Passband Transmission MCQ || Passband Transmission Questions and Answers

Answer.2. 800

Explanation

Bandwidth = 2Rb/log₂M

Analysis:

Transmission bandwidth

$= \frac{{2{R_b}}}{{{{\log }_2}4}} = {R_b}$

Maximum required bandwidth = Rb/2

40% of Rb/2 = 0.2 Rb

Total = (0.2 + 0.5) R_b = 0.7 R_b

No. of channels required

$= \frac{{36 \times {{10}^6}}}{{0.7 \times 64 \times {{10}^3}}} \approx 800$

Answer.4. Scheme where same carrier frequency is used for two different signals

Explanation

QAM is a mixture of ASK and PSK. Hence, amplitude and the phase of the carrier frequency both vary with the message signal. QAM uses two carrier signals which are in quadrature Quadrature here means out phase by 90°. B-ASK, B-PSK all use single carrier. B-FSK use two carriers of very high-frequency carriers which are closely related

Answer.2. ASK and PSK

Explanation

The Digital to Analog Modulation technique is as shown. As shown QAM is the mixture of both ASK and PSK. Hence, amplitude and the phase of the carrier frequency both vary with the message signal.

Answer.3. 2

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 FSK, the carrier frequency is switched between 2 extremes.

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. 4

Explanation

In a BPSK system, each symbol represents 1 bit; in a QPSK system, each symbol represents 2 bits. The Baud Rate (BR) is given by:

$B.R. = \frac{{{R_b}}}{{{{\log }_2}M}}$

Calculation:

Given modulation technique is 16 QAM

The number of QAM states is 2^N, as determined by the number of binary bits per symbol.

Thus a 16-QAM system (N = 4) is one for which the (microwave) carrier is modulated into any one of 16 different amplitude and phase states

Answer.2. 9 and 720 KHz

Explanation

Resolution is given as:

Resolution = Range/2^m ——(1)

n – number of bits

Transmission B.W = R_b       —(2)

Bit Rate(R_b) = m ⋅ n ⋅ f_s

m – number of message signals

n – number of bits

f_s – sampling frequency

Calculation:

Given:

m = 8,

f_m = 5 kHz,

Range = 0 to +2 V

From equation (1)

2/2n = 5 mV

2^{n −1} = 200

n – 1 = 8

n = 9

from equation (2);

BW = R_b = m ⋅ n ⋅ f_s

∴ f_s = 2 f_m = 10 kHz

BW = 8 × 9 × 10

BW = 720 kHz

Answer.1. 0.5 bits-sec^-1/Hz, 2f_b

Explanation

Spectral efficiency, spectrum efficiency, or bandwidth efficiency refers to the information rate that can be transmitted over a given bandwidth in a specific communication system. It is given as:

S.E = f_o/B.W

Bandwidth:

The bandwidth of M-ary PSK:

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

fb = bit rate

α = Roll-off factor

For minimum transmission bandwidth α = 0

$BW = \frac{{2{f_b}}}{{{{\log }_2}M\:}}$

Analysis:

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

$BW = \frac{{2{f_b}}}{{{{\log }_2}M\:}} = 2{f_b}$

Spectral efficiency is given as:

$S.E. = \frac{{{f_b}}}{{BW}} = \frac{{{f_b}}}{{2{f_b}}} = 0.5{f_b}$

Answer.2. Phase

Explanation

The BPSK Demodulator Baseband block demodulates a signal that was modulated using the binary phase-shift keying method. The input is a baseband representation of the modulated signal. This block accepts a scalar or column vector input signal. The input signal must be a discrete-time complex signal.

In binary phase-shift keying (BPSK), the phase of a constant amplitude carrier signal is switched between two possible values m1 and m2. These two values correspond to binary 1 and 0 respectively.

Answer.3. 8,000

Explanation

Baud rate is defined as the number of symbols transmitted per second. It is also known as the symbol rate.

For M -array PSK baud rate is given as:

Baud Rate = Bit Rate/Log₂M

Calculation:

Given: For 8-PSK, M = 8, bit rate (R_b) = 24 kbps

Putting on the respective values, we get:

Baud Rate = 24/Log₂8

Baud Rate = 8000 symbol/sec

Answer.3. Phase Shift Keying

Explanation

ASK, PSK and FSK schemes 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 f_ct

For binary ‘0’ → S₂ (t) = 0

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 f_ct

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

The constellation diagram for binary PSK is given as: