2.1 Represent Relations and Functions Objective: Represent relations and graph linear functions.
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Transcript of 2.1 data represent on cpu
Chapter 2.0:
Data Representation on CPU
FP203 : Computer Organization
2.1 Number System (decimal, binary, octal, and
hexadecimal)
2.2 Arithmetic Operation in number system.
2.3 Convert Decimal, Binary, Octal and Hexadecimal
Numbers to different bases.
2.4 Coding system:Sign and magnitude,
1’s Complement and 2’s Complement
Binary Coded Decimal (BCD system)
ASCII and EBCDIC
Topic Cover
Examples
INTRODUCTION
Real World
Data
Computer
DataInput device
Dear Mom: Keyboard 10110010…
Digitalcamera
10110010…
• Many number system are in use in digital technology.
• Most common are:– Decimal, N10
– Binary, N2
– Octal, N8
– Hexadecimal, N16
2.1 Number System
• Arithmetic operation in number system consist of:– Addition– Subtraction– Multiplication– Division
2.2 ARITHMETIC OPERATION
Only cover this 2 topics
• Decimal system is composed of 10 numerals or symbol.
• Symbol: 0,1,2,3,4,5,6,7,8,9
• Example: 23410
Decimal number system
10 Symbol
103 102 101 100 . 10-1
= 1000 = 100 = 10 = 1 . = 0.1
Multiplier:
2746.210 This number is came from this calculation:
2746.210 = (2x1000) + (7x100) + (4x10) + (2x1) + (2x0.1)
= 2000 + 700 + 40 + 2 + 0.2
= 2746.2
Decimal number = Natural Number
Example:
2 7 4 2 . 2103 102 101 100 . 10-1
= 1000 = 100 = 10 = 1 . = 0.1
Arithmetic Operation
Decimal
+ -
Decimal AdditionExample:
a. 89310 + 32110 =
89310
+32110
121410
Try this : 73310 + 79910 = ?
b. 75710 + 24510 = 75710
+ 24510
100210
Decimal Subtraction
Example:
a. 5410 - 1710 =
5410
- 1710
3710
Try this : 533310 - 3710 = ?
b. 15710 - 8910 = 15710
- 8910
6810
• Octal system is composed of 8 numerals or symbol.
• Symbol: 0,1,2,3,4,5,6,7
• Example: 658
Octal number system
8 Symbol
83 82 81 80 . 8-1
= 512 = 64 = 8 = 1 . = 0.125
Multiplier:
107.158 This number can be convert to decimal value using this calculation:
107.158 = (1x64)+(0x8)+(7x1)+(1x0.1250)+(5x0.0156)
= 64 + 0 + 7 + 0.1250 + 0.078
= 71.20310
Example:
1 0 7 . 1 582 81 80 . 8-1 8-2
= 64 = 8 = 1 . = 0.1250 = 0.0156
Arithmetic Operation
Octal
+ -
Octal AdditionSekiranya setiap hasil perjumlahan yang melebihi atau sama dengan 8 mestilah ditolak dengan 8.
Example:
a. 1238 + 3218 =
1238
+3218
4448
Try this : 7338 + 748 = ?
b. 4578 + 2458 = 4578
+ 2458
7248
Octal Subtraction Sekiranya terdapat peminjam, nombor peminjam mestilah dijumlahkan dengan 8.
Example:
a. 5248 - 1678 =
5248
- 1678
3358
Try this : 15238 - 3648 = ?
b. 1678 - 248 = 1678
- 248
1438
• Binary system is composed of 2 numerals or symbol.
• Symbol: 0,1
• Example: 1012
Binary number system
2 Symbol
25 24 23 22 21 20
= 32 = 16 = 8 = 4 = 2 = 1
Multiplier:
10.1012 This number can be convert to decimal value using this calculation:
10.1012 = (1x2)+(0x1)+(1x0.5)+(0x0.25)+(1x0.125)
= 2 + 0 + 0.5 + 0 + 0.125
= 2.62510
Example:
1 0 . 1 0 121 20 . 2-1 2-2 2-3
= 2 = 1 . = 0.5000 = 0.2500 = 0.1250
Arithmetic Operation
Binary
+ -
Binary AdditionThe four basic rules for adding binary digits are as follows:
Example:
110112 + 100012 =
110112
+ 100012
1011002
Try this : 101112 + 1112 = ?
