Chapter 2 Data Representation on CPU (part 1)
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Transcript of Chapter 2 Data Representation on CPU (part 1)
Chapter 2:
DATA REPRESENTATION
IN COMPUTER MEMORY
SUMMARY: This topic introduces the numbering systems: decimal, binary, octal and hexadecimal. The topic covers the conversion between numbering systems, binary arithmetic, one's complement, two's complement, signed number and coding system. This topic also covers the digital logic components.
CLO 2:apply appropriate method to solve arithmetic problem in numbering system (C3).
RTA: (08 : 08)
2.1 Understand data representation
on CPU.2.1.1 Define decimal, binary, octal and hexadecimal number.
2.1.2 Perform arithmetic operation (addition and subtraction) in different number bases.
2.1.3 Convert decimal, binary, octal and hexadecimal numbers to different bases and vice-versa
INTRODUCTION The binary system and decimal system is
most important in digital system. Decimal - Universally used to represent
quantities outside a digital system. Its means, there will be situations decimal
values must be converted to binary values before entered to digital system.
Example : Calculator / Computer
DECIMAL NUMBERING SYSTEM Decimal system is composed of 10 numerals
or symbols.
These 10 sysmbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Using these symbols as digits of a number, it can express any quantity.
Base number = 10 Basic number = 0,1,2,3,4,5,6,7,8,9
23410
Basic number
Base number
Positional Values (weights)
2 7 4 6 . 2
102103 101 100 10-1
Positional values (weights)
MSD LSD
2746.210 is from calculation below:
2746.2 = (2x103) + (7x102) + (4x101) + (6x100) + (2x10-1) = 2000 + 700 + 40 +6 +0.2 = 2746.2
Most significant digit Least significant digit
BINARY NUMBERING SYSTEM Define Binary numbers
Binary numbers representing number in which only digits 0 or 1.
ADDITION BINARY NUMBERSBasic binary addition rule :
0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 10
1 + 1 + 1 = 11
Example : 101 + 101 = 10101011 + 1011 = ?
Ex 1:
110112 + 100012 = 1011002
Ex 2: 101112 + 1112 = ________
Exercise
Subtraction
Four conditions in binary subtraction 0 - 0 = 0 0 - 1 = 1 borrow 1 1 - 0 = 1 1 - 1 = 0 10 - 1 = 1 If a 10 being borrow a 1, what’s left with
that 10 is a 1
Ex 1:
10012 – 102 = 1112
Ex 2: 1010112 – 11112 =__________
Conversions of Binary NumbersBinary to Decimal conversions Example : 1 1 0 1 12
24 + 23 + 22 + 21 + 20 = 16 + 8 + 2 + 1
= 2710
Decimal to Binary conversions
Convert 2510 to binary number
Exercise: Convert 3010 to binary number
OCTAL NUMBERING SYSTEM The octal number system has a base of eight,
meaning that it has eight possible digits: 0,1,2,3,4,5,6 and 7.
The digit positions in an octal number have weights as follows :
84 83 82 81 80 •. 8-1 8-2 8-3
Ex: 5248 – 1678 = 3358
1678 – 248 = _________
Octal number - (Subtraction - Pengurangan)
Octal –to-decimal conversion
Convert 3728 to decimal number
3728 = 3 x (82) + 7 x (81) + 2 x (80)
= (3 x 64) + (7x 8) + (2 x 1)
= 25010
Octal number - Addition (Penambahan)
Ex:
1238
+ 3218
4448
Ex:
4578
+ 2458
Decimal integer can be converted to octal by using the same repeated-division method with a division factor of 8.
Decimal-to-Octal Conversion
Octal –to- Binary conversion
Binary to Octal conversion
The hexadecimal number system uses base 16.
It has 16 possible digit symbols. It uses the digits 0 through 9 plus the letters
A, B, C, D, E and F as the 16 digit symbols.
HEXADECIMAL NUMBERING SYSTEM
7A16
Basic number
Base number
Hexadecimal number - Addition (Penambahan)
Ex:
3 316
+ 4 716
Ex:
2 0 D 316
+ 1 2 B C16
7 A16
Hexadecimal number - Subtraction (Pengurangan)
Ex:
4 416
- 1 716
Ex:
3 2 5 516
- 3 1 8 216
2 D16
Hexadecimal-To-Decimal Conversion A hexadecimal number can be converted to
its decimal equivalent by using the fact that each hex digit position has a weight that is a power of 16.
Ex 1:
1416 = (1 x 161) + (4 x 160)
= 16 + 4
= 2010
Ex 2:
ABC16 = (10 x 162) + (11 x 161)
= (12 x 160)
= 2560 + 176 + 12
= 274810
Decimal to hex conversion can be done using repeated division by 16.
