Pulse Modulation Techniques Questions and Answers

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.

Q: A \( 10 \)-bit coder is used. Its signal-to-quantisation-noise ratio is about

Applying the standard result for a full-scale sinusoid:\[ \begin{aligned} \mathrm{SNR} &= 6.02n + 1.76 \\ &= 60.2 + 1.76 \approx 62 \ \mathrm{dB} \end{aligned} \]so about 62 dB. Sixty decibels drops the constant term, which is the usual approximation and close enough for most purposes. 48 dB corresponds to eight bits and 30 dB to five. The six-decibels-per-bit slope is the part worth remembering; the additive constant is a refinement. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: A delta modulator has a step of \( 20 \ \mathrm{mV} \) and a clock of \( 64 \ \mathrm{kHz} \). The maximum slope it can follow is

The staircase can climb one step per clock period:\[ 0.02 \times 64000 = 1280 \ \mathrm{V/s} \]so 1280 volts per second, and any input steeper than that causes slope overload. The remaining options misplace a decade or divide where they should multiply. Expressing a coder's limit as a slope rather than as an amplitude is what makes it directly comparable with the signal it has to follow.

Q: A \( 1 \ \mathrm{kHz} \) sinusoid of \( 1 \ \mathrm{V} \) peak is applied to that modulator. Slope overload

The sinusoid's steepest slope is at the zero crossing:\[ \begin{aligned} \text{slope} &= 2\pi f A \\ &= 2\pi(1000)(1) = 6283 \ \mathrm{V/s} \end{aligned} \]which far exceeds 1280, so overload does occur. The first option computes the slope correctly and then draws the wrong conclusion, which is the trap here. The comparison is always peak input slope against step times clock rate, and either can be adjusted to satisfy it.

31.

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.

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32.

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

33.

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

34.

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

35.

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

36.

A \( 10 \)-bit coder is used. Its signal-to-quantisation-noise ratio is about

A.48 dB
B.30 dB
C.62 dB
D.60 dB

Answer: C

Applying the standard result for a full-scale sinusoid:

\[ \begin{aligned} \mathrm{SNR} &= 6.02n + 1.76 \\ &= 60.2 + 1.76 \approx 62 \ \mathrm{dB} \end{aligned} \]

so about 62 dB. Sixty decibels drops the constant term, which is the usual approximation and close enough for most purposes. 48 dB corresponds to eight bits and 30 dB to five. The six-decibels-per-bit slope is the part worth remembering; the additive constant is a refinement. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Discuss in Forum

37.

A delta modulator has a step of \( 20 \ \mathrm{mV} \) and a clock of \( 64 \ \mathrm{kHz} \). The maximum slope it can follow is

A.320 V/s
B.12.8 V/s
C.3200 V/s
D.1280 V/s

Answer: D

The staircase can climb one step per clock period:

\[ 0.02 \times 64000 = 1280 \ \mathrm{V/s} \]

so 1280 volts per second, and any input steeper than that causes slope overload. The remaining options misplace a decade or divide where they should multiply. Expressing a coder's limit as a slope rather than as an amplitude is what makes it directly comparable with the signal it has to follow.

Discuss in Forum

38.

A \( 1 \ \mathrm{kHz} \) sinusoid of \( 1 \ \mathrm{V} \) peak is applied to that modulator. Slope overload

A.Does not occur, since the peak slope is 6283 V/s
B.Occurs, because the peak slope exceeds what the steps allow
C.Occurs only on the negative half cycles of the waveform
D.Cannot be determined without knowing the input impedance

Answer: A

The sinusoid's steepest slope is at the zero crossing:

\[ \begin{aligned} \text{slope} &= 2\pi f A \\ &= 2\pi(1000)(1) = 6283 \ \mathrm{V/s} \end{aligned} \]

which far exceeds 1280, so overload does occur. The first option computes the slope correctly and then draws the wrong conclusion, which is the trap here. The comparison is always peak input slope against step times clock rate, and either can be adjusted to satisfy it.

Discuss in Forum

39.

A pulse train has pulses \( 1 \ \mathrm{\mu s} \) wide. The bandwidth needed is roughly

A.250 kHz
B.500 kHz
C.1 MHz
D.2 MHz

Answer: B

The usual rule takes the bandwidth as the reciprocal of twice the pulse width:

\[ \begin{aligned} B \approx \frac{1}{2\tau} &= \frac{1}{2 \times 10^{-6}} \\ &= 500 \ \mathrm{kHz} \end{aligned} \]

so about 500 kHz. One megahertz is the reciprocal of the width alone, a more conservative figure some texts prefer. The other values follow from a factor of four either way. Narrow pulses demand wide channels, which is the resource every pulse scheme spends and the reason bandwidth rather than power usually limits them.

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40.

Twenty-four channels each band-limited to \( 4 \ \mathrm{kHz} \) are time division multiplexed. The minimum pulse rate is

A.384 kHz
B.48 kHz
C.192 kHz
D.96 kHz

Answer: C

Each channel needs its Nyquist rate and the slots share one train:

\[ 24 \times 8000 = 192 \ \mathrm{kHz} \]

giving 192 kHz, before any allowance for synchronisation. 96 kHz uses the signal bandwidth rather than twice it. 384 kHz doubles unnecessarily. And 48 kHz is far short. Multiplying the per-channel rate by the channel count is the first line of every multiplex calculation, and overhead is then added to it.

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