0 + 0 = 00 + 1 = 11 + 0 = 1
1 + 1 = 0 carry 1
Binary SubtractionThe four basic rules for subtracting binary digits are as follows:-
Example:
10012 -102 =
10012
- 102
1112
Try this : 1010112 – 11112 =?
0 - 0 = 0 0 - 1 = 1 borrow 1 1 - 0 = 1 1 - 1 = 0
Have previously looked at the subtraction operation. A quick review.
Just like subtraction in any other base
10110
-10010
00100
• And when a borrow is needed. Note that the borrow gives us 2 in the current bit position.
.
Binary Subtraction
Example
• When there is no borrow into the msb position, then the subtrahend in not larger than the minuend and the result is
positive and correct.
• If a borrow into the msb does occur, then the subtrahend is larger than the minuend.
In General
• Now do the operation 4 – 6
• Correct difference is -2 or -0010• Different because 2n was brought in and made the operation M-
N+2n
Consider
• Hexadecimal system is composed of 16 numerals or symbol.
• Symbol: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
• Example: 7A16
Hexadecimal number system
16 Symbol
163 162 161 160 . 16-1
= 4096 = 256 = 16 = 1 . = 0.0626
Multiplier:
10 11 12 13 14 15
B6F.7C16 This number can be convert to decimal value using this calculation:
B6F.7C16 = (11x256) + (6x16) + (15x1) + (7x0.0625) + (12x0.0039)
= 2816 + 96 + 15 + 0.4375 + 0.0468
= 2927.484310
Example:
B 6 F . 7 C162 161 160 . 16-1 16-2
= 256 = 16 = 1 . 0.0625 = 0.0039
Arithmetic Operation Hexadecimal
+ -
Hexadecimal AdditionSekiranya setiap hasil perjumlahan yang melebihi atau sama dengan 16 mestilah ditolak dengan 16.
Example:
a. 3316 + 4716 =
3316
+ 4716
7A16
Try this : DF16 + AB16 = ?
b. 20D316 + 12BC16 = 20D316
+ 12BC16
338F16
Hexadecimal SubtractionNilai yang kecil daripada 16 boleh dipinjam dari sebelah dengan nilai 16.
Example:
a. 4416 - 1716 =
4416
- 1716
2D16
Try this : DF16 - AB16 = ?
b. 20D316 - 12BC16 = 20D316
- 12BC16
0E1716
2.3: Convert Decimal, Binary, Octal and Hexadecimal Numbers to different bases
1111012 This number can be convert to decimal value using this calculation:
1111012 = (1x32)+(1x16)+(1x8)+(1x4)+(0x2)+(1x1)
= 32 + 16 + 8 + 4 + 0 + 1
= 6110
Try this: Convert 1100.10112 to decimal?
Convert 100.10112 to decimal?
Example:
1 1 1 1 0 125 24 23 22 21 20
= 32 = 16 =8 = 4 = 2 = 1
Convert Binary to Decimal (N2 – N10)
1111012
1111012 = [(1x4)+(1x2)+(1x1)] [(1x4)+(0x2)+(1x1)]
= [4 + 2 + 1][ 4 + 0 + 1]
= 758
Try this: Convert 110010112 to Octal?
Convert Binary to Octal adalah dengan membahagikan nombor Binary tersebut kepada 3 bit bermula dari sebelah kanan (LSB)
1 0 122 21 20
= 4 = 2 = 1
Convert Binary to Octal (N2 - N8)
1 1 122 21 20
= 4 = 2 = 1
LSB
01012
01012 = (0x8)+(1x4)+(0x2)+(1x1)
= 0 + 4 + 0 + 1
= 516
Try this: Convert 101111012 to Hexadecimal?
