1. Instructor: Yuzhuang Hu yhu1@cs.sfu.ca

2. Another Design Example: PIG (Chapter 7-10) PIG is a single dice game. Two players roll the dice in turns. When 1 is rolled, the current total becomes 0. The first player to reach or exceed 100 wins. HOLD ROLL NEW GAME RESET Player 1Player 2 Turn Total

3. Inputs, Outputs, and Registers of PIG

4. State-Machine Diagram for PIG Default: P1=CP, P2=CP TR1<-0, TR2<-0, CP<-FP SUR<-0 If (DIE=110) DIE<-001 Else DIE<-(DIE+1) RESET ROLL ROLL·HOLD ROLL INIT BEGIN ROL ROH TEST WIN ONE DIE  000, FP<-0 ROLL CP<-CP DIE=1 DIE<>1 SUR<-SUR+DIE ROLL·HOLD CP/(TR1<-TR1+SUR), CP/(TR1<-TR1+SUR) CP<-CP CP(TR1<1100100) +CP(TR2<-1100100) CP(TR1>=1100100)+CP(TR2>=1100100) NEWGAME FP<-FP NEWGAME CP/P1=BLINK, CP/P1=BLINK ROLL

5. Algorithmic State Machine (ASM) The ASM is like state diagrams but less formal and thus easier to be understood. An ASM chart consists of a set of blocks. Each block can be viewed as a directed graph with three types of nodes.  State Box (node).  Binary Decision Box (node).  Conditional Action Box (node).

6. ASM contd. State box: represented by a labeled rectangle. It may contain several register transfer statements or variables. Binary decision box: represented by a hexagon. It indicates that a condition needs to be tested. It is similar to the input condition defined for State-Machine Diagrams. Conditional output action box: represented by an oval box. It contains several register transfer statements or variables. It is similar to the output condition defined for State- Machine Diagrams.

7. Boxes in ASM Charts Register transfer statements (Moore type) State name Condition expression 0 (False) 1 (True) Conditional outputs or actions (Mealy type) (a) State box (b) Binary Decision box (c) Conditional output action box

8. A Design Example using ASM Problem: find the sum for N numbers. algorithm sum_n(S) input: a list S consisting N numbers. output: the sum of the N numbers in S. [1] sum = 0; [2] N = get_input(); [3] while ( N > 0 ) [4] sum = sum + get_input(); [5] N = N – 1; [6] endwhile;

9. Interface of the Sum Machine

10. ASM Diagram for the Sum Machine rdy S0 data 0 Sum <- 0 N <- in_bus ack s1 S1 N=0 1 1 0 rdy S2 data 0 N <- N-1 Sum <- sum+in_bus ack s1 1

11. Digital System Design In most digital system designs, we partition the system into two types of modules: a datapath, and a control unit. Control Unit Data Path Control inputs Data outputs Control outputs Data inputs Control Signals Status Signals

12. Design from ASM processor statuscontrol pts SEQ CTRL PTS SELECTOR External control inputs Data inData out Clock

13. Datapath of the Sum Machine SUM FA N N=0 in_bus out_bus overflow eq0 ls cs ln dn

14. ASM Design Guidelines Write an algorithm for the problem. Translate the algorithm to a sequence of register transfer statements. Group adjacent independent register transfer statements. Draw the ASM diagram, and introduce control signals.

15. Datapath Definition (Chapter 9) The datapath is defined by three basic components:  A set of registers.  The micro-operations performed on data stored in the registers.  The control interface.

16. A Generic Datapath Four parallel-load registers Two mux-based register selectors Register destination decoder Mux B for external constant input Buses A and B with external address and data outputs ALU and Shifter with Mux F for output select Mux D for external data input Logic for generating status bits V, C, N, Z

17. Datapath Examples What to do for R1 <- R2 + R3?  A select, choose R2.  B select, choose R3.  G select, choose A+B.  MF select, choose the ALU output.  MD select, choose MUX F ouput.  Destination select, choose R1.  Load enable, to enable R1.

18. Other Micro-operation Alternatives MF=1: shift operation. MB=1: using a constant. Load enable=0: no register loading, e.g. when providing an address out or data out. MD=1: read from memory.

19. The Arithmetic/Logic Unit ALU performs arithmetic/logic micro-operations.

20. The Arithmetic Circuit The arithmetic circuit consists of a parallel n-bit adder and a selection logic.

21. Function Table for Arithmetic Circuit It is easy to see that Y i =B i S 0 +B i S 1.

22. Function Table for ALU

23. More on ALU The ALU has a fairly high number of logic levels and contributes to propagation delay in the circuit. However simple ripple-carry adders can incur large propagation delays.

24. Carry Look-Ahead Carry look-ahead is designed to reduce the carry propagation delay in the ALU. For a single bit full adder:  Generate a carry out when x=y=1: g=x·y.  Propagate the carry in through the carry out when x or y is 1: p=x xor y.  In terms of p and g, the carry out co=g+ p·ci.

25. Full Adder With Ports p And q

26. A Slight Optimization Redefine p to be x+y. We can do this because of the following reasoning: we only need to consider the case when x=y=1. However when x=y=1, g=1, therefore no matter p takes 0 or 1, co is always equal to 1.

27. Computing the Carry In for Each Bit ci(1)=co(0)=g(0)+p(0)·ci. ci(2)=co(1)=g(1)+p(1) ·g(0)+p(1) ·p(0) ·ci(0). ci(3)=co(2)=g(2)+p(2) · g(1)+p(2) · p(1) ·g(0)+p(2) · p(1) · p(0) ·ci(0). ci(4)=co(3)=g(3)+p(3) · g(2)+p(3) · p(2) · g(1)+p(3) · p(2) · p(1) ·g(0)+p(3) · p(2) · p(1) ·p(0) ·ci(0).

28. Faster Four-bit Addition p(3:0) and g(3:0) are available after 1 gate delay. co(3:0) are available after 2 more gate delays. s(3:0) are available after 1 more gate delay. In total 1+2+1=4 gate delays.

29. A 4-Bit CLA Adder

30. Explanation of P and G Consider the msb position of a bit vector (3:0). Under what condition will a carry be generated out of that position? Under what condition will a carry be propagated through that position? Define  G=g(3)+p(3) · g(2)+p(3) · p(2) · g(1)+p(3) · p(2) · p(1) · g(0)  P= p(3) · p(2) · p(1) · p(0)

31. A 16-bit CLA Adder Use the 4-bit CLA adder as a building box and design a second level CLA logic to build a 16-bit CLA adder.

32. Delay of the 16-bit CLA Adder p(15:0) and g(15:0) are available after one gate delay. It takes 2 more gate delays for the P and G signals for each of the 4-bit box. It takes 2 more gate delays for the second layer to produce ci(12), ci(8) and ci(4). It takes 2 more gate delays for the first layer to produce the rest carry in values. It takes one more gate delay for the sum. In total 1+2+2+2+1=8 gate delays. In general the total delay is 1+2+4( (log n)/2 )+1.

33. Thanks!