AMI line coding with Binary 8 with zero substitution (B8ZS) is shown below. Is that line coding encoded correctly? Discuss the answer by referring the bit position. Bit position 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Data 1101 01 00000000 AMI(B8ZS) Bit position 1 2 3 4 5 6 7 8 9 10 11 12 13 14
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- An 8-bit byte with binary value 10101111 is to be encoded using an even-parity Hamming code. What is the binary value after encoding? ANS: The encoded value is 101001001111 Why is this answer longer from binary value? Please solve this question with details. Please explain why you use that method.Determinethe Decimal Value (Base 10) of the following IEEE-754 32-bit floating-point representation. Show your working. (Hint: identify the sign bit, the biased exponent and the trailing significand /mantissa). 0 01111111 110 0000 0000 0000 0000 0000IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.5625 * 10-2 assuming a version of this format. Calculate the sum of 2.6125*102 and 4.150390625 * 10-1 by hand,assuming both numbers are stored in the 16-bit half precision described above. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps.
- IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.625 * 10-2 assuming a version of this format. Calculate the sum of 2.625*102 and 4.150390625 * 10-1 by hand, assuming both numbers are stored in the 16-bit half precision described above. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps.Show the IEEE 754 bit-pattern for the following numbers (assume excess 127): a. 1.02384375 x 103 b. 5.859375 x 10-3There is an error in the following 7-bit Hamming code with EVEN parity bits. P1 P6 P7 0 0 0 P2 P3 1 1 0 0 P4 1 1 a) What is the position of the erroneous bit? b) What is the correct code? P5 1 1
- Manchester encoding guarantees frequent clock synchronization by changing signal values. What is the maximum number of bits which may be encoded without a signal change? (Note: The way this question has been asked. you are not to include the bit which includes a signal change in your count)Determine the Decimal Value (Base 10) of the following IEEE-754 32-bit floating-point representation. Show your working. (Hint: identify the sign bit, the biased exponent and the trailing significand /mantissa).0 01111111 110 0000 0000 0000 0000 0000IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent 0.2 assuming a version of this format.
- QUESTION 1 We are multiplying 111 1101 and 0000 1110, and are part way through the calculation. Insert the row that is missing and do the next step by filling in the next digit of the result, the next carry one over and the next carry two over. 1111 1101 0000 1110 11111 1010 111 1110 1000 0 1000 0000 carries two over 1011 0000 carries one over 101 0110 result 11 1111 0100 1 1111 1010 111 1110 1000Using even parity, add parity bits to the following bit patterns 0110 100 1011 011 0000 000 Using odd parity, add parity bits to the following bit patterns 0110 100 1011 011 0000 0000The following bit patterns were originally encoded using even parity. In which of them have errors definitely occurred? 10100101 01111010 10001010 01010000 11110011