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Radix Conversion Given a value X represented in source system with radix s, represent the same number in a destination system with radix d Consider the integral part of the number, XI, in the d system: X I xk 1 d k 1 xk 2 d k 2 x1 d x0 d {[( xk 1 d xk 2 ) d 1 0 x2 ] d x1} d x0 0 xi d If XI is divided by d , we obtain x0 as a remainder and quotient Q {[( xk 1 d xk 2 ) d x2 ] d x1} R x0 R is the Desired digit (LSD) – Can Repeatedly Divide to Obtain Converted Value Radix Conversion Example XI = 34610 s=10 d =3 Fixed-point Decimal to Ternary Integer Conversion XI = 1102113 Check by evaluating the radix polynomial 1 35 1 34 0 33 2 32 1 31 1 30 [243 81 18 3 1]10 34610 Radix Conversion (fractional) Consider the fractional part of the value in d Fixed point system X F x1 d 1 x2 d 2 x ( m 1) d ( m 1) x m d m d 1{x1 d 1[ x2 d 1 ( x3 )]} d X F PI PF PI x1 PF d 1[ x2 d 1 ( x3 )] Thus, PI is the Desired Digit We can Repeatedly Multiply by the d Value Radix Conversion Example XI = 0.29110 s=10 d =5 Fixed-point Decimal to Pentary Fractional Conversion 0.291 5 1.455 1 0.455 5 2.275 2 0.275 5 1.375 1 0.375 5 1.875 1 0.875 5 4.375 4 0.375 5 1.875 1 0.875 5 4.375 4 0.29110 (0.12114141414 )5 0.29110 is Finite Fraction for s=10, but infinite fraction for d =5 Fixed Point Negative Numbers Two Common Forms: 1. Signed-magnitude Form 2. Complement Forms Signed Magnitude • • • • First Digit is Sign Digit, Remaining n-1 are the Magnitude Convention (binary) – 0 is a Positive Sign bit – 1 is a Negative Sign bit Convention (non-binary) – 0 is a Positive Sign digit – -1 is a Negative Sign digit Only 2• n-1 Digit Sequences are Utilized Signed Magnitude Example n7 k 4 k 1 3 m3 s 2 d 10 1101.1102 1 (1 22 1 20 1 2 1 1 2 2 )10 1 1 1 4 1 5.7510 2 4 10 Largest Representable Value is: 0111.1112 1 (1 22 1 21 1 20 1 21 1 22 1 23 )10 1 1 1 7 1 4 2 1 7 2 4 8 10 8 10 Signed Magnitude Example (cont) n7 k 4 k 1 3 m3 s 2 d 10 0111.111 1 (1 22 1 21 1 20 1 2 1 1 2 2 1 2 3 )10 1 1 1 7 1 4 2 1 7 2 4 8 10 8 10 1 ulp 1 2 3 1 8 k 1 3, s k 1 23 8 1 1 ulp ( s ) 8 X MAX k 1 ulp X MIN 1 ( k 1 ulp ) X [ X MIN , X MAX ] m 1 Signed Magnitude Ternary Example n7 k 4 k 1 3 m 3 xi {0,1,2} s 3 d 10 2102.120 1 (1 32 2 30 1 31 2 32 )10 1 2 5 1 9 2 11 3 9 10 9 10 (11.555555 )10 Notice that fractional part is infinite in =10 but finite in =3 1 m 1 1 ulp min{xi 0} ( s ) 27 X MAX k 1 ulp X MIN 1 ( k 1 ulp ) X [ X MIN , X MAX ] Signed Magnitude Ternary Bounds n7 k 4 k 1 3 m 3 xi {0,1,2} s 3 d 10 X MAX 0( 1)( 1)( 1).( 1)( 1)( 1) 0222.2223 1 (2 32 2 31 2 30 2 31 2 32 2 33 )10 2 2 2 26 1 18 6 2 26 3 9 27 10 27 1 k 1 ulp 27 27 k 1 [ 0, ulp] Positive Numbers: k 1 Negative Numbers: [( ulp), 0] Range: [( k 1 ulp), k 1 ulp] Signed Magnitude Comments • Two Representations for zero, +0 and –0 • Addition of +K and –K is not zero EXAMPLE 10001010.002 +00001010.002 10010100.002 -1010+1010 Yields a Sum of –2010!!!!! Complement Representations • Two Types of Complement Representations 1. radix complement (binary – 2’s complement) 2. diminished radix complement (binary – 1’s complement) • Positive Values Represented Same Way as Signed Magnitude for Both Types • Negative Value, -Y, Represented as R-Y Where R is a Constant • Obeys the Identity: (Y ) R ( R Y ) R R Y Y • Advantage is No Decisions Needed Based on Operand Sign Before Operations are Applied Complement Representation Example • X is Positive, Y is Negative, Compute X+Y Using Complement Representation X ( R Y ) [ X ( R Y )] [ R Y X ] R (Y X ) • If |Y|>X, Then the Answer is R-(Y-X) • If X>|Y|, Then the Answer Should be X-Y – But X+(R-Y)=R+(X-Y), Thus R Must be Discarded! • Solution is to Choose the Value of R Carefully Requirements for Complementation Value, R • Select R to Simplify (or Eliminate) Correction for the X>|Y| Case • Calculation of Complement of Y, (R-Y) Should be Simple and Fast • Definition of Complement for Single Digit, xi xi ( 1) xi • Definition of Complement for a Word, X X ( xk 1xk 2 x m ) Complementation Value, R • Add Word and Complement Together: X ( xk 1 xk 2 x m ) X ( xk 1 xk 2 x m ) ( 1)( 1) ulp 0 0 1 Now Add 1 ulp • 0 ( 1) 1 0 Therefore, we see that: X X ulp k k X X ulp 0 Answer to Addition Radix Complement Form • The Radix Complement Form is Defined When: R k R X k X X ulp • Using k is Convenient Since Storing Result in Register of Length n Causes MSD of 1 to be Discarded due to Finite Register Length • Therefore, it is Easy to Compute the Complement of X by: 1. Take the Complement of X 2. Add 1ulp to Complement X X ulp k k X X ulp Radix Complement Form (cont) • No Correction is Needed When We have Positive X and Negative Y Such That: X (R Y ) 0 • Since R= k X (R Y ) R ( X Y ) k X Y • And k is discarded Due to Finite Register Length Radix Complement Example 2 k n4 • Since n = m + k m=0 • Therefore 1 ulp = 20=1 • Given X, the radix complement (2’s complement) is: R X k X X ulp X 1 • Range of Positive Numbers is [0000,0111] • 2’s Complement of Largest, 0111: X 1000; • X 1 10012 710 In Radix Complement, There is a Single Representation of Zero (0000) and Each Positive Number has Corresponding Negative Number With MSB=1 Radix Complement Example 2 • k n4 In Radix Complement, There is a Single Representation of Zero (0000) and Each Positive Number has Corresponding Negative Number With MSB=1 • Accounts for 1(zero)+7(pos.)+7(neg.), But Extra Bit Pattern Left • One Additional Negative Number, 10002=-810, 0010 (210 ) 0111 (710 ) 1001 ( 710 ) 1110 ( 210 ) 0 1011 ( 510 ) 1 0101 (510 ) -810X+710 1011 0100 0100 1 0101 ( decode) Same to Encode Diminished Radix Complement 2 • k n4 In Diminished Radix Complement, the Complementation Process is Easier Since the Addition of 1 ulp is Avoided R k ulp R X k ulp X X • Range of Positive Numbers is: [00002,01112]=[010,710] • 1’s Complement of Largest is 10002= -710 • 1’s Complement of Zero is 11112 • Two Representations of Zero! 7 X 7 • In All Cases MSB is Sign Bit Comparison of Two’s Complement, One’s Complement and Signed-Magnitude Sequence 011 010 001 000 111 110 101 100 Two’s One’s Complement Complement 3 3 2 2 1 1 0 0 -1 -0 -2 -1 -3 -2 -4 -3 SignedMagnitude 3 2 1 0 -3 -2 -1 -0