aligned} V_{max} - V_{min} &= 2mV_{c} \\ V_{max} + V_{min} &= 2V_{c} \\ m &= \frac{V_{max} - V_{min}}{V_{max} + V_{min}} \end{aligned} \]

The inverted ratio would always exceed 1, and the product forms have the units of voltage, not a pure number.

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