IEEE Floating Point The IEEE Floating Point Standard

IEEE Floating Point The IEEE Floating Point Standard and execution units for it 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 1

Lecture overview o o o The standard Floating Point Basics A floating point adder design 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 2

The floating point standard o Single Precision o Value of bits stored in representation is: n n If e=255 and f /= 0, then v is Na. N regardless of s s If e=255 and f = 0, then v = (-1) ¥ s If 0 < e < 255, then v = (-1) 2 e-127 (1. f) – normalized number s If e = 0 and f /= 0, the v = (-1) 2 -126 (0. f) o n Denormalized numbers – allow for graceful underflow s If e = 0 and f = 0 the v = (-1) 0 (zero) 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 3

The floating point standard o Double Precision o Value of bits in word representation is: n n n If e=2047 and f /= 0, then v is Na. N regardless of s s If e=2047 and f = 0, then v = (-1) ¥ s e-1023 If 0 < e < 2047, then v = (-1) 2 (1. f) o n s If e = 0 and f /= 0, the v = (-1) 2 -1022 (0. f) o n – normalized number Denormalized numbers – allow for graceful underflow s If e = 0 and f = 0 the v = (-1) 0 (zero) 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 4

The floating point standard o Notes on single and double precision n The leading 1 of the fractional part is not stored for normalized numbers Representation allows for +0 and -0 indicating direction of 0 (allow determination that might matter if rounding was used) Denormalized numbers allow graceful underflow towards 0 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 5

Conversion Examples o Converting from base 10 to the representation Single precision example Covert 10010 Step 1 – convert to binary - 0110 0100 o In a binary representation form of 1. xxx have o o o n 0110 0100 = 1. 100100 x 26 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 6

Conversion Example Continued o o 1. 1001 x 26 is binary for 100 Thus the exponent is a 6 n n n o Biased exponent will be 6+127=133 = 1000 0101 Sign will be a 0 for positive Stored fractional part f will be 1001 Thus we have n n s e f 0 100 0 010 1 1 00 1000…. 4 2 C 8 0 0 in hexadecimal $42 C 8 0000 is representation for 100 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 7

Another example o Representation for -175 n n n n 175 = 128 + 32 + 8 + 4 + 2 +1 = 1010 1111 Or 1. 0101111 x 27 S = 1 Exponent is 7 +127 = 134 = 1000 0110 Fractional part f = 0101111 Representation 1100 0011 0010 1111 0000 …. Or in Hex $C 32 F 0000 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 8

Converting back o o Convert $C 32 F 0000 into decimal Extract components from n n n n 1100 0011 0010 1111 S = 1 Exponent = 1000 0110 = 128+4+2 = 134 unbias 134 – 127 =7 f = 0101111 so mantissa is 1. 0101111 Adjust by exponent 1010 1111 (move binary pt 7 places) Or 128+32+15 = 175 Sign is negative so -175 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 9

Another example o Convert $41 C 8 0000 to decimal n n n 0100 0001 1100 1000 0000 …. S is 0 so positive number Exponent 1000 0011 = 128+3=131 -127=4 f = 1001 so mantissa is 1. 1001 With 4 binary positions have 11001 as final number or a decimal 25 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 10

Arithmetic with floating point numbers o o Add op 1 $42 C 8 0000 and op 2 $41 C 8 0000 First divide into component parts n Op 1 $42 C 8 0000 =0100 0010 1100 1000 0000 …. o o o n S = 0 E = 1000 0101 = 133 – 127 = 6 Mop 1 = 1. 10010000… Op 2 $41 C 8 0000 =0100 0001 1100 1000 0000 …. o o o 1/8/2007 - L 24 IEEE Floating Point Basics S = 0 E = 1000 0011 = 131 – 127 = 4 Mop 2 = 1. 10010000… Copyright 2006 - Joanne De. Groat, ECE, OSU 11

Now add the mantissas o But first align the mantissas n n n Op 1 1. 1001000…. Op 2 1. 1001000…. Which is the smaller number and needs to be aligned Exponent difference between op 1 and op 2 is 2 So shift op 2 by 2 binary places or Op 2 becomes 0. 0110010000… 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 12

Add op 1 mantissa with the aligned op 2 mantissa n n n o o o 1. 1001000000… 0. 0110010000… 1. 1111010000 Result exponent is 6 Value is 1111101 or 64+32+16+8+4+1=125 Values added were 100 and 25 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 13

Constructing Result Value o Sign 0 Exponent 6 E = 1000 0101 = 133 – 127 = 6 Mantissa of Result 1. 1111010000 Fractional Part 1111010000…. o Constructed Value o o o n n 0 100 0010 1 111 1010 0000 $4 2 F A 0 0 (125) 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 14

Floating point representation of 125 o o o Positive so s is 0 Exponent is 6 + 127 = 133 = 1000 0101 Fractional part from mantissa of n o 1. 111101 or 111101 Constructed value n n 0 1000 0101 111101 000000000 $42 FA 0000 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 15

