Chapter 7

- 7.1 -

Chapter 7Problem Solutions

7.1. Excitation and output expressions:

J1 = xQ-2 + y

K1 = yQ-2 + x

-Q2J2 = x

-y-

K2 = xy + Q1z = Q1Q

-2 + x

-Q-1

Excitation table:

Present state(Q1Q2)

Excitation(J1K1,J2K2)

Output(z)

Inputs (xy) Inputs (xy)

00 01 10 11 00 01 10 11

00 00,10 11,00 10,00 11,01 1 1 0 0

01 01,10 11,00 00,00 10,01 1 1 0 0

10 00,11 11,01 10,01 11,01 1 1 1 1

11 01,11 11,01 00,01 10,01 0 0 0 0

Transition table:

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)


00 01 10 11 00 01 10 11

00 01 10 10 10 1 1 0 0

01 01 11 01 10 1 1 0 0

10 11 00 10 00 1 1 1 1

11 00 00 10 10 0 0 0 0

- 7.2 -

7.1. (continued)

State table:

Present state Next state Output(z)


00 01 10 11 00 01 10 11

00 A B C C C 1 1 0 0

01 B B D B C 1 1 0 0

10 C D A C A 1 1 1 1

11 D A A C C 0 0 0 0

State diagram:

- 7.3 -


D1 = x-Q2

D2 = x-Q-1Q2 + xQ

-1Q-2

z = Q1Q2

Excitation table:

Present state(Q1Q2)

Excitation(D1D2)

Output(z)

Input (x)

0 1

00 00 01 0

01 11 00 0

10 00 00 0

11 10 00 1

Transition table (Same as excitation table):

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)

Input (x)

0 1

00 00 01 0

01 11 00 0

10 00 00 0

11 10 00 1

- 7.4 -

7.2. (continued)

State table:


Input (x)

0 1

00 A A B 0

01 B D A 0

10 C A A 0

11 D C A 1

State diagram:

- 7.5 -


T1 = xQ2 + Q-1Q2

T2 = x + Q-1Q2

z1 = xQ-1

z2 = x-Q2

Excitation table:

Present state(Q1Q2)

Excitation(T1T2)

Output(z1z2)

Input (x) Input (x)

0 1 0 1

00 00 01 00 10

01 11 11 01 10

10 00 01 00 00

11 00 11 01 00

Transition table:

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z1z2)

Input (x) Input (x)

0 1 0 1

00 00 01 00 10

01 10 10 01 10

10 10 11 00 00

11 11 00 01 00

- 7.6 -

7.3. (continued)

State table:

Present state Next state Output(z1z2)

Input (x) Input (x)

0 1 0 1

00 A A B 00 10

01 B C C 01 10

10 C C D 00 00

11 D D A 01 00

State diagram:

- 7.7 -


J1 = Q2K1 = x

D2 = x(Q-1+Q

-2)

z = Q2 + x-Q1

Excitation table:

Present state(Q1Q2)

Excitation(J1K1,D2)

Output(z)

Input (x) Input (x)

0 1 0 1

00 00,0 01,1 0 0

01 10,0 11,1 1 1

10 00,0 01,1 1 0

11 10,0 11,0 1 1

Transition table:

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)

Input (x) Input (x)

0 1 0 1

00 00 01 0 0

01 10 11 1 1

10 10 01 1 0

11 10 00 1 1

- 7.8 -

7.4. (continued)

State table:


Input (x) Input (x)

0 1 0 1

00 A A B 0 0

01 B C D 1 1

10 C C B 1 0

11 D C A 1 1

State diagram:

- 7.9 -

7.5.

A: A multiple of three 1's have occurred (where none is a

multiple of three).

B: One greater than a multiple of three 1's have occurred.

C: Two 1's greater than a multiple of three 1's have

occurred.


Input (x) Input (x)

0 1 0 1

* A A B 0 0

B B C 0 0

C C A 0 1

7.6.

A: Last input was 0.

B: One 1 has occurred.

C: Two consecutive 1's have occurred.

D: Three consecutive 1's have occurred.

E: More than three consecutive 1's have occurred.


Input (x) Input (x)

0 1 0 1

* A A B 0 0

B A C 1 0

C A D 0 0

D A E 1 0

E A E 0 0

- 7.10 -

7.7.

A: Waiting for a single 0 (or, equivalently, 1's occurring

after more than one 0 has occurred).

B: Exactly one 0 has occurred.

C: Two or more consecutive 0's have occurred.

D: 1's occurring after exactly one 0.


Input (x) Input (x)

0 1 0 1

* A B A 0 0

B C D 0 1

C C A 0 0

D B D 0 1

- 7.11 -

7.8.

State definitionPresentstate

Next state Output(z)

Numberof inputs

Numberof 1's

Numberof 0's

Input (x) Input (x)

0 1 0 1

0 0 0 * A B C 0 0

1 0 1 B D E 0 0

1 1 0 C E F 0 0

2 0 2 D G H 0 0

2 1 1 E H I 0 0

2 2 0 F I J 0 0

3 0 3 G M K 0 0

3 1 2 H K L 0 0

3 2 1 I L M 0 0

3 3 0 J M M 0 0

4 1 3 K A A 1 0

4 2 2 L A A 0 1

Invalid code group M A A 1 1

- 7.12 -

7.9.

