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# Sistem bilangan

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### Sistem bilangan

1. 1. Sistem Bilangan Oleh : Moch Nur Purnama
2. 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. 3. Analogue vs Digital Analogue Digital
4. 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. 5. Analoge vs Digital Digital D to A Analogue Analogue A to D Processing 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. 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. 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. 8. Positional Number System Decimal point 7x10-1 7x10-2 8x10-3 3578.778 8 x 100 7 x 101 Base 10, weigthing are powers of 10 5 x 102 3 x 103
9. 9. Unsigned binary numbers Binary point 1 x 2-1 = 0.500 0 x 2-2 = 0.000 1 x 2-3 = 0.125 1100.101 Each bit of the Number may be Representaed by A Boolean value 0 x 20= 0.000 0 x 21= 0.000 1 x 22= 4.000 Binary, weightings are powers of 21 x 2 = 8.000 3
10. 10. Multi-precision Arithmatic Additional of A and B A1 B1 A2 B2 + A2 B3 Carry Carry Flag Carry Carry Flag Carry Out In Out
11. 11. Multi-precision Arithmatic Carry Carry Flag Carry Out In A1 B1 A2 B2 - A2 B3
12. 12. Hexadecimal Numbers 660 0 0 4 : 16 1 1 2 2 41 3 3 9 : 16 4 4 2 5 5 6 6 Hexadecimal : 294 Hex 7 7 8 8 215 9 9 13 : 16 10 A 11 B 7 12 C 13 D 14 E Hexadecimal : 7D Hex 15 F
13. 13. Hexadecimal Numbers 0 0000 660 0010 1001 0100 1 0001 2 0010 3 0011 4 0100 2 9 4 5 0101 6 0110 7 0111 8 1000 215 0000 1101 0111 9 1001 A 1010 B 1011 C 1100 0 D 7 D 1101 E 1110 F 1111
14. 14. Decimal to Binary Number = 36.37 Generetee each digit by successive division 5 Or multiplication. Base = 2 There is no guarantee the fraction will be Decimal Binary Converter Number finite Number Digits 0 0 0100100.0110 0.5 1 0100100.011 Fractional part – Multiplication by base 0.75 1 0100100.01 0.375 0 0100100.0 36 0 0100100 18 0 010010 9 1 01001 Whole part – divition by base 4 0 0100 2 0 010 1 1 01 0 0 0
15. 15. Binary Additional 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 carry 1 Easy Layaou ?
16. 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. 17. Binary Subtraction A borrow-out of 1 from This column becomes a borrow in 229 – 46 = 183 of 2 in this column 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 0 1 1 1 Both rows subtracted
18. 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. 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 4 – bit Binary 11 - 1 = 10 Number Circle11 1110-1 1 10 1010
20. 20. Binary Number Circle Subtracting across the boundary Still “works” if you think of result As the distance on the number Circle. (Module arithmetic – ignore The borrow /carry) 4 – bit 8 1000 Binary - 14 - 1110 Number Circle 10 (-1)1010
21. 21. Representing –ve Number  Several choices for natation * sign + magnitude notation * 1’s complement * 2’s complement notation * various ‘excess codes ‘
22. 22. Sign Number – sign + magnitudeNotation Sign Bit Magnitude 0  +ve Simple binary number 1  - ve How about Null or Zero Problem ? + 0  0000 - 0  1000
23. 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. 24. Sign + Magnitude Examples 4-bit sign + Value 8-bit sign + magnitude magnitude +7 0111 00000111 +6 0110 00000110 …… …… …… +1 0001 00000001 +0 0000 00000000 -0 1000 10000000 -1 1001 10000001 -2 1010 10000010 …… …… …… -7 1111 10000111
25. 25. Sign Numbers – 2’sComplement  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. 26. Sign Numbers – 2’sComplement Binary point 1 x 2-1 = 0.500 0 x 2-2 = 0.000 1 x 2-3 = 0.125 1100.101 Sign Bit 0 x 20= 0.000 0 x 21= 0.000 1 x 22= -4.375 4.000 Binary, weightings are powers of 21 x 2 = -8.000 3
27. 27. 2’s Complement Examples 4-bit sign 2’s Value 8-bit sign complement complement +7 0111 00000111 +6 0110 00000110 …… …… …… +1 0001 00000001 +0 0000 00000000 -1 1111 11111111 -2 1110 11111110 …… …… …… -7 1001 11111001 -8 1000 11111000
28. 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. 29. Exercise Converse decimal number above into negative (2’s complement) : 1. -7 ( 4 digit ) 6. 6 (4 digit) 2. -7 (8 digit) 7. 10 (8 digit) 3. -12 (8 digit) 8. 30 (8 digit) 4. -20 (8 digit) 9. 98 (digit) 5. -100 (8 digit) 10. 126 (digit)
30. 30. Addition 2’s Complement For 4 digit : 4 0100 3 + 0011 + 7 0111 22+21+20 = 4+2+1 =7
31. 31. Addition 2’s Complement For 4 digit -1 1111 -2 + 1110 + -3 11101 Carry out -(8)+4 +0 + 1 = -3
32. 32. Exercise For 4 Digit : 1. 7 + (-5) 2. -6 + -1 3. 3 + 4 4. 2 + 3 5. -4 + 7 Converse all item to digital and addition. And then Converse to decimal again
33. 33. Subtraction 2’s Complement +7 0111 + 3 (0011)- 1101 + +4 10100 Discard
34. 34. Subtraction 2’s Complement (-8) 1000 (-3) = 1101 - 0011 + -5 1011
35. 35. Exercise for 4 digit . Converse decimal above to digit and subtraction. After that converse to decimal again : 1. (+3) – (-3) 2. (-4) – (+2) 3. (-8)- (+4) 4. (-3) – (-4) 5. (7) – (5)
36. 36. 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 Signed Numbers The same hardware OP OP C=Carry V=overflow C=Carry Signed Numbers Signed Numbers V=overflow Arithmetic Flags in Condition code register (CCR)
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