1. The IEEE Floating Point Standard and execution units for it
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

2. Copyright 2006 - Joanne DeGroat, ECE, OSU
Lecture overview
The standard
Floating Point Basics
A floating point adder design
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

3. The floating point standard
Single Precision
Value of bits stored in representation is:
If e=255 and f /= 0, then v is NaN regardless of s
If e=255 and f = 0, then v = (-1)s ¥
If 0 < e < 255, then v = (-1)s 2e-127 (1.f) – normalized number
If e = 0 and f /= 0, the v = (-1)s (0.f)
Denormalized numbers – allow for graceful underflow
If e = 0 and f = 0 the v = (-1)s 0 (zero)
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

4. The floating point standard
Double Precision
Value of bits in word representation is:
If e=2047 and f /= 0, then v is NaN regardless of s
If e=2047 and f = 0, then v = (-1)s ¥
If 0 < e < 2047, then v = (-1)s 2e-1023 (1.f)
– normalized number
If e = 0 and f /= 0, the v = (-1)s (0.f)
Denormalized numbers – allow for graceful underflow
If e = 0 and f = 0 the v = (-1)s 0 (zero)
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

5. The floating point standard
Notes on single and double precision
The leading 1 of the fractional part is not stored for normalized numbers
Mantissa ….
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/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

6. Copyright 2006 - Joanne DeGroat, ECE, OSU
Conversion Examples
Converting from base 10 to the representation
Single precision example
Covert 10010
Step 1 – convert to binary
In a binary representation form of 1.xxx have
= x 26
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

7. Conversion Example Continued
x is binary for 100
Thus the exponent is a 6
Biased exponent will be 6+127=133 =
Sign will be a 0 for positive
Stored fractional part f will be 1001
Thus we have
s e f ….
C in hexadecimal
$42C is representation for 100
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

8. Copyright 2006 - Joanne DeGroat, ECE, OSU
Another example
Representation for -175 (sign magnitude rep)
175 = =
Or x 27
S = 1
Exponent is = 134 =
Fractional part f =
Representation ….
Or in Hex $C32F 0000
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

9. Copyright 2006 - Joanne DeGroat, ECE, OSU
A fractional example
Decimal value 0.25
Convert to binary
In power form x 2-2
Sign is + so 0
Exponent is = 125 =
Fractional part is 00000…
Representation is …
And in Hex $3E
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

10. Copyright 2006 - Joanne DeGroat, ECE, OSU
Converting back
Convert $C32F 0000 into decimal
Extract components from
S = 1
Exponent = = = 134
unbias – 127 =7
f = so mantissa is
Adjust man by exponent (move binary pt 7 places)
Or = 175
Sign is negative so -175
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

11. Copyright 2006 - Joanne DeGroat, ECE, OSU
Another example
Convert $41C to decimal ….
S is 0 so positive number
Exponent = 128+3= =4
f = 1001 so mantissa is
With 4 binary positions have as final number in binay which is 25
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

12. Arithmetic with floating point numbers
Add op1 $42C and op2 $41C8 0000
First divide into component parts
Op1 $42C = ….
S = 0
E = = 133 – 127 = 6
Mop1 = …
Op2 $41C = ….
E = = 131 – 127 = 4
Mop2 = …
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

13. Copyright 2006 - Joanne DeGroat, ECE, OSU
Now add the mantissas
But first align the mantissas
Op ….
Op …. Which is the smaller number and needs to be aligned
Exponent difference between op1 and op2 is 2
So shift op2 by 2 binary places or
Op2 becomes …
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

14. Copyright 2006 - Joanne DeGroat, ECE, OSU
Add
Add op1 mantissa with the aligned op2 mantissa … …
Result exponent is 6
Value is or =125
Values added were 100 and 25
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

15. Constructing Result Value
Sign 0
Exponent 6 E = = 133 – 127 = 6
Mantissa of Result
Fractional Part ….
Constructed Value
$4 2 F A (125)
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

16. Floating point representation of 125
Positive so s is 0
Exponent is = 133 =
Fractional part from mantissa of or
Constructed value
$42FA 0000
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

17. Multiplication example
Multiply op1 $42C & op2 $41C8 0000
First divide into component parts
Op1 $42C = ….
S = 0
E = = 133 – 127 = 6
Mop1 = …
Op2 $41C = ….
E = = 131 – 127 = 4
Mop2 = …
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

18. Multiplication basics
Base 10 example
3x102 * 1.1x102 = 3.3 x 104
Have 2 numbers A x 2ea and B x 2eb
Multiply and get
result = A*B x 2ea+eb
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

19. Copyright 2006 - Joanne DeGroat, ECE, OSU
So here
Have sign of both is + so result is +
Exponent addition
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 = 137 =
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

20. Copyright 2006 - Joanne DeGroat, ECE, OSU
The mantissas
Do a binary multiplication
1.1001
1 1001
1100 1
and add
Adjusting for binary point have
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

21. Copyright 2006 - Joanne DeGroat, ECE, OSU
Final result
Exponent is 137 or 10
Mantissa is
Adjusted for exponent
Value is
Or = = 2500
And we were multiplying 100 * 25
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

22. Copyright 2006 - Joanne DeGroat, ECE, OSU
Specification of a FPA
Floating Point Add/Subtract Unit
Specification
Inputs in IEEE 754 Double Precision
Must perform both addition and subtraction
Must handle the full floating point standard
Normalized numbers
Not a Numbers – NaNs
+/- Infinity
Denormalized numbers
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

23. Specifications continued
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
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

24. Specifications continued
Outputs
The correctly represented result
Flags that are output are
Zero result
Overflow to infinity from normalized numbers as inputs
NaN result
Overshift (result is the larger of the two operands)
Denormalized result
Inexact (result was rounded)
Invalid operation for addition
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

25. High level block diagram
Basic architecture interface
Data – 64 bit A,B,& C Busses
Control signals – Latch, Add/Sub, Asel, Drive
Condition Flags Output – 7 Flag signals
Clocks – Phi1 and Phi2 (a 2 phase clocked architecture
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

26. Copyright 2006 - Joanne DeGroat, ECE, OSU
Start the VHDL
The entity interface
In the next lecture
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

27. Copyright 2006 - Joanne DeGroat, ECE, OSU
Denormalized Example
Denormalized example(multiplication by 100)
$ is ……
f =
*$42C (100)
Change Denormalized to 2e-127form(was 2e-126)
S = 0
E = 2-127
M = x.f = ……
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

28. Copyright 2006 - Joanne DeGroat, ECE, OSU
Do multiplication
Had the FP representation of 100
S = 0
E = (133 – 127)
M = 1.f =
Multiply and get a result with
E = = -121
M = *0.01 =
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU

29. Copyright 2006 - Joanne DeGroat, ECE, OSU
Renormalize
The values to store
S = 0
E = for a mantissa of (stored e 6)
Adjust to normailized form
E = for a mantissa of (stored e 4)
Construct value to store
S E F
……… = $
1/8/ L24 IEEE Floating Point Basics
Copyright Joanne DeGroat, ECE, OSU