1. What do you mean by buffer? (બફર એટલે શું ?)
3. What do you mean by redundant bit? (રીડદંડ બીટ એટલે શું ?)
4. What is the Hamming distance for d(000, 011)? d (000, 011) માટે હેમિંગ અંતર કેટલું છે?
5. What is the Hamming distance for d(10101, 11110)? (d (10101, 11110) માટે હેમિંગ અંતર કેટલું છે?)
6. What do you mean by Hamming Distance? (હમિંગ ડિસ્ટન્સનો તમારો મતલબ શું છે?)
Hamming distance is the number of bit positions in which the two bits are different.
In order to calculate the Hamming distance between two strings, and , we perform their XOR operation, (a⊕ b), and then count the total number of 1s in the resultant string.
7. For Data frame 1001101 write the frame with even parity. (ડેટા ફ્રેમ 1001101 માટે ઇવન પેરીટી ફ્રેમ લખો.)
8. For Data frame 1001101 write the frame with odd parity. (ડેટા ફ્રેમ 1001101 માટે ઓડ પેરીટી ફ્રેમ લખો.)
The corresponding binary pattern is obtained as-
Thus, for the given Polynomial the corresponding binary pattern is 11011011.
Buffer is used to temporarily store data while it is being moved from one place to another. The purpose of a buffer is to hold data right before it is used.
2. Write the full name of ARP, DLC and MAC. (એઆરપી, ડીએલસી અને એમએસીનું સંપૂર્ણ નામ લખો.)
MAC: Media Access Control
DLC: Data Link Control
ARP: Address Resolution Protocol
To detect or correct error some extra bit is added is known as redundant bit. Redundant bits are added by the sender and removed by the receiver.
The Hamming distance d(000, 011) is 2 because modulo summation is
000
⊕ 011
---------
011 (here in this answer "011" total number of 1 is two).
The Hamming distance d(10101, 11110) is 3 because modulo summation is
10101
⊕ 11110
------------
01011 (here is this answer "01011", total number of 1 is three)
The Hamming distance between two words is number of differences between corresponding bits.
7. For Data frame 1001101 write the frame with even parity. (ડેટા ફ્રેમ 1001101 માટે ઇવન પેરીટી ફ્રેમ લખો.)
( Procedure: In the case of even parity, for a given set of bits, the number of 1’s are counted. If that count is odd, the parity bit value is set to 1, making the total count of occurrences of 1’s an even number. If the total number of 1’s in a given set of bits is already even, the parity bit’s value is 0. )
here data frame : "1001101"
count the number of "1" = 4 ( which is already even, so add value "0" parity)
Answer: Even parity Frame is : "10011010"
( Procedure: In the case of odd parity, for a given set of bits, the number of 1’s are counted. If that count is even, the parity bit value is set to 1, making the total count of occurrences of 1’s an odd number. If the total number of 1’s in a given set of bits is already odd, the parity bit’s value is 0. )
here data frame : "1001101"
count the number of "1" = 4 ( which is even, so add value "1" parity)
Answer: Even parity Frame is : "10011011"
9. write the binary pattern for the polynomial x7 + x6 + x4 + x3 + x + 1.
(બહુપદી x7 + x6 + x4 + x3 + x + 1.. માટે બાયનરી પેટર્ન લખો.)
Thus, for the given Polynomial the corresponding binary pattern is 11011011.
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