i) 0 1110 1111111 sign bit is 0(+ve) exp bits are 1110 => 1110 => 1x2^3+1x2^2+1x2^1+0x2^0 => 1x8+1x4+1x2+0x1 => 8+4+2+0 => 14 in decimal it is 14 so, exponent/bias is 14-7 = 7 frac bits are 1111111 IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1.1111111 * 2^7 1.1111111 in decimal is 1.9921875 => 1.1111111 => 1x2^0+1x2^-1+1x2^-2+1x2^-3+1x2^-4+1x2^-5+1x2^-6+1x2^-7 => 1x1+1x0.5+1x0.25+1x0.125+1x0.0625+1x0.03125+1x0.015625+1x0.0078125 => 1+0.5+0.25+0.125+0.0625+0.03125+0.015625+0.0078125 => 1.9921875 so, 1.9921875 * 2^7 in decimal is 255.0 so, 011101111111 in IEEE-754 format is 255.0 Answer: 255.0 ii) 0 0001 0000000 sign bit is 0(+ve) exp bits are 0001 => 0001 => 0x2^3+0x2^2+0x2^1+1x2^0 => 0x8+0x4+0x2+1x1 => 0+0+0+1 => 1 in decimal it is 1 so, exponent/bias is 1-7 = -6 frac bits are IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1. * 2^-6 1. in decimal is 1 => 1. => 1x2^0 => 1x1 => 1 => 1 so, 1 * 2^-6 in decimal is 0.015625 so, 000010000000 in IEEE-754 format is 0.015625 Answer: 0.015625 iii) 0 0000 1111111 sign bit is 0(+ve) exp bits are 0000 => 0000 => 0x2^3+0x2^2+0x2^1+0x2^0 => 0x8+0x4+0x2+0x1 => 0+0+0+0 => 0 in decimal it is 0 so, exponent/bias is 0-7 = -7 frac bits are 1111111 IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1.1111111 * 2^-7 1.1111111 in decimal is 1.9921875 => 1.1111111 => 1x2^0+1x2^-1+1x2^-2+1x2^-3+1x2^-4+1x2^-5+1x2^-6+1x2^-7 => 1x1+1x0.5+1x0.25+1x0.125+1x0.0625+1x0.03125+1x0.015625+1x0.0078125 => 1+0.5+0.25+0.125+0.0625+0.03125+0.015625+0.0078125 => 1.9921875 so, 1.9921875 * 2^-7 in decimal is 0.01556396484375 so, 000001111111 in IEEE-754 format is 0.01556396484375 Answer: 0.01556396484375 iv) 0 0000 0000001 sign bit is 0(+ve) exp bits are 0000 => 0000 => 0x2^3+0x2^2+0x2^1+0x2^0 => 0x8+0x4+0x2+0x1 => 0+0+0+0 => 0 in decimal it is 0 so, exponent/bias is 0-7 = -7 frac bits are 0000001 IEEE-754 Decimal value is 1.frac * 2^exponent IEEE-754 Decimal value is 1.0000001 * 2^-7 1.0000001 in decimal is 1.0078125 => 1.0000001 => 1x2^0+0x2^-1+0x2^-2+0x2^-3+0x2^-4+0x2^-5+0x2^-6+1x2^-7 => 1x1+0x0.5+0x0.25+0x0.125+0x0.0625+0x0.03125+0x0.015625+1x0.0078125 => 1+0.0+0.0+0.0+0.0+0.0+0.0+0.0078125 => 1.0078125 so, 1.0078125 * 2^-7 in decimal is 0.00787353515625 so, 000000000001 in IEEE-754 format is 0.00787353515625 Answer: 0.00787353515625
Assignment (show how you calculate these): 1) According to the IEEE 754 protocols, compute the following...
To write a C program (not C++) that converts numbers between Decimal and IEEE-754 format and vice versa. Inputs: Number in Decimal format (including special case of 0) Number in IEEE-754 format (including special cases) Output: Equivalent number in IEEE-754 format Equivalent number in Decimal Specification: The program converts a number based on choosing from a menu of choices, where each choice calls the appropriate procedure, where the choices are: Decimal to IEEE-754 conversion IEEE-754 to Decimal conversion Quit program...
Watching a YouTube tutorial on how to convert decimal to floating point numbers (IEEE 754) and normalisation may prove to be beneficial. Watching a YouTube tutorial on how to convert decimal to floating point numbers (IEEE 754) may prove to be beneficial Convert the decimal number to 32 bits I Decimal number 18 to its binary equivalent I. 18 normalized in binary: 1.-2刈2n) II Biased exponent: 10 IV. Conversion to EE 754 16 I: 10, For ii please normalize the...
1 please IEEE-754 Floating point conversions problems (assume 32 bit machine): 1. For IEEE 754 single-precision floating point, write the hexadecimal representation for the following decimal values: a. 27.1015625 b.-1 2. For IEEE 754 single-precision floating point, what is the decimal number, whose hexadecimal representation is the following? a. 4280 0000 b. 7FE4 0000 c. 0061 0000 3. For IEEE-754 single-precision floating point practice the following problem: Suppose X and Y are representing single precision numbers as follows: X 0100...
This problem covers floating-point IEEE format. (a) Assuming single precision IEEE 754 format, what is the binary pattern for decimal number -6.16? (b) Assuming single precision IEEE 754 format, what decimal number is represented by this word: 0 01111100 01100000000000000000000 (Hint: remember to use the biased form of the exponent.)
2. Represent 25.28255 in 32 bit IEEE-754 floating point format as shown in the following format discussed in class. Sign Bit BIT 31 Exponent BITS 30:23 Mantissa BITS 22:0 BYTE 3+1 bit 7 Bits BYTE 1 BYTE O
5, [points] This problem covers floating-point IEEE format. (a) Assuming single precision IEEE 754 format, what is the binary pattern for decimal number -6.16? (b) Assuming single precision IEEE 754 format, what decimal number is represented by this word: 0 01111100 01100000000000000000000 (Hint: remember to use the biased form of the exponent.)
Question 19 In representing the decimal number 11/8 in IEEE 754 representation, the 8-bit binary value of the exponent plus bias for single precision is 01111111 01111110 01111101 10000000 Question 13 In representing the decimal number 11/8 in IEEE 754 representation, the 23-bit binary value of the fraction for single precision is 110 0000 0000 0000 0000 0000 111 0000 0000 0000 0000 0000 100 0000 0000 0000 0000 0000 011 0000 0000 0000 0000 0000 Question 12 The point...
Consider the following 32 bit binary representation of the value using IEEE 754 single precision floating point representation. Show the corresponding signed number in decimal. 01000001001010100000000000000000
Can you write process of the question? A fictional floating-point encoding scheme uses 1 bit for sign followed by 1 bit for the exponent and 2 bits for the mantissa. It otherwise behaves exactly like the IEEE 754 encoding scheme. List down all decimal values that can be represented by this scheme along with their binary representation.
What would be the IEEE 754 double precision floating point representation of 1.32487359893280124981233898124124 times 10^-17. For explanation, I want you to document the steps you perform, in this order: (1) What is n in decimal fixed point form (ddd.ddd,dd); (2) What is n in binary fixed point form (bbb.bbbb), storing the first 110 bits following the binary point); (3) What is the normalized binary number, written in the form 1.bbbbb...bbb times 2^e, storing 54 bits following the binary point) (4)...