Ex: Convert 2010 to hex
Decimal-To-Hexadecimal Conversion
16 20
16 1 4
2010 = 1416
Like the octal number system, the hexadecimal number system is used primarily as a “shorthand” method for representing binary numbers.
It is a relatively simple matter to convert a hex number to binary .
Each hex digit is converted to its four-bit binary equivalent.
Hexadecimal-to-Binary Conversion
Ex:
111010012 = __________
00101101001111002 =___________
The binary number is grouped into groups of four bits, and each group is converted to its equivalent hex digit.
Zero are added, as needed to complete a four-bit group.
Ex:
1012 = 0101
= 516
Binary-to-Hexadecimal Conversion
Summary
Hexadecimal Decimal Binary
0 0 0000
1 1 0001
2 2 0010
3 3 0011
4 4 0100
5 5 0101
6 6 0110
7 7 0111
Summary
Hexadecimal Decimal Binary
8 8 1000
9 9 1001
A 10 1010
B 11 1011
C 12 1100
D 13 1101
E 14 1110
F 15 1111
2.1.4 Describe the coding system
a. Sign and magnitudeb. 1’s Complement and 2’s Complementc. Binary Coded Decimal (BCD system)d. ASCII and EBCDIC
Describe the coding system
Sign and Magnitude
Positive sign, ( 0 ) orNegative sign ( 1 )
Magnitude =Size
Or value
Describe the coding system
Sign and Magnitude
Example 1 : Represent -9 in sign and magnitude.
-9 sign magnitude 1 1001 11001 -9 in sign &magnitude value is 11001
Describe the coding system
Sign and Magnitude
Example 2 : Represent 25 in sign and magnitude.
+25 sign magnitude 0 11001 011001 25 in sign & magnitude value is 011001
One’s Complements and Two’s Complements
One’s Complements One’s complements is used in binary number. The one’s complement of a binary number is
obtained by changing each 0 to 1 and 1 to a 0. Only change negative number In other words, change each bit in the number to
its complement.
Exp:
10011001 – original binary number
01100110 – complement each bit to form 1’s complement
Thus, we say that the 1’s complement of 10011001 is 01100110.
Exp:
Convert -2710 to 1’s complement
a) -2710 = 001002
Convert -4510 to 1’s complement
-------------
Two’s Complement The 2’s complement of a binary number is formed
by taking the 1’s complement of the number and adding 1 to the least-significant-bit (LSB) position.
Exp: 101101 binary equivalent of 45
010010 complement each bit to form 1’s complement
+ 1 add 1
010011 2’s complement of original binary
number
Two’s complement = One’s Complement + 1
Exercise
1. Convert the number below to 1’s complement 2’s complement
------------------- -------------------
-101110012 0100 0110 0100 0111
-5768 01000 0001 01000 0010
-124516
-4510
Addition in 1’s complement
Exp 1 : 810 + (-310)1000 8 change to binary number
+ 1100 -3 change to 1’s complement 10100
1 add carry to LSB 0101
Exercise 2 : 510 + (-210) ----------------
Exp 1: 810 + (-310)1100 -3 change to 1’s complement
+ 1
1101 -3 change to 2’s complement
+ 1000 8 change to binary number
10101
Addition in 2’s complement
This carry is disregarded, the result is 0101 (sum=5)
Exp 2: -810 + 310
0111 -8 change to 1’s complement
+ 1
1000 -8 change to 2’s complement
+ 0011 3 change to binary number
1011negative sign bit
Only binary number which have –ve sign need to change to 1’s complement. If the number is decimal number, change the number to binary number.
The –ve number that already change to 1’s complement, means that the number already change to +ve. So that, the subtraction process have change to addition process.
Overflow bit in addition process need to carry to LSB and add with the number.
Subtraction in 1’s complement
Exp 1: 2510 – 1310
Step 1 : Convert 2510 and -1310 to binary number.
2510 = 110012 1310 = 11012
Step 2 : Change 1310 = 1101 to 1’s complement
1310 = 11012 change to 1’s complement = 10010
11001 25 change to binary number
+ 10010 -13 change to 1’s complement
101011
+ 1 add 1
01100 total=12
Change the number are given to binary number.
For each –ve binary number, we must change to 1’s complement (change 0 to 1 and 1 to 0).
Then, add the number with 1.
Subtraction in 2’s complement
Exp 1: 2510 – 1310
Step 1 : Convert 2510 and -1310 to binary number.
