Pulse Modulation Techniques Questions and Answers

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.

Q: Why is a narrower slot not always an improvement?

Narrow pulses need wide bandwidth, so squeezing more channels into a frame is paid for in channel bandwidth - and beyond some point the pulses spread and cause crosstalk anyway. It increases rather than reduces the channel count, quantisation is unaffected, and the sampling rate is set by the signals. Capacity and bandwidth are two views of the same resource in a pulse system, which is why one cannot be improved without spending the other. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Q: What is the function of a signalling slot in a multiplex frame?

A signalling slot carries the control information a call needs - dialling, answer, clear - separately from the speech itself, which is why the European primary rate has thirty-two slots for thirty channels. Slot boundaries come from the frame alignment pattern in the other overhead slot, guard time is unallocated margin rather than a slot, and amplitude equalisation is not a framing matter. Overhead in a frame is always doing a job, and identifying which job each slot does explains why the numbers come out as they do.

Q: Why is the frame alignment pattern chosen to be unlikely in traffic?

The pattern must be distinctive enough that a run of traffic bits does not imitate it and pull the receiver out of alignment, which is why standard patterns look nothing like speech samples. Brevity, error tolerance and circuit simplicity are all secondary considerations that a good pattern also satisfies. A marker that has to be found in a stream must be improbable in that stream, which is a requirement common to every framing and synchronisation scheme.

Q: What does an adaptive delta modulator vary?

It varies the step size, enlarging it when the signal is changing fast and shrinking it when the signal is quiet, which addresses slope overload and granular noise together instead of compromising between them. The clock is fixed, there is no word length in a one-bit coder, and the sampling instants are regular. Where a fixed parameter is squeezed by two opposing requirements, making it adaptive is the natural resolution, and this is the textbook example.

Q: Why is differential pulse code modulation more efficient than plain coding?

Because consecutive samples are correlated, the difference between them is smaller than the samples themselves and needs fewer bits to describe to the same accuracy. The quantiser's level count is unchanged, anti-alias filtering is still mandatory, and the sampling rate must still satisfy Nyquist. Exploiting the redundancy a signal already contains is the cheapest possible saving, since it costs nothing but the arithmetic to predict the next sample.

Q: What does a predictor contribute to a differential coder?

The predictor forms an estimate of the coming sample from the history, and only the difference between estimate and reality is transmitted - the better the prediction, the smaller that difference and the fewer bits it needs. An error record, a recovered clock and a companding law are all other things. The decoder runs the identical predictor, which is why both ends must agree on it exactly and why a bit error can take a moment to die away. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Q: A signal is sampled at \( 20 \ \mathrm{kHz} \). Components at \( 12 \ \mathrm{kHz} \) will appear at

Folding about half the sampling rate:\[ \begin{aligned} f_{\text{alias}} &= f_{s} - f \\ &= 20 - 12 = 8 \ \mathrm{kHz} \end{aligned} \]so the component appears at 8 kHz, inside the band and indistinguishable from a genuine signal there. Twelve kilohertz assumes no folding at all, and the remaining options are the sampling rate and the sum. Every component above half the rate has an image below it, which is precisely why the filter must remove them before sampling.

101.

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.

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102.

What limits the number of channels a time division multiplex can carry?

A.The frame period divided by the slot width
B.The amplitude the transmitter is able to deliver
C.The number of quantisation levels chosen in the coder
D.The bandwidth of each individual channel being carried

Answer: A

The frame lasts one sampling interval and each channel needs a slot, so the channel count is the frame period divided by the slot width, less the overhead slots. Transmitter amplitude affects range rather than capacity, quantisation levels affect quality, and the individual channel bandwidth fixes the sampling rate and hence the frame period rather than the count directly. Capacity in a time division system is always a division of one time by another, which makes it easy to compute and easy to check.

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103.

Why is a narrower slot not always an improvement?

A.It requires a lower sampling rate at every channel input
B.It demands more channel bandwidth
C.It reduces the number of channels that can be carried
D.It increases the quantisation noise the coder produces

Answer: B

Narrow pulses need wide bandwidth, so squeezing more channels into a frame is paid for in channel bandwidth - and beyond some point the pulses spread and cause crosstalk anyway. It increases rather than reduces the channel count, quantisation is unaffected, and the sampling rate is set by the signals. Capacity and bandwidth are two views of the same resource in a pulse system, which is why one cannot be improved without spending the other. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Discuss in Forum

104.

What is the function of a signalling slot in a multiplex frame?

