Passband Transmission MCQ Quiz - Objective Question with Answer for Passband Transmission - Download Free PDF

Explanation:

Numerical Solution for Eb/N0 Calculation

Given Data:

• CNDR (Carrier-to-Noise Density Ratio): 100 dBHz
• Modulation Scheme: 256-QAM
• Symbol Rate: 12.5 M Symbols/s (or 12.5 × 10⁶ Symbols/s)

Step-by-Step Solution:

To calculate the energy-per-bit to noise density ratio (Eb/N0), we can use the relationship between CNDR, symbol rate, and the modulation scheme. The formula is as follows:

Formula:

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Where:

• CNDR: Carrier-to-Noise Density Ratio (in dBHz)
• R: Symbol rate (in Symbols/s)
• M: Modulation order (number of distinct symbols in the modulation, for 256-QAM, M = 256)

Step 1: Convert CNDR to dB Scale

In this question, CNDR is already provided in dBHz (100 dBHz), so no additional conversion is necessary.

Step 2: Calculate the Logarithmic Term for M

The modulation order (M) for 256-QAM is 256. The logarithmic term log₂(M) is calculated as:

log₂(256) = log₂(2⁸) = 8

Step 3: Calculate the Logarithmic Term for Symbol Rate

The symbol rate (R) is given as 12.5 M Symbols/s, which is equivalent to:

R = 12.5 × 10⁶ Symbols/s

The logarithmic term 10 × log₁₀(R × log₂(M)) is calculated as:

10 × log₁₀(12.5 × 10⁶ × 8)

First, calculate the product:

12.5 × 10⁶ × 8 = 100 × 10⁶ = 10⁸

Now, calculate the logarithm:

log₁₀(10⁸) = 8

Finally, multiply by 10:

10 × log₁₀(R × log₂(M)) = 10 × 8 = 80 dB

Step 4: Subtract the Logarithmic Term from CNDR

Eb/N0 = CNDR - 10 × log₁₀(R × log₂(M))

Substitute the values:

Eb/N0 = 100 dB - 80 dB

Eb/N0 = 20 dB

The value of Eb/N0 is 20 dB.

Concept:

For an M-ary PSK (M-PSK) system, the symbol rate (baud rate) is:

\(R_s = \frac{R_b}{\log_2 M}\)

Where:

• R_b is the total bit rate
• M is the modulation order

The minimum bandwidth required (Nyquist Bandwidth) is: B = R_S for ideal conditions

Given:

• M = 8 (8-PSK)
• \(R_b = 3 \times 200\text{ Mbps} = 600\text{ Mbps}\) (since I, Q, and C streams each toggle at 200 Mbps)
   

Calculation:

\(R_s = \frac{600}{\log_2 8} = \frac{600}{3} = 200 \text{ Mega-symbols/sec}\)

\(B = 2 \times R_s = 2 \times 200 = 400 \text{ MHz} \quad\)(accounting for both sidebands)

Hence, the correct option is C

QPSK modulation represents symbols by a constellation of four-phase angles of the carrier signal, orthogonal to each other.

There are two bits per symbol. So for example 00 = 0°, 01 = 90° , 10 = 180° and 11= 270°. 

QPSK transmits twice the data rate in a given bandwidth compared to BPSK, at the same BER. 

The constellation diagram for 4 different symbols is as shown:

In Frequency Shift Keying (FSK) modulation, the carrier frequency is switched between two frequencies based on:

The correct answer is option 2: whether a binary 1 or 0 is sent.

Additional Information 

Frequency Shift Keying (FSK) is a method of digital signal modulation in which the frequency of a carrier signal is varied in accordance with the digital signal data. In simpler terms, FSK uses different frequencies to represent different data values. This method is particularly useful for transmitting digital data over analog communication channels. The primary reason for using FSK modulation is its robustness in the presence of noise compared to other modulation techniques like amplitude modulation (AM) or phase modulation (PM).

In FSK modulation, the carrier frequency is switched between two or more discrete frequencies based on whether a binary '1' or '0' is being transmitted. This means:

• When the digital signal is a binary '1', the carrier frequency shifts to a higher frequency, typically referred to as the mark frequency.
• When the digital signal is a binary '0', the carrier frequency shifts to a lower frequency, typically referred to as the space frequency.

This frequency variation allows the transmission of digital data over analog systems efficiently. The receiver must be capable of distinguishing between the different frequencies to decode the transmitted binary information. One of the most common applications of FSK is in modem technology, where it is used to transmit data over telephone lines.

FSK modulation has several advantages, including:

• FSK signals are less susceptible to noise and distortion compared to amplitude-based modulation techniques.
• The implementation of FSK modulation and demodulation is straightforward and does not require complex hardware.
• FSK provides reliable data transmission, especially in environments where signal integrity is a concern.

The correct answer is 1) QPSK (Quadrature Phase Shift Keying).

Explanation

• QPSK (Quadrature Phase Shift Keying):
  • QPSK is a multilevel modulation technique that uses four distinct phase shifts to represent four different symbols.
  • These four phases are typically separated by 90 degrees (e.g., 45°, 135°, 225°, and 315°).
  • Each symbol represents two bits of information, allowing for twice the data rate compared to BPSK.
• Other options:
  • BFSK (Binary Frequency Shift Keying): Uses two different frequencies.
  • BPSK (Binary Phase Shift Keying): Uses two phase shifts.
  • 8-PSK: Uses eight phase shifts.

Analogue modulation techniques:

1. Amplitude Modulation
2. Frequency Modulation
3. Phase Modulation

Digital Modulation Techniques:

1. Amplitude shift key
2. Frequency shift key
3. Phase Shift key
4. QAM

Notes:

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 . The constellation diagram is as shown:

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

The Constellation Diagram Representation is as shown:

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

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

The Constellation Diagram Representation is as shown:

Concept:

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.

In ASK modulation each symbol is transmitted using a single bit

Hence, the bit rate = baud rate

Baud rate = 800 baud

If the question would have asked QPSK in case of ASK

Then QPSK transfers 2 bit for 1 symbol

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.

Constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown.

Concept:

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).

\(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

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

Concept:

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).

\(Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}\)

Analysis:

Transmission bandwidth

\(= \frac{{2{R_b}}}{{{{\log }_2}4}} = {R_b}\)

Maximum required bandwidth

\(= \frac{{{R_b}}}{2}\)

40% of \(\frac{{{R_b}}}{2} = 0.2\;{R_b}\)

Total = (0.2 + 0.5) R_b = 0.7 R_b

No. of channels required

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

Concept:

Probability of symbol error for coherently detected M-ary FSK:

\({P_{se}} \le \left( {M - 1} \right)Q\left( {\sqrt {\frac{{{E_s}}}{{{N_0}}}} \;} \right)\)

Where

\({E_s} = {\log _2}M.\;{E_b}\)

Probability of bit error for coherently detected M-ary FSK-

\(BER = {P_{be}} = \frac{M}{2} \times Q\left( {\sqrt {\frac{{{E_s}}}{{{N_0}}}} \;} \right)\)

\({P_{be}} = \frac{{M/2}}{{M - 1}} \times {P_{se}}\)

Calculation:

Given: M = 32

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

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 . The constellation diagram is as shown:

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

The Constellation Diagram Representation is as shown:

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

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

The Constellation Diagram Representation is as shown:

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:

Concept:

Resolution is given as:

Resolution \(= \frac{{Range}}{{{2^n}}}\)       ---(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)

\(Resolution = \frac{{Range}}{{{2^n}}}\)

\(\frac{2}{{{2^n}}} = 5\;mV\)

\({2^{n - 1}} = 200\)

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