Systems IEEE 754 Format Shankar Balachandran* Associate Professor, - - PowerPoint PPT Presentation

Shankar Balachandran* Associate Professor, CSE Department Indian Institute of Technology Madras

*Currently a Visiting Professor at IIT Bombay

Digital Circuits and Systems

Spring 2015 Week 9 Additional Module

IEEE 754 Format

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Arithmetic Circuits 3

Floating-Point Number Representation

 Sign bit: S is the sign of the floating point number  Exponent: p-bit exponent (E) in excess-B code.  Mantissa: m-bit unsigned mantissa (M).  Radix: R is the radix for the representation.  Actual value of the number represented above is:  F = (-1)S × 1.M × RE-B … (if normalized)  F = (-1)S × M × RE-B … (if unnormalized)

S Exponent (E) Mantissa (M)

1 p bits m bits

Arithmetic Circuits 4

IEEE Floating-Point Format

 Single Precision Format: (32 bits)

 F = (-1)S × 1. M × 2E-127  Special reserved values:

 E = 0 with M = 0 represents ZERO  E = 255 with M = 0 represents ±  E = 255 with M ≠ 0 represents NaN

 Double Precision Format: (64 bits)

 F = (-1)S × 1. M × 2E-1023

• S Exponent (E) unsigned Mantissa (M)

1 8 bits 23 bits 22 23 30 31

S Exponent (E) unsigned Mantissa (M)

1 11 bits 52 bits

Arithmetic Circuits 5

Examples

 Convert 4.62×102 to IEEE single precision format:

4.62×102 = 462 = 111001110.0 = 1.110011100 × 28  Mantissa = 110011100 Exponent = 8+127 = 135 = 10000111 0 1000 0111 1100 1110 0000 0000 0000 000 = 0 87 CE0000

 Convert -0.456×2-3 to IEEE single precision format:

• 0.456×2-3 = - 0.0111 0100 1011 1100 0110 1010 0111 1101 × 2-3

= -1.1101 0010 1111 0001 1010 101 × 2-5

 Exponent = -5+127 = 122 = 01111010 1 0111 1010 1101 0010 1111 0001 1010 101 = 1 7A D2F1AA

 Convert 1 81 99999A to decimal representation:

Mantissa = -1.1001 1001 1001 1001 1001 1010 Exponent = 8116 - 12710 = -1.1001 1001 1001 1001 1001 1010 × 22 = -0110.0110 0110 0110 0110 0110 1 = -6.4

Arithmetic Circuits 6

Floating-Point Addition



How to add two floating-point numbers? (Ma,Ea) + (Mb,Eb)

1.

Place number with larger exponent in register 1 and the other in 2.

2.

d = E1 – E2

3.

Right-shift mantissa M2 by d bits (i.e., left-shift radix point by d bits).

4.

MSUM = M1 + M2; ESUM=E1

5.

If (MSUM) ≥ 2, renormalize (post-normalization) by dividing by 2 (shifting right) and incrementing ESUM .

6.

Rounding may be required to store the result in the same number of bits (precision) as the inputs.

7.

Result = (MSUM, ESUM).

Arithmetic Circuits 7

Example: Addition

 Add (11000000, 011) to (10101100, 100) (input numbers are in the

normalized form with excess-4 exponent).

 That is, (1.11000000 × 2011-100) + (1.10101100 × 2100-100) = ?  Since (100 > 011), M1 = 1.10101100 E1 = 100 and M2 = 1.11000000 E2 =

011

 d = E1 – E2 = 100 – 011 = 001  Right-shift M2 by 001 (1) bits  M2 = 0.11100000 E2 = 100  MSUM = 1.10101100 + 0.11100000 = 10.10001100 and ESUM= E1 =

100

 Post-normalize: MSUM = 1.01000110 and ESUM= 101  Therefore,

(1.11000000 × 2011-100) + (1.10101100 × 2100-100) = (1.01000110 × 2101-100)

 Or,

(11000000, 011) + (10101100, 100) = (01000110, 101)

Arithmetic Circuits 8

Floating-Point Multiplication



How to multiply two floating-point numbers? (Ma, Ea) × (Mb, Eb)

1.

MPROD = M1 × M2

2.

EPROD = E1 + E2- bias

3.

Post-normalize MPROD by shifting by an appropriate amount and then updating EPROD by the same amount.

4.

Rounding may be required to store the result in the same number of bits (precision) as the inputs.

5.

If necessary, normalize and update EPROD

6.

Result = (MPROD, EPROD).

Arithmetic Circuits 9

Example: Multiplication

 Multiply (10101100, 0101) and (11000000, 0110) (input

numbers are in the normalized form excess-8 exponent).

 That is, (1.10101100 × 20101-1000) × (1.11000000 × 20110-1000) = ?  MPROD = 1.10101100 × 1.11000000 = 10.1110110100000000 and

EPROD= E1 + E2 - 8 = 0011

 Post-normalize: MPROD = 1.01110110100000000 and

EPROD= 0100

 Rounding : MPROD = 1.01110111  Therefore,

(1.11000000 × 20101-1000) × (1.10101100 × 20110-1000) = (1.01110111 × 20100-1000)

 Or,

(11000000, 0101) × (10101100, 0110) = (01110111, 0100)

End of Week 9: Additional Module

Thank You

Multipliers+Others 10