Pulse Modulation Techniques Questions and Answers

Q: A pulse position modulated system uses a \( 10 \ \mathrm{\mu s} \) slot with a maximum displacement of \( \pm 2 \ \mathrm{\mu s} \). The fraction of the slot used for modulation is

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.

Q: Which pulse scheme is the natural precursor of pulse code modulation?

Amplitude modulation comes first, because coding begins by taking amplitude samples and then quantising them; the width and position schemes encode into other parameters and lead nowhere digital. Delta modulation is an alternative digital scheme rather than a precursor. Tracing which analogue scheme a digital one grew out of explains why sample-and-hold circuits sit at the front of every coder.

Q: Which scheme carries information in a parameter that noise cannot easily corrupt?

Position modulation encodes into timing while holding amplitude constant, so a receiver can threshold the pulses and discard amplitude noise entirely. All three amplitude variants carry the message in the very quantity noise adds to, and whether the sampling is natural or flat-top makes no difference to that. Noise immunity always comes from encoding into a parameter the noise does not share.

Q: Why does pulse code modulation need more bandwidth than pulse amplitude modulation?

Coding replaces one amplitude sample with a group of bits, multiplying the pulse rate by the word length, and bandwidth follows the pulse rate. The sampling rate is the same for both, and amplitude is irrelevant to bandwidth. Guard bands belong to frequency division multiplexing. Coding buys noise immunity and regeneration by spending bandwidth, which is the same trade wideband FM makes by other means.

Q: Assertion (A): Assertion: Aliasing cannot be removed by filtering after the sampler. Reason (R): Reason: An aliased component occupies the same frequency as a genuine signal component. Select the correct answer from the options below.

Both are true and the second is the reason for the first: once folded down, the alias is at a frequency the wanted signal also occupies and no filter can tell them apart. That is why the anti-alias filter must precede the sampler rather than follow it. The structure of the argument is identical to the receiver's image response, and recognising the shared shape saves learning the two facts separately.

Q: Assertion (A): Assertion: Regeneration removes quantisation noise from a coded signal. Reason (R): Reason: A regenerator reconstructs each pulse cleanly regardless of the noise on it. Select the correct answer from the options below.

The reason is true but the assertion is false: quantisation noise is built into the coded values themselves, so a perfect regenerator reproduces it faithfully. What regeneration removes is noise the channel added. The distinction is the whole reason coder resolution is specified separately from link quality: one impairment can be cleaned up along the route and the other cannot.

Q: Assertion (A): Assertion: Pulse width modulation has better noise immunity than pulse amplitude modulation. Reason (R): Reason: Its pulse amplitude is constant, so the receiver may limit the pulses. Select the correct answer from the options below.

Both hold and the second explains the first: because amplitude carries nothing, clipping the pulses to a standard height discards amplitude noise without losing information. It is exactly the argument that gives frequency modulation its advantage over amplitude modulation, applied to a pulse train instead of a carrier. The same asymmetry between signal and noise is being exploited in both cases.

Q: Consider the following statements about the sampling theorem. 1. The sampling rate must be at least twice the highest signal frequency 2. An anti-alias filter is needed before the sampler 3. Aliasing can be corrected by a filter after the sampler Which of the statements given above is or are correct?

The first two are correct and the third is not: aliasing is irreversible, because the folded component lands on a frequency the signal itself occupies. That single fact is what makes the anti-alias filter mandatory rather than optional, and it is why the sampling rate and the filter's cut-off are chosen as one decision rather than two.

Q: Consider the following statements about delta modulation. 1. It transmits one bit per sample 2. It needs a much higher sampling rate than pulse code modulation 3. Its step size must satisfy two opposing requirements Which of the statements given above is or are correct?

All three are correct. One bit per sample is the definition, the high rate is the price of that simplicity, and the step size is caught between slope overload if it is too small and granular noise if it is too large. A parameter bracketed by two opposing failures always has an optimum rather than a direction of improvement, and adaptive step sizing is the engineering answer to it.

Q: Statement I: Time division multiplexing requires good linearity in the transmission path. Statement II: Frequency division multiplexing requires good linearity in the transmission path. In the light of the above statements, choose the correct answer.

The second is true and the first is not: time division keeps only one channel present at any instant, so non-linearity has nothing to mix, whereas frequency division has every channel coexisting and intermodulation puts them into one another. Time division's own demand is on timing accuracy instead. Whether channels coexist or take turns is what decides which kind of imperfection a system can tolerate.

41.

A pulse position modulated system uses a \( 10 \ \mathrm{\mu s} \) slot with a maximum displacement of \( \pm 2 \ \mathrm{\mu s} \). The fraction of the slot used for modulation is

A.20 per cent
B.80 per cent
C.10 per cent
D.40 per cent

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.

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42.

Which pulse scheme is the natural precursor of pulse code modulation?

