1. Section 1.3 Asynchronous Circuits Exercises Alfredo Benso Politecnico di Torino, Italy Alfredo.benso@polito.it

2. #1 Find a minimal race free state coding for the following transition table.Find a minimal race free state coding for the following transition table. Also, find the boolean functions representing the future state.Also, find the boolean functions representing the future state. 0001111000011110 AAABCDAAB BA-CBADCB CCCCCBBDC DAC-ADDAD ab c=0c=1

3. #1: stable states 0001111000011110 AAABCDAAB BA-CBADCB CCCCCBBDC DAC-ADDAD c=0c=1

4. #1: Vicinity graph 0001111000011110 AAABCDAAB BA-CBADCB CCCCCBBDC DAC-ADDAD AB DC c=0c=1

5. C #1: Minimize the vicinity graph 0001111000011110 AAABCDAAB BA-CBADCB CCCCCBBDC DAC-ADDAD AB D c=0c=1

6. C #1: Minimize the vicinity graph 0001111000011110 AAABCBAAB BA-CBDDAB CCCCCDBAC DBC-CDDBD AB D 10 0001 11 c=0c=1

7. #2 Design the reduced transition table of an asynchronous circuit functioning in fundamental mode. The circuit has three inputs a, b, and c, and one output U, which is 1 only if all inputs are equal to ‘1’, and they went to ‘1’ in the order a, b, and c. U returns to 0 when all inputs returned to ‘0’.Design the reduced transition table of an asynchronous circuit functioning in fundamental mode. The circuit has three inputs a, b, and c, and one output U, which is 1 only if all inputs are equal to ‘1’, and they went to ‘1’ in the order a, b, and c. U returns to 0 when all inputs returned to ‘0’. Also, find the minimum number of state variables required to have a race free state assignment, and find a possible state assignment.Also, find the minimum number of state variables required to have a race free state assignment, and find a possible state assignment.

8. #2: Transition table We have to recognize the following sequence:We have to recognize the following sequence: 000  100  110  111 A 0 - - B 0 - - A 0 - - B 0 - - C 0 - - D - C 0 A - D 1 A 0 E 0 0001111000011110 c=0c=1 Waiting for 100 Wrong sequence. Waiting for 110 Waiting for 111 OK! Now waiting for all 0 A B C D E

9. #2: Transition table We have to recognize the following sequence:We have to recognize the following sequence: 000  100  110  111 A 0 E 0 - - E 0 B 0 - - A 0 - - B 0 E 0 - - C 0 - - E 0 - - D - C 0 A - D 1 A 0 E 0 0001111000011110 c=0c=1 Waiting for 100 Wrong sequence. Waiting for 110 Waiting for 111 OK! Now waiting for all 0 A B C D E

10. #2: Minimize the transition table A 0 E 0 - - E 0 B 0 - - A 0 - - B 0 E 0 - - C 0 - - E 0 - - D - C 0 A - D 1 A 0 E 0 0001111000011110 c=0c=1 Waiting for 100 Wrong sequence. Waiting for 110 Waiting for 111 OK! Now waiting for all 0 A B C D E States A and B are compatible, so I can collapse them in one state

11. #2: Minimize the transition table A 0 E 0 - - E 0 A 0 E 0 - - C 0 - - E 0 - - D - C 0 A - D 1 A 0 E 0 0001111000011110 c=0c=1 Wrong sequence. Waiting for 100 & 110 Waiting for 111 OK! Now waiting for all 0 A C D E States A and B are compatible, so I can collapse them in one state

12. #2: Vicinity graph A 0 E 0 - - E 0 A 0 E 0 - - C 0 - - E 0 - - D - C 0 A - D 1 A 0 E 0 0001111000011110 c=0c=1 A C D E E AC D A-C and C-E are essential, so we can try to eliminate A-E and C-D

13. #2: Vicinity graph A 0 C 0 - - C 0 A 0 C 0 D 0 C 0 A 0 E 0 - - E 0 E - A - C 0 A - D 1 C 0 E 0 0001111000011110 c=0c=1 A C D E E AC D 00 1101 10

