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EECC550 - Shaaban #1 Lec # 5 Winter 2000 12-20-2000 CPU Design Steps 1. Analyze instruction set operations using independent RTN => datapath requirements.

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Presentation on theme: "EECC550 - Shaaban #1 Lec # 5 Winter 2000 12-20-2000 CPU Design Steps 1. Analyze instruction set operations using independent RTN => datapath requirements."— Presentation transcript:

1 EECC550 - Shaaban #1 Lec # 5 Winter 2000 12-20-2000 CPU Design Steps 1. Analyze instruction set operations using independent RTN => datapath requirements. 2. Select set of datapath components & establish clock methodology. 3. Assemble datapath meeting the requirements. 4. Analyze implementation of each instruction to determine setting of control points that effects the register transfer. 5. Assemble the control logic.

2 EECC550 - Shaaban #2 Lec # 5 Winter 2000 12-20-2000 CPU Design & Implantation Process Bottom-up Design: –Assemble components in target technology to establish critical timing. Top-down Design: –Specify component behavior from high-level requirements. Iterative refinement: –Establish a partial solution, expand and improve. datapath control processor Instruction Set Architecture => Reg. FileMuxALURegMemDecoderSequencer CellsGates

3 EECC550 - Shaaban #3 Lec # 5 Winter 2000 12-20-2000 Single Cycle MIPS Datapath: CPI = 1, Long Clock Cycle

4 EECC550 - Shaaban #4 Lec # 5 Winter 2000 12-20-2000 Drawback of Single Cycle Processor Long cycle time. All instructions must take as much time as the slowest: –Cycle time for load is longer than needed for all other instructions. Real memory is not as well-behaved as idealized memory –Cannot always complete data access in one (short) cycle.

5 EECC550 - Shaaban #5 Lec # 5 Winter 2000 12-20-2000 Abstract View of Single Cycle CPU PC Next PC Register Fetch ALU Reg. Wrt Mem Access Data Mem Instruction Fetch Result Store ALUctr RegDst ALUSrc ExtOp MemWr Equal nPC_sel RegWr MemWr MemRd Main Control ALU control op fun Ext

6 EECC550 - Shaaban #6 Lec # 5 Winter 2000 12-20-2000 Single Cycle Instruction Timing PCInst Memory mux ALUData Mem mux PCReg FileInst Memory mux ALU mux PCInst Memory mux ALUData Mem PCInst Memorycmp mux Reg File Arithmetic & Logical Load Store Branch Critical Path setup

7 EECC550 - Shaaban #7 Lec # 5 Winter 2000 12-20-2000 Reducing Cycle Time: Multi-Cycle Design Cut combinational dependency graph by inserting registers / latches. The same work is done in two or more fast cycles, rather than one slow cycle. storage element Acyclic Combinational Logic storage element Acyclic Combinational Logic (A) storage element Acyclic Combinational Logic (B) =>

8 EECC550 - Shaaban #8 Lec # 5 Winter 2000 12-20-2000 Clock Cycle Time & Critical Path Critical path: the slowest path between any two storage devices Cycle time is a function of the critical path must be greater than: –Clock-to-Q + Longest Path through the Combination Logic + Setup Clk........................

9 EECC550 - Shaaban #9 Lec # 5 Winter 2000 12-20-2000 Instruction Processing Cycles Obtain instruction from program storage Determine instruction type Obtain operands from registers Compute result value or status Store result in register/memory if needed (usually called Write Back). Update program counter to address of next instruction } Common steps for all instructions Instruction Fetch Instruction Decode Execute Result Store Next Instruction

10 EECC550 - Shaaban #10 Lec # 5 Winter 2000 12-20-2000 Partitioning The Single Cycle Datapath Add registers between smallest steps PC Next PC Operand Fetch Exec Reg. File Mem Access Data Mem Instruction Fetch Result Store ALUctr RegDst ALUSrc ExtOp MemWr nPC_sel RegWr MemWr MemRd

11 EECC550 - Shaaban #11 Lec # 5 Winter 2000 12-20-2000 Example Multi-cycle Datapath PC Next PC Operand Fetch Ext ALU Reg. File Mem Acces s Data Mem Instruction Fetch Result Store ALUctr RegDst ALUSrc ExtOp nPC_sel RegWr MemWr MemRd IR A B R M Reg File MemToReg Equal Registers added: IR: Instruction register A, B: Two registers to hold operands read from register file. R: or ALUOut, holds the output of the ALU M: or Memory data register (MDR) to hold data read from data memory

