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Trang 1- The base is ten – there are 10 different symbols, the digits 0, 1, 2, etc upto 9
- To represent value less than ten involves only one digit larger values need two or more digits
Binary system
- The base must be two, with only the digits 0 and 1 available
- To show values of two or ever require two or more binary digits
- Octal system has eight as its base, it uses the symbol 0, 1, 2 up to 7 only
- Two or more digits are needed for values of eight and above
- Hexadecimal system has sixteen as its base, it use the symbols 0, 1, 2 ,9
& A, B, C, D, E, F, to stand for the “digits” ten, eleven, twelve, thirteen, fourteen, fifteen
Question 2 Converting from Bases To Bases?
1 Change the decimal
- Binary:
Eg (2559) 10
2559 1
1279 1
639 1
319 1
159 1
79 1 (2559)10 = (10111111111)2
39 1
19 1
9 1
4 0
2 1
0 0
- Octal:
7690 8
49 96,1 8
10 16 120 8
40 15 8
- Hexadecimal:
6396 16
159 399 16
156 79 24 16
Trang 2
C F (6369)10 = (CF81)16 2 Convert to others from binary - To decimal (101010)2 (?)10
1.25 + 0.24 + 1.23 + 0.22 + 1.21 + 0.20 = 42
(101010)2 = (42)10
- To octal 100101101 1st step change into denary = 1.28 + 1.25 + 1.23 + 1.22 + 1.20 = 256 + 32 + 8 + 4 + 1
=(301)10 2nd step: convert to octal 301 8
61 37 8
(301)10 = (455)8 (100101101)2 = (455)8 - To hexadecimal 110111011011 1st step = 1.211 + 1.210 + 1.28 + 1.27 + 1.26 + 1.24 + 1.23 + 1.21 + 1.20 = 2048+ 1024 + 256 + 158 + 64 + 16 + 8 + 2 + 1 = (3547)10 2nd step 3547 16
384 221 16
27 61
(3547)10 = (CCA)16 (110111011011)2 = (CCA)16 3 Convert into binary and display the answer in normalized exponential form 247 1
123 1
61 1
30 1
13
Trang 315 1
7 1
3 1
1 1
0 1 (247)10 = (11110111)2
= 0 1111011 x 2 normalized exponential form
Question 3 Integer and Floating – point arithmetic?
1 Floating – point Addition
a (0.1011 x 25
) + (0.1001 x 25
) = (0.1011 + 0 1001) x 25
= 1.0100 x 25
= 0.10100 x 26
b (0.1001 x 23 ) + (0.1110 x 25 ) = (0.001001 x 25 ) + (0.1110 x 25 ) = (0.001001 + 0.111000) x 25 = 1.000001 x 25
= 0.1000 x 26
(here have truncation) (0.1000001 x 26
)
2 Floating – point subtraction
a (0.1110 x 27
) – (0.1100 x 27
) = 0.0010 x 27
= 0 10 x 25
b (0.1001 x 28 ) – ( 0.1000 x 25 ) = (0.1001 x 28 ) – ( 0.0001 x 28 ) = 0.1000 x 28
3 Floating – point multiplication
a (0.1010 x 23
) x (0.1100 x 23
) = (0.1010 x 0.1100) x 26
= 0.01111 x 26
= 0.1111 x 25
b (0.11110 x 23
) x ((0.01011) x 24
) = (0.11110 x 0 01011) x 27
= 0.001111 x 27 = 0.1111 x 25
4 Floating – point division
a (0.11010 x 26) : (0.001 x 26) = (0.11010 x 26
) : (1 x 23
) = 0.1101 x 26 : 1x 23 = 0.1101 x 23
b (0.110111 x 26
) : (0.1001 x 24
)
= (0.110111 : 0.1001) x 22
= (1101.11 : 1001) x 22
= 1.100001 x 22
= 0.1100001 x 23
Chapter 3: TYPES OF INSTRUCTION AND ADDRESSING
Trang 4Question 1 Types of instructions used in CS?
1 Arithmetic instructions
Arithmetic instructions include directives to the computers to perform additions, subtraction, multiplications, divisions and exponentiations
2 Input/ output instructions
They direct the computer to read data values from the specified input devices into the main store for processing
They also include instructions to write the contents of memory locations holding the result of processing to a specified output device
3 Decision or control instructions
Most data processing application will contain situations where alternative calculations
or procedures will have to be executed based on the result of condition tests carried out
4 Data handling instructions They include the copying of the content of one memory location to another or setting
a memory locations to an initial value
Also include the management or insertion of characters into data items Examples of such instructions include branch instructions, jump instruction & stop instruction
Question 2 Types of addressing?
1 Direct addressing The operands of each machine instructions is used to retrieve the data
2 Indirect addressing The operands is used to specify the memory address which contains the address of the data to be processed
Op – code
12345 Data item
12345 Data item Main storage
Direct addressing
Indirect addressing
3 Indexed addressing
- The main applications of this type of addressing technique is to enable to access of sequential locations in memory that are adjacent to each other
- Each adjcent memory address has value n+1, where n is the address of the previous location
- When the first of the location have been accessed, the next memory location in sequence is accessed by simply increasing the add of the present location by 1 & using accessing it
Trang 5- The starting address of the series of locations is specified in the operand of the instruction
- In order to access the next location in sequence, the content of the index register is increased by 1 a added to the opreand address
- This is done repeatedly until the last memory location in the series is processed
Indexed addressing
Index Register
Chapter 4: PROGRAMMING LANGUAGES
Question 1 Program and level of language?
Program is group of constructions that is linked together to perform specific task It’s necessary for a computer program to be written in a “PL” because at a computer program is created by a programing using a sys analyst’s specification of the job in the hand
1 Machine language
- ML is the set of bit(0,1) that can performed considered by CPU
- Ads
fast
short prog
store in small memory
- Dis
difficult to understand & remember its code
takes a lot of time to programming
difficult to use
2 Low level language
- LLL is used to dercribe exactly procedure of performance of CPU at certain time
- Features:
Instruction is written by natrural English or natural language
More powerful and so the prog is shortest
Need less instruction
Is a one to one relationship between the written instruction and the machine instructions
It’s instruction tend to be machine It runs in OS