Pulse Modulation Techniques Questions and Answers

Q: Which coder design choice improves quality without raising the bit rate?

Non-uniform spacing, which companding implements, puts fine steps where the signal spends most of its time and coarse steps where it rarely goes, improving the perceived ratio at no cost in bits. Doubling the levels and raising the sampling rate both increase the rate. And a regenerator addresses path noise rather than coder quality. Matching a quantiser to the signal's own statistics is free in bandwidth terms and is the first place to look for an improvement.

Q: Why does companding help speech in particular?

Speech has a large dynamic range but the amplitude distribution is heavily weighted towards small values, so fine steps down there are used far more often than fine steps at the top. Its bandwidth is modest, silence suppression is a separate technique, and no signal tolerates aliasing. A coder tuned to the statistics of the particular signal always beats a general-purpose one, and speech's amplitude statistics are unusually well suited to the trade.

Q: A delta modulator's step is halved while its clock is doubled. Its maximum followable slope

The slope limit is the product of step and clock rate:\[ \begin{aligned} \text{before} &: \delta f_{s} \\ \text{after} &: \tfrac{\delta}{2} \times 2f_{s} = \delta f_{s} \end{aligned} \]so it is unchanged, while the granular noise improves because the steps are smaller. That is precisely the trade a faster clock buys: the same tracking ability with finer resolution, at the cost of bit rate. Recognising which combination of parameters a limit depends on shows immediately which changes cancel.

Q: Sixteen channels of \( 3 \ \mathrm{kHz} \) each are multiplexed, each coded to \( 7 \) bits and sampled at \( 8 \ \mathrm{kHz} \). The aggregate bit rate is

Rate per channel times the number of channels:\[ \begin{aligned} 8000 \times 7 &= 56 \ \mathrm{kbit/s} \\ 16 \times 56 &= 896 \ \mathrm{kbit/s} \end{aligned} \]giving 896 kbit/s before overhead. The 3 kHz figure is a distractor: the sampling rate is given directly and the signal bandwidth plays no further part. Spotting which of the quoted numbers the calculation does not need is half the work in a question of this shape. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Q: Assertion (A): Assertion: Pulse position modulation is more efficient in power than pulse width modulation. Reason (R): Reason: All its pulses are of the same short duration. Select the correct answer from the options below.

Both are true and the second is the reason: identical short pulses mean the average transmitted power is low and independent of the modulation, whereas a width-modulated train's power rises with its average duty cycle. The efficiency is bought with a stricter timing requirement, which is the standard shape of these trades. Where a scheme's average power does not depend on the message, the transmitter can be sized for the average rather than the worst case.

Q: A coder is specified as follows. ParameterValue Sampling rate8 kHzWord length8 bitsCompandingA-law Which figure does the companding chiefly affect?

Companding redistributes the eight bits so that small signals get finer steps, which improves the ratio on quiet passages without touching the bit rate, the bandwidth limit or the frame capacity - those follow from the sampling rate and word length alone. Reading a specification table means asking which line each other line depends on, and here the companding line stands apart from the two that set the rate.

71.

Which coder design choice improves quality without raising the bit rate?

A.Raising the sampling rate above twice the signal bandwidth
B.Adding a second regenerator midway along the transmission path
C.Non-uniform spacing of the quantisation levels
D.Doubling the number of uniformly spaced levels available

Answer: C

Non-uniform spacing, which companding implements, puts fine steps where the signal spends most of its time and coarse steps where it rarely goes, improving the perceived ratio at no cost in bits. Doubling the levels and raising the sampling rate both increase the rate. And a regenerator addresses path noise rather than coder quality. Matching a quantiser to the signal's own statistics is free in bandwidth terms and is the first place to look for an improvement.

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72.

Why does companding help speech in particular?

A.Speech occupies a wider bandwidth than most signals do
B.Speech contains long silences that need no coding at all
C.Speech is more tolerant of aliasing than other signals are
D.Speech spends most of its time at low amplitudes

Answer: D

Speech has a large dynamic range but the amplitude distribution is heavily weighted towards small values, so fine steps down there are used far more often than fine steps at the top. Its bandwidth is modest, silence suppression is a separate technique, and no signal tolerates aliasing. A coder tuned to the statistics of the particular signal always beats a general-purpose one, and speech's amplitude statistics are unusually well suited to the trade.

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73.

A companded coder gives \( 38 \ \mathrm{dB} \) of signal-to-noise ratio over a \( 40 \ \mathrm{dB} \) input range. A uniform coder of the same word length would give \( 38 \ \mathrm{dB} \) only

A.At full amplitude
B.Over the same forty decibel range of input
C.At the lowest amplitude within that range
D.At no amplitude within the range at all

Answer: A

A uniform quantiser's step is fixed, so its ratio falls decibel for decibel as the signal falls:

\[ \begin{aligned} \text{at full scale} &: 38 \ \mathrm{dB} \\ \text{40 dB below} &: -2 \ \mathrm{dB} \end{aligned} \]

so it reaches 38 dB only at full amplitude. Companding's achievement is holding the ratio roughly constant across the range instead, which is what a listener actually notices.

