Dr. Saied Abd El Wahab By: Shery Boushra Tamer 13/3/2010 9:00 to 12:00 Lecture 2

1. Multi Media Dr. Saied Abd El Wahab By: Shery Boushra Tamer 13/3/2010 9:00 to 12:00 Lecture 2

2. Bit : is the basic unit of information in computing and telecommunications; it is the amount of information that can be stored by a device or other physical system that can normally exist in only two distinct states (0 or 1). • 1 Bit • 2 Bit 2 Bit = 2² = 4 (Alternatives) Multi Media 1.Text 0 1 Truth Table 0 1 0 1

3. Multi Media 3 Bit 3Bit = 2³= 8(Alternatives) Truth Table 0 1 0 1 0 1

4. Multi Media • 4 Bit 4 Bit = 2 = 16 (Alternatives) • 5 Bit 5 Bit = 2=32 (Alternatives) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

5. Multi Media • 6 Bit = 64 (Alternatives) • 7Bit= 128 (Alternatives) • 8Bit = 256(Alternatives) = 1 Byte 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 ASCII Code Changed to 1 Byte to be understood by the computer system

6. Multi Media ASCII Code : Is The American Standard Code for Information Interchange Look to Lec11MM.doc

7. 1 Character ------ ASCII Code ------- 1 Byte ----- 8 Bit • Every character has its own ASCII Code = 1 Byte Multi Media …….… …….… ………. ………. ASCII Code 8 Bit 1 Byte

8. Universal Code: • In computer architecture, 16-bit integers, memory addresses, or other data units are those that are at most 16 bits (2 octets) wide. • A 16-bit file format is a binary file format for which each data element is defined on 16 bits (or 2 Bytes). An example of such a format is UTF-16 and the Windows Metafile Format Multi Media

9. The 16-bit Intel architecture allows for different memory models ways to access a particular memory location. The reason for using a specific memory model is the size of the assembler instructions or required storage for pointers. Multi Media

10. Compilers of the 16-bit era generally had the following type width characteristic: • “Tiny” Code and data will be in the same segment (especially, the registers CS,DS,ES,SS will point to the same segment);“near” pointers are always used. Code, data and stack together cannot exceed 64K. Multi Media

11. “Small” Code and data will be in different segments, and “near” pointers are always used. There will be 64K of space for code and 64K for data/stack. • “Medium” Code pointers will use “far” pointers, enabling access to 1MB. Data pointers remain to be of the “near” type. • “Compact” Data pointers will use “far” and code will use “near” pointers. • “Large”/“Huge” Code and data pointers will be “far”. Multi Media

12. Case study: An archive CONSISTS of A4 Paper (40 Lines and each Line have 80 characters) AN average rate for data entry is 60 Lines (free of errors)per one hour • Calculate: Page contains = amount of lines * amount of character in each line =40*80 = 3200 characters Multi Media

13. 3200 characters ----- 3200 Click ----- 3200 save in buffer ------ 3200 ASKII code ---- 3200 Byte • Then will enter 60 correct line * 80 character = 80 character/minute • It means 80 character/minute ÷ 60 second = 1.3 character/second • The person will enter the paper in = 3200 character / page ÷80 character/minute= 40minte Multi Media

14. We need to take care for : 1.Time established 2.Cost 3.Amount finished Multi Media

15. Sources of composing an image are : 1. Camera 2. Scanner 3.Mobile phone • OCR Software: Optical Character Recognition • Input Image ------- Output Character • Paper ------ Scanned ----- OCR ------Text • 1 minute to enter text using scanner and OCR • 40 minute to enter text using person and keyboard Multi Media 2.Image

16. Value of work , time Information service , Economic service , QoS Between at least two ( A & B ) : Machine & machine or machine & person or person & person Service ( Economy , Time , Cost , Quality ) Service doesn’t depend on infrastructure Information service providers ( ISD ) Multi Media Project (A)

17. SQL 2000 SQL 2005 SQL 2008 Multi Media Project (B)

18. Interactive multimedia service Multi Media Project (C)

19. Usage of multimedia Digital sensors Digital Camera+ Soft ware ( Security ) Movie maker usage Multi Media Project (D)