The minimum bandwidth of 2-array PSK (BPSK) will be:

$W = \frac{{2{R_b}}}{{{{\log }_2}M\:}} = 2{R_b}$

For ASK: BW_{ASK} = 2Rb

For PSK: BW_{PSK} = 2Rb

For FSK: BW_{FSK} = 2 Rb + |f1 – f2|

f1 and f2 are the two lower and upper frequencies.

Observation:

BW_{ASK} = BW_{PSK} < BW_{FSK}

Hence compared to ASK and PSK, FSK requires a large bandwidth.

Note:

The probability of error is maximum in ASK.

Hence PSK is the most widely used.

3. ASK, PSK, FSK, and QAM is the examples of:

Digital communication

Analog communication

Ionosphere communication

RF communication

Answer.1. Digital communication

Explanation

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S1 (A) = Acos 2π f_{H}t

For binary ‘0’ → S2 (t) = A cos 2π f_{L}t .

ASK(Amplitude Shift Keying):

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

PSK(Phase Shift Keying):

In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier

For binary ‘1’ → S1 (A) = Acos 2π fct

For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = – A cos 2π fct

4. ASK stands for:

Amplitude Shift Keying

Amplification Shift Keying

Altitude Shuffle Keying

Amplitude Shuffle Keying

Answer.1. Amplitude Shift Keying

Explanation

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space.

In these schemes, bit-by-bit transmission through free space occurs.

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S_{1} (t) = Acos 2π fct

For binary ‘0’ → S_{2} (t) = 0

5. What is the maximum data rate that can be transmitted using a QPSK modulation with a roll-off factor of 0.2 for a 36 MHz transponder?

Bandwidth for QPSK $= \frac{{{R_b}\left( {1 + 0.2} \right)}}{{{{\log }_2}4}}$

2 × 36 × 10^{6} = 1.2 R_{b}

R_{b} = 72/1.2 Mbps

= 60 Mbps

6. For a total transmit power (Pt) of 1000 W and for a transmission rate of 50 Mbps, the energy per bit (Eb) will be

10 μJ

20 μJ

30 μJ

40 μJ

Answer.2. 20 μJ

Explanation

The transmitted power and bit energy are related by the relation:

E_{b} = P_{t} × T_{b}

P_{t} = Total carrier power (Watts)

T_{b} = Time of a single bit (sec) given by the inverse of transmission bit rate (f_{b}), i.e.

T_{b} = 1/f_{b}

Calculation:

Given f_{b} = 50 Mbps and P_{t} = 1000 W

The energy per bit will be:

E_{b} = P_{t}/f_{b}

E_{b} = 1000/50 × 10^{6} J

E_{b} = 20 μJ

7. The bandwidth required for QPSK modulated channel is

Half of the bandwidth of BPSK

Twice of the bandwidth of BPSK

Equal to BPSK

Equal to FSK

Answer.1. Half of the bandwidth of BPSK

Explanation

BPSK transfers one bit per symbol. It either transmits 0 or 1 at a time.

QPSK is a modulation scheme that allows one symbol to transfer two bits of data. There are four possible two-bit numbers (00, 01, 10, 11)

$Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}$

Calculation:

For BPSK M = 2 bandwidth = 2Rb

For QPSK M = 4 bandwidth = Rb

The bandwidth required for QPSK modulated channel is half of the bandwidth of BPSK.

8. A video camera generates data at a rate of 5 Mbps. The data is channel coded at rate 1/3 and 8 PSK modulated. Which of the following statements is correct?

Information rate: 15 Mbps; Symbol rate: 5 Msps

Information rate: 5 Mbps; Symbol rate: 15 Msps

Information rate: 15 Mbps; Symbol rate: 15 Msps

Information rate: 5 Mbps; Symbol rate: 5 Msps

Answer.1. Information rate: 15 Mbps; Symbol rate: 5 Msps

Explanation

Given: m = 8, code rate = 1/3, data rate = 5 Mbps

The information rate (R) will be = Data code/Code Rate

R = 5/1/3 = 15 Mbps

And the symbol rate will be = Information rate/log_{2}m

Symbol Rate = 15/log_{2}8

Symbol rate = 5 Msps

9. For a coherent FSK with 32 levels, the probability of bit error is ______ the probability of symbol error.

31 times

1 / 31 times

2 times

0.5 times

Answer.4. 0.5 times

Explanation

Probability of bit error for coherently detected M-ary FSK is given by

10. In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 kHz and 25 kHz respectively. These waveforms will be orthogonal for a bit interval of

250 μsec

200 μsec

50 μsec

45 μsec

Answer.2. 200 μsec

Explanation

Given, In FSK modulation, let f1, f2 be frequencies for bit ‘0’