A.Pulse amplitude modulation
B.Pulse width modulation
C.Pulse position modulation
D.Delta modulation

Answer: A

Amplitude modulation comes first, because coding begins by taking amplitude samples and then quantising them; the width and position schemes encode into other parameters and lead nowhere digital. Delta modulation is an alternative digital scheme rather than a precursor. Tracing which analogue scheme a digital one grew out of explains why sample-and-hold circuits sit at the front of every coder.

Discuss in Forum

43.

Which scheme carries information in a parameter that noise cannot easily corrupt?

A.Flat-top sampled pulse amplitude modulation
B.Pulse position modulation
C.Pulse amplitude modulation
D.Natural sampled pulse amplitude modulation

Answer: B

Position modulation encodes into timing while holding amplitude constant, so a receiver can threshold the pulses and discard amplitude noise entirely. All three amplitude variants carry the message in the very quantity noise adds to, and whether the sampling is natural or flat-top makes no difference to that. Noise immunity always comes from encoding into a parameter the noise does not share.

Discuss in Forum

44.

Why does pulse code modulation need more bandwidth than pulse amplitude modulation?

A.Its pulses must be transmitted at a greater amplitude
B.It requires guard bands between the multiplexed channels
C.Each sample becomes several bits
D.Its sampling rate must be higher to avoid aliasing

Answer: C

Coding replaces one amplitude sample with a group of bits, multiplying the pulse rate by the word length, and bandwidth follows the pulse rate. The sampling rate is the same for both, and amplitude is irrelevant to bandwidth. Guard bands belong to frequency division multiplexing. Coding buys noise immunity and regeneration by spending bandwidth, which is the same trade wideband FM makes by other means.

Discuss in Forum

45.

Assertion (A): Assertion: Aliasing cannot be removed by filtering after the sampler.

Reason (R): Reason: An aliased component occupies the same frequency as a genuine signal component.

Select the correct answer from the options below.

A.Both A and R are true, but R is not 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, and R is the correct explanation of A

Answer: D

Both are true and the second is the reason for the first: once folded down, the alias is at a frequency the wanted signal also occupies and no filter can tell them apart. That is why the anti-alias filter must precede the sampler rather than follow it. The structure of the argument is identical to the receiver's image response, and recognising the shared shape saves learning the two facts separately.

Discuss in Forum

46.

Assertion (A): Assertion: Regeneration removes quantisation noise from a coded signal.

Reason (R): Reason: A regenerator reconstructs each pulse cleanly regardless of the noise on it.

Select the correct answer from the options below.

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

Answer: A

The reason is true but the assertion is false: quantisation noise is built into the coded values themselves, so a perfect regenerator reproduces it faithfully. What regeneration removes is noise the channel added. The distinction is the whole reason coder resolution is specified separately from link quality: one impairment can be cleaned up along the route and the other cannot.

Discuss in Forum

47.

Assertion (A): Assertion: Pulse width modulation has better noise immunity than pulse amplitude modulation.

Reason (R): Reason: Its pulse amplitude is constant, so the receiver may limit the pulses.

Select the correct answer from the options below.

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

Answer: B

Both hold and the second explains the first: because amplitude carries nothing, clipping the pulses to a standard height discards amplitude noise without losing information. It is exactly the argument that gives frequency modulation its advantage over amplitude modulation, applied to a pulse train instead of a carrier. The same asymmetry between signal and noise is being exploited in both cases.

Discuss in Forum

48.

Consider the following statements about the sampling theorem.

1. The sampling rate must be at least twice the highest signal frequency

2. An anti-alias filter is needed before the sampler

3. Aliasing can be corrected by a filter after the sampler

Which of the statements given above is or are correct?

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

Answer: C

The first two are correct and the third is not: aliasing is irreversible, because the folded component lands on a frequency the signal itself occupies. That single fact is what makes the anti-alias filter mandatory rather than optional, and it is why the sampling rate and the filter's cut-off are chosen as one decision rather than two.

Discuss in Forum

49.

Consider the following statements about delta modulation.

1. It transmits one bit per sample

2. It needs a much higher sampling rate than pulse code modulation

3. Its step size must satisfy two opposing requirements

Which of the statements given above is or are correct?

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

Answer: D

All three are correct. One bit per sample is the definition, the high rate is the price of that simplicity, and the step size is caught between slope overload if it is too small and granular noise if it is too large. A parameter bracketed by two opposing failures always has an optimum rather than a direction of improvement, and adaptive step sizing is the engineering answer to it.

Discuss in Forum

50.

Statement I: Time division multiplexing requires good linearity in the transmission path.

Statement II: Frequency division multiplexing requires good linearity in the transmission path.

In the light of the above statements, choose the correct answer.

A.Statement I is incorrect but Statement II is correct
B.Both Statement I and Statement II are correct
C.Both Statement I and Statement II are incorrect
D.Statement I is correct but Statement II is incorrect

Answer: A

The second is true and the first is not: time division keeps only one channel present at any instant, so non-linearity has nothing to mix, whereas frequency division has every channel coexisting and intermodulation puts them into one another. Time division's own demand is on timing accuracy instead. Whether channels coexist or take turns is what decides which kind of imperfection a system can tolerate.

Discuss in Forum

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