Passband Transmission MCQ Quiz - Objective Question with Answer for Passband Transmission - Download Free PDF
Last updated on Jun 13, 2025
Latest Passband Transmission MCQ Objective Questions
Passband Transmission Question 1:
Given that CNDR of 256 QAM signal operating at 12.5 M Symbols/s is 100 dBHz. The value of Eb/N0 (Energy-per-bit to Noise Density ratio) is:
Answer (Detailed Solution Below)
Passband Transmission Question 1 Detailed Solution
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.
Passband Transmission Question 2:
The bandwidth occupied by an 8 PSK modulated signal, where each of the I, Q and C data streams which form the n-tuple IQC toggle at a rate 200 Mbps, is:
Answer (Detailed Solution Below)
Passband Transmission Question 2 Detailed Solution
Concept:
For an M-ary PSK (M-PSK) system, the symbol rate (baud rate) is:
Where:
- Rb is the total bit rate
is the modulation order
The minimum bandwidth required (Nyquist Bandwidth) is: B = RS for ideal conditions
Given:
- M = 8 (8-PSK)
(since I, Q, and C streams each toggle at 200 Mbps)
Calculation:
Hence, the correct option is C
Passband Transmission Question 3:
In which modulation technique is the carrier modulated using phase shifts of 0°, 90°, 180°, and 270°?
Answer (Detailed Solution Below)
Passband Transmission Question 3 Detailed Solution
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:
Passband Transmission Question 4:
In Frequency Shift Keying (FSK) modulation, the carrier frequency is switched between two frequencies based on _______.
Answer (Detailed Solution Below)
Passband Transmission Question 4 Detailed Solution
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.
Passband Transmission Question 5:
__________ is a multilevel modulation in which four phase shift are used for representing four different symbols.
Answer (Detailed Solution Below)
Passband Transmission Question 5 Detailed Solution
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.
Top Passband Transmission MCQ Objective Questions
ASK, PSK, FSK, and QAM are the examples of:
Answer (Detailed Solution Below)
Passband Transmission Question 6 Detailed Solution
Download Solution PDFAnalogue modulation techniques:
- Amplitude Modulation
- Frequency Modulation
- Phase Modulation
Digital Modulation Techniques:
- Amplitude shift key
- Frequency shift key
- Phase Shift key
- 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:
The bandwidth required for QPSK modulated channel is
Answer (Detailed Solution Below)
Passband Transmission Question 7 Detailed Solution
Download Solution PDFConcept:
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).
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.
If the bit rate for an ASK signal is 800 bps, the baud rate is:
Answer (Detailed Solution Below)
Passband Transmission Question 8 Detailed Solution
Download Solution PDFIn 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
Baud rate = 800/2 = 400 baud
Quadrature Amplitude Modulation (QAM) is a combination of:
Answer (Detailed Solution Below)
Passband Transmission Question 9 Detailed Solution
Download Solution PDFDigital 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.
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?
Answer (Detailed Solution Below)
Passband Transmission Question 10 Detailed Solution
Download Solution PDFConcept:
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).
For bandpass transmission,
Bandwidth for M-ary PSK
Given:
BW = 36 × 106 Hz
Roll-off factor = 0.2
Calculation:
Bandwidth for QPSK
2 × 36 × 106 = 1.2 Rb
Rb = 72/1.2 Mbps
= 60 Mbps
For Baseband |
For Passband |
Binary: 1) B.W. = Rb |
Binary: 1) BW = 2 Rb |
Raised cosine (α) : 2) |
Raised cosine (α) : = Rb (1 + α) |
M-ary: 1) |
M-ary: 1) |
Raised cosine (α): 2) |
Raised cosine (α) : 2) |
A satellite system employs QPSK modulation with 40% excess bandwidth per carrier including guard band. The voice channels use 64 kbps PCM coding. The no. of channels supported by 36 MHz bandwidth of the transponder in bandwidth limited case will be
Answer (Detailed Solution Below)
Passband Transmission Question 11 Detailed Solution
Download Solution PDFConcept:
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).
Analysis:
Transmission bandwidth
Maximum required bandwidth
40% of
Total = (0.2 + 0.5) Rb = 0.7 Rb
No. of channels required
For a coherent FSK with 32 levels, the probability of bit error is ______ the probability of symbol error.
Answer (Detailed Solution Below)
Passband Transmission Question 12 Detailed Solution
Download Solution PDFConcept:
Probability of symbol error for coherently detected M-ary FSK:
Where
Probability of bit error for coherently detected M-ary FSK-
Calculation:
Given: M = 32
In FSK, the carrier frequency is switched between ______ extremes.
Answer (Detailed Solution Below)
Passband Transmission Question 13 Detailed Solution
Download Solution PDFASK, 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:
Consider a binary transmission system with the symbol waveforms s1(t) = A cos(ωct), s2(t) = - A cos(ωct). This modulation format is termed as
Answer (Detailed Solution Below)
Passband Transmission Question 14 Detailed Solution
Download Solution PDFASK, 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 fct
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 fct
For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = - A cos 2π fct
The constellation diagram for binary PSK is given as:
8 Channel multiplex system has 5 kHz BW speech channels of the analog voltage range 0 to +2V. For 5 mV resolution, the minimum number of bits per sample in each channel and total BW of the system are _____ and _____ respectively.
Answer (Detailed Solution Below)
Passband Transmission Question 15 Detailed Solution
Download Solution PDFConcept:
Resolution is given as:
Resolution
n - number of bits
Transmission B.W = Rb ---(2)
Bit Rate(Rb) = m ⋅ n ⋅ fs
m – number of message signals
n – number of bits
fs – sampling frequency
Calculation:
Given:
m = 8,
fm = 5 kHz,
Range = 0 to +2 V
From equation (1)
n – 1 = 8
n = 9
from equation (2);
BW = Rb = m ⋅ n ⋅ fs
∴ fs = 2 fm = 10 kHz
BW = 8 × 9 × 10
BW = 720 kHz