Pulse Modulation Techniques Questions and Answers

Q: Which change would cure that distortion?

A larger step raises the slope the staircase can follow, and a faster clock would do the same; adaptive delta modulation varies the step automatically for exactly this reason. A narrower filter would dull the output without helping the tracking. A longer hold does not enter into it. And a slower clock would make matters worse by reducing the slope limit further. Once a limit is expressed as a product, every factor in it is a candidate remedy, and the cheapest one wins.

Q: What is the trade-off in choosing a delta modulator's step size?

A step too small causes slope overload on fast signals, and a step too large causes granular noise on steady ones, so the parameter has an optimum rather than a direction of improvement. The other pairings are real trades elsewhere in the system but not the one the step size decides. Whenever a parameter is squeezed from both sides, adaptive control of it is the natural next step, and that is exactly how adaptive delta modulation came about. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Q: Which pulse scheme requires the least receiver complexity?

Width modulation needs only a low-pass filter, since its average value is the message - which is why it appears in the simplest control applications. Code modulation needs a decoder and framing, adaptive delta modulation needs a matched adaptive decoder, and position modulation needs slot timing. Ranking schemes by receiver complexity often reverses the ranking by performance, and which ranking matters depends entirely on where the equipment has to sit.

Q: Which scheme would suit a link where the receiving equipment must be extremely simple?

Width modulation is the answer, for the same reason as before: a filter is the whole demodulator. The three alternatives all need decoders, timing recovery or adaptation logic at the far end. Where one end of a link must be cheap and the other may be expensive, the scheme chosen should put the complexity at the expensive end, and this is the clearest example of that principle in the pulse family.

Q: Which scheme would suit a route with many amplifiers in cascade?

Code modulation, because each regenerator rebuilds the pulses and noise does not accumulate along the route, whereas every analogue pulse scheme degrades a little at each hop. The analogue schemes differ among themselves in noise immunity but all of them accumulate. Once a route is long enough, the ability to regenerate outweighs every other consideration, which is why long-haul transmission went digital before anything else did. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Q: Arrange these events in the order they occur along a pulse code link. 1. The message is band-limited by a filter 2. Samples are quantised and encoded 3. Pulses are regenerated at intermediate points 4. A reconstruction filter recovers the waveform Select the correct sequence.

The order is 1, 2, 3, 4. Each step depends on the one before it: filtering must precede sampling because aliasing is irreversible, coding must precede transmission, regeneration happens along the route, and reconstruction is the last operation before the output. A chain whose order follows from the reasons for it can be reconstructed rather than memorised, which is more reliable under pressure.

Q: Match each quantity with the parameter that sets it. List I A. Highest frequency carried B. Quantisation noise C. Aggregate bit rate List II P. Sampling rate Q. Word length R. Both, times the channel count Select the correct match.

The pairing is A-P, B-Q, C-R. Sampling rate and word length are the two independent levers, one governing bandwidth and the other governing resolution, and the bit rate is their product multiplied by the number of channels. Seeing which quantities are independent and which are derived is what makes a coder's specification tractable rather than a list of numbers to be learned.

Q: Statement I: Quantisation noise can be reduced by using more bits per sample. Statement II: Quantisation noise can be reduced by regenerating the pulses along the route. In the light of the above statements, choose the correct answer.

The first is true and the second is not: quantisation noise is created in the coder and carried in the coded values, so a regenerator reproduces it exactly. Regeneration removes only what the channel added. The distinction is worth being firm about, because it explains why coder resolution has to be chosen for the quality wanted at the far end and cannot be improved anywhere along the way.

Q: Which of the following is not a source of impairment in a sampled system? Regeneration along the route Aliasing at the sampler Quantisation in the coder Crosstalk between channel slots

Regeneration is a remedy, not an impairment: it removes accumulated path noise by rebuilding each pulse cleanly. The other three are genuine impairments arising at three different stages, and each has its own remedy - a better filter, more bits, and more bandwidth or guard time respectively. Sorting the mechanisms into causes and cures is what makes a fault list usable rather than merely long.

91.

Which change would cure that distortion?

A.A longer holding time in the sampler
B.A slower clock at the coder
C.A larger step size
D.A narrower reconstruction filter

Answer: C

A larger step raises the slope the staircase can follow, and a faster clock would do the same; adaptive delta modulation varies the step automatically for exactly this reason. A narrower filter would dull the output without helping the tracking. A longer hold does not enter into it. And a slower clock would make matters worse by reducing the slope limit further. Once a limit is expressed as a product, every factor in it is a candidate remedy, and the cheapest one wins.

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92.

What is the trade-off in choosing a delta modulator's step size?

A.Bit rate against the number of channels carried
B.Aliasing against the reconstruction filter's steepness
C.Bandwidth against the transmitted average power
D.Slope overload against granular noise

Answer: D

A step too small causes slope overload on fast signals, and a step too large causes granular noise on steady ones, so the parameter has an optimum rather than a direction of improvement. The other pairings are real trades elsewhere in the system but not the one the step size decides. Whenever a parameter is squeezed from both sides, adaptive control of it is the natural next step, and that is exactly how adaptive delta modulation came about. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Discuss in Forum

93.