Convert Binary to Hexadecimal adalah dengan membahagikan nombor binary kepada 4 bit bermula dari LSB. Sekiranya bit tersebut tidak mencukupi, maka digit ‘0’ perlu ditambah pada MSB
0 1 0 123 22 21 20
= 8 = 4 = 2 = 1
Convert Binary to Hexadecimal (N2 – N16)
LSB
181
1810 = 100102
Try this: Convert 32.20210 to binary?
Convert 8910 to binary?
Example: Convert 1810 to binary
Convert Decimal to Binary (N10 – N2)
2 18
2 9 0
2 4 1
2 2 0
2 1 0
0 1
0.3410 =
300.3410 = 454.256058
Try this: Convert 32.20210 to Octal?
Example: Convert 300.3410 to Octal i. 300 Divide by 8 ii. 0.34 Multiply by 8
Convert Decimal to Octal (N10 – N8)
8 300
8 37 4
8 4 5
8 0 4
0.34 x 8 = 2.72 ( 2+0.72 )
0.72 x 8 = 5.76 ( 5+0.76 )
0.76 x 8 = 6.08 ( 6+0.08 )
0.08 x 8 = 0.64 ( 0+0.64 )
0.64 x 8 = 5.12 ( 5+0.12 ) 0.25605
454
2010= 1416
Try this: Convert 343410 to hexadecimal?
Example: Convert 2010 to Hexadecimal
Convert Decimal to Hexadecimal (N10 – N16)
16 20 Balance
16 1 4
0 1
4 58
1001018 = [(1x4)+(0x2)+(0x1)] [(1x4)+(0x2)+(1x1)]
= [4 + 0 + 0][ 4 + 0 + 1]
= 100 1012
Try this: Convert 110010112 to Octal?
Convert Octal to Binary adalah dengan menukar setiap digit oktal kepada nilai 3 bit binary nya
1 0 122 21 20
= 4 = 2 = 1
Convert Octal to Binary (N8 – N2)
1 0 022 21 20
= 4 = 2 = 1
MSB LSB
3 A16
3A16 = [(0x8)+(0x4)+(1x2)+(1x1)][(1x8)+(0x4)+(1x2)+(0x1)]
= [0 + 0 + 2 + 1][ 8 + 0 + 2 + 0]
= 0011 10102
Try this: Convert EFA16 to Binary?
Convert Octal to Binary adalah dengan menukar setiap digit hexadecimal kepada nilai 4 bit binary nya
1 0 1 023 22 21 20
= 8 = 4 = 2 = 1
Convert Hexadecimal to Binary (N16 – N2)
0 0 1 123 22 21 20
= 8 = 4 = 2 = 1
MSB LSB
2.4: Coding System Sign and magnitude, 1’s Complement & 2’s
Complement
Apply what you have learned to the binary number systems. How do you represent negative numbers in this 8-bit binary system?
Cut the number system in half.
Use 00000001 – 01111111 to indicate positive numbers.
Use 10000000 – 11111111 to indicate negative numbers.
Notice that 00000000 is not positive or negative.
8-Bit Binary Number System
01111111
01111110
01111101
00000001
00000000
11111111
11111110
10000001
10000000
pos(+)
neg(-)
+127
+126
+125
+1
0
-1
-2
-127
-128
• As there is no third symbol available to store a negative symbol explicitly we must use a bit to show if a number is negative or not.– We name this bit the ‘Sign Bit’– We use the leftmost bit.– If the ‘Sign Bit’ is 1 then the number is
negative, if it is 0 then it is positive.
Representing Negative Numbers
• What did do you notice about the most significant bit of the binary numbers?
• The MSB is (0) for all positive numbers.
• The MSB is (1) for all negative numbers.
• The MSB is called the sign bit.
• In a signed number system, this allows you to instantly determine whether a number is positive or negative.
Sign Bit
01111111
01111110
01111101
00000001
00000000
11111111
11111110
10000001
10000000
pos(+)
neg(-)
+127
+126
+125
+1
0
-1
-2
-127
-128
• This is just inverting each bit.
