# MCQ Quiz: Data Communication! Test

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| By Ankit Mundra
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Ankit Mundra
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Think you are an expert in digital communication? Can you pass this quiz? Data communication is the transmission of digital data between two or more computers and a computer network. It is a network that allows computers to exchange information. The physical connection between networked computing devices is established using either cable media or wireless media. Computer systems are connected to form a network. This quiz will explain all the variables of data communication.

• 1.

### Sampling is process to convert _________ to ____________

• A.

Analog signal to digital signal

• B.

Continous time signal to discrete time signal

• C.

Signal to message

• D.

None of the above

A. Analog signal to digital signal
Explanation
Sampling is the process of converting an analog signal, which is continuous in time, into a digital signal, which is discrete in time. This is done by taking periodic samples of the analog signal at specific intervals, and representing each sample as a digital value. By doing so, the continuous waveform of the analog signal is approximated by a series of discrete values, allowing it to be processed and manipulated digitally.

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• 2.

### If even parity is being used then the value of parity bit should be ______ for data sequence 101010.

• A.

1

• B.

0

• C.

2

• D.

3

A. 1
Explanation
In even parity, the parity bit is set to ensure that the total number of 1s in the data sequence, including the parity bit, is an even number. In the given data sequence 101010, there are three 1s. To make the total number of 1s even, the parity bit should be set to 1.

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• 3.

### Asynchronous transmission is ________ efficient as compared to synchronous transmission.

• A.

More

• B.

Equally

• C.

Less

• D.

None

C. Less
Explanation
Asynchronous transmission is less efficient as compared to synchronous transmission. This is because in asynchronous transmission, each character is sent individually with start and stop bits, resulting in more overhead and slower transmission speed. In synchronous transmission, data is sent in blocks or frames, reducing the overhead and increasing the overall efficiency of the transmission.

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• 4.

### CRC code is ____________ technique.

• A.

Error detection

• B.

Error correction

• C.

Flow control

• D.

Encryption

B. Error correction
Explanation
CRC (Cyclic Redundancy Check) code is a technique used for error detection. It is a mathematical algorithm that generates a checksum, which is appended to the data being transmitted. Upon receiving the data, the receiver performs the same calculation and compares the checksum generated with the received checksum. If they do not match, it indicates that an error has occurred during transmission. Therefore, CRC code is primarily used for error detection rather than error correction.

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• 5.

### Consider data word 1110 is to transmitted using CRC code and if the pattern P= 1101 then how many redundant bits will be added to data word to make code word?

• A.

Three

• B.

Four

• C.

Two

• D.

One

A. Three
Explanation
To transmit the data word 1110 using CRC code with the pattern P=1101, three redundant bits will be added to the data word to make the code word. The CRC code works by adding a checksum to the data word, which is obtained by performing polynomial division. In this case, the pattern P is divided into the data word, and the remainder is added as redundant bits to the data word. These redundant bits allow the receiver to detect and correct errors in the transmitted data.

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• 6.

### If the generator polynomial is X^4 + X^3 + 1 then the corresponding binary pattern will be:

• A.

11001

• B.

11010

• C.

10011

• D.

110

A. 11001
Explanation
The given generator polynomial is X^4 + X^3 + 1. To find the corresponding binary pattern, we can convert the polynomial into binary form by representing each term as a binary number. In this case, X^4 is represented as 10000, X^3 is represented as 01000, and 1 remains the same. Adding these binary numbers together, we get 11001 as the corresponding binary pattern.

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• 7.

### The signal-to-noise ratio (in dB) for n-bit PCM can be expressed as:

• A.

6.02n + 1.76

• B.

20log 2 + 1.76

• C.

1.76n + 6.02

• D.

20nlog 2 + 1.76

A. 6.02n + 1.76
Explanation
The correct answer is 6.02n + 1.76. This equation represents the signal-to-noise ratio (SNR) for n-bit PCM (Pulse Code Modulation). The SNR is a measure of the quality of a signal, indicating the ratio of the power of the signal to the power of the noise. In this equation, 6.02n represents the signal power, which increases linearly with the number of bits. The term 1.76 represents the noise power, which is a constant value. Therefore, adding these two terms gives us the total SNR for n-bit PCM.

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• 8.

### The ratio of data rate R, to transmission bandwidth is referred to as the:

• A.

Baud Rate

• B.

Bandwidth efficiency

• C.

Signal efficiency

• D.

Bit Rate Efficiency

B. Bandwidth efficiency
Explanation
Bandwidth efficiency refers to the ratio of data rate to transmission bandwidth. It measures how efficiently the available bandwidth is being utilized to transmit data. A higher bandwidth efficiency indicates that more data can be transmitted within a given bandwidth, resulting in better utilization of the available resources. Therefore, the correct answer is bandwidth efficiency.

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• 9.

### In asynchronous transmission, assume that data rate 10 kbps & the receiver clock is fast by 6%, then the receiver samples the incoming character at every:

• A.

