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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

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

Analogue

Digital

Processing

D to A

A to D

Why Digital Only by using in Processing?

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

8 x 100

7 x 101

5 x 102

3 x 103

3578.778

Base 10, weigthing are powers of 10

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

A1

B1

A2

B2

+

A2

B3

Carry

Flag

Carry

Flag

Carry

Out

Carry

In

Carry

Out

Multi-precision Arithmatic

Multi-precision Arithmatic

Carry

Flag

Carry

Out

Carry

In

A1

B1

A2

B2

-

A2

B3

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

215

13

: 16

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

660

0010 1001 0100

2 9 4

215

0000 1101 0111

0 D 7

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

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0 carry 1

Easy Layaou ?

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

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

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

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

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

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

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 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

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

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) :

• -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)

For 4 digit :

4 0100

3 + 0011 +

7 0111

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

For 4 digit

-1 1111

-2 + 1110 +

-3 11101

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

Carry out

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

Subtraction 2’s Complement

+ 7 0111

+ 3 (0011)- 1101 +

+4 10100

Subtraction 2’s Complement

(-8) 1000

(-3) = 1101 - 0011 +

-5 1011

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)

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)