1. Sequential Design Part II

2. Output

3. A(t+1) =D A = AX + BX B(t+1) =D B = AX Y = AX + BX

4. 2-D State Table

5. Even Easier, State Diagrams:

7. Using State Diagrams

8. Using J-K Flip Flops J A = BK A = BX J B = XK B = AX + AX What about designing a state machine using J-K Flip Flops... Given this state table

9. Recall

10. Sequential Circuit Design Obtain either the state diagram of the state table from the statement of the problem If only a state diagram is available from set 1, obtain the state table Assign binary codes to the states Derive the flip-flop input equations from the next-state entries in the encoded state table Derive the output equations from the output entries in the state table Simplify the flip-flop input equations and output equations Draw the logic diagram with D flip-flops and combinational gates, as specified by the flip-flop input equations and output equations

11. Mealy Machine State Register C1 x(t) s(t+1) s(t) z(t) clk init present state present input next state C2 Output based on state and present input

12. Moore Machine State Register C1 x(t) s(t+1) s(t) z(t) clk init present state present input next state C2 Output base on state only

13. Example: A Sequence Recognizer Let’s detect the sequence “1101” in a bit sequence We need to “remember” what bits have passed by. If the input is a ‘1’ then move to state B and the output is a 0 (have not yet detected the “1101” sequence

14. If we are at state B (which means that we have read a ‘1’ immediately beforehand) and the next input is a ‘1’ then we are making our way towards a successful “1101” read so move to state C. Means first bit was a ‘1’

15. The next bit we would like to read along our “1101” sequence is a ‘0’. So if we read a 0, go to State D -- notice output is still 0, we have not yet read the entire sequence. After state D we have succeeded if a ‘1’ is read so we will proceed and the output will now be ‘high’ or ‘1’

16. We don’t want to proceed to an E state, instead, if we have detected “1101”, we have not only detected the bit sequence but we also are on our way to detecting another “1101” sequence. Consider “1101101”. Two Sequences We set the output ‘high’ and go to State B.

17. We must also fill in the “unsuccessful” states, ones in which we have not read a “1101” sequence. Begins with ‘0’ Second bit is ‘0’ Third bit is a ‘1’ which means we have read a “111” seq. This puts us waiting for a ‘0’ A ‘0’ is the last bit (“1100”) back to the beginning.

18. Given the following: Designing with D Flip-Flops A(t + 1) = D A (A,B,X) =  m(2,4,5,6) B(t + 1) = D B (A,B,X) =  m(1,3,5,6) Y(A,B,X) =  m(1,5)

20. A(t + 1) = D A (A,B,X) =  m(2,4,5,6)

21. B(t + 1) = D B (A,B,X) =  m(1,3,5,6)

23. Using k-maps to reduce the equations: A(t + 1) = D A (A,B,X) =  m(2,4,5,6) B(t + 1) = D B (A,B,X) =  m(1,3,5,6) Y(A,B,X) =  m(1,5) D A = AB + BX A BX A D B = AX + BX + ABX A BX Y = BX

24. Logic Diagram for Circuit with D Flip-Flops

25. Don’t cares: A Design Advantage

26. J A = BXJ B = X K A = BXK B = AX + AX Significantly Reduces Logic Equations