1’s compliment of 00000010 is 1111101
1’s Complement
1 1 1 1 1 0
flip the number.
flip the number.
1
0 0 0 0 0 1 0
The steps in the 2’s Complement process
First, complement all of the digits in a number.
– A digit’s complement is the number you add to the digit to make it equal to the largest digit in the base (i.e., 1 for binary). In binary language, the complement of 0 is 1, and the complement of 1 is 0.
Second, add 1.
– Without this step, our number system would have two zeroes (+0 & -0), which no number system has.
2’S Complement Process
2’s Complement Examples
Example #1
Example #2
Complement Digits
Add 1
5 = 00000101
-5 = 11111011
11111010
+1
Complement Digits -13 = 11110011
13 = 00001101
00001100
+1
Using The 2’s Compliment Process
9 + (-5)
4
(-9) + 5
- 4
(-9)+ (-5)
- 4
9 + 5
14
POS + POS
POS
POS + NEG
POS
NEG + POS
NEG
NEG + NEG
NEG
Use the 2’s complement process to add together the following numbers.
POS + POS → POS Answer
If no 2’s complement is needed, use regular binary addition.
00001001 9 + 5
14
00001110
00000101 +
POS + NEG → POS Answer
Take the 2’s complement of the negative number and use regular binary addition.
00001001 9 + (-5)
4
11111011+
00000101
11111010+1
11111011
2’s Complement
Process
1]000001008th Bit = 0: Answer is Positive
Disregard 9th Bit
POS + NEG → NEG Answer
Take the 2’s complement of the negative number and use regular binary addition.
11110111 (-9) + 5
-4
00000101+
00001001
11110110+1
11110111
2’s Complement
Process
111111008th Bit = 1: Answer is Negative
11111100
00000011+1
00000100
To Check:Perform 2’s ComplementOn Answer
NEG + NEG → NEG Answer
Take the 2’s complement of both negative numbers and use regular binary addition.
11110111 (-9) + (-5)
-14
11111011 +
2’s ComplementNumbers, See Conversion ProcessIn Previous Slides
1]111100108th Bit = 1: Answer is Negative
Disregard 9th Bit
11110010
00001101+1
00001110
To Check:Perform 2’s ComplementOn Answer
2.4: Coding System Binary Coded Decimal (BCD
System) ASCII and EBCDIC
Four bits per digit
Binary-Coded Decimal (BCD)
Digit Bit pattern
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
Note: the following bit patterns are not used:
101010111100110111101111
• 709310 = ? (in BCD)
Example
7 0 9 3
0111 0000 1001 0011
• ASCII = American National Standard Code for Information Interchange
• 7-bit code• 8th bit is unused (or used for a parity bit)• 27 = 128 codes• Two general types of codes:
– 95 are “Graphic” codes (displayable on a console)
– 33 are “Control” codes (control features of the console or communications channel)
ASCII
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
ASCII Chart
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
Most significant bit
Least significant bit
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
e.g., ‘a’ = 1100001
95 Graphic codes
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
33 Control codes
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
Alphabetic codes
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
Numeric codes
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 $ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB ' 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L \ l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
Punctuation, etc.
• Representing text strings, such as “Hello, world”, in a computer
The Problem
“Hello, world” Example
============
Binary010010000110010101101100011011000110111100101100001000000111011101100111011100100110110001100100
Hexadecimal48656C6C6F2C207767726C64
Decimal72
1011081081114432
119103114108100
Hello,
world
============
============
Extended BCD Interchange Code (pronounced ebb’-se-dick)• 8-bit code• Developed by IBM• Rarely used today• IBM mainframes only
EBCDIC
EBCDIC “Extended Binary Coded Decimal Interchange Code” code
table
EBCDIC “Extended Binary Coded Decimal Interchange Code” code
tableExample:
1111 1111 1110 1001 1111 0111 1101 0111EBCDIC CODE
Z 6
P
Message below are represented in EBCDIC code. What is the message? Please convert by using EBCDIC Code table given:
i) 1111 1100 1011 0101 1101 1001 EBCDIC CODE
LSBMSB