Which pulse scheme requires the least receiver complexity?

A.Pulse width modulation
B.Pulse code modulation
C.Delta modulation with adaptive stepping
D.Pulse position modulation

Answer: A

Width modulation needs only a low-pass filter, since its average value is the message - which is why it appears in the simplest control applications. Code modulation needs a decoder and framing, adaptive delta modulation needs a matched adaptive decoder, and position modulation needs slot timing. Ranking schemes by receiver complexity often reverses the ranking by performance, and which ranking matters depends entirely on where the equipment has to sit.

Discuss in Forum

94.

Which scheme would suit a link where the receiving equipment must be extremely simple?

A.Pulse position modulation with framing
B.Pulse width modulation
C.Pulse code modulation with companding
D.Adaptive delta modulation with variable stepping

Answer: B

Width modulation is the answer, for the same reason as before: a filter is the whole demodulator. The three alternatives all need decoders, timing recovery or adaptation logic at the far end. Where one end of a link must be cheap and the other may be expensive, the scheme chosen should put the complexity at the expensive end, and this is the clearest example of that principle in the pulse family.

Discuss in Forum

95.

Which scheme would suit a route with many amplifiers in cascade?

A.Pulse width modulation with a low duty cycle
B.Pulse position modulation with tight framing
C.Pulse code modulation
D.Pulse amplitude modulation

Answer: C

Code modulation, because each regenerator rebuilds the pulses and noise does not accumulate along the route, whereas every analogue pulse scheme degrades a little at each hop. The analogue schemes differ among themselves in noise immunity but all of them accumulate. Once a route is long enough, the ability to regenerate outweighs every other consideration, which is why long-haul transmission went digital before anything else did. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Discuss in Forum

96.

Arrange these events in the order they occur along a pulse code link.

1. The message is band-limited by a filter

2. Samples are quantised and encoded

3. Pulses are regenerated at intermediate points

4. A reconstruction filter recovers the waveform

Select the correct sequence.

A.2, 1, 3, 4
B.1, 3, 2, 4
C.1, 2, 4, 3
D.1, 2, 3, 4

Answer: D

The order is 1, 2, 3, 4. Each step depends on the one before it: filtering must precede sampling because aliasing is irreversible, coding must precede transmission, regeneration happens along the route, and reconstruction is the last operation before the output. A chain whose order follows from the reasons for it can be reconstructed rather than memorised, which is more reliable under pressure.

Discuss in Forum

97.

Match each quantity with the parameter that sets it.

List I

A. Highest frequency carried

B. Quantisation noise

C. Aggregate bit rate

List II

P. Sampling rate

Q. Word length

R. Both, times the channel count

Select the correct match.

A.A-P, B-Q, C-R
B.A-Q, B-P, C-R
C.A-R, B-Q, C-P
D.A-P, B-R, C-Q

Answer: A

The pairing is A-P, B-Q, C-R. Sampling rate and word length are the two independent levers, one governing bandwidth and the other governing resolution, and the bit rate is their product multiplied by the number of channels. Seeing which quantities are independent and which are derived is what makes a coder's specification tractable rather than a list of numbers to be learned.

Discuss in Forum

98.

Statement I: Quantisation noise can be reduced by using more bits per sample.

Statement II: Quantisation noise can be reduced by regenerating the pulses along the route.

In the light of the above statements, choose the correct answer.

A.Statement I is incorrect but Statement II is correct
B.Statement I is correct but Statement II is incorrect
C.Both Statement I and Statement II are correct
D.Both Statement I and Statement II are incorrect

Answer: B

The first is true and the second is not: quantisation noise is created in the coder and carried in the coded values, so a regenerator reproduces it exactly. Regeneration removes only what the channel added. The distinction is worth being firm about, because it explains why coder resolution has to be chosen for the quality wanted at the far end and cannot be improved anywhere along the way.

Discuss in Forum

99.

Statement I: Aliasing is prevented by filtering before the sampler.

Statement II: Aliasing is prevented by filtering after the sampler.

In the light of the above statements, choose the correct answer.

A.Both Statement I and Statement II are incorrect
B.Statement I is incorrect but Statement II is correct
C.Statement I is correct but Statement II is incorrect
D.Both Statement I and Statement II are correct

Answer: C

The first is correct and the second is not, for the reason that runs through this whole topic: once a component has folded down it occupies a frequency the signal itself uses, and nothing downstream can separate them. The filter after the sampler is the reconstruction filter and does a different job entirely. Where two inputs are mapped onto one output, the discrimination must happen before the mapping.

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100.

Which of the following is not a source of impairment in a sampled system?

Regeneration along the route

Aliasing at the sampler

Quantisation in the coder

Crosstalk between channel slots

A.Aliasing at the sampler
B.Quantisation in the coder
C.Crosstalk between channel slots
D.Regeneration along the route

Answer: D

Regeneration is a remedy, not an impairment: it removes accumulated path noise by rebuilding each pulse cleanly. The other three are genuine impairments arising at three different stages, and each has its own remedy - a better filter, more bits, and more bandwidth or guard time respectively. Sorting the mechanisms into causes and cures is what makes a fault list usable rather than merely long.

Discuss in Forum

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