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0. Introduction to Computer

Dec. Binary. Hex. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17. 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001. 0 1 2 3 4 5 6 7 8

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0. Introduction to Computer

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  1. Dec Binary Hex 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 0 1 32 4 2 32 16 8 4 2 1 + N S S N 1001102 = 3810 0. Introduction to Computer 0.1 Binary Number System Physical Representation Conversion(Bin Dec) binary digit Units: bit Byte(8 bit) word(2 or 4 B) KB(1,024 B) MB(1,024 KB) GB(1,024 MB) 28 = 256 Chap. 0 Introd. to Computer

  2. 0.2 Organization of a Hypothetic-Computer CPU CU ALU CPU: Cent’l Proc. Unit bus: CU: Control Unit ALU: Arith./ Logic Unit PC: Program Counter IR: Instruction Register ID: Instruction Decoder Flag Reg: o, s, z, c Registers: AX - DX clock: clock generator Interface Card R: Ready bit data: buffer register PC AX BX IR CX DX ID clock o s z c 00 01 02 03 04 05 06 07 08 09 0A 0B MEMORY R data R data OUTPUT INPUT Chap. 0 Introd. to Computer

  3. 0.3 Instruction Set Group Machine code Mnemonic code Operation Transfer 0001RdRs MOVRd, Rs Rd (Rs) 0010RdMm LOADRd, Mm Rd (Mm) 0011MmRs STOREMm, Rs Mm (Rs) Arithmetic 0100RdRs ADDRd, Rs Rd (Rd) + (Rs), set Flag 0101RdRs SUBRd, Rs Rd (Rd) - (Rs), set Flag Branch 0110Mm JMPMm PCMm 0111Mm JLMm PCMm, if S=1 1000Mm JZMm PCMm, if Z=1 Input/Output 1001RdPt INRd, Pt Rd (Pt) 1010PtRs OUTPt, Rs Pt (Rs) Control 0001 HALT PC (PC) Rd: destination register Rs: source register Mm: memory address PC: program counter Pt: port number Chap. 0 Introd. to Computer

  4. CU ALU PC AX BX ALU IR CX DX ID clock Assembly L. Prog. o s z c .... LOAD AX, A LOAD BX, B SUB AX, BX JL ITSB LOAD BX, A ITSB: STORE C, BX HALT A: DS 2 B: DS 2 C: DS 2 Fetch 00 0010000111 01 0010011000 Execute 02 01010001 03 01110101 04 0010000111 Assemble 05 0011100101 06 10110000 Machine Lang. Prog. Stored Prog. 07 00000011 08 00000101 09 00000000 von Neumann 0A 0.4 Machine Language Program vs Assembly Language Program To store the bigger of A and B to C Chap. 0 Introd. to Computer

  5. Low-level PL Machine Language Assembly Language • High-level PL 10 C = A 20 if (B>A) then C = B Interpreter L Basic APL Smalltalk ... if (A>B) then C := A else C := B; compiler L Fortran, Cobol, PL/1 Pascal, Ada C, C++, ... C = (A>B)? A : B; 0.5 Programming Languages Chap. 0 Introd. to Computer

  6. 개인용 컴퓨터의 규격 • CPU • Memory • Secondary Storage • I/O Device Chap. 0 Introd. to Computer

  7. CPU • Microprocessor • 8088, 80286, 80386, 80486 • Pentium II, III, IV • Celeron • CPU의 속도 • MHz (650 MHz, 800 MHz, 1 GHz) Chap. 0 Introd. to Computer

  8. Memory • Cache Memory • 용량: 256 KB, 512 KB (L1 cache, L2 cache) • DRAM • Dynamic Random Access Memory - 휘발성 • 용량: 128 MB, 256 MB • DIMM, SIMM • ROM • Read Only Memory (ROM BIOS) - 비휘발성 Chap. 0 Introd. to Computer

  9. Secondary Storage • Hard Disk • Magnetic Storage • Seek time + Rotational delay + Transfer time • 용량: 20 GB, 30 GB • CD-ROM • 속도: 48배속, 용량: 650 MB (DVD: 4 GB) • Floppy Disk • 3.5” 2HD = 1.44 MB, ZIP: 100 MB • Magnetic Tape • Sequential Storage (DAT: 4GB ~ 8GB) Chap. 0 Introd. to Computer

  10. I/O Device • Graphics Card Video Adapter • Pixel: 12801024 (640480, 1024768) • Frame Buffer: 1028  1024  3B  8  32 MB • Sound Card • Analog to Digital Converter • Modem: 56 Kbps • Keyboard, Mouse Chap. 0 Introd. to Computer

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