V 0.1 1Fixed Point Numbers• The binary integer arithmetic you are used to is known by the more general term of Fixed Point arithmetic.– Fixed Point means that we view the decimal point being in the same place for all numbers involved in the calculation.– For integer interpretation, the decimal point is all the way to the right$C0+ $25--------$E5192.+ 37.--------229.Unsigned integers, decimal point to the right.A common notation for fixed point is ‘X.Y’, where X is the number of digits to the left of the decimal point, Y is the number of digits to the right of the decimal point.V 0.1 2Fixed Point (cont).• The decimal point can actually be located anywhere in the number -- to the right, somewhere in the middle, to the right$11+ $1F--------$30Addition of two 8 bit numbers; different interpretations of results based on location of decimal point17+ 31--------48xxxxxxxx.0decimal point to right. This is 8.0 notation.4.25+ 7.75--------12.00xxxxxx.yytwo binary fractional digits. This is 6.2 notation.0.07+ 0.12--------0.190.yyyyyyyy decimal point to left (all fractional digits). This is 0.8 notation.V 0.1 3Algorithm for converting X.Y binary to decimal, decimal to X.Y format.Convert the number 34.7 to its closest representation in 8.4 (12bit) format:Four bits to right of decimal point, so multiply by 24= 16, truncate34.7 * 16 = 555.2 = 555 = 0x22BAssume that 0x73A represents a 8.4 (12 bits) fixed point number, what is its decimal representation?Four bits to right of decimal point, so divide by 24= 160x73A/16 = 1850 /16 = 115.625.V 0.1 4Signed X.Y format (2’s complement)Follow the same conversion rules!For –3.25 to 8.4 signed format:-3.25 * 16 = -52 = 0xFCC as a 2’s complement numberWhat is the decimal value of 0xA27 if 8.4 signed format?This is a negative number. Magnitude is:0x000 – 0xA27 = 0x5D9, so number is –1497.Decimal value is –1497/16 = 93.5625V 0.1 5Unsiged Overflow• Recall that a carry out of the Most Significant Digit is an unsigned overflow. This indicates an error - the result is NOT correct!$FF+ $01--------$00255+ 1--------063.75+ 0.25-----------0 xxxxxxxx.0decimal point to rightxxxxxx.yytwo binary fractional digits (6.2 notation)0.yyyyyyyy decimal point to left (all fractional digits). This 0.8 notation0.99600+ 0.00391-----------0 Addition of two 8 bit numbers; different interpretations of results based on location of decimal pointV 0.1 6Saturating Arithmetic• Saturating arithmetic means that if an overflow occurs, the number is clamped to the maximum possible value.– Gives a result that is closer to the correct value– Used in DSP, Graphic applications.– Requires extra hardware to be added to binary adder. – Pentium MMX instructions have option for saturating arithmetic.$FF+ $01--------$FF255+ 1--------25563.75+ 0.25-----------63.75 xxxxxxxx.0decimal point to rightxxxxxx.yytwo binary fractional digits.0.yyyyyyyy decimal point to left (all fractional digits)0.99600+ 0.00391-----------0.99600V 0.1 7Saturating ArithmeticThe Intel Xx86 MMX instructions perform SIMD operations between MMX registers on packed bytes, words, or dwords. The arithmetic operations can made to operate in Saturation mode. What saturation mode does is clip numbers to Maximum positive or maximum negative values during arithmetic.In normal mode: FFh + 01h = 00h (unsigned overflow)In saturated, unsigned mode: FFh + 01 = FFh (saturated to maximum value, closer to actual arithmetic value)In normal mode: 7fh + 01h = 80h (signed overflow) In saturated, signed mode: 7fh + 01 = 7fh (saturated to max value)V 0.1 8Saturating Adder: Unsigned and 2’Complement• For an unsigned saturating adder, 8 bit:– Perform binary addition– If Carryout of MSB =1, then result should be a $FF.– If Carryout of MSB =0, then result is binary addition result.• For a 2’s complement saturating adder, 8 bit:– Perform binary addition– If Overflow = 1, then:• If one of the operands is negative, then result is $80• If one of the operands is positive, then result is $7f– If Overflow = 0, then result is binary addition result.V 0.1 9Saturating Adder: Unsigned, 4 Bit exampleA[3:0]B[3:0]T[3:0]+CO011111SUM[3:0]01S2/1 MuxV 0.1 10Saturating Adder: Signed, 4 Bit exampleA[3:0]B[3:0]T[3:0]+0A3 A3’A3’A3’1SUM[3:0]Vflag = T3 A3’ B3’ + T3’ A3 B3Vflag is true if sign of both operands are the same (both negative, both positive) and different from Sum (overflow if add two positive numbers, get a negative or add two negative numbers and get a positive number. Can’t get overflow if add a postive and a negative).Saturated value has same sign as one of the operands, with other bits equal to NOT (sign) : 0111 (positive saturation), 1000 (negative saturation).V 0.1 11Saturating ArithmeticThe MMX instructions perform SIMD operations between MMX registers on packed bytes, words, or dwords. The arithmetic operations can made to operate in Saturation mode. What saturation mode does is clip numbers to Maximum positive or maximum negative values during arithmetic.In normal mode: FFh + 01h = 00h (unsigned overflow)In saturated, unsigned mode: FFh + 01 = FFh (saturated to maximum value, closer to actual arithmetic value)In normal mode: 7fh + 01h = 80h (signed overflow) In saturated, signed mode: 7fh + 01 = 7fh (saturated to max value)V 0.1 12Why Saturating Arithmetic?• In case of integer overflow (either signed or unsigned), many applications are satisfied with just getting an answer that is close to the right answer or saturated to maximum result• Many DSP (Digital Signal Processing) algorithms depend on this feature– Many DSP algorithms for audio data (8 to 16 bit data) and Video data (8-bit R,G,B values) are integer based, and need saturating arithmetic.• This is easy to implement in hardware, but slow to emulate in software. A nice feature to have.V 0.1 13Binary Coded DecimalBinary coded decimal (BCD) encodes decimal digits as their equivalent 4-bit values:73 (decimal) = 0x73 (BCD)Makes conversion of binary-to-decimal and decimal-to-binary trivial.Arithmetic of binary coded values requires special help.V 0.1 14BCD Addition, SubtractionV 0.1 15PIC24 Code for BCD Addition/SubtractionV 0.1 16Floating Point Representations• The goal of floating point representation is represent a large range of numbers• Floating point in decimal representation looks like:+3.0 x 10 3, 4.5647 x 10 -20 , etc• In binary, sample
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