Transcript Intro to Computer - Naresuan University

305171 Computer Programming
Rattapoom Waranusast
Department of Electrical and Computer Engineering
Faculty of Engineering, Naresuan University

A COMPUTER is an electronic device that
can:
– Receive information
– Perform processes
– Produce output
– Store info for future use.
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
Personal computers (PC)
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
Mobile devices
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
Game consoles
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
Servers
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
Mainframe computers
IBM zEnterprise
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Super computers
Fujitsu’s K - Computer
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
Embedded systems
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
Robots
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PC
Mainframe
Mobile
Etc.
Hardware
Programmer
System Analyst
User
Administrator
Etc.
Software
OS
Applications
Firmware
Content
Etc.
Peopleware
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Feedback
Input
(Input Devices)
Process
(CPU)
Output
(Output Devices)
Storage
(Memory)
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
Arithmetic and logic unit (ALU)
– performs arithmetic and logical operations

Control unit
– controls the flow of data through the
processor, and coordinates the activities of
the other units within it

Registers
– small amount of storage available on the CPU
whose contents can be accessed more
quickly than storage available elsewhere
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
Main (primary) memory
– Stores programs and data while computer is
running. Main memory is fast and limited in
capacity. The CPU can only directly access
information in main memory.
– Random Access Memory (RAM)
– Read-Only Memory (ROM)
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Memory mapping
Address
Values
1 0 1 1 1 0 0 1
0000
0 0 0 0 1 1 1 1
0001
0002
..
..
..
..
FFFF
1 1 1 0 0 0 1 1
• Size :
•
•
•
•
•
•
•
•
Kilobyte
Megabyte
Gigabyte
Terabyte
Petabyte
Exabyte
Zettabyte
Yottabyte
KB
MB
GB
TB
PB
EB
ZB
YB
210 ~103
220 ~106
230 ~109
240 ~1012
250 ~1015
260 ~1018
270 ~1021
280 ~1024
0 0 1 0 0 1 0 0
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
External (secondary) memory
– Holds information too large for storage in
main memory.
– Information on external memory can only be
accessed by the CPU if it is first transferred to
main memory.
– It retains information when the computer is
switched off.
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CPU
ALU
Main
Memory
CU
VDO
controller
Registers
BUS
Clock
Disk
controller
I/O ports
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

A clock signal is a signal generated by quartz
crystal circuit that oscillates between a high and
a low state and is utilized to coordinate actions
of circuits.
Clock rate is measured in cycles/sec (Hertz ).
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User
Level 5
Problem oriented language level
Translation (compiler)
Level 4
Assembly Language
Translation (assembler)
Level 3
Operating System
Partial interpretation (operating system)
Level 2
Instruction Set Architecture
Interpretation (microprogram) or direct execution
Level 1
Micro-architecture level
Hardware
Level 0
Digital Logic Level
Circuits
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
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
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The interesting objects at this level are gates;
Each gate has one or more digital inputs
(signals representing a 0 or 1) and computes
as output some simple function of these
inputs, such as AND or OR;
A small number of gates can be combined to
form a 1-bit memory, which can store a 0 or 1;
The 1-bit memories can be combined in
groups of, for example, 16, 32 or 64 to form
registers.
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A collection of 8-32 registers that form a
local memory and ALU
 The registers are connected to the ALU to
form a data path over which the data
flow;
 The operation of the data path may be
controlled by a microprogram, directly by
hardware, or even by software.

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The ISA level is defined by the machine’s
instruction set
 This is a set of instructions carried out
interpretively by the microprogram or
hardware execution sets;
 Machine languages

– Strings of numbers giving machine specific
instructions
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

The Operating System Machine (OSM) level
is a complete set of instructions available to
application programmers. This includes all
ISA level instructions and a new set of
instructions that the operating system adds
called system calls.
The OSM level is always interpreted. For
instance, if reading data from file, the
operating system carries it out step by step.
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




This level is really a symbolic form.
This level provides a method for people to write programs
for levels 1, 2 and 3 in a form that is more readable.
Programs in assembly language are first translated to level
1, 2 or 3 language and then interpreted by the
appropriate virtual or actual machine.
The program that performs the translation is called an
assembler.
Example:
LOAD
ADD
MOV
R1, #FFFF
R2,R1
R3,R2
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



This level usually consists of languages designed
to be used by applications programmers.
These languages are generally called higher
level languages, for examples: C, C++, Java,
BASIC, LISP, Prolog, Pascal, FORTRAN, COBOL,
etc.
Programs written in these languages are
generally translated to Level 3 or 4 by
translators known as compilers, although
occasionally they are interpreted instead.
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
Computers use binary numbers internally
because storage devices like memory and disk
are made to store 0s and 1s. A number or a
text inside a computer is stored as a sequence
of 0s and 1s. Each 0 and 1 is called a bit, short
for binary digit. The binary number system has
two digits, 0 and 1.

