DSB and SSB Modulation Questions and Answers

Q: In the vestigial sideband amplitude modulation (AM-VSB), the transmitted signal has a bandwidth of ..............., where B is bandwidth of original analog signal.

The transmitted bandwidth of VSB is B + Δ.Vestigial sideband modulation sends one sideband in full (bandwidth B) plus a small vestige of the other sideband, of width Δ. The total bandwidth is therefore B + Δ, more than single sideband (B) but much less than double sideband (2B). Television broadcasting used VSB for this reason.B − Δ would be less than one full sideband, and the ± forms do not describe a single bandwidth.

Q: A double sideband suppressed carrier signal is generated by

A balanced modulator produces the product of message and carrier, and its symmetry cancels the carrier at the output so only the two sidebands remain. An envelope detector is a demodulator, not a modulator. A saturated class C stage generates harmonics rather than a clean product. And a reactance modulator with multipliers is the classic frequency modulation chain, which produces an angle-modulated signal with a constant envelope, quite unlike DSB-SC.

Q: The envelope of a DSB-SC signal for a tone message is

With no carrier to bias it, the envelope follows the magnitude of the message, so it looks like a full-wave rectified tone repeating at twice the message frequency, and the carrier phase reverses at each null. That is exactly why a simple envelope detector recovers the wrong waveform. The tone itself appears as the envelope only when a large carrier is added. A constant envelope belongs to angle modulation. And a square envelope would need a square-wave message.

Q: Why does an envelope detector fail on DSB-SC?

The detector responds to magnitude only, so it misses the phase reversals and delivers the rectified message instead of the message, which is severe distortion rather than a small error. Coherent detection is therefore mandatory. The time constant can be set correctly and still not help. The sidebands are certainly present, indeed they are all there is. And the diode does not saturate; it simply rectifies the wrong quantity. The three schemes differ on exactly two axes, power and bandwidth, and every question on the topic reduces to one of them.

Q: Coherent detection of DSB-SC requires a local carrier matched in

Both must match: a frequency error produces a beat and a phase error scales the output by its cosine, falling to zero when the reference is in quadrature with the signal. That quadrature null is the reason a phase-tracking loop is needed rather than a free-running oscillator. Amplitude of the reference affects only the output level. And no choice of intermediate frequency removes the need for a coherent reference. Keeping the carrier is a decision about the receiver, and keeping the second sideband a decision about the channel.

Q: A Costas loop recovers the carrier phase by

The Costas arrangement runs two multipliers in quadrature and uses the product of their outputs as an error signal, which vanishes only when the reference is aligned with the signal. It locks without needing any transmitted carrier. Squaring and dividing is the squaring-loop alternative, which works but needs a divider and has a phase ambiguity. A pilot carrier is a different scheme entirely. And zero-crossing counting is a frequency measurement, not a phase-recovery method.

In the vestigial sideband amplitude modulation (AM-VSB), the transmitted signal has a bandwidth of ..............., where B is bandwidth of original analog signal.

A.B − Δ
B.B + Δ
C.B ± Δ
D.B ± 2Δ

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The transmitted bandwidth of VSB is B + Δ.

Vestigial sideband modulation sends one sideband in full (bandwidth B) plus a small vestige of the other sideband, of width Δ. The total bandwidth is therefore B + Δ, more than single sideband (B) but much less than double sideband (2B). Television broadcasting used VSB for this reason.

B − Δ would be less than one full sideband, and the ± forms do not describe a single bandwidth.

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The balanced modulator essentially is

A.an oscillator
B.a mixer
C.a product modulator
D.an amplifier

Exam Year(s)

UGC NET - 2013,

Answer: C

A balanced modulator multiplies the carrier with the modulating signal, and its output contains the sum and difference frequencies, which are the two sidebands, with the carrier suppressed. It is therefore essentially a product modulator, and it is used to generate DSB-SC and, with a filter, SSB. An oscillator generates a signal rather than processing one, a mixer is a related product device but is meant for frequency conversion, and an amplifier only increases the level. The answer is a product modulator.

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Assertion-Reason type questions :

The following items consist of two statements, one labelled as ‘Assertion A’ and the other labelled the ‘Reason (R)’. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : A SSB system is used for broadcasting applications.
Reason (R) : The saving of power in SSB system is ≥ 75%.
Codes :

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: D

Assertion (A) is false, since single sideband is used for point-to-point and long distance communication, not for broadcasting, which uses AM or FM because of the simpler receivers. Reason (R) is true, since SSB saves at least 75 percent of the power (83.3 percent for 100 percent modulation) compared with full AM. So (A) is false and (R) is true.

