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Sistem Bilangan

Sistem Bilangan. Oleh : Mukhlidi Muskhir. Analogue vs Digital. Analogue * Continuous range of value * Precision limited by Noise Digital * Discrete range of values * Precision limited by number of “Bit”. Analogue vs Digital. Analogue. Digital. Analogue vs Digital.

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Sistem Bilangan

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  1. Sistem Bilangan Oleh : Mukhlidi Muskhir

  2. Analogue vs Digital • Analogue * Continuous range of value * Precision limited by Noise • Digital * Discrete range of values * Precision limited by number of “Bit”

  3. Analogue vs Digital Analogue Digital

  4. Analogue vs Digital • The real world is analogue ( by because all signal in world be shape analogue) • But in controlling, Digital one had using for process. • Both of signal had been converter each other

  5. Analoge vs Digital Analogue Analogue Digital Processing D to A A to D Why Digital Only by using in Processing? ^ Adventure in integrated Circuit has made the complex processing of digital data. ^ Digital Control processing has made easier than analogue ^ Digital circuits are inherently more noise resistant

  6. Digital and Boolean • Digital represented by boolean logic • Boolean is the name of mathematician’s expert • Now boolean is called by conventional logic because there is new logic that called by fuzzy logic • But all electronic still using boolean logic to processing the controlling system

  7. Why Boolean • It is convenient in electrical system to use a two-value system to represent value true/false, on/off, yes/no and 1/0 * Two voltage or current levels can be used * Easier to process and distribute reliably (diandalakan) * Don’t think of them as numbers (even though we often represent them as 0/1 for brevity(ketangkasan)) • The need for binary numbers * Multi-value quantities need to be represented in the digital system. Therefore need numbers made up from the simple two value system

  8. Positional Number System Decimal point 7x10-1 7x10-2 8x10-3 8 x 100 7 x 101 5 x 102 3 x 103 3578.778 Base 10, weigthing are powers of 10

  9. Unsigned binary numbers Binary point 1 x 2-1 = 0.500 0 x 2-2 = 0.000 1 x 2-3 = 0.125 0 x 20= 0.000 0 x 21= 0.000 1 x 22= 4.000 1 x 23= 8.000 1100.101 Each bit of the Number may be Representaed by A Boolean value Binary, weightings are powers of 2

  10. A1 B1 A2 B2 + A2 B3 Carry Flag Carry Flag Carry Out Carry In Carry Out Multi-precision Arithmatic Additional of A and B

  11. Multi-precision Arithmatic Carry Flag Carry Out Carry In A1 B1 A2 B2 - A2 B3

  12. Hexadecimal Numbers 660 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 8 9 A B C D E F 4 : 16 41 9 : 16 2 Hexadecimal : 294 Hex 215 13 : 16 7 Hexadecimal : 7D Hex

  13. Hexadecimal Numbers 0 1 2 3 4 5 6 7 8 9 A B C D E F 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 660 0010 1001 0100 2 9 4 215 0000 1101 0111 0 D 7

  14. Decimal to Binary Generetee each digit by successive division Or multiplication. There is no guarantee the fraction will be finite Fractional part – Multiplication by base Whole part – divition by base

  15. Binary Additional 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 carry 1 Easy Layaou ?

  16. Binary Addition 190 + 141 =331 Carry out of Each column 1 1 1 1 1 1 0 1 1 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 1 1 Carry out of 8-bit number

  17. Binary Subtraction A borrow-out of 1 from This column becomes a borrow in of 2 in this column 229 – 46 = 183 2 2 2 2 2 1 1 1 0 0 1 0 1 Borrow in from Left column 0 0 1 0 1 1 1 0 1 1 1 1 1 Borrow out 1 0 1 1 1 0 1 1 Both rows subtracted

  18. Exercise • Convert to 8-bit binary and do the arithmetic operation * 120 + 54 * 110 + 100 * 224 – 134 * 200 + 20 * 112 – 89 * 111 – 25 • Convert back to decimal and check the result

