6-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 -6.633 assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit floating point format compares to the single precision IEEE 754 standard.
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- IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmostbit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and themantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern torepresent -4.573 assuming a version of this format, which uses an excess-16format to store the exponent. Comment on how the range and accuracy of this16-bit floating point format compares to the single precision IEEE 754 standard.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.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 -1.5625 * 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. Th e 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 1.5625 101 assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit fl oating point format compares to the single precision IEEE 754 standard.When calculating error detection code FCS, you may use a technique called cyclic redundancy checking (CRC). What is the preset divisor P's length in bits for a 16-bit bit block with a total of 20 bits (total length of bit block and FCS combined)?In ___________ there is always a transition at the middle of the bit, but the bit values are determined at the beginning of the bit. If the next bit is 0, there is a transition; if the next bit is 1, there is none. a. Manchester b. differential Manchester c. both (a) and (b) d. neither (a) nor (b)
- implement bitMask(x,y) Generate a mask consisting of all 1’s from lowbit to highbit /* * bitMask - Generate a mask consisting of all 1's * lowbit and highbit* Examples: bitMask(5,3) = 0x38* Assume 0 <= lowbit <= 31, and 0 <= highbit <= 31* If lowbit > highbit, then mask should be all 0's* Legal ops: ! ~ & ^ | + << >>* Max ops: 16* Rating: 3*/int bitMask(int highbit, int lowbit) {return 2;}In ___________ there is always a transition at the middle of the bit, but the bit values are determined at the beginning of the bit. If the next bit is 0, there is a transition; if the next bit is 1, there is none.Using a 32-bit bitfield, assume two's complement encoding with a binary point between bits 23 and 24. | What is the largest possible number that can be expressed? What is the numeric precision of this format?
- One way of detecting errors is to transmit data as a block of n rows of k bits per row and adding parity bits to each row and each column. The lower-right corner is a parity bit that checks its row and its column, Will this scheme detect all single errors? Double errors? Triple errors?Represent -53.625 using 32-bit IEEE-754 floating point standard. Indicate Sign bit (1 bit), Biased Exponent bits (8 bits), and Fractional bits (23 bits). For the fractional bits, indicate just the first 9 bits out of 23. For example for 228 (slide 55 of chapter 5) your answer for the fractional bits would be 110010000. sign bit --> biased exponent --> first 9 fractional bits -->Consider a SEC code that protects 8-bit words with 4 parity bits. If we read the value0xABC, is there an error? If so, correct the error.Note: 0xABC = 1010 1011 1100Hint: Decode the 12-bit encoded data and find if there is an error.