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A double sideband suppressed carrier signal is generated by

A.A balanced modulator
B.A diode envelope detector
C.A class C amplifier driven into saturation
D.A reactance modulator followed by a frequency multiplier

Answer: A

A balanced modulator produces the product of message and carrier, and its symmetry cancels the carrier at the output so only the two sidebands remain. An envelope detector is a demodulator, not a modulator. A saturated class C stage generates harmonics rather than a clean product. And a reactance modulator with multipliers is the classic frequency modulation chain, which produces an angle-modulated signal with a constant envelope, quite unlike DSB-SC.

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The envelope of a DSB-SC signal for a tone message is

A.A square wave at twice the message frequency
B.The rectified tone
C.The tone itself
D.A constant, with the message carried entirely in the phase

Answer: B

With no carrier to bias it, the envelope follows the magnitude of the message, so it looks like a full-wave rectified tone repeating at twice the message frequency, and the carrier phase reverses at each null. That is exactly why a simple envelope detector recovers the wrong waveform. The tone itself appears as the envelope only when a large carrier is added. A constant envelope belongs to angle modulation. And a square envelope would need a square-wave message.

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Why does an envelope detector fail on DSB-SC?

A.The signal has no sidebands to rectify
B.The diode saturates because there is no carrier to bias it
C.It cannot follow the phase reversal at each null
D.Its time constant is always too long

Answer: C

The detector responds to magnitude only, so it misses the phase reversals and delivers the rectified message instead of the message, which is severe distortion rather than a small error. Coherent detection is therefore mandatory. The time constant can be set correctly and still not help. The sidebands are certainly present, indeed they are all there is. And the diode does not saturate; it simply rectifies the wrong quantity. The three schemes differ on exactly two axes, power and bandwidth, and every question on the topic reduces to one of them.

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Coherent detection of DSB-SC requires a local carrier matched in

A.Amplitude only
B.Frequency only, phase being irrelevant to the product
C.Neither, provided the intermediate frequency is high enough
D.Frequency and phase

Answer: D

Both must match: a frequency error produces a beat and a phase error scales the output by its cosine, falling to zero when the reference is in quadrature with the signal. That quadrature null is the reason a phase-tracking loop is needed rather than a free-running oscillator. Amplitude of the reference affects only the output level. And no choice of intermediate frequency removes the need for a coherent reference. Keeping the carrier is a decision about the receiver, and keeping the second sideband a decision about the channel.

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A coherent detector's local carrier is in error by a phase angle \( \phi \). The recovered output is scaled by

A.Cosine of phi
B.Sine of phi
C.Phi itself in radians
D.Cosine squared of phi

Answer: A

Multiplying the received signal by a reference at phase error \( \phi \) gives an output proportional to \( \cos\phi \), so a small error costs little but a quarter-cycle error nulls the output completely. That is the quadrature null, and it is what a Costas loop exists to avoid. A sine dependence would null at zero error, which is the reverse of the truth. The angle itself has the wrong form, and a squared cosine would describe a power rather than a voltage.

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A Costas loop recovers the carrier phase by

A.Limiting the signal and counting zero crossings
B.Comparing two quadrature detector outputs
C.Squaring the signal and dividing the result by two
D.Filtering a residual pilot carrier out of the received spectrum

Answer: B

The Costas arrangement runs two multipliers in quadrature and uses the product of their outputs as an error signal, which vanishes only when the reference is aligned with the signal. It locks without needing any transmitted carrier. Squaring and dividing is the squaring-loop alternative, which works but needs a divider and has a phase ambiguity. A pilot carrier is a different scheme entirely. And zero-crossing counting is a frequency measurement, not a phase-recovery method.

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10.

A squaring loop suffers from which limitation?

A.It requires a transmitted pilot
B.It works only for single sideband
C.A phase ambiguity of half a cycle
D.It cannot track frequency drift once the loop has acquired lock

Answer: C

Squaring destroys the sign, so the recovered carrier carries a half-cycle ambiguity and the recovered message may come out inverted; for analogue speech that hardly matters, but for data it must be resolved by a known reference or by differential coding. It tracks drift perfectly well once locked. It needs no pilot, which is its attraction. And it is a suppressed-carrier technique applicable to DSB rather than restricted to SSB.

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DSB and SSB Modulation Questions and Answers

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