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CSE 102 Introduction to Computer Engineering Central Processing Unit.

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Presentation on theme: "CSE 102 Introduction to Computer Engineering Central Processing Unit."— Presentation transcript:

1 CSE 102 Introduction to Computer Engineering Central Processing Unit

2 Classes of Computers Supercomputer$5-20 million Mainframe$0.5-4 million Server$10-200 thousand PC/Workstation$1-10 thousand Game console$300-$1000 Embedded device$1-$100 Future disposable1-100 cents

3 Building Computer Chips Complex multi-step process –Slice ingots into wafers –Process wafers into patterned wafers –Dice patterned wafers into dies –Test dies, select good dies –Bond to package –Test parts –Ship to customers and make money

4 Building Computer Chips

5 Microprocessors and Memory Microprocessor Performance Factors: What makes one microprocessor perform better than another? CPU speed is influenced by several factors: –clock speed – Megahertz, Gigahertz –word size 8-bit, 16-bit, 32-bit or 64-bit word sizes –Cache – Level 1, Level 2 caches –instruction set size

6 Performance vs. Design Time Time to market is critically important E.g., a new design may take 3 years –It will be 3 times faster –But if technology improves 50%/year –In 3 years 1.5 3 = 3.38 –So the new design is worse! (unless it also employs new technology)

7 Moore’s Law Considering the rate of technological development, the complexity of an integrated circuit, with respect to minimum component cost will double in about 24 months (18-24).

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9 Input Devices Processing Data into Information Output Devices Secondary Storage Devices Central Processing Unit Computer System Components Keyboard Mouse Touch Screen Voice... Monitor Printer Disks, Tapes, Optical Disks

10 CPU (Central Processing Unit) The CPU executes computer instructions Popular CPU’s: Intel-Pentium, AMD, Power PC It is on a chip called the microprocessor System clock –Megahertz Has three parts: 1.Control Unit 2.ALU (Arithmetic Logic Unit) 3.Registers

11 How Computers Work CD on CPU simulation Fetch the Instruction Interpret the Instruction Prepare Circuitry to Execute Instruction Manipulate the Data Store the Result Fetch the Data The Instruction CycleThe Execution Cycle 1. Fetch 2. Decode 3. Execute 4. Store

12 Code Translation (C to Assembly Language) #include int main( ) { int a=10; int b=20; int c; c=a+b*2; return 0; } MOVR1, #10 MOVR2, #20 MOVR3, #2 MULR4, R2, R3 ADDR5, R1, R4 BREAK

13 Code Translation (C to Machine Code) #include int main( ) { int a=10; int b=20; int c; c=a+b*2; return 0; } MOVR1, #10 MOVR2, #20 MOVR3, #2 MULR4, R2, R3 ADDR5, R1, R4 BREAK Adress Data ============== 1000 110A 1002 1214 1004 1302 1006 3423 1008 2514 100A 0000 MEMORY

14 CPU Control Unit Adress Data ============== 1000 110A 1002 1214 1004 1302 1006 3423 1008 2514 100A 0000 MEMORY Fetch from PC 110A DECODE 0001 Opcode MOV 0001 Register 0110 Data1 0100 Data2 CPU PC = 1000Program Counter

15 Registers and MOV operation 1 1 0 A 0001 0001 0000 1010 MOV CPU PC = 1000 REGISTERSDATA R0 R1 R2 R3 R4 R5 R6 R7 00001010 = 10 in decimal 00001010

16 Registers and MOV operation 1 2 1 4 0001 0010 0001 0100 MOV CPU PC = 1002 REGISTERSDATA R0 R1 R2 R3 R4 R5 R6 R7 00010100 = 20 in decimal 00001010 00010100

17 Registers and MOV operation 1 3 0 2 0001 0011 0000 0010 MOV CPU PC = 1004 REGISTERSDATA R0 R1 R2 R3 R4 R5 R6 R7 00000010 = 2 in decimal 00001010 00010100 00000010

18 Registers and MUL operation 3 4 2 3 0011 0100 0010 0011 MUL CPU PC = 1004 REGISTERSDATA R0 R1 R2 R3 R4 R5 R6 R7 00001010 00010100 00000010 Mul ALU 20 2 40 = 00101000 00101000

19 Registers and ADD operation 2 5 1 4 0010 0101 0001 0100 ADD CPU PC = 1006 REGISTERSDATA R0 R1 R2 R3 R4 R5 R6 R7 00001010 00010100 00000010 Add ALU 10 40 50 = 00110010 00101000 00110010


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