12 EECC550 - Shaaban #12 Lec # 5 Winter 2000 12-20-2000 Operations In Each Cycle Instruction Fetch Instruction Decode Execution Memory Write Back R-Type IR  Mem[PC] A  R[rs] B  R[rt] R  A + B R[rd]  R PC  PC + 4 Logic Immediate IR  Mem[PC] A  R[rs] R  A OR ZeroExt[imm16] R[rt]  R PC  PC + 4 Load IR  Mem[PC] A  R[rs] R  A + SignEx(Im16) M  Mem[R] R[rd]  M PC  PC + 4 Store IR  Mem[PC] A  R[rs] B  R[rt] R  A + SignEx(Im16) Mem[R]  B PC  PC + 4 Branch IR  Mem[PC] A  R[rs] B  R[rt] If Equal = 1 PC  PC + 4 + (SignExt(imm16) x4) else PC  PC + 4

13 EECC550 - Shaaban #13 Lec # 5 Winter 2000 12-20-2000 Finite State Machine (FSM) Control Model State specifies control points for Register Transfer. Transfer occurs upon exiting state (same falling edge). State X Register Transfer Control Points Depends on Input Control State Next State Logic Output Logic inputs (conditions) outputs (control points)

14 EECC550 - Shaaban #14 Lec # 5 Winter 2000 12-20-2000 Control Specification For Multi-cycle CPU Finite State Machine (FSM) IR   MEM[PC] R-type A  R[rs] B  R[rt] R  A fun B R[rd]  R PC  PC + 4 R  A or ZX R[rt]  R PC  PC + 4 ORi R  A + SX R[rt]  M PC  PC + 4 M  MEM[R] LW R  A + SX MEM[R]  B PC  PC + 4 BEQ & Equal BEQ & ~Equal PC  PC + 4 PC  PC + SX || 00 SW “instruction fetch” “decode / operand fetch” Execute Memory Write-back To instruction fetch

15 EECC550 - Shaaban #15 Lec # 5 Winter 2000 12-20-2000 Traditional FSM Controller State 6 4 11 next State op Equal control points stateopcond next state control points Truth or Transition Table datapath State To datapath

16 EECC550 - Shaaban #16 Lec # 5 Winter 2000 12-20-2000 Traditional FSM Controller datapath + state diagram => control Translate RTN statements into control points. Assign states. Implement the controller.

17 EECC550 - Shaaban #17 Lec # 5 Winter 2000 12-20-2000 Mapping RTNs To Control Points Examples & State Assignments IR  MEM[PC] 0000 R-type A  R[rs] B  R[rt] 0001 R  A fun B 0100 R[rd]  R PC  PC + 4 0101 R  A or ZX 0110 R[rt]  R PC  PC + 4 0111 ORi R  A + SX 1000 R[rt]  M PC  PC + 4 1010 M  MEM[S] 1001 LW R  A + SX 1011 MEM[S]  B PC  PC + 4 1100 BEQ & Equal BEQ & ~Equal PC  PC + 4 0011 PC  PC + SX || 00 0010 SW “instruction fetch” “decode / operand fetch” Execute Memory Write-back imem_rd, IRen Aen, Ben ALUfun, Sen RegDst, RegWr, PCen To instruction fetch state 0000 To instruction fetch state 0000

18 EECC550 - Shaaban #18 Lec # 5 Winter 2000 12-20-2000 Detailed Control Specification StateOp fieldEqNext IRPC OpsExec Mem Write-Back en selA B Ex Sr ALU S R W MM-R Wr Dst 0000???????00011 0001BEQ000111 1 0001BEQ100101 1 0001R-typex01001 1 0001orIx01101 1 0001LWx10001 1 0001SWx10111 1 0010xxxxxxx00001 1 0011xxxxxxx00001 0 0100xxxxxxx01010 1 fun 1 0101xxxxxxx00001 00 1 1 0110xxxxxxx01110 0 or 1 0111xxxxxxx00001 00 1 0 1000xxxxxxx10011 0 add 1 1001xxxxxxx10101 0 0 1010 xxxxxxx00001 01 1 0 1011xxxxxxx11001 0 add 1 1100xxxxxxx0000 1 00 1 R ORI LW SW BEQ