14. #3 Find the Future State and Output functions of an asynchronous circuit functioning in fundamental mode. The circuit has three inputs X, Y, and Z and one output U, which goes to ‘1’ only when:Find the Future State and Output functions of an asynchronous circuit functioning in fundamental mode. The circuit has three inputs X, Y, and Z and one output U, which goes to ‘1’ only when: –There is a rising edge of X AND –While the value of X was ‘0’, Y and Z followed the sequence 00 - 10 – 11 (even if BEFORE or AFTER this sequence Y and Z changed to different values). U returns to ‘0’ as soon as X returns to ‘0’.U returns to ‘0’ as soon as X returns to ‘0’. The circuit must have a race free state assignment.The circuit must have a race free state assignment.

15. #3: Transition Table B 0 A 0 B 0 A 0 - - C 0 A 0 - - B 0 - - D 0 C 0 - - A 0 D 0 E - B - A - E 1 0001111000011110 X=0X=1 Wait for 00 Wait for X to return to 0 Wait for 10 Wait for 11 Now wait for rising edge X A B C D E We have to recognize the following sequence on Y and Z with X=0:We have to recognize the following sequence on Y and Z with X=0: 00  10  11

16. #3: Transition Table Minimization B 0 A 0 B 0 A 0 D 0 B 0 A 0 - - A 0 D 0 E - B - A - E 1 0001111000011110 X=0X=1 Wait for 00 Wait for X to return to 0 Wait for 10 & 11 Now wait for rising edge X A B D E B and C are compatible so they can be collapsed in one stateB and C are compatible so they can be collapsed in one state

17. #3: Transition Table Minimization B 0 A 0 B 0 A 0 D 0 B 0 A 0 - - A 0 D 0 E - B - A - E 1 0001111000011110 X=0X=1 A B D E I can remove E-B going trough E - A - B D AB E

18. #3: Transition Table Minimization B 0 A 0 B 0 A 0 D 0 B 0 A 0 - - A 0 D 0 E - A - E 1 0001111000011110 X=0X=1 A B D E D AB E11 0001 10

19. #3: Output function 00000000 00000--- 0000---- ----1111 0001111000011110 X=0X=1 00 01 11 10

20. #3: Output function 00000000 00000000 00001111 00001111 0001111000011110 X=0X=1 00 01 11 10 U = X S 0

21. #3: Future state function 0100000000000000 0100100100000000 1111111110101010 0000000010101010 0001111000011110 X=0X=1 00 01 11 10

22. #3: Future state function 00000000 00100010 11111111 00001111 0001111000011110 X=0X=1 00 01 11 10 S 0 = S 0 S 1 + XS 0 + YZS 1

23. #3: Future state function 10000000 10010000 11110000 00000000 0001111000011110 X=0X=1 00 01 11 10 S 1 = S 0 S 1 X + XYZS 1 + XYZS 0 + XYZS 1

24. #5 Design the reduced transition table of an asynchronous circuit functioning in fundamental mode. The circuit has two inputs X and Y, and one output U, which is 1 only if X=1 and, during the last two rising-edges of X, Y maintained the same value.Design the reduced transition table of an asynchronous circuit functioning in fundamental mode. The circuit has two inputs X and Y, and one output U, which is 1 only if X=1 and, during the last two rising-edges of X, Y maintained the same value. Also, find out the minimum number of state variables required to have a race free state assignment.Also, find out the minimum number of state variables required to have a race free state assignment.

25. #5: Primitive Transition table XY 00011110 AA,0A,0E,0B,0wait for the first transition of X BC,0C, 0B,0B,01st trans. & Y=0 – wait for X to go low CC,0C,0E,0D,-wait for another transition of X DH,-H,-D,1D,12nd trans. & Y=0 – wait for X to go low EF,0F,0E,0E,01st trans. & Y=1 – wait for X to go low FF,0F,0G,-B,0wait for another transition GI,-I,-G,1G,12nd trans. & Y=1 – wait for X to go low HH,0H,0E,0D,-last 2 trans Y=0 – wait for a new transition II,0I,0G,-B,0 last 2 trans Y=1 – wait for a new transition

