Question

2. Convert the following real numbers into single precision IEEE floating point format. Give the final answer in hexadecimal and specify: the sign bit, exponent bits, and significand bits. Show your work. (10 + 10 points) A. 69.625 B. -123.7 the following IEEE single precision floating point numbers. Show your work. (10 + 10 points) A. 0xc1be0000 B. 0x42c68000

Answer 1

2.

A. 69.625

First i am converting decimal to 32 bit single precision IEEE binary floating point.

in 32 bits: 1st bit for sign,next 8 bits for exponent,next 23 bits for mantissa

First converting 69 to binary:

we divide number by 2 and write quotient,remainder a side.

then repeat the process by dividing quotient by 2 and write quotient,remainder a side.

repeat process until quotient is 0.

69 ÷ 2 = 34 (quotient), remainder 1;

34 ÷ 2 = 17 (quotient), remainder 0;

17 ÷ 2 = 8 (quotient), remainder 1;

8 ÷ 2 = 4 (quotient), remainder 0;

4 ÷ 2 = 2 (quotient), remainder 0;

2 ÷ 2 = 1 (quotient), remainder 0;

1 ÷ 2 = 0 (quotient), remainder 1;

quotient is 0. so take the remainders in reverse order (100 0101)_2.

Next convert fractional decimal into binary:

0.625:

Process : multiply fractional part with 2 continuously , write the integer part a side. Next repeat this process with new fractional part until the fractional part becomes 0.

0.625 × 2 = 1 (integer part), new fractional part 0.25;

0.25 × 2 = 0 (integer part), new fractional part 0.5;

0.5 × 2 = 1 (integer part), new fractional part 0;

take the integer parts from top to bottom (101)₂

total number 69.625 = 100 0101.101₂

Normalizing binary number:

shift the decimal point to left so that only one non zero digit will be left before decimal point:

Normalized the binary number = 1.000101101 X 2⁶

up to this we have

sign bit : 0 positive number

exponent : 6 (unadjusted)

mantissa : 1.000101101

But as specified, exponent is of 8 - bits, so we need to adjust the exponent so that it will have 8 bits

We have a formula : Exponent adjusted =  Exponent unadjusted + 2^(8-1) (8 is the no of bits in exponent) - 1

= 6 + 127 -1

= 133₁₀

now convert 133 to binary number :

33 ÷ 2 = 66 (quotient), remainder 1;

66 ÷ 2 = 33 (quotient), remainder 0;

33 ÷ 2 = 16 (quotient), remainder 1;

16 ÷ 2 = 8 (quotient), remainder 0;

8 ÷ 2 = 4 (quotient), remainder 0;

4 ÷ 2 = 2 (quotient), remainder 0;

2 ÷ 2 = 1 (quotient), remainder 0;

1 ÷ 2 = 0 (quotient), remainder 1;

writing in reverse order 133 = 1000 0101₂

Now Normalizing the mantissa, Remove the 1 to left of decimal and adjust the binary number so that it contains 23 bits

Normalized mantissa : 00010110100000000000000₂

Overall binary representation is:

69.625₁₀ = 0 10000101 00010110100000000000000₂ (1st bit is sign bit, next 8 are exponent, next 23 are mantissa)

sign bit 0 means it is positive number

1 means it is negative number

binary	Hexa decimal
0000	0
0001	1
0010	2
0011	3
0100	4
0101	5
0110	6
0111	7
1000	8
1001	9
1010	A
1011	B
1100	C
1101	D
1110	E
1111	F

Converting it into HEXA DECIMAL: ox428B4000 FROM TABLE

b.-123.75₁₀

First i am converting decimal to 32 bit single precision IEEE binary floating point.

in 32 bits: 1st bit for sign,next 8 bits for exponent,next 23 bits for mantissa

First converting |-123| (take mod) to binary:

we divide number by 2 and write quotient,remainder a side.

then repeat the process by dividing quotient by 2 and write quotient,remainder a side.

123 ÷ 2 = 61 (quotient), remainder  1;

61 ÷ 2 = 30 (quotient), remainder 1;

30 ÷ 2 = 15 (quotient), remainder 0;

15 ÷ 2 = 7 (quotient), remainder 1;

7 ÷ 2 = 3 (quotient), remainder  1;

3 ÷ 2 = 1 (quotient), remainder 1;

1 ÷ 2 = 0 (quotient), remainder 1;

123 = 111 1011₂

Next convert fractional decimal into binary:

0.75:

Process : multiply fractional part with 2 continuously , write the integer part a side. Next repeat this process with new fractional part until the fractional part becomes 0.