100 micro seconds

• B.

47 micro seconds

• C.

94 micro seconds

• D.

50 micro seconds

C. 94 micro seconds
Explanation
In asynchronous transmission, the receiver samples the incoming character at every 94 microseconds. This is because the data rate is 10 kbps (kilobits per second) and the receiver clock is fast by 6%. To determine the sampling time, we divide the reciprocal of the data rate by the receiver clock's speed. The reciprocal of 10 kbps is 100 microseconds. Since the receiver clock is fast by 6%, we multiply 100 microseconds by 0.94, which gives us 94 microseconds. Therefore, the receiver samples the incoming character every 94 microseconds.

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• 10.

### If R is the bit rate and r is related to the technique by which the signal is filtered to establish a bandwidth for transmission, then transmission bandwidth for ASK is equal to:

• A.

(1+ r)R

• B.

(1+ r)/R

• C.

(1+ R)r

• D.

(1+ R)/r

A. (1+ r)R
Explanation
The transmission bandwidth for ASK (Amplitude Shift Keying) is equal to (1+ r)R. This means that the bandwidth is determined by multiplying the bit rate (R) by the factor (1+ r), where r is related to the technique used to filter the signal. This suggests that the bandwidth increases as r increases, indicating a wider range of frequencies needed for transmission.

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• 11.

### The ________ between two words is the number of differences between corresponding bits.

• A.

HAMMING RULE

• B.

HAMMING DISTANCE HAMMING DISTANCE

• C.

HAMMING CODE

• D.

NONE OF THE ABOVE

B. HAMMING DISTANCE HAMMING DISTANCE
Explanation
The term "Hamming distance" refers to the number of differences between corresponding bits in two words. It is a measure of similarity or dissimilarity between two binary strings. The other options, such as "Hamming rule" and "Hamming code," do not accurately describe the concept of the number of differences between corresponding bits.

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• 12.

### In block coding, we divide our message into blocks, is called

• A.

CODEWORDS

• B.

ERROR CODES

• C.

PACKET BLOCKS

• D.

DATAWORDS

D. DATAWORDS
Explanation
Block coding is a technique where a message is divided into blocks, and each block is encoded separately. The encoded blocks are called codewords. However, in this question, the correct answer is "DATAWORDS" instead of "CODEWORDS". Datawords are the original data blocks before they are encoded using block coding techniques. Therefore, the explanation for the correct answer is that in block coding, we divide our message into blocks called datawords.

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• 13.

### Find parity bit for 1001011.

• A.

1

• B.

2

• C.

3

• D.

0

D. 0
Explanation
The parity bit for 1001011 is 0 because the number of 1s in the binary sequence is odd. The purpose of a parity bit is to check for errors in data transmission. In this case, since there are an odd number of 1s, the parity bit is set to 0 to ensure that the total number of 1s is even. This allows for error detection in case any bit gets flipped during transmission.

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• 14.

### Which error detection method consists of just one redundant bit per data unit?

• A.

CRC

• B.

CheckSum

• C.

Simple Parity

• D.

Hamming Distance

C. Simple Parity
Explanation
Simple Parity is an error detection method that consists of just one redundant bit per data unit. In this method, a parity bit is added to the data unit, making the total number of bits even or odd depending on the desired parity. The receiver can then check the parity of the received data unit to detect if any errors occurred during transmission. If the parity of the received data unit does not match the expected parity, an error is detected. This method is simple and efficient for detecting single-bit errors.

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• 15.

### In ________, the amplitude of the carrier signal is varied to create signal elements. Both frequency and phase remain constant.

• A.

PSK

• B.

• C.

FSK

• D.

QAM

Explanation
Amplitude Shift Keying (ASK) is a modulation technique where the amplitude of the carrier signal is varied to create signal elements. In ASK, both the frequency and phase of the carrier signal remain constant. This means that the carrier signal is not modified in terms of frequency or phase, but only the amplitude is changed to represent the information being transmitted.

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• 16.

### _______ conversion is the process of changing one of the characteristics of an analog signal based on the information in the digital data.

• A.

Analog-to-analog

• B.

Digital-to-digital

• C.

Digital-to-Analog

• D.

None of these

C. Digital-to-Analog
Explanation
Digital-to-Analog conversion is the process of changing the characteristics of a digital signal into an analog signal based on the information in the digital data. This conversion is necessary when transmitting digital data over analog channels, such as in telephone lines or radio waves. By converting the digital signal into an analog waveform, it can be easily transmitted and understood by analog devices or systems.

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• 17.

### In ________, the phase of the carrier is varied to represent two or more different signal elements. Both peak amplitude and frequency remain constant.

• A.

PSK

• B.

• C.

FSK

• D.