26. #5: Reduced Transition table XW 00011110 AA,0A,0E,0B,0 Equivalent states: C & H  C BC,0C, 0B,0B,0 F & I  F CC,0C,0E,0D,- No compatible states DH,-H,-D,1D,1 EF,0F,0E,0E,0 FF,0F,0G,-B,0 GI,-I,-G,1G,1 HH,0H,0E,0D,- II,0I,0G,-B,0

27. #5: Reduced Transition table XW 00011110 AA,0A,0E,0B,0 Equivalent states: C & H  C BC,0C, 0B,0B,0 F & I  F CC,0C,0E,0D,- No compatible states DH,-H,-D,1D,1 EF,0F,0E,0E,0 FF,0F,0G,-B,0 GI,-I,-G,1G,1 HH,0H,0E,0D,- II,0I,0G,-B,0

28. #5: Reduced Transition table XW 00011110 AA,0A,0E,0B,0 Equivalent states: C & H  C BC,0C, 0B,0B,0 F & I  F CC,0C,0E,0D,- No compatible states DH,-H,-D,1D,1 EF,0F,0E,0E,0 FF,0F,0G,-B,0 GI,-I,-G,1G,1 HH,0H,0E,0D,- II,0I,0G,-B,0

29. #5: Reduced Transition table XW 00011110 AA,0A,0E,0B,0 Equivalent states: C & H  C BC,0C, 0B,0B,0 F & I  F CC,0C,0E,0D,- No compatible states DC,-C,-D,1D,1 EF,0F,0E,0E,0 FF,0F,0G,-B,0 GF,-F,-G,1G,1

30. #5: Vicinity Graph Max outdegree is 3. I need at least three state variablesMax outdegree is 3. I need at least three state variables ABCDEFG

31. #5: Vicinity Graph It seems that there is no solution with three variables.It seems that there is no solution with three variables. Either we need 4 state variablesEither we need 4 state variables A BCDEFG 000 111 100 101 001 010 110 011 ?? ? ??? ? ?

32. #6 Design the gate level model of an asynchronous circuit functioning in fundamental mode. The circuit has two inputs P (pulse) and R (reset) and one output Z, which is normally at ‘0’. Z follows the following rules:Design the gate level model of an asynchronous circuit functioning in fundamental mode. The circuit has two inputs P (pulse) and R (reset) and one output Z, which is normally at ‘0’. Z follows the following rules: –It goes to ‘1’ when a rising edge is detected on P and R=0. –It goes to ‘0’ when R=1.

33. 00011110 A B #6: Transition table P R A 0 B - B 1 C - B 1 C - Let’s start with the correct sequence. R=0 and P=0 and we wait for a transition on P. C - - We can stay in B until R changes back to 1. If R goes to 1, Z goes to 0 but we have to wait for R to go to 0 again before checking for another rising edge on P

34. 00011110 A B #6: Transition table P R A 0 B - B 1 C - B 1 C - We can exit C only when both R and P are ‘0’. C - - If R goes to 1 while we are waiting for a rising edge on P, we go out-of- sequence. C 0 A 0 C 0

35. 00011110 A B #6: Vicinity graph P R A 0 B - B 1 C - B 1 C - C - - C 0 A 0 C 0 B AC2 21 0001

36. 00011110 A B #6: Vicinity graph P R A 0 B - B 1 A 0 B 1 A 0 C C 0 A 0 C 0 B AC2 21

37. 00011110 A B #6: Race free state assignment P R A 0 B - B 1 A 0 B 1 A 0 C C 0 A 0 C 0 B AC 0010 01

38. 00011110 00 01 #6: Output function P R A 0 B - B 1 A 0 B 1 A 0 10 C 0 A 0 C 0 000-1001 ---- 0000 00 01 P R 11 10 Z = S 0 S 1 R S0 S1S0 S1S0 S1S0 S1 00011110

39. 00 01 #6: State function P R A 0 B - B 1 A 0 B 1 A 0 10 C 0 A 0 C 0 0010100101000001 -------- 00101010 00 01 P R 11 10 S0 S1S0 S1S0 S1S0 S1 00011110

