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Unsigned 0 to 2n - 1 2 s Complement -2n-1 to 2n-1 - 1 BCD 0 to 10n/4 - 1 But, what about? very large numbers 9,349,398,989,787,762,244,859,087,678 very small number 0.0000000000000000000000045691 rationals 2/3 Irrationals 2.71828& ., 3.1415926& .., #ZMZZZ#bbjbjbjbjbb b &   2 H-Conceptual Overview: Finite-precision numbers*. f$$b$ . ORepresenting Real Numbers`  How to represent fractional parts to the right of the ``decimal'' point? A number like 0.12 (i.e., (1/2)10) not represented well by integers 0 or 1! Two ways to represent real numbers better: Fixed point Floating pointIN+I``h `h-`+` `   BFixed-Point Data Formatb(   Q Fixed Point `  Pros Add two reals just by adding the integers Much less logic than floating point Faster Often used in digital signal processing Cons The range of numbers is narrow number=400 000 000 000 000 000 000 000 000. 000 It is much more economical to represent as 4*1026} b`}`` ` ~ b  j  b &    3Recall Scientific Notation`  Issues: Arithmetic (+, -, *, / ) Representation, Normal form Range and Precision(Determined by?) Rounding and errors Exceptions (e.g., divide by zero, overflow, underflow) Properties~0FFb)bfb fgb  {Scientific Notation: Normalized ( 12.35 x 10^-9 ? 1.235 x 10^-8 ? scientific notation: has a single digit to the left of the decimal point Normalized scientific notation: such a single digit must be non-zero.T Numerical Form: ( 1)s M 2E Sign bit s determines whether number is negative or positive Significand M normally a fractional value in range [1.0,2.0). Exponent E weights value by power of two Encoding >Ua UU`dedlddl `d?`d9`d` `2Y   f  KIEEE 754 standardd  SThree formats: single/double/extended precision (32,64,80 bits). Single precision:TT` T y  the representation: ( 1)s (1+ Fraction) 2E-bias where E is the exponent representation in the exp field Sign bit s determines whether number is negative or positive Fraction is normally a fractional value in range between 0 and 1 A leading 1 added to the fraction is  implicit Exponent using a  biased notation For single precision  the bias is 127 UlUa'aUb fgfnffnn2&nn n bf>bfnbf( b {  z4Advantage of using the biased notation for exponents55$ Under the single-precision IEEE 754 standard: bias = 127 If the real exponent is +1, what is the biased exponent ? How about if the real exponent is -1 ?WNormalized Encoding Exampleb(  Value: Float F = 15213.010; 1521310 = 1.1101101101101 X 213 Significand M = 1.1101101101101 frac = 11011011011010000000000 (23 bits! With leading 1 hiding!) Exponent E = 1310, Bias = 12710 Exp = 14010 = 10001100 Z!Z ZjZ Z8ZZbcjcbbjbjb b b  b  b   b  b   b   b 5 b  bbbjbbbjbbbjbcbbL=        mDenormalized Values`   Condition exp = 000& 0 Significand value M = 0.xxx& x xxx& x: bits of frac Cases frac = 000& 0 Represents value 0 Note that have distinct values +0 and  0 frac 000& 0 Numbers very close to 0.0 Lose precision as get smaller  Gradual underflow  Z ZZZZ Z<Z ZLZ bbcbcbcbbbbccbc c  b  c  b  c  b bcbcbc<bcb bcbcLbt   #    E  U nSpecial Values`  Condition exp = 111& 1 (infinity) Operation that overflows Both positive and negative E.g., 1.0/0.0 = -1.0/-0.0 = +, 1.0/-0.0 = - Not-a-Number (NaN) Represents case when no numeric value can be determined E.g., sqrt( 1), -  Z Z ZcZZNZ bbcbcbcc  c  $c   bD b  c   b  c   b  c   b   b  c   b  c   c   b bHbc  @  @    xIEEE 754: Summary g$  oIEEE754: Summary (Cont.)b(   aSpecial Properties of Encoding`  FP Zero Same as Integer Zero All bits = 0 Can (Almost) Use Unsigned Integer Comparison Must first compare sign bits Must consider -0 = 0 NaNs problematic Will be greater than any other values Otherwise OK Denorm vs. normalized Normalized vs. infinity -Cab b-bC b ab@  @  ( | FP Addition b(  &Operands ( 1)s1 M1 2E1 ( 1)s2 M2 2E2 Assume E1 > E2 Exact Result: ( 1)s M 2E Exponent E: E1 Sign s, significand M: Result of signed align & add ZZZZDZ bgfnffngfnffnbbbbb b  g  f  n  f  f  n  b  bbbbbbbbbb&h   ! ~Decimal Number Conversion  b(  Convert Binary to Decimal (base 2 to base 10) x x x x x. d d d d d d & 2 / %0 Z -`#b(c (k (c (bb$ h !Decimal Number Conversion (Cont.)" "b$ " Convert Decimal Integer to binary Integer divide the decimal value by 2 and then write down the remainder from bottom to top (Divide 2 and Get the Remainders) 3710 = ?2 + P P)bbVbf*bbjbjbb  !Decimal Number Conversion (Cont.)" "b$ " Convert Decimal Fraction to Binary Fraction multiply the decimal value by 2 and then write down the integer number from top to bottom (Multiply 2 and Get the Integers) . 0.37510 = ?2 , P P7bQb!f*bbjbjb  !Decimal Number Conversion (Cont.)" "b$ " Convert Decimal Number to Binary Binary Put Together: Integer Part . 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GGG___PPT10.+?kDD9' {= @B D' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|Y%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|Y%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =-o6Bdissolve*<3<*|DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* |%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* |,%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* |,A%(+p+0+|0 ++0+ |0 +0  0  X(  Xr X S ?     X 0n @ }<$ 0   X < o  *l (a   X(a ,$D  00N 0P`0  X 1(a  X Z\wawa8c?0P0  Ss" c   X Z|wawa8c? PP 0  Uexp" c    X Zwawa̙8c?` P`0  dfrac" c    X <`7   [ 1 bit 8 bits 23 bits   X <b v5 ,$  0 ^,Double precision: see page 192 in P&H book--H X 0޽h ? @Eff؂o___PPT10.+۔ҧD' = @B D' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*XA%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XAD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*XAD{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*XAT%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XATD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*XATD' =%(D' =%(DG' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =-6B'barn(inHorizontal)*<3<* XD' =%(D' =%(DT' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =-6B'barn(inHorizontal)*<3<* X+p+0+X0 ++0+ X0 +   0L0  S(    S ~(wawa1 ?@ <$ 0  ,4aYi%  Z8۫ 8c 8c1 ?\ /IEEE 754 Standard Floating Point Representation 00g$ 0    $lwawa1 ?Rectangle: Click to edit Master text styles Second level Third level Fourth level Fifth levelD t*0wnUc  4aYiH  0޽h ? GGG___PPT10.+r<DN' {= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*;%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*1%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*1T%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*T{%(+8+0+0 +  0 0Z(  r  S ?    r  S @ 0   #  < ,$  0 / Answer: -1+127 = 126! (Note 126-127 = -1!) 8/ wn/  <Tl,$D  0 Z(How about if the real exponent is -127 ?))  B|} *~ ,$  0 [+Answer: 1+127 = 128! (Note 128-127 = +1!),,H  0޽h ? @Eff؂o  ___PPT10 .`