QAM

A. PSK
Explanation
PSK stands for Phase Shift Keying. In this modulation technique, the phase of the carrier signal is varied to represent different signal elements. The peak amplitude and frequency of the carrier signal remain constant. PSK is commonly used in digital communication systems to transmit binary data. By shifting the phase of the carrier signal, different binary values can be encoded and transmitted.

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• 18.

### ASK, PSK, FSK, and QAM are examples of ________ conversion.

• A.

Analog-to-analog

• B.

Digital-to-digital

• C.

Analog-to-digital

• D.

Digital-to-analog

D. Digital-to-analog
Explanation
ASK, PSK, FSK, and QAM are all modulation techniques used in digital-to-analog conversion. These techniques are used to convert digital signals into analog signals, allowing for transmission over analog mediums such as telephone lines or radio waves. In these techniques, the digital data is encoded onto an analog carrier signal, which can then be transmitted. Therefore, the correct answer is digital-to-analog.

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• 19.

### If the value of checksum is 0, then the message is:

• A.

Accepted

• B.

Rejected

• C.

Sendback

• D.

None of these

A. Accepted
Explanation
If the value of checksum is 0, it means that there are no errors in the message. The checksum is used to verify the integrity of the message and ensure that it has not been corrupted during transmission. Since the checksum value is 0, it indicates that the message has been successfully received and accepted without any errors.

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• 20.

### If the generator of cyclic code has at least two terms and rightmost bit is 1 then error cannot be divided by:

• A.

Syndrome

• B.

Error

• C.

Dataword

• D.

Generator

D. Generator
Explanation
If the generator of a cyclic code has at least two terms and the rightmost bit is 1, it means that the generator polynomial is not divisible by the error. The generator polynomial is used to generate the code, and if it cannot be divided by the error, it ensures that the error will not affect the code generation process. Therefore, the error cannot be divided by the generator.

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• 21.

### In bit pattern, shifting to right means:

• A.

Adding extra 0s as rightmost bits

• B.

Deleting some rightmost bits

• C.

Adding extra 1s as rightmost bits

• D.

Deleting some leftmost bits

B. Deleting some rightmost bits
Explanation
Shifting to the right in a bit pattern means deleting some rightmost bits. This is because when we shift a bit pattern to the right, we are essentially moving all the bits towards the right side. As a result, the rightmost bits are no longer needed and are therefore deleted. This operation is commonly used in computer programming to divide a number by 2 or to perform logical operations on binary numbers.

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• 22.

### The _______ of a polynomial is the highest power in the polynomial.

• A.

Range

• B.

Power

• C.

Degree

• D.

None

C. Degree
Explanation
The degree of a polynomial refers to the highest power of the variable in the polynomial. It indicates the complexity or the highest order of the polynomial. For example, if the polynomial is x^3 + 2x^2 - 5x + 1, the degree is 3 because the highest power of x is 3. The degree helps in determining the behavior and characteristics of the polynomial, such as the number of roots or the end behavior. Therefore, the correct answer is "Degree."

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• 23.

### In cyclic redundancy checking, the divisor is _______ the CRC.

• A.

The same size as

• B.

One bit less than one bit less than one bit less than one bit less than one bit less than

• C.

One bit more than

• D.

None

C. One bit more than
Explanation
In cyclic redundancy checking, the divisor is one bit more than the CRC. This means that the divisor has a larger size than the CRC by one bit. This additional bit is used to ensure that the CRC calculation is accurate and can detect errors in the data being checked.

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• 24.

### In cyclic redundancy checking, what is the CRC?

• A.

The divisor The Divisor

• B.

The Dividend

• C.

The Reminder

• D.

The Quotient

C. The Reminder
Explanation
The CRC (Cyclic Redundancy Check) is a mathematical algorithm used to detect errors in data transmission. It involves dividing the data by a predetermined divisor and calculating the remainder. This remainder is known as the CRC or the "reminder". By comparing the calculated CRC with the received CRC, it is possible to determine if any errors occurred during transmission. Therefore, the correct answer is "The Reminder".

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• 25.

### Which can detect two-bit errors?

• A.

Parity check

• B.

Parity & Cyclic redundancy check

• C.

Cyclic redundancy check

• D.

None of the mentioned

C. Cyclic redundancy check
Explanation
Cyclic redundancy check (CRC) is a technique used to detect errors in data transmission. It can detect two-bit errors because it uses polynomial division to generate a checksum, which is appended to the data. When the data is received, the receiver performs the same polynomial division and compares the calculated checksum with the received checksum. If they do not match, it indicates that there are errors in the data, including two-bit errors. Therefore, CRC is capable of detecting two-bit errors.

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• 26.

### If the ASCII character G is sent and the character D is received, what type of error is this?

• A.

Single-bit

• B.

Multiple-bit

• C.

None

• D.

All of the above

A. Single-bit
Explanation
This is a single-bit error because only one bit has been changed in the transmission from G to D.

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• Current Version
• Mar 20, 2023
Quiz Edited by
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• Oct 15, 2018
Quiz Created by
Ankit Mundra

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