Pulse Modulation Techniques Questions and Answers

Q: What additional step distinguishes pulse code modulation from pulse amplitude modulation?

Quantisation is the step that makes the scheme digital: each sample is rounded to one of a finite set of levels and then encoded as a group of bits. Pre-filtering, multiplexing and synchronisation are all common to the analogue pulse schemes as well. That one step is what buys regeneration - a quantised sample can be recognised and rebuilt exactly, which no analogue pulse can.

Q: What is quantisation noise?

Quantisation noise is the rounding error between the true sample value and the discrete level chosen to represent it, and unlike channel noise it is created inside the coder and travels with the signal. Thermal noise, aliasing and timing jitter are separate and independent impairments. Quantisation noise is the price of the digital step, and it is the one impairment that regeneration cannot remove.

Q: Why is companding used in telephony coders?

Companding compresses before quantising and expands afterwards, so quiet passages get finer effective steps and the ratio holds up across a wide dynamic range. It does not reduce the bit rate, which stays at eight bits per sample. Aliasing is the filter's business and synchronisation is the frame's. Making the quantiser's resolution follow the signal's own statistics is a far better use of a fixed number of bits than spreading them uniformly. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Q: What is delta modulation?

Delta modulation sends a single bit per sample saying whether the signal has risen or fallen since the last one, which makes the coder and decoder extremely simple at the cost of a much higher sampling rate. It has nothing to do with differences between channels, second derivatives or sample reordering. Trading sample rate for word length is the whole idea, and it is worth taking whenever simplicity matters more than efficiency. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Q: What is slope overload in a delta modulator?

Slope overload occurs when the input's rate of change exceeds the step size times the sampling rate, so the staircase cannot keep up. The remedy is a larger step or a faster clock, and adaptive delta modulation varies the step for exactly this reason. Drift and quantiser saturation describe other things, and an inadequate sampling rate is aliasing. A tracking system fails when the target moves faster than the tracker can, which is a limitation shared by every feedback quantiser.

Q: What is granular noise in a delta modulator?

With a steady input the modulator alternates up and down by one step, producing an idling pattern heard as granular noise. The error on a rapidly changing input is slope overload, the opposite failure. Channel noise and filter distortion are external to the coder. The two defects pull the step size in opposite directions - larger to avoid overload, smaller to reduce granularity - which is precisely why adaptive step sizing was invented. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Q: A signal band-limited to \( 3.4 \ \mathrm{kHz} \) is sampled at \( 8 \ \mathrm{kHz} \) and coded to \( 8 \) bits. The bit rate is

Multiplying the sampling rate by the word length:\[ 8000 \times 8 = 64 \ \mathrm{kbit/s} \]which is 64 kbit/s, the standard telephony channel rate. 27.2 kbit/s wrongly uses the signal bandwidth in place of the sampling rate. 32 kbit/s halves it, and 128 kbit/s doubles it. The whole digital telephone hierarchy is built from this one product, so it is worth being able to reproduce it without hesitation.

Q: Thirty such channels are multiplexed with two extra slots for signalling and synchronisation. The aggregate rate is

Thirty-two slots of sixty-four kilobits each:\[ 32 \times 64 = 2048 \ \mathrm{kbit/s} \]giving 2.048 Mbit/s, the European primary rate. 1.920 Mbit/s counts only the thirty traffic channels and forgets the overhead. 1.544 Mbit/s is the North American rate, built from twenty-four channels on a different plan. And 4.096 Mbit/s doubles the count. Overhead slots are part of the aggregate rate even though they carry no traffic, which is the point the wrong answer misses.

31.

What additional step distinguishes pulse code modulation from pulse amplitude modulation?

A.Quantisation of each sample to a discrete level
B.Filtering of the signal before it reaches the sampler
C.Multiplexing of several channels onto one pulse train
D.Synchronisation of the receiver to the frame structure

Answer: A

Quantisation is the step that makes the scheme digital: each sample is rounded to one of a finite set of levels and then encoded as a group of bits. Pre-filtering, multiplexing and synchronisation are all common to the analogue pulse schemes as well. That one step is what buys regeneration - a quantised sample can be recognised and rebuilt exactly, which no analogue pulse can.

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32.

What is quantisation noise?

A.The jitter in the timing of the sampling instants themselves
B.The error between a sample and its assigned level
C.The thermal noise added by the sampler's own circuitry
D.The aliasing produced when the sampling rate is too low

Answer: B

Quantisation noise is the rounding error between the true sample value and the discrete level chosen to represent it, and unlike channel noise it is created inside the coder and travels with the signal. Thermal noise, aliasing and timing jitter are separate and independent impairments. Quantisation noise is the price of the digital step, and it is the one impairment that regeneration cannot remove.

Discuss in Forum

33.

By how much does the signal-to-quantisation-noise ratio improve per extra bit?

