Pulse Modulation Techniques Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

Answer: B

Dividing the slot among its eight bits:

\[ \frac{3.9}{8} = 0.49 \ \mathrm{\mu s} \]

giving 0.49 microseconds, which corresponds to the 2.048 megabit aggregate rate as it must. The other options are the slot, the frame and an unrelated round number. Checking that a computed bit interval agrees with the aggregate rate obtained independently is a cheap and effective way to catch an arithmetic slip in a multiplex calculation. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.