Binary numbers are not intuitive. When you
write a number like 20 in a program, it is
assumed to be a decimal number. Internally,
computer software is used to convert decimal
numbers into binary numbers, and vice versa.
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
The digits in the decimal number system are 0, 1,
2, 3, 4, 5, 6, 7, 8, and 9. The value that each digit
in the sequence represents depends on its
position. A position in a sequence has a value that
is an integral power of 10.
5892 = 5 x 103 + 8 x 102 + 9 x 101 + 2 x 100

We say that 10 is the base or radix of the decimal
number system. Similarly, the base of the binary
number system is 2 since the binary number
system has two digits and the base of the hex
number system is 16 since the hex number
system has sixteen digits.
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
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
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A positional number system
Has only 2 symbols (0 and 1), so that base = 2
The maximum value of a single digit is 1
(base – 1)
The maximum value of N digits is 2N-1
Used mostly in computer system
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5 digits binary can store number up to
25-1 = 32-1 = 31
 Example

101012
=
=
=
(1x24) + (0x23) + (1x22) + (0x21)
+ (1x20)
16 + 0 + 4 + 0 + 1
2110
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
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
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A positional number system
Has only 8 symbols (0, 1, 2, 3, 4, 5, 6, 7), so that
base = 8
The maximum value of a single digit is 7
(base – 1)
The maximum value of N digits is 8N-1
Used mostly in computer system
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
Example
20578
=
=
=
(2x83) + (0x82) + (5x81) + (7x80)
1024 + 0 + 40 + 7
107110
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
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
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
A positional number system
Has only 16 symbols (0-9, A, B, C, D, E, F), so
that its base = 16
The maximum value of a single digit is 15 (F)
(base – 1)
The maximum value of N digits is 16N-1
Used mostly in computer system
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
Example
1AF16
=
=
=
(1x162) + (10x161) + (15x160)
256 + 160 + 15
43110
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
Method
1. Determine the column (positional) value of
each digit.
2. Multiply the obtained column values by the
digits in the corresponding columns.
3. Calculate the sum of these products.
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
Division-Remainder Method
1. Divide the decimal number by the value of the
new base.
2. Record the remainder from step 1 as the right
most digit of the new base number
3. Divide the quotient of the previous divide by the
value of the new base.
4. Record the remainder from step 3 as the next digit
(to the left) of the new base number.
5. Repeat step 3 and 4 until the quotient become 0,
the last remainder will be the most significant digit
of the new base number.
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
Example
95210 = ?8
Solution
8 )952
)119
)14
)1
0
Hence, 95210 = 016708
Remainders
0
7
6
1
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
Example
95210 = ?16
Solution
16)952
)59
)3
0
Hence, 95210 = 3B816
Remainders
8
11
3
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
Method
1. Convert the original number to a decimal
number
2. Convert the decimal number that obtained
on in step 1 to the new base number
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
Method
1. Divide the digits into groups of three
starting from the right.
2. Convert each group of three binary digits to
one octal digit using the method of binary
to decimal conversion.
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
Example
11010102 = ?8
Solution
1. Divide the binary into groups of three digits
001 101 010
2. Convert each group into one octal digit
0012 =
0x22 + 0x21 + 1x20
=
1
1012 =
1x22 + 0x21 + 1x20
=
5
0102 =
0x22 + 1x21 + 0x20
=
2
Hence, 11010102 = 1528
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
Method
1. Convert each octal digit to 3 binary digits.
2. Combine all the resulting binary groups into
a single binary number.
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
Example
5628 = ?2
Solution
1. Convert each octal digit to 3 binary digits.
58 = 1012
68 = 1102
28 = 0102
2. Combine all the resulting binary groups into a
single binary number.
5628 =
101
110
010 2
Hence, 5628 = 1011100102
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
Method
1. Divide the digits into groups of four starting
from the right.
2. Convert each group of four binary digits to
one hexadecimal digit using the method of
binary to decimal conversion.
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
Example
1111012 = ?16
Solution
1. Divide the binary into groups of four digits
0011 1101
2. Convert each group into one hexadecimal digit
00112 = 0x23 + 0x22 + 1x21 + 1x20 = 3
11012 = 1x23 + 1x22 + 0x21 + 1x20 = 13 = D
Hence, 1111012 = 3D16
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
Method
1. Convert each hexadecimal digit to 4 binary
digits.
2. Combine all the resulting binary groups into
a single binary number.
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
Example
2AB16 = ?2
Solution
1. Convert each hex digit to 4 binary digits.
216 = 00102 A16 = 10102 B16 = 10112
2. Combine all the resulting binary groups into a
single binary number.
2AB16 =
0010 1010 1011 2
Hence, 2AB16 = 0010101010112
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
All information that is processed by computers
is converted in one way or another into a
sequence of numbers. This includes
– numeric information
– textual information and
– Pictures, sound, video, etc.

