Q.1 How can you determine the no. of redundant bits in Hamming Code error correction technique? Q.2 Compare line discipline methods ENQ/ACK and Poll/Select? Q.3 What the receiver wil receive if the check sum method is applied to the following bit? 10101001 00111001
http://www.google.co.uk/search?q=5+gigabits+in+megabits 5120 megabits in 5 gigabits
did u get the q papers
1+q
1 Byte is 8 bits
4 bits=1 nybble 2 nybbles=1 byte8 bits in a byte
8 Bits = 1 Byte
1 bit = a 1 or 0 (b) 4 bits = 1 nybble (?) 8 bits = 1 byte (B)
8 bits
Factoring the left and right parts separately gives: q3 - q2 + 2q - 2 = q(q2 - 1) + 2(q - 1) = q(q + 1)(q - 1) + 2(q - 1) Now we have a common factor (q - 1) that we can take out. I'll combine the remaining terms again: (q - 1)[q(q + 1) + 2(q - 1)] = (q - 1)[q2 + q + 2q - 2] = (q - 1)(q2 + 3q - 2) The right part has no obvious factorization; but you can use the quadratic formula to get the factors, which will probably include square roots.
To convert megabits to bits you just have to multiply megabits by 1,048,576 bits. 1 megabit = 1,048,576 bits.
8 Bits = 1 Byte Also, 4 Bits = 1 Nibble = Half Byte