## Saturday, March 30, 2013

## Sunday, March 3, 2013

### IEEE-754 (anon. post)

*Summary of IEEE-754*

This is a summary of what I understand about IEEE-754 notation.

*CMIIW ?*

For the purpose of explanation, single precision is chosen.

*Normalised:*

IEEE-754 is another way to represent the scientific notation.

Example of scientific notation : 1.00 x 10^5

Where 1.00 represents the fraction part (Mantissa), and 5 represent the exponent part. 10 represents the radix.

Note that the explanation below uses binary.

Notation :

Notation :

First bit = Sign of the number, 0 as positive, 1 as negative - the 'sign bit'

The next 8 bits = the Exponent part. Represented by 8-bit (in excess-127 format)

FORBIDDEN à 00000000 and 11111111 are forbidden for normalised single precision, will be explained later.

00000000 as an exponent is ONLY FOR DENORMALISED numbers

11111111 as exponent is ONLY FOR INFINITY.

00000000 as an exponent is ONLY FOR DENORMALISED numbers

11111111 as exponent is ONLY FOR INFINITY.

The
last 23 bits = the Fraction part (aka the Mantissa).

The next 23 bits represent the bits to the right of the binary point. To the left of the binary point is always a 1 (strictly for normalised - and therefore it can be disregarded and assumed to be a one).

All 0 (twenty-three 0's) represents the value of 1.0 while all 1 (twenty-three 1's) represents the value of

Note : Binary point is the dot between the number of the fraction part,

For negative : From -2.0 x 2^(127) to -1.0 x 2^(-126)

For positive : From 1.0 x 2^(-126) to 2.0 x 2^(127)

1.125 (decimal) in IEEE-754 :

In scientific notation of binary: 1.125 = 1.001 x 2^(0).

Sign : Positive (0)

Fraction part : 1.001 (2^0 + 2^(-3))

Exponent part : 0

Solve the Exponent part first; exponent part must be in the form of 8-bit excess-127.

0 in 8-bit excess-127 à 10000000

Fraction part à In IEEE-754, ignore the 1 to the left of the binary point, and COPY the numbers to the right of the binary point (001 in this example) and then fulfill the 23 bits requirement (in this case 001 + twenty more 0s).

Therefore the 1.125 in IEEE-754 notation become:

0 10000000 00100000000000000000000

(Spaces between the sign, exponent and fraction were used to make things more visible)

To make it shorter, convert to Hexadecimal

1.125 in IEEE-754 notation in hexadecimal = 40100000

Basically
the same as normalised. However, exponent part is ALWAYS 00000000.
Although in excess-127 00000000 is considered to be -127, FOR
denormalised form, 00000000 indicates that the exponent is -126, but the
number to the left of the binary point of the fraction part is ALWAYS
0.

For denormalised, all 0 in the fraction part represents the value of 0.0 and all 1 in the fraction part represents the value of approximately 1.0.

Example : 0.001 x 2^(-126)

Fraction part is ALWAYS BELOW 1.0, exponent part is ALWAYS -126.

Range :

Negative : From 0 to -1 x 2^(-126)

Positive : From 0 to 1 x 2^(-126)

Note : Two zeros exist: +0 and -0.

Smallest possible non zero value à -2^(-23) * 2^(-126) and 2^(-23) * 2^(-126)

Basically the same thing except the exponent part is 11111111 with all 0 as fraction part.

Note : There exist positive infinity and negative infinity.

32 bit and 64 bit floating point formats:

The next 23 bits represent the bits to the right of the binary point. To the left of the binary point is always a 1 (strictly for normalised - and therefore it can be disregarded and assumed to be a one).

All 0 (twenty-three 0's) represents the value of 1.0 while all 1 (twenty-three 1's) represents the value of

*approximately*2.0.Note : Binary point is the dot between the number of the fraction part,

*like a decimal point - but in binary*:)

Range :Range :

For negative : From -2.0 x 2^(127) to -1.0 x 2^(-126)

For positive : From 1.0 x 2^(-126) to 2.0 x 2^(127)

Example :Example :

1.125 (decimal) in IEEE-754 :

In scientific notation of binary: 1.125 = 1.001 x 2^(0).

Sign : Positive (0)

Fraction part : 1.001 (2^0 + 2^(-3))

Exponent part : 0

Solve the Exponent part first; exponent part must be in the form of 8-bit excess-127.