Therefore, if we can derive a way to store and
retrieve numbers electronically this method can
be used by computers to store and retrieve any
type of information.
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
Computers Store ALL information using binary numbers

Computers use binary numbers in different ways to
store different types of information.

Common types of information that are stored by
computers are :
– Whole numbers (i.e. Integers).
Examples: 8 97 -732 0
-5 etc.
– Numbers with decimal points.
Examples: 3.5 -1.234 0.765 999.001 etc.
– Textual information (including letters, symbols and digits)
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
Unsigned integers
– Given n bits, it is possible to represent the
range of values from 0 to 2n - 1
– For example an 8-bit representation would
allow representations that range 0 to 255
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
Signed Magnitude
– An extra bit in the most significant position is
designated as the sign bit which is to the left
of an unsigned integer. The unsigned integer
is the magnitude.
S
MMMMMMMMMMMMMMM
– A 0 in the sign bit means positive, and a 1
means negative
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
Signed Magnitude (cont.)
– Given an n bit signed magnitude number the range
of values that it can represent is
-(2n-1-1) to +(2n-1-1)
for example:
8 bits  -127 to 127
16 bits  -32767 to 32767
– Signed magnitude representation associates a sign
bit with a magnitude that represents zero, thus it has
two distinct representation of zero:
00000000 and 10000000
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
1’s Complement representation
– For positive numbers, the representation is
the same as for unsigned integers where the
most significant bit is always zero
– The additive inverse of a 1’s complement
representation is found by inverting each bit.
– Inverting each bit is also called taking the 1’s
complement
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

1’s Complement representation
Examples
0000 0011 (3)
1111 1100 (-3)
0001 0111 (23)
1110 1000 (-23)
0000 0000 (0)
1111 1111 (0)
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
2’s Complement representation
– The additive inverse of a 2’s complement integer
can be obtained by adding 1 to its 1’s
complement
– The 2’s complement representation for a
negative number is the additive inverse of its
positive representation
– An advantage of 2’s complement is that there is
only one representation for zero
– Given an n bit 2’s complement number the range
of values that it can represent is
-(2n-1) to +(2n-1-1)
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2’s Complement representation
 Examples

010001 (17)
take
000000 (0)
the 1’s complement
101110 +
1
101111 (-17)
111111 +
1
1000000 (0)
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mantissa
exponent
6.0210 x 1023
decimal point
radix (base)
 Normalized form: no leadings 0s
(exactly one digit to left of decimal point)
 Alternatives to representing 1/1,000,000,000
–Normalized:
1.0 x 10-9
–Not normalized:
0.1 x 10-8,10.0 x 10-10
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mantissa
exponent
1.02 x 2-1
“binary point”
radix (base)
Computer
arithmetic that supports it called
floating point, because it represents numbers
where the binary point is not fixed, as it is for
integers
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Normalized format: +1.xxxxxxxxxx2 x 2yyyy2
 Multiple of Word Size (32 bits)

31 30
23 22
S Exponent
1 bit
Mantissa
8 bits



0
23 bits
S represents Sign
Exponent represents y’s
Mantissa represents x’s
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


To store letters and symbols, the computer
assigns every character a numerical value.
Computers remember letters and other
symbols by storing the binary number for
the symbol.
For this system to work a standard
numbering system needs to be defined and
consistently used for all symbols that the
computer needs to process.
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
ASCII (American
Standard Code for
Information
Interchange) is the
standard
numbering given to
all characters.

“ASCII values”
range in number
from 1 to 128 (8
bits).
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

Unicode is a computing industry standard for
the consistent encoding, representation and
handling of text expressed in most of the
world's writing systems.
Unicode can be implemented by different
character encodings. The most commonly
used encodings are UTF-8 (which uses one
byte for any ASCII characters, which have the
same code values in both UTF-8 and ASCII
encoding, and up to four bytes for other
characters.
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

A raster image, or
bitmap, consists of
a rectangular array
of pixels.
Each pixel has a
corresponding red,
green, and blue
value that combine
to determine the
color displayed by
that pixel.
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
Audio analog to
digital converters
work by repeatedly
measuring the
amplitude of an
incoming electrical
pressure sound
wave, and
outputting these
measurements as a
long list of binary
bytes.
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