1. Guest Lecture by David Johnston
CprE 281: Digital Logic
Guest Lecture by David Johnston

2. Moore Machines CprE 281: Digital Logic Iowa State University, Ames, IA
Copyright © 2013

3. The machine’s current state decides the current output.
Moore Machine Summary
The machine’s current state and current input are used to decide which next state to transition into.
The machine’s current state decides the current output.

4. Let’s finish implementing our previous example.

5. C: z = 1
Reset
B: z = 0
A: z = 0 w = 1
We need to find both the next state logic and the output logic implied by this machine.

6. Next state Present Output state A B C z w = w = 1 C: z = 1 Reset
B: z = 0
A: z = 0 w = 1
Next state
Present
Output
state z w = w = 1 A B C

7. Next state Present Output state A A B B A C C A C 1 z w = w = 1
Reset
B: z = 0
A: z = 0 w = 1
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1

8. Let’s use two flip flops to hold the state machine’s state.

9. Let Q(t) = y2y1 and Q(t+1) = Y2Y1.w
Next State Logic
Output Logic z Y y 2 2
Clock

10. So, we need to find logic expressions for
Y1(w, y1, y2), Y2(w, y1, y2), and z(y1, y2). Y y 1 1 w
Next State Logic
Output Logic z Y y 2 2
Clock

11. Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Suppose we encoded our states in the same order in which they were labeled:
A ~ 00
B ~ 01
C ~ 10

12. The finite state machine will never reach a state encoded as 11.
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Next state
Present
Output
state w = w = 1 z A 00
The finite state machine will never reach a state encoded as 11. B 01 C 10 11

13. Next state Present Output state A A B B A C C A C 1 z w = w = 1w = 1 A A B B A C C A C 1
Present
Next state
state w = 1
Output y Y 2 z A 00 01 B 10 C 11 dd d
We arbitrarily chose these as our state encodings. We could have used others.

14. Present Next state state w = 1 Output y Y 2 z 00 01 10 11 dd d
q(t) = y2y1 and q(t+1) = Y2Y1 w y2 y1 Y2 Y1 1
Present
Next state
state w = 1
Output y Y 2 z 00 01 10 11 dd d y2 y1 z 1

15. Y 1 y y w y2 y1 Y2 Y1 1 d 2 1 w 00 01 11 10 d 1 1 d
Y 2 y y 2 1 w 00 01 11 10 d 1 1 d 1 y2 y1 z 1 d z y 1 y 2 1 1 1 d

16. Don’t care conditions simplify the combinatorial logic.2 1
Ignoring don't cares
Using don't cares w 00 01 11 10 d Y = wy y Y = wy y 1 1 2 1 1 2 1 1 d
Y 2 y y 2 1 w 00 01 11 10 d Y = wy y + wy y Y = wy + wy 2 1 2 1 2 2 1 2 1 1 d 1 = w ( y + y ) 1 2 z y 1 y 2 1 z = y y z = y 1 2 2 1 1 d
Don’t care conditions simplify the combinatorial logic.

18. Lastly, we add a reset signal, which forces the machine back to its start state, which is state 00 in this case.

19. Summary: Designing a Moore Machine
Obtain the circuit specification.
Derive a state diagram.
Derive the state table.
Decide on a state encoding.
Encode the state table.
Derive the output logic and next-state logic.
Add a reset signal.

20. An Alternative State Encoding

21. Our previous design decisions led us to:
Y 1 y y z 2 1
Y 2 y y y 2 1 1 y w 00 01 11 10 w 2 1 00 01 11 10 d d 1 1 d 1 1 d 1 1 d 1
Can we do better?

22. A Better State Encoding
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Suppose we encoded our states another way:
A ~ 00
B ~ 01
C ~ 11

23. A Better State Encoding
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Next state
Present
Output
state w = w = 1
A ~ 00
B ~ 01
C ~ 11 z

24. A Better State Encoding
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Present
Next state
state w = 1
Output y 2 Y z A 00 01 B 11 C 10 dd d

25. What Are Our Logic Expressions?
Present
Next state
state w = 1
Output y 2 Y z A 00 01 B 11 C 10 dd d

26. What Are Our Logic Expressions?
Present
Next state
state w = 1
Output y 2 Y z A 00 01 B 11 C 10 dd d
Warning:
This table does not enumerate y2y1, in the standard way, so be careful when filling out the K-Map. Y2 y y Y1 y y z 2 1 2 1 y 1 w y w 00 01 11 10 00 01 11 10 2 1 1 1 1

