UNIT 2: Physical Layer
Physical Layer: Transmission media, Signal transmission and encoding, Network performance and transmission impairments, Switching techniques and multiplexing, Overview of PSTN, ISDN, and ATM
Q61 – Maximum data rate of a channel for a noiseless 3-kHz binary channel is
3000 bps
6000 bps
1500 bps
None
(Timothy Williams, McGraw Hill Education)
Ans – (2)
Explanation – The maximum data rate of a noiseless channel can be calculated using the Nyquist formula, which states that the maximum data rate (𝑅) of a noiseless channel is twice the bandwidth (𝐵) of the channel
R = 2B = 2*3000 bps = 6000 bps
Q62 – The maximum data rate of a channel of 3000 Hz bandwidth and SNR of 30 dB is
15000 bps
60000 bps
20000 bps
3000 bps
(Timothy Williams, McGraw Hill Education)
Ans – (1)
Explanation – The maximum data rate of a channel can be calculated using the Shannon-Hartley theorem, which states
C = B*log2 (1+SNR)
C = 3000*log2 (1+30) = 3000*5 = 15000 bps
Q63 – The _________ product defines the number of bits that can fill the link.
bandwidth-period
frequency-amplitude
bandwidth-delay
delay-amplitude
Ans – (3)
Explanation –
The bandwidth-delay product (BDP) is a networking term that represents the maximum amount of data that can be in transit in a network (or link) at any point in time.
It is calculated by multiplying the bandwidth (in bits per second) of the link by the round-trip delay time (in seconds) of the network.
The BDP helps in determining the appropriate size of buffers or the amount of data that can be sent without overflowing the network buffers.
Q64 – The portion of physical layer that interfaces with the media access control sublayer is called
physical signalling sublayer
physical data sublayer
physical address sublayer
none of the mentioned
Ans – (1)
Explanation –
The portion of the physical layer that interfaces with the Media Access Control (MAC) sublayer is called the physical signalling sublayer.
This sublayer of the physical layer is responsible for transmitting raw data bits over the physical medium and includes functions like encoding, modulation, and transmission.
It interfaces directly with the MAC sublayer, which is responsible for controlling access to the physical medium.
Q65 – In asynchronous serial communication the physical layer provides
start and stop signalling
flow control
both (a) and (b)
none of the mentioned
Ans – (3)
Explanation –
Start and Stop Signalling – The physical layer is responsible for generating and interpreting the start and stop bits that frame each byte of data in asynchronous serial communication.
Flow Control – Flow control, which manages the rate of data transmission between sender and receiver to avoid data loss due to buffer overflow, is typically handled at higher layers of the OSI model (e.g., data link layer).
Q66 – Which of the following devices takes data sent from one network device and forwards it to all devices on the network regardless of the intended recipient?
DNS Server
Switch
Hub
Gateway
Ans – (3)
Explanation –
A hub operates at the physical layer (Layer 1) of the OSI model.
It receives data packets from one network device and then broadcasts or forwards them to all other devices connected to the hub, regardless of the intended recipient.
Hubs do not examine the destination address in the data packet; they simply replicate and send the data out on all ports.
Q67 – Which of the following devices takes data sent from one network device and forwards it to the destination node based on MAC address?
Hub
Switch
Gateway
Modem
Ans – (2)
Explanation –
A switch operates at the data link layer (Layer 2) of the OSI model.
It receives data frames from one network device and forwards them only to the specific network device intended to receive them, based on the destination MAC address in the Ethernet frame.
Switches maintain a table (MAC address table or forwarding table) that maps MAC addresses to the physical ports on the switch, enabling efficient data forwarding within the local network.
Q68 – Which of the following devices is used to connect different network segments and manage the traffic between them?
Bridge
Hub
Gateway
Repeater
Ans – (1)
Explanation –
A bridge operates at the data link layer (Layer 2) of the OSI model.
Its primary function is to connect two or more network segments and manage the traffic between them by examining the MAC addresses of data frames and selectively forwarding them based on their destination MAC address.
Bridges help to reduce collision domains and improve overall network performance by filtering traffic and segmenting the network into smaller collision domains.
Q69 – _______ encoding has a transition at the beginning of each 0 bit.
RZ
Manchester
Differential Manchester
All the above
Ans – (3)
Explanation –
Differential Manchester is unique because the transition at the beginning indicates the bit value, specifically for the 0 bit, which makes it distinct from RZ and Manchester encoding.
0 Bit: There is a transition at the beginning of each 0 bit.
1 Bit: There is no transition at the beginning of each 1 bit.
Each bit period has at least one transition in the middle, ensuring synchronization.
It is used in various communication systems for reliable data transmission.
Q70 – The Nyquist theorem specifies the minimum sampling rate to be_______.
equal to the lowest frequency of a signal
equal to the highest frequency of a signal
twice the bandwidth of a signal
twice the highest frequency of a signal
Ans – (4)
Explanation –
The correct answer is twice the highest frequency of a signal.
The Nyquist theorem, also known as the Nyquist-Shannon sampling theorem, states that to accurately reconstruct a continuous signal from its samples, the sampling rate must be at least twice the highest frequency present in the signal. This minimum sampling rate is known as the Nyquist rate.
Q71 – Which of the following encoding methods does not provide for synchronization?
