Passband Transmission MCQ || Passband Transmission Questions and Answers

Answer.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

Answer.1. Higher than 

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 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.

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_Ht

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

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

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’ → S₁ (t) = Acos 2π fct

For binary ‘0’ → S₂ (t) = 0

Answer.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 R_b

R_b = 72/1.2 Mbps

= 60 Mbps

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 × 10⁶ J

E_b = 20 μJ

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.

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

Answer.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

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

Answer.2. 200 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 T_b 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/T_b

35 KHz = m/T_b

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

i.e. T_b = 200 μs. (minimum value)