This set of Network Security Inteview Questions and Answers for freshers focuses on “Layers – II”.

1.

What are the bits transmitted for the NRZ-L system?

a) 10001101

b) 01001110

c) 10110001

d) 10001101

View Answer

Explanation: High – 0, Low -1.

2.

What are the bits transmitted for the return to zero system?

a) 01001

b) 10110

c) 10010

d) 01101

View Answer

Explanation: First half low – 0, First Half High – 1.

3.

What are the bits transmitted in the differential Manchester coding?

a) 001101

b) 101100

c) 110010

d) 010011

View Answer

Explanation: -ve to +ve transition – 1 bit, +ve to –ve – 0 bit.

4.

What are the bits transmitted for the AMI and Pseudoternary System?

a) 101101

b) 101100

c) 010010

d) none of the mentioned

View Answer

Explanation: Pseudoternary Has alternating high and low levels as 0 and neutral level as logic 1.

AMI has Alternating high levels as logic 1 and neutral level as logic 0.

5. Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?

a) 4600 kHz

b) 500 kHz

c) 540 kHz

d) 580 kHz

View Answer

Explanation: For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 × 100 + 4 × 10 = 540 kHz.

6. Find the checksum byte for the fallowing data words: 10110011, 10101011, 01011010, 11010101

a) 10001101

b) 01110010

c) 10110101

d) 01110010

View Answer

Explanation: Do binary addition to compute the result.

7. Generate the CRC codeword for the message x^{3}+1 using the generator polynomial x^{3}+x+1.

a) 1001101

b) 1001001

c) 1001110

d) 1001111

View Answer

Explanation: M(x) = 1001. G(x) = 1011.

x

^{3}*M(x) = 1001000.

On Dividing we get remainder as 110. Therefore, Codeword is 1001110.

8. Generate the CRC codeword for the data word 1101011011 using generator 10011. Also write both in the polynomial form.

a) 11010110110011

b) 11010110110110

c) 11010110111100

d) 11010110111110

View Answer

Explanation: Solve using polynomial division and then appending the remainder to the divisor.

9. What is the hamming distance between these 2 codes: 10010010 and 11011001?

a) 3

b) 4

c) 6

d) 2

View Answer

Explanation: Hamming distance is number of dissimilar bits between 2 streams.

10. What is the hamming code for the data: 1001101?

a) 10011100101

b) 11010000101

c) 10001100101

d) 11111001011

View Answer

Explanation:Find the 1st 2nd 4th and 8th bits using the hamming algorithm and thus proceed to get the hamming code.

11. 10010100101 is the code received. Find the error bit.

a) 7

b) 5

c) 2

d) 3

View Answer

Explanation: Error occurs in bit 7 and can be found via hamming code.

12. The channel capacity is 100Mbps, the frame length is 10000 bits and the arrival rate is 5000 frames/sec. Calculate the mean time delay.

a) 400 µsec

b) 20 msec

c) 2000 µsec

d) 200 µsec

View Answer

Explanation: T = 200 µsec from T= 1/(μC-λ).

**Sanfoundry Global Education & Learning Series – Cryptography and Network Security.**

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