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
Q21 – If the frequency spectrum of a signal has a bandwidth of 500 Hz with the highest frequency at 600 Hz, what should be the sampling rate, according to the Nyquist theorem?
200 samples/s
500 samples/s
1000 samples/s
1200 samples/s
Ans – (4)
Explanation –
According to the Nyquist theorem, the sampling rate must be at least twice the highest frequency present in the signal to avoid aliasing. Given the highest frequency in the signal is 600 Hz, the minimum sampling rate should be 2×600 Hz = 1200 samples/s
Q22 – 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 –
This is to ensure that the signal can be accurately reconstructed without aliasing.
Q23 – One factor in the accuracy of a reconstructed PCM signal is the ________.
Signal bandwidth
Carrier frequency
Number of bits used for quantization
Baud rate
Ans – (3)
Explanation –
The number of bits used for quantization determines the resolution of the signal, affecting the signal’s accuracy and the level of quantization noise.
Q24 – Which encoding type always has nonzero average amplitude
Unipolar
Polar
Bipolar
All the above
Ans – (1)
Explanation –
In unipolar encoding, the signal levels are all positive or all negative, which means the average amplitude is always nonzero.
Q25 – 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 does not inherently provide synchronization because it does not include a mechanism for indicating the beginning or end of a bit, leading to potential issues with distinguishing consecutive bits of the same value.
Q26 – Which encoding method uses alternating positive and negative values for 1s?
NRZ-I
RZ
Manchester
AMI
Ans – (4)
Explanation –
In AMI encoding, positive and negative pulses are alternated to represent consecutive 1s, with a neutral (zero) level used for 0s. This method helps to ensure a sufficient number of transitions for clock recovery and synchronization purposes.
Q27 – If the maximum value of a PCM signal is 31 and the minimum value is -31, how many bits were used for coding?
4
5
6
7
Ans – (3)
Explanation –
To determine the number of bits used for coding in PCM, we can use the formula
Number of bits = log2 Number of quantization levels
In this case, the number of quantization levels is determined by the range of values from the maximum to the minimum value, plus 1 (to include zero). So, the number of quantization levels is 31 – (-31) + 1 = 63 + 1 = 64.
Now, let’s calculate the number of bits
Number of bits = log2(64) = 6
Q28 – RZ encoding involves __________ signal levels.
Two
Three
Four
Five
Ans – (2)
Explanation –
In RZ encoding, the signal can have three different levels: high, low, and zero (or neutral).
During each bit period, the signal starts at zero, transitions to a high or low level to represent a bit, and then returns to zero before the end of the bit period. This return to the zero level distinguishes RZ encoding from NRZ (Non-Return-to-Zero) encoding, which only uses two levels (high and low) without returning to zero within the bit period.
Q29 – Which quantization level results in a more faithful reproduction of the signal?
2
8
16
32
Ans – (4)
Explanation –
The more quantization levels there are, the more faithfully the signal can be reproduced.
Quantization is the process of mapping a large set of input values to a smaller set. More quantization levels mean finer granularity, which allows the signal to be represented more accurately.
With 32 levels, the signal can be captured with more detail compared to using 2, 8, or 16 levels.
Q30 – Block coding can help in _________ at the receiver.
Synchronization
Error detection
Attenuation
(1) and (2)
Ans – (4)
Explanation – Block coding can help in synchronization and error detection at the receiver.
Synchronization – Block coding can help synchronize the receiver with the incoming data stream by providing specific patterns or sequences that facilitate timing recovery and frame alignment.
Error Detection – Block coding techniques often include redundancy in the transmitted data, which allows the receiver to detect errors and potentially correct them using error detection and correction algorithms.
Q31 – In __________ transmission, bits are transmitted over a single wire, one at a time.
Asynchronous serial
Synchronous serial
Parallel
(1) and (2)
Ans – (4)
Explanation –
In asynchronous serial transmission, data is sent one bit at a time over a single wire without a clock signal controlling the timing. Each transmitted character is framed by start and stop bits to synchronize the sender and receiver. This method is commonly used in serial communication protocols like RS-232.
In Synchronous serial transmission, data is sent over a communication link in a synchronized manner using a clock signal. Unlike asynchronous serial transmission, where individual bits are sent with start and stop bits to signify the beginning and end of a data byte, synchronous serial transmission does not rely on start and stop bits.
Q32 – In _________ transmission, a start bit and a stop bit frame a character byte.
Asynchronous serial
Synchronous serial
Parallel
(1) and (2)
Ans – (1)
Explanation –
In asynchronous serial transmission, data is sent one bit at a time over a single wire without a clock signal controlling the timing. Each transmitted character is framed by start and stop bits to synchronize the sender and receiver. This method is commonly used in serial communication protocols like RS-232.
Q33 – In asynchronous transmission, the gap time between is __________.
Fixed
Variable
A function of data rate
Zero
Ans – (2)
Explanation –
In asynchronous transmission, the gap time between characters (bits) can vary. This gap, known as the inter-character gap, is necessary because there is no continuous clock signal to synchronize the sender and receiver.
Q34 – If the baud rate is 400 for a 4-PSK signal, the bit rate is ________ bps.
100
400
1600
800
Ans – (4)
Explanation – If a 4-PSK signal has a baud rate of 400 and each symbol represents 2 bits (because of 2^2 possibilities), then the total bit rate would be 400 * 2 = 800 bits per second or 800 bps.
Q35 – If the bit rate for an ASK signal is 1200 bps, the baud is _________ bps.
300
400
600
1200
Ans – (4)
Explanation – In ASK (Amplitude Shift Keying) modulation, each symbol represents one bit.
Therefore, the baud rate (symbols per second) is equal to the bit rate (bits per second), 1200 bps.
Q36 – If the bit rate for an FSK signal is 1200 bps, the baud rate is ________.
300
400
600
1200
Ans – (4)
Explanation – In FSK (Frequency Shift Keying) modulation, each symbol represents one bit.
Therefore, the baud rate (symbols per second) is equal to the bit rate (bits per second), 1200 bps.
Q37 – The physical layer is concerned with ___________
bit-by-bit delivery
process to process delivery
application to application delivery
port to port delivery
Ans – (1)
Explanation –
The physical layer is primarily concerned with bit-by-bit delivery, handling the transmission and reception of raw data bits over a physical medium.
Q38 – In which OSI layers does the FDDI protocol operate?
Physical
Data link
Network
A and B
Ans – (4)
Explanation –
The FDDI (Fiber Distributed Data Interface) protocol operates in both the Physical and Data Link layers of the OSI model.
Q39 – A simple cabling method, known as the ……………… topology allows about 30 computers on a maximum cable length of about 600 feet.
Ring
Bus
Star
Mesh
Ans – (2)
Explanation –
The cabling method described, which allows about 30 computers on a maximum cable length of about 600 feet, is known as the Bus topology.
In a Bus topology, all devices are connected to a single cable (the backbone), and each device has its own connection point. The maximum cable length is limited in Bus topology to avoid signal degradation.
Q40 – TCP/IP model does not have ______ layer but OSI model have this layer.
session layer
transport layer
application layer
network layer
Ans – (1)
Explanation – The TCP/IP model does not have a session layer, but the OSI model does have this layer.