NRZ-L
RZ
NRZ-I
Manchester
Ans – (1)
Explanation –
NRZ-L (Non-Return-to-Zero-Level) encoding method does not provide inherent synchronization because there are no guaranteed transitions between bits to help the receiver synchronize with the signal. Long sequences of 0s or 1s can lead to loss of synchronization.
Q72 – Which quantization level results in a more faithful reproduction of the signal?
2
8
16
32
Ans – (4)
Explanation – Quantization is the process of mapping a large set of input values to a smaller set, such as rounding values to the nearest integer. In digital signal processing, quantization levels refer to the number of discrete values that a signal can take. A higher number of quantization levels allows for a more precise representation of the signal’s amplitude.
2 Levels: Very coarse, resulting in significant distortion.
8 Levels: Better but still with noticeable distortion.
16 Levels: Improved fidelity but not perfect.
32 Levels: Provides a more faithful reproduction of the signal compared to the other options, as it allows for finer granularity and less quantization error.
Q73 – _______ provides redundancy to ensure synchronization and inherent error detection.
Block coding
Line coding
Scrambling
None of the above
Ans – (1)
Explanation –
Block Coding adds redundancy to data by dividing it into blocks and adding extra bits for synchronization and error detection. This helps ensure that the transmitted data can be correctly interpreted and any errors can be detected and corrected.
Q94 – ________ is the process of converting digital data to a digital signal.
Block coding
Line coding
Scrambling
None of the above
Ans – (2)
Explanation –
Line Coding is the process of converting digital data into a digital signal. It involves representing the binary data (0s and 1s) with specific voltage levels or signal forms suitable for transmission over a communication medium.
Q75 – In decoding a digital signal, the receiver calculates a running average of the received signal power, called the _______.
baseline
base
line
none of the above
Ans – (1)
Explanation –
In decoding a digital signal, the receiver often calculates a running average of the received signal power to determine the baseline. This helps to differentiate between different signal levels and correctly interpret the digital data.
Q76 – The ________ rate defines the number of data elements sent in 1s; the ______ rate is the number of signal elements sent in 1s.
data; signal
signal; data
baud; bit
none of the above
Ans – (1)
Explanation –
The data rate (or bit rate) defines the number of data elements (bits) sent in one second.
The signal rate (or baud rate) defines the number of signal elements (symbols) sent in one second.
Q77 – The idea of RZ and the idea of NRZ-L are combined into the ________ scheme.
Manchester
Differential Manchester
both (a) and (b)
neither (a) nor (b)
Ans – (1)
Explanation –
Manchester encoding combines elements from both Return-to-Zero (RZ) and Non-Return-to-Zero-Level (NRZ-L) schemes to achieve reliable data transmission.
In Manchester encoding, each bit period is divided into two halves. A logical ‘0’ is represented by a transition from high to low in the middle of the bit period, while a logical ‘1’ is represented by a transition from low to high in the middle of the bit period. This approach ensures synchronization between the sender and receiver because there is always a transition in the middle of each bit period, akin to the RZ scheme.
Additionally, Manchester encoding utilizes two signal levels, similar to NRZ-L, but introduces transitions to encode data, improving error detection capabilities compared to NRZ-L.
Therefore, Manchester encoding effectively combines aspects of RZ and NRZ-L to provide reliable and synchronized data transmission in communication systems.
Q78 – In ___________ there is always a transition at the middle of the bit, but the bit values are determined at the beginning of the bit. If the next bit is 0, there is a transition; if the next bit is 1, there is none.
Manchester
differential Manchester
both (a) and (b)
neither (a) nor (b)
Ans – (2)
Explanation –
In differential Manchester encoding, there is always a transition in the middle of each bit period to maintain synchronization. The bit values are determined by the presence or absence of a transition at the beginning of the bit period:
If the next bit to be transmitted is a ‘0’, there will be a transition in the middle of the bit period.
If the next bit to be transmitted is a ‘1’, there will be no transition in the middle of the bit period.
This method ensures synchronization while encoding data, making it distinct from standard Manchester encoding where transitions in the middle of the bit period directly represent the bit values.
Q79 – If the baud rate is 400 for a QPSK signal, the bit rate is ________ bps.
100
400
800
1600
Ans – (3)
Explanation –
To determine the bit rate for a QPSK (Quadrature Phase Shift Keying) signal with a baud rate of 400, we consider that QPSK can encode 2 bits per symbol. Given the baud rate of 400 symbols per second, each symbol represents 2 bits of data. Therefore, to find the bit rate, we multiply the baud rate by the number of bits per symbol
Bit rate = Baud rate × Number of bits per symbol
Bit rate = 400 × 2 = 800 bps
Q80 – If the bit rate for a 16-QAM signal is 4000 bps, what is the baud rate?
300
400
1000
1200
Ans – (3)
Explanation –
To determine the baud rate for a 16-QAM (Quadrature Amplitude Modulation) signal with a bit rate of 4000 bps, we first recognize that 16-QAM encodes 4 bits per symbol (since log2 (16) = 4. The baud rate (N) relates to the bit rate Rb through the formula
𝑅𝑏 = 𝑁 × baud rate
Given the bit rate of 4000 bps and 16-QAM’s 4 bits per symbol, we calculate the baud rate
baud rate = 4000 bps / 4 = 1000
Therefore, the baud rate for the 16-QAM signal is 1000 baud. This calculation shows how the bit rate and the number of bits per symbol determine the necessary baud rate for effective transmission of data using 16-QAM modulation.