40. 00 01 #6: S 0 variable P R A 0 B - B 1 A 0 B 1 A 0 10 C 0 A 0 C 0 01100000 ---- 0111 00 01 P R 11 10 S0 S1S0 S1S0 S1S0 S1 00011110 S 0 = S 0 P + S 1 R

41. 00011110 00 01 #6: S 1 variable P R A 0 B - B 1 A 0 B 1 A 0 10 C 0 A 0 C 0 00011001 ---- 0000 00 01 P R 11 10 S0 S1S0 S1S0 S1S0 S1 00011110 S 1 = S 1 R + S 0 P R

42. #6: Final circuit P R S0S0S0S0 Z S1S1S1S1

43. #10 Design an asynchronous circuit functioning in fundamental mode. The circuit has two inputs A and B, and one output Z, which is normally at ‘0’. Z changes value when it recognize the sequence 00 – 01 – 00 on the inputs. 00 can be considered concurrently the end of a sequence and the beginning of a new one.Design an asynchronous circuit functioning in fundamental mode. The circuit has two inputs A and B, and one output Z, which is normally at ‘0’. Z changes value when it recognize the sequence 00 – 01 – 00 on the inputs. 00 can be considered concurrently the end of a sequence and the beginning of a new one. Show:Show: –A Minimized Transition table –A race free state assignment –The Boolean function for Z

44. #10: Transition Table B 0 A 0 B 0 C 0 - - D - C 0 - - D 1 E 1 - - B - E 1 - - 00011110 Wait for 00 Wait for 01; Wait for 00 Sequence OK; wait for 01 A B C D E F Sequence OK Out of sequence

45. #10: Transition Table B 0 A 0 B 0 C 0 - - A 0 D - C 0 A 0 - - D 1 E 1 - - F 1 B - E 1 F 1 - - D 1 F 1 00011110 Out of sequence; Wait for 00 Wait for 00 Wait for 01 Wait for 00 Sequence OK; wait for 01 A B C D E F Sequence OK Out of sequence; wait for 00 U=0 U=1

46. #10: Vicinity Graph B 0 A 0 B 0 C 0 - - A 0 D - C 0 A 0 - - D 1 E 1 - - F 1 B - E 1 F 1 - - D 1 F 1 00011110 A B C D E F There are no equivalent or compatible states.There are no equivalent or compatible states. B AC E FD

47. #10: Vicinity Graph B 0 A 0 B 0 C 0 - - A 0 D - C 0 A 0 - - D 1 E 1 - - F 1 B - E 1 F 1 - - D 1 F 1 00011110 A B C D E F We can eliminate A – C and F - EWe can eliminate A – C and F - E B AC E FD

48. #10: Vicinity Graph B 0 A 0 B 0 C 0 A 0 D - C 0 B 0 - - D 1 E 1 F 1 B - E 1 D 1 - - D 1 F 1 00011110 A B C D E F We can eliminate A – C and F - EWe can eliminate A – C and F - E B AC E FD

49. #10: Vicinity Graph B 0 A 0 B 0 C 0 A 0 D - C 0 B 0 - - D 1 E 1 F 1 B - E 1 D 1 - - D 1 F 1 00011110 A B C D E F A race free state assignment is:A race free state assignment is: 110 111010 100 001000

50. #10: Output Function 0000 0000 -00- 1111 -11- 1111 00011110 111 110 010 000 100 001 Be careful! To cover the table, you have to order the rows correctly: it has to be a K-map!Be careful! To cover the table, you have to order the rows correctly: it has to be a K-map!

