Question

1. (a) Convert the following decimal numbers into their EEE-754 single-precision (32-bit) representations. Give your answers in hexadecimal form. (12 marks) (1)-3.3125 () (11) 522240 6) Convert the following IEEE 754 single-precision numbers in hexadecimal into their decimal values accurate to 5 significant figures. (8 marks) (1) 0x800E0000 (1) Ox9FACE600

Answer 1

Ans) give direct solutions hoping you know the process of conversion. Gneralisied steps are also enclosed.

1)-3.3125 = 0xc0540000

2)0 =0x00000000

3)52240 =0x48ff0000

genaralised procedure for conversion solving for 1)

1. We start with the positive version of the number:

|-3.312 5| = 3.312 5

2. First, convert to binary (base 2) the integer part: 3. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

• division = quotient + remainder;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

3₍₁₀₎ =

11₍₂₎

4. Convert to binary (base 2) the fractional part: 0.312 5. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

• #) multiplying = integer + fractional part;
• 1) 0.312 5 × 2 = 0 + 0.625;
• 2) 0.625 × 2 = 1 + 0.25;
• 3) 0.25 × 2 = 0 + 0.5;
• 4) 0.5 × 2 = 1 + 0;

5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.312 5₍₁₀₎ =

0.0101₍₂₎

Positive number before normalization:

3.312 5₍₁₀₎ =

11.0101₍₂₎

6. Normalize the binary representation of the number, shifting the decimal mark 1 positions to the left so that only one non zero digit remains to the left of it:

3.312 5₍₁₀₎ =

11.0101₍₂₎ =

11.0101₍₂₎ × 2⁰ =

1.1010 1₍₂₎ × 2¹

Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 1 (a negative number)

Exponent (unadjusted): 1

Mantissa (not normalized): 1.1010 1

7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2:

Exponent (adjusted) =

Exponent (unadjusted) + 2^(8-1) - 1 =

1 + 2^(8-1) - 1 =

(1 + 127)₍₁₀₎ =

128₍₁₀₎

• division = quotient + remainder;
• 128 ÷ 2 = 64 + 0;
• 64 ÷ 2 = 32 + 0;
• 32 ÷ 2 = 16 + 0;
• 16 ÷ 2 = 8 + 0;
• 8 ÷ 2 = 4 + 0;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =

128₍₁₀₎ =

1000 0000₍₂₎

8. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by adding the necessary number of zeros to the right:

Mantissa (normalized) =

1. 1 0101 00 0000 0000 0000 0000 =

101 0100 0000 0000 0000 0000