0 in 8-bit excess-127 à 10000000

Fraction part à In IEEE-754, ignore the 1 to the left of the binary point, and COPY the numbers to the right of the binary point (001 in this example) and then fulfill the 23 bits requirement (in this case 001 + twenty more 0s).

Therefore the 1.125 in IEEE-754 notation become:

0 10000000 00100000000000000000000

(Spaces between the sign, exponent and fraction were used to make things more visible)

To make it shorter, convert to Hexadecimal

1.125 in IEEE-754 notation in hexadecimal = 40100000

*Denormalised*

Note :Note :

For denormalised, all 0 in the fraction part represents the value of 0.0 and all 1 in the fraction part represents the value of approximately 1.0.

*What can be written with denormalised?*

Fraction part is ALWAYS BELOW 1.0, exponent part is ALWAYS -126.

Range :

Negative : From 0 to -1 x 2^(-126)

Positive : From 0 to 1 x 2^(-126)

Note : Two zeros exist: +0 and -0.

Smallest possible non zero value à -2^(-23) * 2^(-126) and 2^(-23) * 2^(-126)

*Infinity*

Note : There exist positive infinity and negative infinity.

32 bit and 64 bit floating point formats:

## Saturday, March 2, 2013

### handy summary of digital Number formats:

*one of my fellow students posted (anon.) this handy summary of*

*digital Number formats:*

-------------------------------------------------------------------

*8-bit unsigned*

Range : 0 to 255

Note : Always positive . Start from 00000000 (0) to 11111111 (255).

Example :

69 = 01000101

*8-bit signed magnitude*

Range : -127 to 127

Note : Two zeros, +0 (00000000) and -0 (-00000000).

MSB is only the positive or negative sign, with 0 as positive and 1 as negative, contain no value.

Start from 10000000 (-127) to 01111111 (127).

Example :

+69 = 01000101

-69 = 11000101

*8-bit One's Complement*

Range : -127 to 127

Note : Two zeros, +0 (00000000) and -0 (11111111).

MSB is always negative and has value of -127. Start from 10000000 (-127) to 01111111 (127).

Example :

+69 = 01000101

-69 = 10111010

How to get this from signed number (!!) :

If MSB of signed number is 0 (positive), same as signed number

If MSB of signed number is 1 (negative), keep MSB as 1, invert the rest of the bits.

NB. on my current practice exam questions there are a number of questions such as the following:

Which of the following statements about the 8 bit binary number 10001111 is true ?

a) the number is odd when considered to be in signed magnitude, 2's complement, 1's complement, or excess 128 bit formats.

*In general a binary number ending in a zero is usually even*

**unless it is in 1's complement format and its negative**, in which case ending in a zero will cause it to be negative and a bin number ending in 1 will be positive, but for all other cases , as per normal - a bin ending in 0 will be even and ending in a 1 will be negative.*8-bit Two's Complement*

Range : -128 to 127

Note : -128 written as 100000000, only one zero; +0 (00000000).

MSB is always negative and has value of -128. Start from 10000000 (-128) to 01111111 (127).

Example :

+69 = 01000101

-69 = 10111011

How to get this from signed number (?) :

If MSB of signed number is 0 (positive), same as signed number

If MSB of signed number is 1 (negative), keep MSB as 1, invert the rest of the bits, add 1 to the LSB.

Addition in 2's compliment format

when adding bit sequences (on paper) and you have a 'carry bit' on the left at the MS end

(most significant end) you can simply

**discard**this carry bit.

*8-bit excess-128*

Range : -128 to 127

Note : From decimal number, add 128 to the value and write the 8-bit unsigned.

Start from 00000000 (-128) to 11111111 (127).

Example :

+69 = 69 + 128 = 197 = 11000101

-69 = -69 + 128 = 59 = 00111011

*Binary to Octal/Hex and vice versa*

Start from the right hand side (or the least significant bit) and group the bits in groups of three for octal, and four for hexadecimal. Then treat each group as an individual binary string and the result is the

*Octal/Hexadecimal representation.*

For Binary strings that don't divide evenly into groups of three or four, just add as many zeros as needed to the left side.

Example :

1000101 can become 001 | 000 | 101 for the purpose of conversion (note the

added zeros to the left side)

Hence 69 (or 1000101) is 105 in Octal.

To convert Octal or Hexadecimal to Binary, all you need to do is expand each character into the appropriate number of binary digits (three for octal or four for Hex)

Also, if you want to convert from Hex to Octal or vise-versa, the easiest

way to do it is use conversion to binary in between.

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