51. #10: Output Function 0000 0000 -00- 1111 -11- 1111 111 110 010 000 100 00111111111 ---- -00- -11- ---- 0000 0000 Z = S 1 00011110 000 001 011 010 100 101 111 110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

52. #16 Design an asynchronous circuit with two inputs (A and B) and one output, functioning in fundamental mode. The output U of the circuit is equal to 1 iff:Design an asynchronous circuit with two inputs (A and B) and one output, functioning in fundamental mode. The output U of the circuit is equal to 1 iff: –A=0 AND –During the last interval in which A=1 there has been a SINGLE transition (1-0-1) on B. U changes back to ‘0’ as soon as A goes back to ‘1’.U changes back to ‘0’ as soon as A goes back to ‘1’. Show:Show: –A Minimized Transition table –A race free state assignment –The Output and State Boolean functions

53. #16: Comment The problem asks to design a sequence recognizer for the following sequence:The problem asks to design a sequence recognizer for the following sequence: 11  10  11  0- A B U

54. #16: transition table A B C D E F A 0 Let’s start to draw the transition table from a state where we wait for the correct beginning of the sequence.Let’s start to draw the transition table from a state where we wait for the correct beginning of the sequence. 00011110 A 0B 0 A correct sequence starts when A goes to 1 while B is already 1A correct sequence starts when A goes to 1 while B is already 1 B 0

55. #16: transition table A B C D E F A 0 Let’s start to draw the transition table from a state where we wait for the correct beginning of the sequence.Let’s start to draw the transition table from a state where we wait for the correct beginning of the sequence. 00011110 A 0B 0 A B

56. #16: transition table A B C D E F A 0 Now we can fill the transition table with the CORRECT sequenceNow we can fill the transition table with the CORRECT sequence 00011110 A 0B 0 C 0 D 0C 0 E - E 1 B 0 D 0 A B

57. #16: transition table A B C D E F A 0 Now we have to complete the table with the out- of-sequence conditionsNow we have to complete the table with the out- of-sequence conditions 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 If A goes to 1 with B=0, I am already out of sequence, because I will have too many transitions of B while A=1.If A goes to 1 with B=0, I am already out of sequence, because I will have too many transitions of B while A=1. Too many transitions! I can now wait in F until A goes back to 0, and then go to state A to wait for the right condition.I can now wait in F until A goes back to 0, and then go to state A to wait for the right condition.

58. #16: transition table A B C D E F A 0 Now we have to complete the table with the out- of-sequence conditionsNow we have to complete the table with the out- of-sequence conditions 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 If A goes to 0 to early, I go back to state A waiting for the beginning of a new sequence.If A goes to 0 to early, I go back to state A waiting for the beginning of a new sequence. Too many transitions! If B goes to 0 again, I have too many transitions on B, and therefore I will have to go to the F stateIf B goes to 0 again, I have too many transitions on B, and therefore I will have to go to the F state - A 0 - F 0

59. #16: Minimize the Transition table A B C D E F A 0 There are no equivalent statesThere are no equivalent states There are no compatible statesThere are no compatible states 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0

60. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0 B AC E FD 3 3 1 1 1 1 1 1 1

61. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0 B AC E FD 3 3 1 1 1 1 1 1 1

62. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- F 0 B AC E FD 3 3 1 1 1 1 1 1 1

63. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- F 0 B AC E FD 3 3 1 1 1 1 1 1 1

64. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 - F 0 D 0 B AC E FD 3 3 1 1 1 1 1 1 1 A 0 B 0

65. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 - F 0 D 0 B AC E FD A 0 B 0

66. #16: Race free state assignment A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 - F 0 D 0 000 100001 010 ???011 A 0 B 0

67. #16: Race free state assignment It seems there is no solution with three state variables.It seems there is no solution with three state variables. Before deciding to add a new variable, let’s try to modify the graph in another wayBefore deciding to add a new variable, let’s try to modify the graph in another way

68. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0 B AC E FD 3 3 1 1 1 1 1 1 1

69. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0 B AC E FD A-C

70. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- F 0 B AC E FD A-C

71. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- F 0 B AC E FD D-F

72. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- E 0 F 0 B AC E FD D-F

73. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0- E 0 F 0 B AC E FD D-E In this case we can try to move the arcIn this case we can try to move the arc

74. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 F 0 B AC E FD D-E In this case we can try to move the arcIn this case we can try to move the arc

75. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 F 0 B AC E FD E-F Again, we can try to remove the arcAgain, we can try to remove the arc

76. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 B AC E FD E-F Again, we can try to remove the arcAgain, we can try to remove the arc

77. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 B AC E FD B-E Again, we can try to remove the arcAgain, we can try to remove the arc

78. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 B AC E FD B-E The last edge we can remove is B-EThe last edge we can remove is B-E

79. #16: Vicinity Graph A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 001 000011 010 100111 Now a race free assignment is possibleNow a race free assignment is possible B AC E FD

80. #16: Output table 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 Let’s start with the output tableLet’s start with the output table00000000 0000 1100 0000 ---- -000 ---- 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

81. #16: Output table 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0 Let’s start with the output tableLet’s start with the output table00000000 0000 1100 0000 0000 0000 1100 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2 U=S 1 S 2 A

82. #16: State variables 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 0000000001100000000001011 001010111011 010010001000 000000100100 ---- -011111010 ---- 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

83. #16: S 0 state variable 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 000010000 0010 0000 0011 ---- -010 ---- 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

84. #16: S 1 state variable 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 000000001 0111 1100 0000 ---- -111 ---- 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

85. #16: S 2 state variable 000 001 011 111 010 100 A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 D 0 E 1 B 0 A 0 F 0 D 0 A 0 B 0E 0 - E 0 A 000100011 1011 0010 0000 ---- -110 ---- 000 001 011 010 100 101 111 110 00011110 S0 S1S2S0 S1 S2S0 S1S2S0 S1 S2

86. #16: Another solution for State coding There is another possible solution for the vicinity graph.There is another possible solution for the vicinity graph. A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 - F 0 B AC E FD

87. #16: Another solution for State coding By removing only A-C and F-D, a solution is possible:By removing only A-C and F-D, a solution is possible: A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 B 0- E 0 F 0 B AC E FD

88. #16: Another solution for State coding By removing only A-C and F-D, a solution is possible:By removing only A-C and F-D, a solution is possible: A B C D E F A 0 00011110 A 0B 0F 0 B 0C 0 D 0C 0 E -D 0 E 1 B 0 A 0 F 0 D 0 - A 0 B 0- E 0 F 0 001 000101 011 010111

89. #9 Design an asynchronous circuit with two inputs (A and B) and one output, functioning in fundamental mode. The output U of the circuit is equal to 1 iff it recognize the following input sequence:Design an asynchronous circuit with two inputs (A and B) and one output, functioning in fundamental mode. The output U of the circuit is equal to 1 iff it recognize the following input sequence: –AB = 00 – 01 – 11 – 10 - 00 U changes back to ‘0’ as soon as A or B change to 1.U changes back to ‘0’ as soon as A or B change to 1. Show:Show: –A Minimized Transition table –A race free state assignment

90. #9: Transition Table B 0 A 0 B 0 C 0 - - A 0 B 0 C 0 D 0 - - A 0 D 0 E 0 F - - - A 0 E 0 F 1 C - - - A - A B C D E 00011110 F

91. #9: Transition Table B 0 A 0 B 0 C 0 - - A 0 B 0 C 0 D 0 - - A 0 D 0 E 0 F - - - A 0 E 0 F 1 C - - - A - A B C D E 00011110 F Compatible States

92. #9: Transition Table B 0 A 0 B 0 D 0 A 0 - - A 0 D 0 E 0 F - - - A 0 E 0 F 1 B - - - A - A B D E 00011110 F D AB EF

93. #9: Transition Table B 0 A 0 B 0 D 0 A 0 - - A 0 D 0 E 0 F - - - A 0 E 0 F 1 B - - - A - A B D E 00011110 F D AB EF

94. #9: Transition Table B 0 A 0 B 0 D 0 A 0 - - E 0 D 0 E 0 F - A 0 E 0 F 1 B - - - B - A B D E 00011110 F D AB EF

95. #9: Transition Table B 0 A 0 B 0 G 0 A 0 - - E 0 D 0 E 0 F - A 0 E 0 F 1 B - - - B - - - H 0 - - D 0 - - A B D E 00011110 F D AB EF H G G H

96. #9: Transition Table B 0 A 0 B 0 G 0 A 0 - - E 0 D 0 E 0 F - A 0 E 0 F 1 B - - - B - - - H 0 - - D 0 - - A B D E 00011110 F 110 000001 010011 111 101 G H