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74.

A delta modulator's step is halved while its clock is doubled. Its maximum followable slope

A.Is quadrupled
B.Is unchanged
C.Is halved
D.Is doubled

Answer: B

The slope limit is the product of step and clock rate:

\[ \begin{aligned} \text{before} &: \delta f_{s} \\ \text{after} &: \tfrac{\delta}{2} \times 2f_{s} = \delta f_{s} \end{aligned} \]

so it is unchanged, while the granular noise improves because the steps are smaller. That is precisely the trade a faster clock buys: the same tracking ability with finer resolution, at the cost of bit rate. Recognising which combination of parameters a limit depends on shows immediately which changes cancel.

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75.

Sixteen channels of \( 3 \ \mathrm{kHz} \) each are multiplexed, each coded to \( 7 \) bits and sampled at \( 8 \ \mathrm{kHz} \). The aggregate bit rate is

A.1024 kbit/s
B.336 kbit/s
C.896 kbit/s
D.384 kbit/s

Answer: C

Rate per channel times the number of channels:

\[ \begin{aligned} 8000 \times 7 &= 56 \ \mathrm{kbit/s} \\ 16 \times 56 &= 896 \ \mathrm{kbit/s} \end{aligned} \]

giving 896 kbit/s before overhead. The 3 kHz figure is a distractor: the sampling rate is given directly and the signal bandwidth plays no further part. Spotting which of the quoted numbers the calculation does not need is half the work in a question of this shape. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

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76.

Assertion (A): Assertion: A higher sampling rate always improves a pulse code system's quality.

Reason (R): Reason: A higher rate reduces the quantisation noise for a given word length.

Select the correct answer from the options below.

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

Answer: D

Both statements are defensible but the reason does not explain the assertion: a higher rate eases the anti-alias filter and pushes the spectral repeats further apart, but quantisation noise depends on the step size and hence on the word length, not on the rate. Rate and resolution are independent levers on quality, and confusing them leads to spending bit rate where word length was wanted.

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77.

Assertion (A): Assertion: Pulse position modulation is more efficient in power than pulse width modulation.

Reason (R): Reason: All its pulses are of the same short duration.

Select the correct answer from the options below.

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 but R is false
D.A is false but R is true

Answer: A

Both are true and the second is the reason: identical short pulses mean the average transmitted power is low and independent of the modulation, whereas a width-modulated train's power rises with its average duty cycle. The efficiency is bought with a stricter timing requirement, which is the standard shape of these trades. Where a scheme's average power does not depend on the message, the transmitter can be sized for the average rather than the worst case.

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78.

Consider the following statements about time division multiplexing.

1. Only one channel occupies the path at any instant

2. Crosstalk arises chiefly from limited channel bandwidth

3. It demands better amplifier linearity than frequency division

Which of the statements given above is or are correct?

A.1, 2 and 3
B.1 and 2 only
C.1 and 3 only
D.2 and 3 only

Answer: B

The first two are correct and the third is the reverse of the truth: because the channels take turns rather than coexisting, non-linearity has nothing to mix and the linearity requirement is relaxed. It is frequency division that needs excellent linearity, because intermodulation there puts every channel into every other. The two schemes trade linearity against timing accuracy, and knowing which way round that trade runs is the practical consequence.

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79.

Consider the following pairs of parameter and what it limits.

Sampling rate - Highest signal frequency carried

Word length - Signal-to-quantisation-noise ratio

Guard time - Tolerance to pulse spreading

Pulse amplitude - Number of channels multiplexed

How many of the pairs given above are correctly matched?

A.Only two
B.All four
C.Only three
D.Only one

Answer: C

The first three are correctly paired and the fourth is not, so three match. The number of channels is limited by the frame length and the slot width, both time quantities; pulse amplitude affects the received signal-to-noise ratio and nothing about channel capacity. Each design parameter governs one property, and pairing them correctly is what lets a specification be traced back to the choice that produced it.

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80.

A coder is specified as follows.

Parameter	Value
Sampling rate	8 kHz
Word length	8 bits
Companding	A-law

Which figure does the companding chiefly affect?

A.The bit rate the channel is required to carry
B.The highest signal frequency the coder can accept
C.The number of channels that can share one frame
D.The ratio achieved on quiet passages

Answer: D

Companding redistributes the eight bits so that small signals get finer steps, which improves the ratio on quiet passages without touching the bit rate, the bandwidth limit or the frame capacity - those follow from the sampling rate and word length alone. Reading a specification table means asking which line each other line depends on, and here the companding line stands apart from the two that set the rate.

Discuss in Forum

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