ASK FSK and PSK Questions and Answers

Q: In order to transmit at 4·8 kbps over a channel with 3·2 kHz bandwidth, which would be most suitable ?

The most suitable choice is QPSK.With Nyquist signalling, a channel of bandwidth B can carry at most 2B symbols per second in theory, and in practice, with filtering, about B symbols per second. The 3.2 kHz channel therefore supports about 3.2 kbaud. Each scheme needs:BPSK, 1 bit per symbol: 4.8 kbaud, more than the channel carries;QPSK, 2 bits per symbol: 2.4 kbaud, which fits with margin;8-PSK, 3 bits: 1.6 kbaud; 16-PSK, 4 bits: 1.2 kbaud.Higher-order PSK would also fit, but its constellation points are closer together and it needs a much higher signal to noise ratio for the same error rate. QPSK is the simplest scheme whose symbol rate fits the band, which makes it the most suitable.

Q: In frequency shift keying :(a) The combined signal in FSK modulator generated from two separate oscillators cannot have discontinuities in amplitude.(b) The combined signal in FSK modulator generated from two separate oscillators cannot have discontinuities in phase.(c) The combined signal in FSK modulator generated from single oscillator method can have discontinuities in amplitude.(d) The combined signal in FSK modulator generated from single oscillator method cannot have discontinuities in amplitude.

Only statement (d) is true, so the answer is (d) only.When the FSK signal is made by switching between two independent oscillators, the two sine waves are not in step at the instant of switching, so the combined waveform jumps in value there: it has a discontinuity in phase, and in general in amplitude too. That makes (a) false, since it says such a signal cannot have amplitude discontinuities, and also (b) false, which claims no phase discontinuities.When one oscillator is used and only its frequency is changed (a voltage-controlled oscillator), the phase carries on from where it was, so the waveform is continuous: no discontinuity in amplitude or phase. That makes (d) true and (c) false.Every option other than (d) only contains (a), (b) or (c), which is why options A, B and D fail.

Q: Data sequence "1101100001" transmitted at a speed of 1,00,000 bits/sec with a carrier frequency of 150 kHz corresponds to ................. two-bit symbols/sec.

The symbol rate is 50,000 two-bit symbols per second.Each symbol carries 2 bits, so the symbol (baud) rate is half the bit rate:\[ \begin{aligned} R_{s} &= \frac{100000\ \text{bits/s}}{2\ \text{bits/symbol}} \\ &= 50000\ \text{symbols/s} \end{aligned} \]The carrier frequency of 150 kHz and the particular data sequence do not affect the symbol rate; they are distractors. 1,00,000 is the bit rate itself, 25,000 would be for four-bit symbols, and 30,000 does not follow from the data.

In order to transmit at 4·8 kbps over a channel with 3·2 kHz bandwidth, which would be most suitable ?

A.BPSK
B.QPSK
C.8-PSK
D.16-PSK

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The most suitable choice is QPSK.

With Nyquist signalling, a channel of bandwidth B can carry at most 2B symbols per second in theory, and in practice, with filtering, about B symbols per second. The 3.2 kHz channel therefore supports about 3.2 kbaud. Each scheme needs:

BPSK, 1 bit per symbol: 4.8 kbaud, more than the channel carries;
QPSK, 2 bits per symbol: 2.4 kbaud, which fits with margin;
8-PSK, 3 bits: 1.6 kbaud; 16-PSK, 4 bits: 1.2 kbaud.

Higher-order PSK would also fit, but its constellation points are closer together and it needs a much higher signal to noise ratio for the same error rate. QPSK is the simplest scheme whose symbol rate fits the band, which makes it the most suitable.

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In frequency shift keying :

(a) The combined signal in FSK modulator generated from two separate oscillators cannot have discontinuities in amplitude.
(b) The combined signal in FSK modulator generated from two separate oscillators cannot have discontinuities in phase.
(c) The combined signal in FSK modulator generated from single oscillator method can have discontinuities in amplitude.
(d) The combined signal in FSK modulator generated from single oscillator method cannot have discontinuities in amplitude.

A.(b) and (d)
B.(a) and (d)
C.(d) only
D.(a) and (c)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

Only statement (d) is true, so the answer is (d) only.

When the FSK signal is made by switching between two independent oscillators, the two sine waves are not in step at the instant of switching, so the combined waveform jumps in value there: it has a discontinuity in phase, and in general in amplitude too. That makes (a) false, since it says such a signal cannot have amplitude discontinuities, and also (b) false, which claims no phase discontinuities.

When one oscillator is used and only its frequency is changed (a voltage-controlled oscillator), the phase carries on from where it was, so the waveform is continuous: no discontinuity in amplitude or phase. That makes (d) true and (c) false.

Every option other than (d) only contains (a), (b) or (c), which is why options A, B and D fail.

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Data sequence "1101100001" transmitted at a speed of 1,00,000 bits/sec with a carrier frequency of 150 kHz corresponds to ................. two-bit symbols/sec.

A.30,000
B.50,000
C.1,00,000
D.25,000

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The symbol rate is 50,000 two-bit symbols per second.

Each symbol carries 2 bits, so the symbol (baud) rate is half the bit rate:

\[ \begin{aligned} R_{s} &= \frac{100000\ \text{bits/s}}{2\ \text{bits/symbol}} \\ &= 50000\ \text{symbols/s} \end{aligned} \]

The carrier frequency of 150 kHz and the particular data sequence do not affect the symbol rate; they are distractors. 1,00,000 is the bit rate itself, 25,000 would be for four-bit symbols, and 30,000 does not follow from the data.

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