SInce the number system used is a 16-bit binary system(also referred as 16 base 2 format,where 16 represents number of bits used and 2 represents base of the number system:binary)
In general the sign of the number systems will be represented in the Most Significant Bit (MSB)position on the left most position of the number that is represented just like representing an integer
for signed numbers , the value of MSB signifies the positive or negative symbol for the number
MSB=1 means negative
MSB=0 means positive
for the given number system , the number format is like :
1 bit (MSB) for representing the sign +/-
6 bits to represent the value of exponent
9 bits for representing the mantissa
making a total of 16 bits as shown below
IT WAS ALSO MENTIONED THAT THE NUMBER SYSTEM IS SIGNED AND BIASED .This means the value that represents the exponent is an offset from the actual value.this helps higher values to be stored in less number of bits in 2's complement form instead of direct representation of binary values
Here the numbee system uses 6 bit exponent which is capable of representing values from 0- 26 (0 to 31) making a total of 32 values with a maximum bias value : 31
in order to get the true exponent the offset of 15(011112) has to be subtracted from the stored exponent.
The stored exponents
(00000)2 used to represent zero and
(11111)2 used to represent infinity
are interpreted specially in floating point number system.
since an infinity value cannot be represented numerically with a limited number of bits in the mantissa,all the bits representing the mantissa value remain zeroes
the floating point 16-bit representation of + infinity is shown below:
bits 0-8 = 000000 (mantissa all bits reset)
bits 10-14=111111 (representing special interpretation of value infinity)
bit 15(MSB) = 0 (representing + symbol)
I need all steps with the reason and detail Thanks on 15 0 out of 1...
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please help Problem 4 (10 points): 1. Consider the numbers 23.724 and 0.3344770219. Please normalize both 2. Calculate their sum by hand. 3. Convert to binary assuming each number is stored in a 16-bit register. Half-precision binary floating-point has: sign bit: lbit, exponent width: 5bits and a bias of 15, and significand 10 bits (16 bits total) 4. Show cach step of their binary addition, assuming you have one guard, one round, and one sticky bit, rounding to the nearest...
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Please show work, thanks. Consider the following two 16-bit floating-point representations 1. Format A. There is one sign bit There are k 6 exponent bits. The exponent bias is 31 (011111) There are n 9 fraction/mantissa bits 2. Format B There is one sign bit There are k 5 exponent bits. The exponent bias is 15 (01111) There are n 10 fraction/mantissa bits Problem 1 (81 points total /3 points per blank) Below, you are given some bit patterns in...
Calculate 1.666015625 x 10° (1.9760 x 104 + - 1.9744 x 10^) by hand, assuming each of the values are stored in the 16-bit half precision format IEEE 754-2008. IEEE 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. Assume 1 guard, 1 round bit,...
The answer below is correct, but can you guys show me how to get it? I need conplete solution. thanks Consider a 16-bit floating-point representation based on the IEEE floating-point format, with one sign bit, seven exponent bits (k=7), and eight fraction bits (n:8). The exponent bias is 27-1-1-63. Fill in the table that follows for each of the numbers given, with the following instructions for each column: Hex: the four hexadecimal digits describing the encoded form. M the value...
I need help with this please. Calculate: 3758 + 7728 = 101110001012 + 101110001012 = ADF16 – 12316 = 10100111012 - 6368 = 16 For the following do the subtraction in 2’s compliment arithmetic in binary. Use 8 bit size binary representation for each of the values below. All the following values are base 10 numbers. Show all work please. 13 – 9 = 29 – 45 = 81 – 81 = ...