0.75 × 2 = 1 (integer part), new fractional part 0.5

0.5 × 2 = 1 (integer part), new fractional part 0

0.75₁₀  = 11₂

total number 123.75₁₀ = 111 1011.11₂

Normalizing the binary number:

shift the decimal point to left so that only one non zero digit will be left before decimal point:

Normalized binary number = 1.11 101111 X 2⁶

up to this we have

sign bit : 1 so it is negative number

exponent : 6 (unadjusted)

mantissa : 1.11 101111

But as specified exponent is of 8 - bits, we need to adjust the exponent so that it will have 8 bits

We have a formula : Exponent adjusted =  Exponent unadjusted + 2^(8-1) (8 is the no of bits in exponent) - 1

= 6 + 127 -1

= 133₁₀

now convert 133 to binary number :

33 ÷ 2 = 66 (quotient), remainder 1;

66 ÷ 2 = 33 (quotient), remainder 0;

33 ÷ 2 = 16 (quotient), remainder 1;

16 ÷ 2 = 8 (quotient), remainder 0;

8 ÷ 2 = 4 (quotient), remainder 0;

4 ÷ 2 = 2 (quotient), remainder 0;

2 ÷ 2 = 1 (quotient), remainder 0;

1 ÷ 2 = 0 (quotient), remainder 1;

writing in reverse order 133₁₀ = 1000 0101₂

Now Normalizing the mantissa, Remove the 1 to left of decimal and adjust the binary number so that it contains 23 bits.

Normalized mantissa : 11 101111000000000000000₂

Overall binary representation is:

-123.75₁₀ = 1 10000101 11101111000000000000000₂ (1st bit is sign bit, next 8 are exponent, next 23 are mantissa)

sign bit 0 means it is positive number

1 means it is negative number

Converting it into HEXA DECIMAL: oxC2F78000 (for every 4 bits)

3.

a. 0xc1be0000

Converting it to binary from table :

1 10000011 01111100000000000000000₂

1 - sign bit (so negative number)

10000011 -exponent

01111100000000000000000 - mantissa

Converting the binary to decimal for exponent:

multiply the bits from left side with their 2^{position number} (position number starts from 0), add the all numbers

10000011₂ : 1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰ = 128 + 0 + 0 + 0 + 0 + 0 + 2 + 1=131₁₀

As in the previous example we have adjusted the exponent, here also same thing ,

we need to subtract 131 from (2^(8-1) (8 is the no of bits in exponent) - 1) (reverse process to the 2nd problems)

131-127 = 4 (adjusted exponent)

Convert mantissa to decimal :

01111100000000000000000₂ = 0 × 2^-1 + 1 × 2^-2 + 1 × 2^-3 + 1 × 2^-4 + 1 × 2^-5 + 1 × 2^-6 + 0 × 2^-7 + 0 × 2^-8 + 0 × 2^-9 + 0 × 2^-10 + 0 × 2^-11 + 0 × 2^-12 + 0 × 2^-13 + 0 × 2^-14 + 0 × 2^-15 + 0 × 2^-16 + 0 × 2^-17 + 0 × 2^-18 + 0 × 2^-19 + 0 × 2^-20 + 0 × 2^-21 + 0 × 2^-22 + 0 × 2^-23 =  

0 + 0.25 + 0.125 + 0.062 5 + 0.031 25 + 0.015 625 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =

= 0.484 375₍₁₀₎

Formula for converting binary to decimal : (-1)^{Sign bit} × (1 + mantissa) × 2^{(adjusted Exponent )}

= (-1)¹ × (1 + 0.484 375) × 2^{4 =} - 23.75₁₀

final answer is 0xc1be0000 = - 23.75₁₀

b. 0x42c68000

Converting it to binary :

0 10000101 10001101000000000000000

0 - sign bit (so positive number)

10000101 -exponent

10001101000000000000000 - mantissa

Converting the binary to decimal for exponent:

multiply the bits from left side with their 2^{position number} (position number starts from 0), add the all numbers

10000101₂ : 1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰ = 133₁₀

As in the previous example (2nd problem) we have adjusted the exponent, here also same thing ,

we need to subtract 133 from (2^(8-1) (8 is the no of bits in exponent) - 1) (reverse process to the 2nd problems)

133-127 = 6 (adjusted exponent)

Convert mantissa to decimal :

1 × 2^-1 + 0 × 2^-2 + 0 × 2^-3 + 0 × 2^-4 + 1 × 2^-5 + 1 × 2^-6 + 0 × 2^-7 + 1 × 2^-8 + 0 × 2^-9 + 0 × 2^-10 + 0 × 2^-11 + 0 × 2^-12 + 0 × 2^-13 + 0 × 2^-14 + 0 × 2^-15 + 0 × 2^-16 + 0 × 2^-17 + 0 × 2^-18 + 0 × 2^-19 + 0 × 2^-20 + 0 × 2^-21 + 0 × 2^-22 + 0 × 2^-23

= 0.5 + 0 + 0 + 0 + 0.031 25 + 0.015 625 + 0 + 0.003 906 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

10001101000000000000000₂ =0.550 781 25₁₀

Formula for converting binary to decimal : (-1)^{Sign bit} × (1 + mantissa) × 2^{(adjusted Exponent )}

(-1)⁰ × (1 + 0.550 781 25) × 2⁶

=1.550 781 25 × 2⁶

=99.25₁₀

final answer is 99.25₁₀