A.Providing the guard time between adjacent channels
B.Equalising the amplitudes of the multiplexed channels
C.Carrying call control rather than speech
D.Marking the boundary of each individual channel slot

Answer: C

A signalling slot carries the control information a call needs - dialling, answer, clear - separately from the speech itself, which is why the European primary rate has thirty-two slots for thirty channels. Slot boundaries come from the frame alignment pattern in the other overhead slot, guard time is unallocated margin rather than a slot, and amplitude equalisation is not a framing matter. Overhead in a frame is always doing a job, and identifying which job each slot does explains why the numbers come out as they do.

Discuss in Forum

105.

Why is the frame alignment pattern chosen to be unlikely in traffic?

A.So it occupies as few bits of the frame as possible
B.So it survives bit errors on the transmission path
C.So it can be generated by the simplest possible circuit
D.So the receiver cannot mistake data for the pattern

Answer: D

The pattern must be distinctive enough that a run of traffic bits does not imitate it and pull the receiver out of alignment, which is why standard patterns look nothing like speech samples. Brevity, error tolerance and circuit simplicity are all secondary considerations that a good pattern also satisfies. A marker that has to be found in a stream must be improbable in that stream, which is a requirement common to every framing and synchronisation scheme.

Discuss in Forum

106.

What does an adaptive delta modulator vary?

A.The step size, according to the signal's slope
B.The clock rate, according to the signal's amplitude
C.The word length, according to the channel's quality
D.The sampling instant, according to the frame timing

Answer: A

It varies the step size, enlarging it when the signal is changing fast and shrinking it when the signal is quiet, which addresses slope overload and granular noise together instead of compromising between them. The clock is fixed, there is no word length in a one-bit coder, and the sampling instants are regular. Where a fixed parameter is squeezed by two opposing requirements, making it adaptive is the natural resolution, and this is the textbook example.

Discuss in Forum

107.

Why is differential pulse code modulation more efficient than plain coding?

A.Its sampling rate can be lower than the Nyquist rate
B.Successive samples of a real signal are similar
C.Its quantiser has more levels for the same word length
D.It requires no anti-alias filtering at the transmitter

Answer: B

Because consecutive samples are correlated, the difference between them is smaller than the samples themselves and needs fewer bits to describe to the same accuracy. The quantiser's level count is unchanged, anti-alias filtering is still mandatory, and the sampling rate must still satisfy Nyquist. Exploiting the redundancy a signal already contains is the cheapest possible saving, since it costs nothing but the arithmetic to predict the next sample.

Discuss in Forum

108.

What does a predictor contribute to a differential coder?

A.A clock recovered from the incoming pulse stream
B.A companding characteristic matched to the signal
C.An estimate of the next sample from previous ones
D.A record of the errors made on earlier samples

Answer: C

The predictor forms an estimate of the coming sample from the history, and only the difference between estimate and reality is transmitted - the better the prediction, the smaller that difference and the fewer bits it needs. An error record, a recovered clock and a companding law are all other things. The decoder runs the identical predictor, which is why both ends must agree on it exactly and why a bit error can take a moment to die away. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Discuss in Forum

109.

A signal is sampled at \( 20 \ \mathrm{kHz} \). Components at \( 12 \ \mathrm{kHz} \) will appear at

A.12 kHz
B.20 kHz
C.32 kHz
D.8 kHz

Answer: D

Folding about half the sampling rate:

\[ \begin{aligned} f_{\text{alias}} &= f_{s} - f \\ &= 20 - 12 = 8 \ \mathrm{kHz} \end{aligned} \]

so the component appears at 8 kHz, inside the band and indistinguishable from a genuine signal there. Twelve kilohertz assumes no folding at all, and the remaining options are the sampling rate and the sum. Every component above half the rate has an image below it, which is precisely why the filter must remove them before sampling.

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110.

A frame carries \( 32 \) slots of \( 8 \) bits at \( 8000 \) frames per second. The slot duration is

A.3.9 microseconds
B.125 microseconds
C.31.25 microseconds
D.0.49 microseconds

Answer: A

The frame lasts one sampling interval, shared among the slots:

\[ \begin{aligned} T_{f} &= \tfrac{1}{8000} = 125 \ \mathrm{\mu s} \\ T_{s} &= \tfrac{125}{32} = 3.9 \ \mathrm{\mu s} \end{aligned} \]

giving 3.9 microseconds. 125 microseconds is the whole frame, 31.25 would follow from eight slots, and 0.49 is the duration of a single bit. Keeping frame, slot and bit intervals distinct is essential, since all three are quoted in microseconds and differ only by the factors that define them.

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