DSB and SSB Modulation Questions and Answers

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.

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.

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.

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.

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.

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.

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.

Answer: D

Suppressing the carrier removes a spectral line but no bandwidth, so DSB-SC occupies the same twice-the-message width as full-carrier AM. The saving is entirely in power, since the carrier consumed at least two thirds of it. Halving the bandwidth requires removing a sideband, which is single sideband and the next step. Doubling or quartering has no basis in the spectrum, which contains exactly the same two sidebands in both schemes.

Answer: A

A small pilot, typically twenty decibels below full carrier, gives the receiver something to lock its local oscillator to while costing little power. It is the compromise between full-carrier simplicity and suppressed-carrier efficiency. It adds power but that is a cost rather than the purpose. Bandwidth is unchanged. And it is far too small to support envelope detection, which needs a carrier larger than the sidebands. That is the trade the suppressed-carrier family exists to make explicit.

Answer: B

It needs a sharp filter, and it must act at a low frequency where the required fractional bandwidth is achievable; the signal is then translated up to the operating frequency. Crystal, ceramic and mechanical filters all serve. Quadrature networks belong to the phasing method. A limiter would destroy the amplitude information. And multiplication would multiply the frequency errors and distort the sideband, which is why SSB uses mixing rather than multiplication for translation.