27. What Are Our Logic Expressions?
Present
Next state
state w = 1
Output y 2 Y z A 00 01 B 11 C 10 dd d
Warning:
The present state variables, y1y2, do not follow our standard enumeration, so be careful when filling out the K-Map. Y2 Y1 y y y y z y 2 1 2 1 1 w y 2 00 01 11 10 w 00 01 11 10 1 d d 1 1 1 1 d 1 1 1 1 d d 1
Y2(w, y2, y1) = wy1
Y1(w, y2, y1) = w
z(y2, y1) = y2

28. Original State Encodings
New State Encodings: Y1
Y 1 y y y y 2 1 2 1 w 00 01 11 10 w 00 01 11 10 d d 1 1 d 1 1 1 1 d Y2
Y 2 y y y y 2 1 2 1 w w 00 01 11 10 00 01 11 10 d d 1 1 1 d 1 1 1 d z y z y 1 1 y y 2 1 2 1 1 1 d 1 d 1

29. Improved Result Y1(w, y2, y1) = w Y2(w, y2, y1) = wy1 z(y2, y1) = y2 YQ z
Y2(w, y2, y1) = wy1 Q
z(y2, y1) = y2 Y y 1 1 w D Q
Clock Q
Resetn

30. Some may be better than others.
Main Idea
Different state assignments of the same Moore machine generally lead to different circuits.
Some may be better than others.

31. Another Moore Machine: Register Swap Controller

33. Design a Moore machine control circuit for swapping the contents of registers R1 and R2 by using R3 as a temporary.

34. If an output is not given, then assume that it is 0.D: R 3
out 1 = in
Done , w C: 2 B:
A: No Transfer
Reset
If an output is not given, then assume that it is 0.

35. How many logic expressions do we need to find?
How many flip-flops are we going to use?

36. , D: R 3 1 = Done w C: 2 B: A: No Transfer Reset W Z Next State Logic
out 1 = in
Done , w C: 2 B:
A: No Transfer
Reset W Z
Next State Logic
Q(t+1)
Flip-Flop Array
Q(t)
Output Logic
Clock

37. , D: R 3 1 = Done w C: 2 B: A: No Transfer Reset R1out win
Done , w C: 2 B:
A: No Transfer
Reset
R1out w
Next State Logic
Y[1:2]
Flip-Flop Array
y[1:2]
Output Logic
R1in
R2out
R2in
R3out
Clock
R3in
Done

38. Present Next state Outputs state w = 0 w = 1 A B C D , D: R 3 1 = Donein
Done , w C: 2 B:
A: No Transfer
Reset
Present
Next state
Outputs
state
Done
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in A B C D

39. Present Next state Outputs state w = 0 w = 1 A A B B C C 1 1 C D D 1 13
out 1 = in
Done , w C: 2 B:
A: No Transfer
Reset
Present
Next state
Outputs
state
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in
Done A A B B C C 1 1 C D D 1 1 D A A 1 1 1

40. Consider two encoding schemes.
As we saw before, we can expect that some state encodings will be better than others.
Consider two encoding schemes.

41. Present Next state Outputs state w = 0 w = 1 A A B B C C 1 1 C D D 1 1
Naïve State Encoding
Present
Next state
Outputs
state
Done
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in A A B B C C 1 1 C D D 1 1 D A A 1 1 1
Present
Next state
state
Outputs A 00 1 B 01 10 C 11 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done

42. Present Next state state Outputs A 00 1 B 01 10 C 11 D w = 0 w = 11 B 01 10 C 11 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done w y2 y1 Y2 Y1 1

43. Present Next state state Outputs A 00 1 B 01 10 C 11 D w = 0 w = 11 B 01 10 C 11 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done w y2 y1 Y2 Y1 1

44. Present Next state state Outputs A 00 1 B 01 10 C 11 D w = 0 w = 11 B 01 10 C 11 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done

45. Alternative State Encoding
Present
Next state
Outputs
state
Done
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in A A B B C C 1 1 C D D 1 1 D A A 1 1 1
Present
Next state
state
Outputs A 00 1 B 01 11 C 10 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done

46. Present Next state state Outputs A 00 1 B 01 11 C 10 D Y Y y y y y w1 B 01 11 C 10 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done Y Y 1 y y 2 y y 2 1 2 1 w w 00 01 11 10 00 01 11 10 1 1

47. Present Next state state Outputs A 00 1 B 01 11 C 10 D Y Y y y y y w1 B 01 11 C 10 D
w = 0
w = 1
y2y1
Y2Y1
Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done Y Y 1 y y 2 y y 2 1 2 1 w w 00 01 11 10 00 01 11 10 1 1 1 1 1 1 1 1 1 Y = y Y = wy + y y 1 2 1 2 2 1

48. One-Hot State Encoding

49. So far, we have been encoding states in a way that minimizes flip-flops.
But we can sometimes decrease the complexity of our combinational logic if we encode states more sparsely.

