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  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

Analoge vs Digital

Analogue

A to D

Digital Processing

D to A

Analogue

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

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

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

Positional Number System Decimal point

7x10-1 7x10-2 8x10-3

3578.778

Base 10, weigthing are powers of 10

8 x 100 7 x 101 5 x 102 3 x 103

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 3 Binary, weightings are powers of 21 x 2 = 8.000

Multi-precision Arithmatic Additional of A and B A1

B1

A2

B2 +

A2 Carry Flag

Carry Out

B3 Carry In

Carry Flag

Carry Out

Multi-precision Arithmatic Carry Out

Carry Flag

Carry In

A1

B1

A2

B2 -

A2

B3

Hexadecimal Numbers 660

4

41

9

: 16 : 16

2 Hexadecimal :

294 Hex 13

215 7

Hexadecimal :

7D Hex

: 16

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

Hexadecimal Numbers 660

0010 1001 0100

2

215

9

4

0000 1101 0111

0

D

7

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

Decimal to Binary Number =

36.375

Base =

2

Decimal Number

Binary Digits

Generetee each digit by successive division Or multiplication. Converter Number

0

0

0100100.0110

0.5

1

0100100.011

0.75

1

0100100.01

0.375

0

0100100.0

36

0

0100100

18

0

010010

9

1

01001

4

0

0100

2

0

010

1

1

01

0

0

0

There is no guarantee the fraction will be finite Fractional part – Multiplication by base

Whole part – divition by base

Binary Additional

0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 carry 1

Easy Layaou ?

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

Carry out of 8-bit number

0 1 0 0 1 0 1 1

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

0

0

1

0

1

1

1

0

1

1

1

1

1

1

1

0

1

1

1

1

0

Both rows subtracted

Borrow in from Left column

Borrow out

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

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

11 -1

1110 1 10

1010

4 – bit Binary Number Circle

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)

8 - 14

1000 - 1110 10

(-1)1010

4 – bit Binary Number Circle

Representing –ve Number  Several

choices for natation * sign + magnitude notation * 1’s complement * 2’s complement notation * various ‘excess codes ‘

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

?

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!

Sign + Magnitude Examples Value

4-bit sign + magnitude

8-bit sign + 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

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

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

1100.101 Sign Bit

0 x 20= 0.000 0 x 21= 0.000 1 x 22= -4.375 4.000 3 Binary, weightings are powers of 21 x 2 = -8.000

2’s Complement Examples Value

4-bit sign 2’s complement

8-bit sign 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

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

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)

Addition 2’s Complement For 4 digit : 4 3 + 7

0100 0011 0111

+

22+21+20 = 4+2+1 =7

Addition 2’s Complement For 4 digit -1 -2 -3

+

1111 1110 11101

+

Carryout

-(8)+4 +0 + 1 = -3

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

Subtraction 2’s Complement +7 + 3 (0011)+4

Discard

0111 1101 + 10100

Subtraction 2’s Complement (-8) (-3) = 1101 -5

1000 0011 + 1011

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)

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 C=Carry

V=overflow Signed Numbers

Signed Numbers

Arithmetic Flags in Condition code register (CCR)

V=overflow