Pulse Modulation Techniques Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

Answer: D

The displacement spans four microseconds in total:

\[ \frac{2 \times 2}{10} = 0.4 \]

so 40 per cent, with the rest given to the pulse itself and to guard time. Twenty per cent counts only one direction of displacement, which is the commonest slip on this calculation. The remaining options double or quarter the result. A budget expressed as a fraction of the slot makes it obvious how much room is left for guard time.