CS Free MCQ Bank

[ Option D ]

Industrial, Scientific, and Medical (ISM) bands are portions of the radio spectrum reserved internationally for non-commercial purposes like industrial heating, medical equipment, and scientific experimentation. Over time, they have also been widely used for unlicensed wireless communication, such as Wi-Fi and Bluetooth.

[ Option A ]

A scrambler is used in line coding schemes to modify the original data before transmission.

• To remove long sequences of 0s or 1s which can cause synchronization problems.
• To make the transmitted data appear random, even if the original data has patterns.
• Helps in maintaining clock synchronization between sender and receiver.

[ Option B ]

In digital communication, Modulation is the process of varying a carrier signal (sine wave) to transmit data. Digital modulation schemes often use two or more carriers, which can vary in amplitude, phase, or frequency to represent binary information.

A Constellation Diagram is a graphical representation of these modulated signals on a 2D plane, where the x-axis represents the in-phase component (I) and the y-axis represents the quadrature component (Q). Each point in the diagram corresponds to a unique amplitude and phase combination, allowing engineers to visualize the transmitted symbols and detect errors.

[ Option B ]

[ Option B ]

[ Option C ]

A Useless Channel is a communication channel where there is no correlation between input and output symbols. This means the output is completely independent of the input, the received symbol gives no information about what was sent. In this, the channel capacity is zero because the output does not carry any meaningful information about the input.
A noiseless channel, output is exactly the same as input, so there is a perfect correlation between input and output.
A lossless channel, input can be uniquely determined from the output, so correlation still exists.
A deterministic channel, each input symbol corresponds to exactly one output symbol, hence strong correlation.

[ Option C ]

In Manchester Encoding, each bit period must have a transition in the middle, Low to High for bit 1 and High to Low for bit 0. This transition helps maintain synchronization between sender and receiver. If the line stays high throughout a bit period (High to High), it means no transition occurred, violating the Manchester rule. Therefore, the receiver interprets it as a bit error or loss of synchronization.

[ Option C ]

To digitize a human voice signal, the Pulse Code Modulation (PCM) method is used, which requires sampling the analog signal. 
According to the Nyquist theorem, the sampling rate must be at least twice the highest frequency present in the signal. 
Since the human voice ranges up to 4000 Hz, the signal must be sampled at 8000 samples per second. If each sample is represented by 8 bits, the total bit rate becomes 8000 × 8 = 64,000 bits per second, or 64 kbps.

[ Option C ]

According to Nyquist’s theorem, the maximum data rate of a noiseless channel depends on its bandwidth (B) and the number of discrete signal levels (V) used in transmission.
Maximum Data Rate = 2B log₂V bits per second

• B is bandwidth of the channel in hertz (Hz).
• V is number of discrete signal levels.

[ Option C ]

In computer networks, binary data cannot always be transmitted directly over text-based protocols like email (SMTP) or web (HTTP). To solve this problem, Base64 encoding is used. Base64 takes every 24 bits of input data (3 bytes) and divides it into four groups of 6 bits each. Each 6-bit value is then mapped to a printable ASCII character.

Since each Base64 output is an ASCII character, it requires 8 bits of storage. Therefore, four characters × 8 bits = 32 bits of output are generated from the original 24 bits. This explains why Base64 increases the size of the encoded data by about 33%.

When encoding the string "SRK" into Base64, the process begins by converting each character into its ASCII binary form. The character S has an ASCII value of 83, which is 01010011 in binary. The character R has an ASCII value of 82, represented as 01010010. Finally, the character K has an ASCII value of 75, represented as 01001011. Together, the string "SRK" produces a 24-bit binary stream:

01010011 01010010 01001011

Next, these 24 bits are divided into groups of 6 bits each:

010100 110101 001001 001011

Each 6-bit group is then converted into its decimal equivalent:

• 010100 = 20
• 110101 = 53
• 001001 = 9
• 001011 = 11

Using the Base64 index table, these decimal values are mapped to their respective characters:

• 20 = U
• 53 = 1
• 9 = J
• 11 = L

Thus, the final Base64 encoded output of the string "SRK" is: U1JL

[ Option A ]

For a Noiseless communication channel, the theoretical maximum bit rate is determined by the Nyquist bit rate formula. The Nyquist formula calculates the highest possible data rate based on the channel bandwidth (B) and the number of discrete signal levels (M) used in transmission.
Maximum bit rate = 2 × B × log₂(M)

In contrast, the Shannon bit rate formula considers noisy channels and accounts for signal-to-noise ratio (SNR), making it applicable when noise is present.

[ Option A ]

Quantizing noise occurs during the quantization process in Pulse Code Modulation (PCM). PCM converts an analog signal into a digital signal by sampling the amplitude of the analog signal at regular intervals and then quantizing these samples to the nearest value among a finite set of levels. The difference between the actual analog value and the quantized digital value is called quantization error or quantizing noise.

• PCM (Pulse Code Modulation): Quantization is a necessary step, hence quantizing noise appears.
• TDM (Time Division Multiplexing): No quantization; it only multiplexes signals in the time domain.
• FDM (Frequency Division Multiplexing): Signals are separated by frequency bands, no quantization noise.
• PPM (Pulse Position Modulation): The noise mainly comes from timing errors, not quantization.

[ Option B ]

The maximum data rate of a noiseless channel is determined by the Nyquist Theorem. It defines the theoretical upper limit for data transmission rate in an noiseless channel.
Maximum bit rate = 2×B×log₂(M)
Where:
B = Bandwidth of the channel.
M = Number of discrete signal levels (symbols) used.
Given Values:
B=3000 Hz, M=2 (says binary signal and binary means two levels, 0 and 1)
Maximum bit rate = 2 × 3000 × log₂(2) and log₂(2)=1
So, Maximum bit rate = 2 × 3000 × 1 = 6000 bps.