1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU1 IEEE Floating Point Adder Using the IEEE Floating Point Standard for an.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU3 Adder is double precision  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 2 e-1023 (1.f)  – normalized number If e = 0 and f /= 0, the v = (-1) s 2 -1022 (0.f)  Denormalized numbers – allow for graceful underflow If e = 0 and f = 0 the v = (-1) s 0 (zero)

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU4 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

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU5 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

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU6 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

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU7 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

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU13 And what goes in between?  In the final design lots goes in between but  You first want to make sure that the latches are working properly  So just pass one input to the output and check  And once this works properly can move on with the design

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU15 The exponent unit portion  Must get the larger exponent  And the difference between the exponents which is the shift distance  Also several control signals Exponent all 0s and all 1s Exponent A>B, A<B, =

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU16 Mantissa Processing Logic  Need to examine the two fractional parts and generate several control signals that are required to prepare the operands  Need relational signals M>, M=, M< Needed to know which operand to shift  Need to know if stored fractional part if all 0’s or not Needed for NaN, 0,  and determination

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU17 After generating control signals  Step 1 is to select between a normalized mantissa and a denormalized mantissa  For normalized – Prepend NOT(Ex0) If Ex0 is a 1 then the exponent if all 0s and you have a denormalized number or 0 When Ex0 is a 0 you have a NaN, infinity, or a normalized number  Other selection is the factional part shifted left by 1 and postpended by a 0 For denormalized numbers Taking it from 2 -126 to 2 -127 and can now treat it like a normalized number

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU18 Now select between these two  Select the denormalized WHEN Ex0 * (NOT Mx0) When Ex0 is a 1 you have a denormalized number or 0 When Mx0 is a 0 there is a least 1 bit of the fractional part that is a 1 and thus you have a denormalized number  Select the NaN, infinity, 0, normalized number Select this case when Ex0 is a 0 or Mx0 is a 1 When Mx0 is a 1 have infinity, 0, or a normalized number When Ex0 is a 0 have a normalized number, infinity, or NaN

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU19 Shown in table form  Selection table to also point out this relationship  Note that for a 0 have NOT(Ex0) prepended to the fractional part or a 0.00000…000

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU20 Selections are input to a crossbar  The crossbar switch place the larger value on the right path and the small onto the left path  The small is the operand to shift if any shifting to align the binary point is needed  The equation for exchange on the crossbar is E> + (E=*M>) or shift the A input to the right side if the exponent of A is the larger OR the exponents are equal and the fractional part of A is larger

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU21 The next multiplexers  Now have the smaller on the left path and the larger on the right path.  On the left path if either exponent is all 1s then that operand is NaN or infinity and has been crossbarred, or is equal, to the right path operand. In this case want to simply pass it through to the output by adding 0 to it. So a 0 is one choice of the left path mux.  On the right path select the right path value or mux in a hardwired NaN for an illegal operation

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU22 Linear shifting  Next step is to linear shift the left operand  The exponent generates the exponent > signals by subtracting the exponents ExpA-ExpB and ExpB-ExpA  Then with the help of the all control signals the exponent difference is known and this value is sent to the shifter.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU23 One last mulitplexer  The right path operand, the larger is simply input to the ADDER.  On the left path the output of the linear shifter is sent to the ADDER for a + operation  OR  The one’s complement of the value is sent to the ADDER for a – operation. In this case the input carry is handled appropriately.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU25 Xbar code highlight  Code swap <= expgt OR (expeq AND mangt); xbar_r <= lxbarin when (swap = ‘1’) else rxbarin; xbar_l <= rxbarin when (swap = ‘1’) else lxbarin;

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU26 Hard code NaN VHDL code  The code -- Control equation for mux in_mux_r_man <= expa1 AND mana0 and expb1 AND manb0 and (signa XOR signb); in_mux_r <= nan_man WHEN (in_mux_r_man = ‘1’) ELSE xbar_r;

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU27 Now add the mantissas  Simply add the two mantissas.  As the sign of the B input was XORed with the operation, i.e., inverted if it was a subtract operation, the carry in the the XOR of the two signs. If the signs are different then a subtract is being performed and a ‘1’ if being input to the carry in of the adder.  The adder does two’s complement addition.  Inputs are of the form x.xxxxx…xx or 54 bits.  The output is of the form xx.xxxx…xxx or 58 bits

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU28 On to the next challenge  This is perhaps the hardest part – renormalization of the result  Have a result exponent (the exponent of the larger) and a mantissa in the form xx.xxxxxx…xxxx  The following slide shows the processing needed

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU30 Many choices to deal with  May need to shift the mantissa 1 position to the right on a fixed binary point.  May be OK as is  May have to shift left – then need to know the position of the leading 1. In a behavioral model can simply shift left once, increment a counter and then check. In hardware need a leading 1 detector that give the position of the leading 1 so that the mantissa can be shifter left.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU31 Interactions  All shifts of mantissa result in exponent adjustment.  There are 4 choices on the exponent As is Incremented by 1 Adjusted down by some amount depending on shift Zero

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU32 Interactions  There are 5 choices on the mantissa As is Right shifted by 1 – increment exp by 1 Left shifted for leading 1 Left shifted and then right shifted by 1 Hardwired 0  This part is the same for both addition and multiplication. Easy to do algorithmically.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU33 Rounding Unit  Once done with renormalization will look at the guard bits to determine rounding.  Standard specifies several rounding modes.  Can also just truncate.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU34 Rounding  Can result in changes to both the mantissa and the exponent.  After rounding final result is output in normalized form.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU35 And don’t forget the flags  Any arithmetic unit output flags on the status and validity of the result.  The flags to be generated are output from various control signals or combinations of various control signals.

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU36 To test (verify) the design  Must test for normal operation and boundary conditions  Will check A by B NaN NaN +/- infinity +/- infinity +/- 0 +/- 0 Denorm Denorm Norm Norm  For both direct and all crossed pairings

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU37 Boundary conditions  Wish to check several boundary conditions Denorm + Denorm = Max Denorm Denorm + Denorm = Min Norm Norm – Norm = Max Denorm … Rounding using first guard bit Rounding using 1 st and 2 nd guard bits …

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU38 Testing  Testing of the design code is not necessarily the same as the testing the would be done on the chip.  The “testing” of the design is call verification and must insure that all possible input combinations produce the specified output.

Result cases in verification  Input classes – NaN, Inf, 0, Norm, Dnorm  Each class by other NaN by all the others gives NaN  For the multiplier design this is easy since all results are NaN and NaN is generated directly by unit. Inf – possible results – Inf or NaN when illegal op of +Inf * -Inf – result generated directly 0 – result is 0 – also generated directly 1/8/2007 - L2 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU41

Result cases 2  Norm and Denormalized cases Norm by Norm  3 possible results  Overflow to infinity  Normalized  Underflow to denormalized or 0 Denorm by Denorm  Denormalized  Underflow to 0 1/8/2007 - L2 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU42

1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU1 IEEE Floating Point Adder Using the IEEE Floating Point Standard for an.

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1/8/2007 - L25 Floating Point Adder Copyright 2006 - Joanne DeGroat, ECE, OSU1 IEEE Floating Point Adder Using the IEEE Floating Point Standard for an.

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