A.About 12 dB
B.About 1.5 dB
C.About 6 dB
D.About 3 dB

Answer: C

Each additional bit doubles the number of levels and so halves the step size, improving the ratio by about 6 dB - the familiar rule that a sixteen-bit coder betters an eight-bit one by some forty-eight decibels. 3 dB corresponds to a doubling of power rather than of amplitude resolution. And the remaining figures follow from neither. A rule expressed per bit makes the whole design trade between bit rate and quality immediately visible. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

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34.

Why is companding used in telephony coders?

A.To reduce the bit rate the channel has to carry
B.To prevent aliasing of components above half the sampling rate
C.To synchronise the coder and decoder at the two ends of the link
D.To give small signals finer effective resolution

Answer: D

Companding compresses before quantising and expands afterwards, so quiet passages get finer effective steps and the ratio holds up across a wide dynamic range. It does not reduce the bit rate, which stays at eight bits per sample. Aliasing is the filter's business and synchronisation is the frame's. Making the quantiser's resolution follow the signal's own statistics is a far better use of a fixed number of bits than spreading them uniformly. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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35.

Which impairment does regeneration in a pulse code system remove?

A.Noise accumulated on the transmission path
B.Quantisation error introduced in the coder
C.Aliasing caused by an inadequate sampling rate
D.Aperture distortion from the flat-top sampling process

Answer: A

A regenerator recognises each pulse and transmits a clean replacement, so path noise is discarded rather than accumulated - the decisive advantage of digital transmission over long routes. Quantisation error, aliasing and aperture distortion are all created before transmission and are carried along faithfully by every regenerator. Regeneration removes what the channel added and nothing that the coder built in, which is exactly why coder design matters so much.

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36.

What is delta modulation?

A.Transmitting the samples in a different order each frame
B.Transmitting one bit per sample for the direction of change
C.Transmitting the difference between two adjacent channels
D.Transmitting the second derivative of the sampled waveform

Answer: B

Delta modulation sends a single bit per sample saying whether the signal has risen or fallen since the last one, which makes the coder and decoder extremely simple at the cost of a much higher sampling rate. It has nothing to do with differences between channels, second derivatives or sample reordering. Trading sample rate for word length is the whole idea, and it is worth taking whenever simplicity matters more than efficiency. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Discuss in Forum

37.

What is slope overload in a delta modulator?

A.The quantiser saturating on the loudest passages of the signal
B.The sampling rate falling below twice the highest frequency
C.The signal changing faster than the steps can follow
D.The accumulated output drifting away from the input's mean

Answer: C

Slope overload occurs when the input's rate of change exceeds the step size times the sampling rate, so the staircase cannot keep up. The remedy is a larger step or a faster clock, and adaptive delta modulation varies the step for exactly this reason. Drift and quantiser saturation describe other things, and an inadequate sampling rate is aliasing. A tracking system fails when the target moves faster than the tracker can, which is a limitation shared by every feedback quantiser.

Discuss in Forum

38.

What is granular noise in a delta modulator?

A.The error introduced when the input changes too rapidly
B.The noise added by the channel between coder and decoder
C.The distortion caused by an inadequate reconstruction filter
D.Hunting about a constant input by one step

Answer: D

With a steady input the modulator alternates up and down by one step, producing an idling pattern heard as granular noise. The error on a rapidly changing input is slope overload, the opposite failure. Channel noise and filter distortion are external to the coder. The two defects pull the step size in opposite directions - larger to avoid overload, smaller to reduce granularity - which is precisely why adaptive step sizing was invented. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Discuss in Forum

39.

A signal band-limited to \( 3.4 \ \mathrm{kHz} \) is sampled at \( 8 \ \mathrm{kHz} \) and coded to \( 8 \) bits. The bit rate is

A.64 kbit/s
B.27.2 kbit/s
C.32 kbit/s
D.128 kbit/s

Answer: A

Multiplying the sampling rate by the word length:

\[ 8000 \times 8 = 64 \ \mathrm{kbit/s} \]

which is 64 kbit/s, the standard telephony channel rate. 27.2 kbit/s wrongly uses the signal bandwidth in place of the sampling rate. 32 kbit/s halves it, and 128 kbit/s doubles it. The whole digital telephone hierarchy is built from this one product, so it is worth being able to reproduce it without hesitation.

Discuss in Forum

40.

Thirty such channels are multiplexed with two extra slots for signalling and synchronisation. The aggregate rate is

A.4.096 Mbit/s
B.2.048 Mbit/s
C.1.920 Mbit/s
D.1.544 Mbit/s

Answer: B

Thirty-two slots of sixty-four kilobits each:

\[ 32 \times 64 = 2048 \ \mathrm{kbit/s} \]

giving 2.048 Mbit/s, the European primary rate. 1.920 Mbit/s counts only the thirty traffic channels and forgets the overhead. 1.544 Mbit/s is the North American rate, built from twenty-four channels on a different plan. And 4.096 Mbit/s doubles the count. Overhead slots are part of the aggregate rate even though they carry no traffic, which is the point the wrong answer misses.

Discuss in Forum

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