50. Vehicle Speed Controller
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Let’s use three flip-flops and using the following one- hot state encoding scheme:
A ~ 001
B ~ 010
C ~ 100

51. Vehicle Speed Controller
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1 A
001 B
010 C
100 1

52. Vehicle Speed Controller
Next state
Present
Output
state z w = w = 1 A A B B A C C A C 1
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1 A
001
001
010 B
010
001
100 C
100
001
100 1

53. Next State Present state Output w = w = 1 z y y y Y Y Y Y Y Y 001 0011
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1
001
001
010
010
001
100
100
001
100 1

54. Next State Present state Output w = w = 1 z y y y Y Y Y Y Y Y 001 001d 1
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1
001
001
010
010
001
100
100
001
100 1

55. Next State Present state Output w = w = 1 z y y y Y Y Y Y Y Y 001 0011
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1
001
001
010
010
001
100
100
001
100 1

56. w = y3 Y3 = wy1 Y2 = wy1 Y1 = w Next State Present state Output w = w1
Next State
Present
state
Output w = w = 1 z y y y 3 2 1 Y Y Y Y Y Y 3 2 1 3 2 1
001
001
010
010
001
100
100
001
100 1
w = y3
Y3 = wy1
Y2 = wy1
Y1 = w

57. Register Swap Controller
Present
Next state
Outputs
state
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in
Done A A B B C C 1 1 C D D 1 1 D A A 1 1 1

58. Register Swap Controller
Present
Next state
Outputs
state
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in
Done A A B B C C 1 1 C D D 1 1 D A A 1 1 1
Let’s use four flip-flops and the following one-hot state encoding scheme:
A ~ 0001
B ~ 0010
C ~ 0100
D ~ 1000

59. Register Swap Controller
Present
Next state
Outputs
state
Done
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in A A B B C C 1 1 C D D 1 1 D A A 1 1 1
Present
Next State
Outputs
State
w = 0
w = 1
y4y3y2y1
Y4Y3Y2Y1
Y4Y3Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done A
001 B
010 1 1 C
100 1 1 D 1
000 1 1 1

60. Register Swap Controller
Present
Next state
Outputs
state
w = 0
w = 1
R1out
R1in
R2out
R2in
R3out
R3in
Done A A B B C C 1 1 C D D 1 1 D A A 1 1 1
Present
Next State
Outputs
State
w = 0
w = 1
y4y3y2y1
Y4Y3Y2Y1
Y4Y3Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done A
001
0001
0010 B
010
0100
0100 1 1 C
100
1000
1000 1 1 D 1
000
0001
0001 1 1 1

61. What next state logic do we need?
Y4(w, y4, y3, y2, y1)
Y3(w, y4, y3, y2, y1)
Y2(w, y4, y3, y2, y1)
Y1(w, y4, y3, y2, y1)
Present
Next State
Outputs
State
w = 0
w = 1
y4y3y2y1
Y4Y3Y2Y1
Y4Y3Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done A
001
0001
0010 B
010
0100
0100 1 1 C
100
1000
1000 1 1 D 1
000
0001
0001 1 1 1

62. What output logic do we need?
R1out(y4, y3, y2, y1)
R1in(y4, y3, y2, y1)
R2out(y4, y3, y2, y1)
R2in(y4, y3, y2, y1)
R3out(y4, y3, y2, y1)
R3in(y4, y3, y2, y1)
Done(y4, y3, y2, y1)
Present
Next State
Outputs
State
w = 0
w = 1
y4y3y2y1
Y4Y3Y2Y1
Y4Y3Y2Y1
R1out
R1in
R2out
R2in
R3out
R3in
Done A
001
0001
0010 B
010
0100
0100 1 1 C
100
1000
1000 1 1 D 1
000
0001
0001 1 1 1

63. Questions?

64. THE END