MAT1101 Assignment 1 2020 MAT1101 ASSIGNMENT 1 SEMESTER 1, 2020 WEIGHT: 30% TOTAL MARKS: 30 Due date: Monday 5th April XXXXXXXXXX:55pm AEST1 SUBMISSION INSTRUCTIONS The assignment is to be...

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Answer To: MAT1101 Assignment 1 2020 MAT1101 ASSIGNMENT 1 SEMESTER 1, 2020 WEIGHT: 30% TOTAL MARKS: 30 Due...

Valupadasu answered on Apr 05 2021
146 Votes
Answers:
Question 1 answers start==================
1. The sign bit is 1.
Convert 515 to binary: 515 = 010 0000 0011
Find the 2’s complement: 101 1111 1101

The 12-bit representation for signed integer 515 is: 1101 1111 1101
2. The Sign bit is 0
Convert 51 to binary = 110011
Convert 0.25 to binary = 01
The representation for 51.25 is: 0 110011.01
Scientific notation is 1.100101 * 2^5(2 power 5)
Bias value is 31, since the exponent value is positive. The decimal value of exponent is 35.
Final value to be stored is
0 100011 100101
(As per the instruction provided in the question 6-bits of from the 12-bits are reserved for
the mantissa and 1 for sign so will left with 5 bits for exponent part so wrong value will be
saved)
3. No, the actual 12-bit floating point stored value number is 0 00011 100101
Again, this is a positive number (the first bit, the sign, is 0), the exponent is 100011 and
the mantissa is 1.100101.
Converting the exponent to decimal:
The conversion is a basic binary to decimal conversion.
1 + 2 = 3
Since the bias value is 15(as per formula 2 ̂k-1 − 1 as the exponent bias (k is 5). 15 must be
subtracted from the converted value:
3-15=-12
The resulting exponent is -12.
The whole mantissa can now be converted to decimal:
Converting the mantissa does not need the normalization to be undone . The whole number
can be calculated as follows:
1.100101* 2^-12(2 power -12)
either with negative exponents: (1+ 0.5 + 0.0625 + 0.015625) * 2^-12 = 0.000244
or fractions: (1 + 1/2 + 1/16 + 1/64) * 2^-12 = = 0.000244


4. Hexadecimal value of §USQ© is: A7 55 53 51...
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