ASK FSK and PSK Questions and Answers

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.

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.

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.

Answer: A

Phase shift keying is used for transmitting digital data over an analog channel.

PSK is a digital modulation (keying) scheme: the bits select the phase of a sinusoidal carrier, for example 0° for a 0 and 180° for a 1 in BPSK, or four phases for two bits at a time in QPSK. The result is a band-pass analog waveform that can travel over a radio link, telephone line or other channel that cannot carry the baseband pulses directly.

Sending analog data over a digital channel is the job of pulse code modulation. PSK is not restricted to binary signals, since QPSK and 8-PSK carry several bits per symbol, and it is not used to transmit FM signals.

Answer: C

On-off keying is Amplitude shift keying and Uni-polar PAM.

In on-off keying a 1 is sent as a burst of carrier and a 0 as no carrier. That is binary amplitude shift keying with one of the two amplitudes set to zero. The baseband form of the same idea, a pulse for 1 and nothing for 0, is unipolar (on-off) pulse amplitude modulation, and OOK is in effect a unipolar PAM signal multiplied by a carrier. So both names describe on-off keying.

Option B names only one of the two forms. FSK switches between two frequencies and PSK between phases, and in both the carrier is always on, so neither is on-off keying.

Answer: A

A modem carries digital data on analog lines.

A modem (modulator-demodulator) converts digital data into an analog signal suitable for an analog channel such as a telephone line, by modulating a carrier with techniques such as FSK, PSK or QAM. At the other end it demodulates the analog signal back into digital data. Because data must flow in both directions, both functions are needed.

A modulator alone or a demodulator alone does only half the job. A multiplexer combines several signals onto one channel; it does not convert digital data to an analog form.