9-6 The Control Word

1. 9-6 The Control Word The selection variables for the datapath control the microoperations executed within datapath for any given clock pulse Fig. 9-11

2. Table 9-5 Example: The binary control word for this operation: 001_010_011_0_0101_0_1

3. Example of microoperation -- using symbolic notation

4. Example of microoperation -- using binary control word

5. SimulationFig. 9-12 Status_bits: vector (V, C, N, Z)

6. 9-7 A Simple Computer Architecture • In programmable system, a portion of the input to the processor consists of a sequence ofinstructions. • Instructions are usually stored in memory (ROM or RAM) • The address for the instruction to be executed is stored program counter(PC) • In non-programmable system, the control unit determines the (sequence of) operations to be perform based on only input and status bit. • In non-programmable system, the control unit is not responsible to any instruction.

7. 9-7 A Simple Computer Architecture • Instruction set architecture • Program for specifying the operation to be performed • Program is a list of instruction • Control unit reads an instruction from memory • decode and execute the instruction by issuing a sequence of one or more microoperations

8. Instruction set architecture • Instruction set: a collection of instructions • Instruction set architecture (ISA): a thorough description of instruction set • Three major components: • Storage resource • Instruction formats • Instruction specifications

9. Storage resources The instruction memory and data memory can be the same or be different memory depend on different standpoint of the CPU.

10. Instruction formats • If a branch occurs, new address is formed by PC contents + AD (6bits) • The 6 bits AD is referred to address offset (in signed 2s complement) • Sign extension is usedto preserving 2s complement representation

11. Instruction specifications Mnemonic: symbolic representation for opcode Assembler: a program converts the opcode symbol to binary zf: zero fill se: sign extension

12. In memory Suppose R4 contains 70 and R5 contains 80 Why PC <－ PC – 20 in address 55?

13. 9-8 Single-Cycle Hardwired Control • A control unit that fetches and executes an instruction in a single clock cycle. • We refer to this computer as the single-cycle computer.

14. Fig. 9-15 Block diagram for a single-cycle computer

15. Instruction decoder 1. PL=0, PC is incremented (no jump or branch) PL=1, JB=1 for jump, JB=0 for branch 2. BC: for branch condition (Z or N) 3. For Branch on Zero, FS=0000 to pass reg. A . But bit 9 (FS0) is 1. (a contradiction) How to solve?? (See text) Fig. 9-16

16. Truth table for instruction decoder You should refer to Table 9-8 for realizing this table.

17. Six instruction

18. A program for 83-(2+3) • Suppose • R3 contains 248 • Memory in location 248 contains 2 • Memory in location 249 contains 83 • The result will be placed in location 249

19. A program for 83-(2+3) • LD R1, R3 • ADI R1, R1, 3 • NOT R1,R1 • INC R1, R1 • INC R3, R3 • LD R2, R3 • ADD R2, R2, R1 • INC R3, R3 • ST R3, R2

20. Single-Cycle Computer Issues • Shortcomings of single-cycle computer • When performing complex operation, f.g. binary multiplication, it can’t be accomplished by a microoperation that can be executed in a single clock cycle. • The control organization that provides multiple clock cycle is needed • If single memory for instruction and data is used (at present, there are two distinct memory), at least two clocks, are required for obtaining and executing the instruction. First one is used to fetch instruction the second one is used to read/write data. • This can easily be accomplished by multiple-cycle control.

21. Single-Cycle Computer Issues • Long worst-case delay path – lower limit on the clock period • Example Fig. 9-17 • 9.8ns delay => 102MHz limited

22. End this lecture • We neglect the section 9-9 multiple-cycle hardwired control due to less of time.