Multiplication example o o Multiply op 1 $42 C 8 0000 & op 2 $41 C 8 0000 First divide into component parts n Op 1 $42 C 8 0000 =0100 0010 1100 1000 0000 …. o o o n S = 0 E = 1000 0101 = 133 – 127 = 6 Mop 1 = 1. 10010000… Op 2 $41 C 8 0000 =0100 0001 1100 1000 0000 …. o o o 1/8/2007 - L 24 IEEE Floating Point Basics S = 0 E = 1000 0011 = 131 – 127 = 4 Mop 2 = 1. 10010000… Copyright 2006 - Joanne De. Groat, ECE, OSU 16

Multiplication basics o Base 10 example n o o o 3 x 102 * 1. 1 x 102 = 3. 3 x 104 Have 2 numbers A x 2 ea and B x 2 eb Multiply and get result = A*B x 2 ea+eb 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 17

So here o o Have sign of both is + so result is + Exponent addition n n Both exponents are biased as stored If you add stored binary exponents you need to subtract the extra bias or 127 Or using pencil and paper (or powerpoint) can just add the unbiased exponent of one operand to the other biased exponent Here have 133 + 4 = 137 = 1000 1001 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 18

The mantissas o Do a binary multiplication n n n o 1. 1001 1100 1 1100111 0001 and add Adjusting for binary point have 10. 01110001 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 19

Final result o o o Exponent is 137 or 10 Mantissa is 10. 01110001 Adjusted for exponent 1001 1100 0100 Value is 2048+256+128+64+4 Or 2304+128+68 = 2432 + 68 = 2500 And we were multiplying 100 * 25 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 20

More Examples o o o A = 100 $42 C 8 0000 0100 0010 1100 1000 0000 S = 0 E = 1000 0101= 133 – 127 = 6 F = 1001 0000 --Man. A = 1. 100100000 B = 25 $41 C 8 0000 0100 0001 1100 1000 0000 S = 0 E = 1000 0011 = 131 – 127 = 4 F = 1001 0000 --Man. B = 1. 100100000 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 21

Example Continued o For A + B need to align binary pt by 2 places n n n Man. A = 1. 10010000000 Shf. Man. B = 0. 01100100000 Sum is 1. 1111010000 with a bin exp of 6 0 100 0010 1 111 1010 0000 ---$4 2 F A 0 0 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 22

Subtraction example o B = -25 $C 1 C 8 0000 n n o 1100 0001 1100 1000 0000 S = 1 E = 10000011 = 131 -127 = 4 F = 1001 0000 ---Man. B = 1. 100100000 C = 10 $41 C 8 0000 n n 0100 0001 0010 0000 0000 S = 0 E = 1000 0010 = 130 -137 = 3 F =0100 0000 ---Man. C = 1. 010000000 -- 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 23

Subtraction ex concluded o For B+C need to subtract aligned mantissa of C from B n n n o Man. B = 1. 1001000000 Man. Cshftd = 0. 1010000000 result 0. 111100000 and exp of 4 Normalized mantissa is 1. 111 exponent of 3 n n n Result sign =1 Exp = 130 = 1000 0010 Result Man = 1. 111000000 --Result 1100 0001 0111 0000 0000 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 24

Specification of a FPA o o Floating Point Add/Subtract Unit Specification n Inputs in IEEE 754 Double Precision Must perform both addition and subtraction Must handle the full floating point standard o o 1/8/2007 - L 24 IEEE Floating Point Basics Normalized numbers Not a Numbers – Na. Ns +/- Infinity Denormalized numbers Copyright 2006 - Joanne De. Groat, ECE, OSU 25

Specifications continued n n Result will be a IEEE 754 Double Precision representation Unit will correctly handle the invalid operation of adding + ¥ and - ¥ = Nan per the standard Unit latches it inputs into registers from parallel 64 -bit data busses. There is a separate signal line that indicates the operation add or subtract 9/25/08 – ECE 764 L 2 a IEEE Floating Point Basics Copyright 2008 - Joanne De. Groat, ECE, OSU 26

Specifications continued o Outputs n n The correctly represented result Flags that are output are o o o o 1/8/2007 - L 24 IEEE Floating Point Basics Zero result Overflow to infinity from normalized numbers as inputs Na. N result Overshift (result is the larger of the two operands) Denormalized result Inexact (result was rounded) Invalid operation for addition Copyright 2006 - Joanne De. Groat, ECE, OSU 27

High level block diagram o Basic architecture interface n n Data – 64 bit A, B, & C Busses Control signals – Latch, Add/Sub, Asel, Drive Condition Flags Output – 7 Flag signals Clocks – Phi 1 and Phi 2 (a 2 phase clocked architecture 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 28

Start the VHDL o The entity interface o VHDL code covered in next lecture 1/8/2007 - L 24 IEEE Floating Point Basics Copyright 2006 - Joanne De. Groat, ECE, OSU 29