Kimia

1. Sistem Bilangan
Oleh :
Moch Nur Purnama

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
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. 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
3578.778
8 x 100
7 x 101
Base 10, weigthing are powers of 10 5 x 102
3 x 103

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

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. 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. 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. 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
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. 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
4 – bit
Binary
11 - 1 = 10
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.
(Module arithmetic – ignore
The borrow /carry)
4 – bit
8 1000 Binary
- 14 - 1110 Number Circle
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
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. 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
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. 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. 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) :
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. 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
Carry out
-(8)+4 +0 + 1 = -3

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. 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 :
1. (+3) – (-3)
2. (-4) – (+2)
3. (-8)- (+4)
4. (-3) – (-4)
5. (7) – (5)

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)