# Passband Transmission MCQ || Passband Transmission Questions and Answers

1. If the bit rate for an ASK signal is 800 bps, the baud rate is:

1. 1600 baud
2. 400 baud
3. 600 baud
4. 800 baud

Explanation

In ASK modulation each symbol is transmitted using a single bit

Hence, the bit rate = baud rate

Baud rate = 800 baud

2. The bandwidth of BFSK is ______ BPSK

1. Higher than
2. The same as
3. Lower than
4. None of the other options

Explanation

If Rb is the bit rate, then:

The bandwidth of M-ary PSK:

$W = \frac{{2{R_b}}}{{{{\log }_2}M\:}}\left( {1 + \alpha } \right)$

Rb = Bitrate

α = Roll-off factor

For minimum transmission bandwidth α = 0

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

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

For PSK: BWPSK = 2Rb

For FSK: BWFSK  = 2 Rb + |f1 – f2|

f1 and f2 are the two lower and upper frequencies.

Observation:

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:

1. Digital communication
2. Analog communication
3. Ionosphere communication
4. RF 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 fHt

For binary ‘0’ → S2 (t) = A cos 2π fLt .

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

1. Amplitude Shift Keying
2. Amplification Shift Keying
3. Altitude Shuffle Keying
4. Amplitude Shuffle 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’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (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?

1. 7.2 Mbps
2. 30 Mbps
3. 43.2 Mbps
4. 60 Mbps

Explanation

For bandpass transmission,

Bandwidth for M-ary PSK $= \frac{{{R_b}\left( {1 + \alpha } \right)}}{{{{\log }_2}M}}$

Given:

BW = 36 × 106 Hz

Roll-off factor = 0.2

Calculation:

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

2 × 36 × 106 = 1.2 Rb

Rb = 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

1. 10 μJ
2. 20 μJ
3. 30 μJ
4. 40 μJ

Explanation

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

Eb = Pt × Tb

Pt = Total carrier power (Watts)

Tb = Time of a single bit (sec) given by the inverse of transmission bit rate (fb), i.e.

Tb = 1/fb

Calculation:

Given fb = 50 Mbps and Pt = 1000 W

The energy per bit will be:

Eb = Pt/fb

Eb = 1000/50 × 106 J

Eb = 20 μJ

7. The bandwidth required for QPSK modulated channel is

1. Half of the bandwidth of BPSK
2. Twice of the bandwidth of BPSK
3. Equal to BPSK
4. 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?

1. Information rate: 15 Mbps; Symbol rate: 5 Msps
2. Information rate: 5 Mbps; Symbol rate: 15 Msps
3. Information rate: 15 Mbps; Symbol rate: 15 Msps
4. 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/log2m

Symbol Rate = 15/log28

Symbol rate = 5 Msps

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

1. 31 times
2. 1 / 31 times
3. 2 times
4. 0.5 times

Explanation

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

$\begin{array}{l} BER = {P_{be}} = \frac{M}{2} \times Q\left( {\sqrt {\frac{{{E_s}}}{{{N_0}}}} \:} \right)\\ \\ {P_{be}} = \frac{{M/2}}{{M – 1}} \times {P_{se}} \end{array}$

Calculation:

Given: M = 32

${P_{be}} = \frac{{32/2}}{{31}} \times {P_{se}} \approx \frac{1}{2} \times {P_{se}}$

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

1. 250 μsec
2. 200 μsec
3. 50 μsec
4. 45 μsec

Explanation

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

s(t) = A sin (2πf1t)

For bit ‘1’

s(t) = A sin (2πf2t)

Let Tb be bit duration

For both waveforms to be orthogonal,

$\left| {{f_1} – {f_2}} \right| = \frac{n}{{{T_b}}}\:and\:\left| {{f_1} + {f_2}} \right| = \frac{m}{{{T_b}}}$

Application:

Given, f1 = 10 kHz and f2 = 25 kHz

15 KHz = n/Tb

35 KHz = m/Tb

It is possible for 1/Tb = 5 KHz with minimum m and n value, i.e.

i.e. Tb = 200 μs. (minimum value)

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