19 EECC550 - Shaaban #19 Lec # 5 Winter 2000 12-20-2000 Alternative Multiple Cycle Datapath (In Textbook) Miminizes Hardware: 1 memory, 1 adder

20 EECC550 - Shaaban #20 Lec # 5 Winter 2000 12-20-2000 Alternative Multiple Cycle Datapath (In Textbook) Shared instruction/data memory unit A single ALU shared among instructions Shared units require additional or widened multiplexors Temporary registers to hold data between clock cycles of the instruction: Additional registers: Instruction Register (IR), Memory Data Register (MDR), A, B, ALUOut

21 EECC550 - Shaaban #21 Lec # 5 Winter 2000 12-20-2000 Operations In Each Cycle Instruction Fetch Instruction Decode Execution Memory Write Back R-Type IR  Mem[PC] PC  PC + 4 A  R[rs] B  R[rt] ALUout  PC + (SignExt(imm16) x4) ALUout  A + B R[rd]  ALUout Logic Immediate IR  Mem[PC] PC  PC + 4 A  R[rs] B  R[rt] ALUout  PC + (SignExt(imm16) x4) ALUout  A OR ZeroExt[imm16] R[rt]  ALUout Load IR  Mem[PC] PC  PC + 4 A  R[rs] B  R[rt] ALUout  PC + (SignExt(imm16) x4) ALUout   A + SignEx(Im16) M  Mem[ALUout] R[rd]  Mem Store IR  Mem[PC] PC  PC + 4 A  R[rs] B  R[rt] ALUout  PC + (SignExt(imm16) x4) ALUout  A + SignEx(Im16) Mem[ALUout]  B Branch IR  Mem[PC] PC  PC + 4 A  R[rs] B  R[rt] ALUout  PC + (SignExt(imm16) x4) If Equal = 1 PC  ALUout

22 EECC550 - Shaaban #22 Lec # 5 Winter 2000 12-20-2000 High-Level View of Finite State Machine Control First steps are independent of the instruction class Then a series of sequences that depend on the instruction opcode Then the control returns to fetch a new instruction. Each box above represents one or several state.

23 EECC550 - Shaaban #23 Lec # 5 Winter 2000 12-20-2000 Instruction Fetch and Decode FSM States

24 EECC550 - Shaaban #24 Lec # 5 Winter 2000 12-20-2000 Load/Store Instructions FSM States

25 EECC550 - Shaaban #25 Lec # 5 Winter 2000 12-20-2000 R-Type Instructions FSM States

26 EECC550 - Shaaban #26 Lec # 5 Winter 2000 12-20-2000 Jump Instruction Single State Branch Instruction Single State

27 EECC550 - Shaaban #27 Lec # 5 Winter 2000 12-20-2000

28 EECC550 - Shaaban #28 Lec # 5 Winter 2000 12-20-2000 Finite State Machine (FSM) Specification Finite State Machine (FSM) Specification IR  MEM[PC] PC  PC + 4 R-type ALUout  A fun B R[rd]  ALUout ALUout  A op ZX R[rt]  ALUout ORi ALUout  A + SX R[rt]  M M  MEM[ALUout] LW ALUout  A + SX MEM[ALUout]  B SW “instruction fetch” “decode” Execute Memory Write-back 0000 0001 0100 0101 0110 0111 1000 1001 1010 1011 1100 BEQ 0010 If A = B then PC  ALUout A  R[rs] B  R[rt] ALUout  PC +SX To instruction fetch

29 EECC550 - Shaaban #29 Lec # 5 Winter 2000 12-20-2000 MIPS Multi-cycle Datapath Performance Evaluation What is the average CPI? –State diagram gives CPI for each instruction type –Workload below gives frequency of each type TypeCPI i for typeFrequency CPI i x freqI i Arith/Logic 440%1.6 Load 5 30%1.5 Store 410%0.4 branch 320%0.6 Average CPI: 4.1 Better than CPI = 5 if all instructions took the same number of clock cycles (5).


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