Question

What are the largest positive representable numbers in 32-bit IEEE 754 single precision floating point and double precision floating point? Show the bit encoding and the values in base 10. a) Single Precision

b) Double Precision

link to circuit:http://i.imgur.com/7Ecb2Lw.png

Answer 1

In IEEE 754 standard Single Precision floating point will be represented with 32 bit and double precision point using 64 bit.

Initially we have to convert real number(ex: 3.5) into binary (11.1)

-3.5 will be represented as (-11.1)

11.1 is represented as 1.11*2^1

Therefore any number can be wriiten in the form of 1.xxxxx.... * 2^y

bit '1' before decimal will be common and we need to store bits after the decimal and exponent of y to decode again into respective real number.

Single Precison floating point will be represented as below

First bit will be sign bit...( 0 means positive,1 means negative real number)

Next 8 bits will be used to store the exponent of 2 in binary
Remainging 29 bits will be used to store the bits after decimal point

so -3.5 will be represented as

-11.1 = - 1.11*2^1

so sign bit will become '1'

23 bits after decimal point will become "110000000..."

Since exponent can be negative number... there are just 8 bits to represent both negative and positive numbers

using 8 bits we can store 0-255

We add bias number so that y should fall into above range

-1.11 * 2^1

1 will be stored as 1+127 = 128

if exponent is negative we have to add 127

lets say exponent is -3 then it is stored as -3 + 127 = 124

Bias value is 127 which in binary is 01111111

For double precision exponent is represented using 11 bits and bias value is 01111111111= 1023

Largest value in single precision

0.11111110 11111111111111111111111

(2-2^-23)*2*127 = 3.403*10^38

0.00000001 00000000000000000000000

2–126 = 1.1755 × 10–38 .

In double precision

Largest would be

0 11111111110 1111………………………………………..1111

(2 – 2–52)×2 (2046 – 1023) = (2 – 2–52)×2^ 1023 =10 ^3083

minimum would be

0 00000000001 0000………………………………….. 00000

2 1–1023 = 2–1022 .