  19. Binary Number Circle In real hardware there is a fixed number Of bits available. We often ignore leading zeros But they are still there! Examlpe : If we only use 4 bits then the binary Counting sequence “wraps around” At 15 ↔ 0 11 - 1 = 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 4 – bit Binary Number Circle 11 1110 - 1 1 10 1010

  20. Binary Number Circle Subtracting across the boundary Still “works” if you think of result As the distance on the number Circle. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 4 – bit Binary Number Circle (Module arithmetic – ignore The borrow /carry) 8 1000 - 14 - 1110 10 (-1)1010

  21. Representing –ve Number • Several choices for natation * sign + magnitude notation * 1’s complement * 2’s complement notation * various ‘excess codes ‘

  22. Sign Number – sign + magnitude Notation Sign Bit Magnitude 0  +ve Simple binary number 1  - ve How about Null or Zero Problem ? + 0  0000 - 0  1000

  23. Signed Numbers – Sign + magnitude Notation Arithmetic • Difficult to do – have to work out that operation to perform • 5 + -6 actually calculate –(6-5) i.e. exchange the operands and do subtraction! • -5+ -6 actually calculate –(5+6) i.e. negate the addition of the negated numbers ! • Required action depends the signs of the numbers and on which has the large magnitude. Natural for us –a bit hard for the computer since the only way it can work out the bigger number is to do a subtraction!

  24. Sign + Magnitude Examples

  25. Sign Numbers – 2’s Complement • As for straight binary numbers but with the weighting of the most significant bit being negative • Example * 4 bit – weights are -8, 4,2,1 * 8 bit – weights are -128, 64,32,16,8,4,2,1 • Need to know how many bits are being used to work out the value of the number – don’t omit leading zeroes

  26. Sign Numbers – 2’s Complement Binary point 1 x 2-1 = 0.500 0 x 2-2 = 0.000 1 x 2-3 = 0.125 0 x 20= 0.000 0 x 21= 0.000 1 x 22= 4.000 1 x 23= -8.000 1100.101 Sign Bit -4.375 Binary, weightings are powers of 2

  27. 2’s Complement Examples

  28. 2’s Complement Examples Example : -4 (decimal) Become 4 = 0100 ( binary) = 1x22 = 4 2’s Complement -4= 1100 (binary) = -(23) + 22 = -8 + 4 = -4

  29. Exercise Converse decimal number above into negative (2’s complement) : • -7 ( 4 digit ) 6. 6 (4 digit) • -7 (8 digit) 7. 10 (8 digit) • -12 (8 digit) 8. 30 (8 digit) • -20 (8 digit) 9. 98 (digit) • -100 (8 digit) 10. 126 (digit)

  30. Addition 2’s Complement For 4 digit : 4 0100 3 + 0011 + 7 0111 22+21+20 = 4+2+1 =7

  31. Addition 2’s Complement For 4 digit -1 1111 -2 + 1110 + -3 11101 -(8)+4 +0 + 1 = -3 Carry out

  32. Exercise For 4 Digit : • 7 + (-5) • -6 + -1 • 3 + 4 • 2 + 3 • -4 + 7 Converse all item to digital and addition. And then Converse to decimal again

  33. Subtraction 2’s Complement + 7 0111 + 3 (0011)- 1101 + +4 10100 Discard

  34. Subtraction 2’s Complement (-8) 1000 (-3) = 1101 - 0011 + -5 1011

  35. Exercise for 4 digit . Converse decimal above to digit and subtraction. After that converse to decimal again : • (+3) – (-3) • (-4) – (+2) • (-8)- (+4) • (-3) – (-4) • (7) – (5)

  36. Signed Numbers OP C=Carry V=overflow Signed Numbers 2’s Complement ALU • Addition and subtraction use the same rules as unsigned binary. • Same hardware may be used for both • Carry (C) is used for unsigned, overflow (v) for signed Signed Numbers The same hardware OP C=Carry V=overflow Signed Numbers Arithmetic Flags in Condition code register (CCR)

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