2510 = 110012 1310 = 11012
Step 2 : Change -1310 to 2’s complement
-1310 = 011012 change to 1’s complement = 10010
10010 change to 2’s complement = 10011
11001 25 change to binary
+ 10010 -13 change to 2’s complement
101100 total=12Carry disregard
Solve this arithmetic with 2’s complement.
i. 428 – 158
ii. 101110 - 56910
Exercise
Signed Number
Addition of Signed Number Addition number same sign Exp: +4 + (+8) = +12
+4 00000100
+8 00001000
+12 00001100
Addition number different sign Exp 1:
(-4) + (+8) = +4
-4 11111100
+ (+8) 00001000
+4 1 00000100
This carry is disregarded
00000100 +411111011 1’s complements
+ 111111100 2’s complements
Exp 1:
(-12) + (+5) = -7
-12 11110100
+ +5 00000101
-7 11111001
00001100 +1211110011 1’s complements
+ 111110100 2’s complements
Negatif sign bit
Subtraction of Signed Number
Positive Number (-) Negative Number Exp:
+10 – (-5) = +10 + (+5)
= + 15
+10 00001010
+ 5 00000101
+15 00001111
Negative Number (-) Positive Number Exp:
-10 – (+5) = - 15
- 10 11110110 11110110
- (+5) 00000101 + 11111011
-15 111110001
Negative Number (-) Negative Number Exp:
-10 – (-5) = -10 + (5)
= - 5
-10 11110110
- (- 5) + 00000101
- 5 11111011
BCD Code
BCD – Binary Coded Decimal
It’s contain BCD 8421, 2421, 3321, 5421, 5311, 4221 and etc.
The common used is BCD 8421 and BCD 2421.
4 bit BCD CodeDesimal 5421 5311 4221 3321 2421 8421 7421
0 0000 0000 0000 0000 0000 0000 0000
1 0001 0001 0001 0001 0001 0001 0001
2 0010 0011 0010 0010 0010 0010 0010
3 0011 0100 0011 0011 0011 0011 0011
4 0100 0101 1000 0101 0100 0100 0100
5 1000 1000 0111 1010 1011 0101 0101
6 1001 1001 1100 1100 1100 0110 0110
7 1010 1011 1101 1101 1101 0111 1000
8 1011 1100 1110 1110 1110 1000 1001
9 1100 1101 1111 1111 1111 1001 1010
Binary-Coded-Decimal Code
If each digit of a decimal number is represented by its binary equivalent, the result is a code called binary-coded-decimal.
Decimal digit can be as large as 9, four bits are required to code each digit (the binary code for 9 is 1001)
Exp: 87410
8 7 4 (decimal)
1000 0111 0100 (BCD)
BCD 8421 Code – to – Binary Number
Exp:
Convert 1001 0110BCD 8421 to binary number.
Step 1: Change BCD 8421 code to decimal number.
1001 0110
9 6
Step 2 : Change decimal number to binary number.
1001 0110BCD 8421 = 11000002
Exp:
Convert 10010102 to BCD 8421
code.Step 1: Change binary number to decimal
number.
10010102 = 7410
Step 2: Change decimal number to BCD
8421 code.
10010102 = 01110111BCD
8421
Binary Number – to – BCD 8421 Code
The most widely used alphanumeric code is the American Standard Code for Information Interchange (ASCII).
The ASCII code is a seven-bit code and so it has 27
= 128 possible code groups.
ASCII Code
MSBLSB
Binary 000 001 010 011 100 101 110 111
Binary Hex 0 1 2 3 4 5 6 7
0000 0 Nul Del sp 0 @ P p
0001 1 Soh Dc1 1 1 A Q a q
0010 2 Stx Dc2 “ 2 B R b r
0011 3 Etx Dc3 # 3 C S c s
0100 4 Eot Dc4 $ 4 D T d t
0101 5 End Nak % 5 E U e u
0110 6 Ack Syn & 6 F V f v
0111 7 Bel Etb ‘ 7 G W g w
1000 8 Bs Can ( 8 H X h x
1001 9 HT Em ) 9 I Y i y
1010 A LF Sub . : J Z j z
1011 B VT Esc + ; K k
1100 C FF FS , < L l
1101 D CR GS - = M m
1110 E SO RS . > N n
1111 F SI US / ? O o
ASCII code
Exp:
An operator is typing in a BASIC program
at the keyboard of a certain
microcomputer. The computer converts
each keystroke into its ASCII code and
stores the code as a byte in memory.
Determine the binary strings that will be
entered into memory when operator types
in the following BASIC statement:
GOTO 25
Solution:
Locate each character (including the
space) and record ASCII code.
G 01000111
O 01001111
T 01010100
O 01001111
(space) 00100000
2 00110010
5 00110101
*0 was added to the leftmost bit of each ASCII code because theCodes must be stored as bytes (eight bits).
Exercise :
1.The following message encode in ASCII code. What the meaning of this code ?
a) 54 4F 4C 4F 45 47 b) 48 45 4C 4C 4F c) 41 50 41 4B 48 41 42 41 52
EBCDIC Stand for Extended Binary Coded Decimal
Interchange Code. was first used on the IBM 360 computer,
which was presented to the market in 1964. Used with large computer such as
mainframe.