Presentstate


State definitionInput (x) Input (x)

0 1 0 1

Awaiting 1st input ofsequence

* A B C 0 1

1st bit applied, retainapplied input

B D E 0 1

1st bit applied, complementapplied input

C E E 1 0

2nd bit applied, retainapplied input

D F G 0 1

2nd bit applied, complementapplied input

E G G 1 0

3rd bit applied, retainapplied input

F A A 0 1

3rd bit applied, complementapplied input

G A A 1 0

- 7.13 -

7.10. (a)

Presentstate



0 1 0 1

Awaiting 1st bit ofsequence (b4=g4)

* A B C 0 1

b4 bit was 0 (b3=g3) B D E 0 1

b4 bit was 1 (b3=g-3) C E D 1 0

b3 bit was 0 (b2=g2) D F G 0 1

b3 bit was 1 (b2=g-2) E G F 1 0

b2 bit was 0 (b1=g1) F H I 0 1

b2 bit was 1 (b1=g-1) G I H 1 0

b1 bit was 0 (b0=g0) H A A 0 1

b1 bit was 1 (b0=g-0) I A A 1 0

- 7.14 -

7.10. (continued)

(b) In problem 3.34 it was established that the algorithm

to convert an n-bit binary number bn-1bn-2...b1b0 into

its equivalent n-bit Gray code gn-1gn-2...g1g0 is

gn-1 = bn-1 gk-1 = bk bk-1 for n-1k1

Presentstate



0 1 0 1

Awaiting 1st bit ofsequence (g4=b4)

* A B C 0 1

g4 bit was 0 (g3=b4 b3) B D E 0 1

g4 bit was 1 (g3=b4 b3) C D E 1 0

g3 bit was 0 (g2=b3 b2) D F G 0 1

g3 bit was 1 (g2=b3 b2) E F G 1 0

g2 bit was 0 (g1=b2 b1) F H I 0 1

g2 bit was 1 (g1=b2 b1) G H I 1 0

g1 bit was 0 (g0=b1 b0) H A A 0 1

g1 bit was 1 (g0=b1 b0) I A A 1 0

- 7.15 -

7.11.


Next state Output(QAQBQC)

QA QB QCInput (x)

0 1

0 0 0 * A A B 000

1 0 0 B C D 100

0 1 0 C E F 010

1 1 0 D G H 110

0 0 1 E A B 001

1 0 1 F C D 101

0 1 1 G E F 011

1 1 1 H G H 111

7.12.

A: Even number of 1's have occurred, current output should

be 0.

B: Odd number of 1's have occurred, current output should

be 0.

C: Even number of 1's have occurred, current output should

be 1.

D: Odd number of 1's have occurred, current output should

be 1.


Input (x) Input (x)

0 1 0 1

* A A B 0 0

B B C 0 0

C C D 1 1

D D A 1 1

- 7.16 -

7.13. A possible initial state set is:

A: In a sequence of an even number of 0's.

B: In a sequence of an odd number of 0's.

C: In a sequence of 1's having a parity disagreeing with

the previous sequence of 0's.

D: In a sequence of 1's having a parity agreeing with the

previous sequence of 0's.


Input (x) Input (x)

0 1 0 1

* A B C 0 0

B A D 0 0

C B D 1 0

D B C 0 0

It is immediately noted that AD. The minimal state table

is:


Input (x) Input (x)

0 1 0 1

(A,D): * 1 2 3 0 0

(B): 2 1 1 0 0

(C): 3 2 1 1 0

- 7.17 -

7.14.

A: Waiting for the first 1 in a sequence of 1's.

B: One 1 has occurred.

C: Two 1's have occurred.

D: Three or more 1's have occurred.


Input (x) Input (x)

0 1 0 1

* A A B 0 0

B A C 0 0

C A D 0 0

D A D 0 1

7.15. Let the present state denote the previous three inputs,

i.e., xt-3, xt-2, and xt-1 where xt-i is the input at time

i.



xt-3 xt-2 xt-1Input (x) Input (x)

0 1 0 1

0 0 0 * A A B 0 0

0 0 1 B C D 1 0

0 1 0 C E F 1 1

0 1 1 D G H 0 0

1 0 0 E A B 1 0

1 0 1 F C D 0 0

1 1 0 G E F 0 0

1 1 1 H G H 1 0

- 7.18 -

7.16. Using the same approach as the previous problem, the state

table is:



xt-3 xt-2 xt-1Input (x) Input (x)

0 1 0 1

0 0 0 * A A B 0 1

0 0 1 B C D 0 0

0 1 0 C E F 0 0

0 1 1 D G H 1 0

1 0 0 E A B 0 1

1 0 1 F C D 0 0

1 1 0 G E F 0 0

1 1 1 H G H 1 0

For this table, however, AE, BF, CG, and DH. The

minimal state table is:


Input (x) Input (x)

0 1 0 1

(A,E): * 1 1 2 0 1

(B,F): 2 3 4 0 0

(C,G): 3 1 2 0 0

(D,H): 4 3 4 1 0

- 7.19 -

7.17. (a)

Equivalence classes of states:

{(A,E,H), (B,F), (C,G,I), (D)}

Minimal state table:


Input (x) Input (x)

0 1 0 1

(A,E,H): * 1 2 3 1 1

(B,F): 2 4 2 0 0

(C,G,I): 3 2 1 0 1

(D): 4 4 1 0 0

- 7.20 -

7.17. (continued)

(b)


{(A,C,E), (B,G), (D), (F), (H)}



Input (x) Input (x)

0 1 0 1

(A,C,E): * 1 2 1 0 0

(B,G): 2 1 3 1 0

(D): 3 4 1 0 0

(F): 4 4 5 1 0

(H): 5 3 4 1 1

- 7.21 -

7.18.


{(A,E,F), (B,G), (C,H), (D)}



Input (x)

0 1

(A,E,F): * 1 2 3 1

(B,G): 2 4 1 0

(C,H): 3 1 1 1

(D): 4 1 3 0

- 7.22 -

7.19. Transition table:

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)

Input (x) Input (x)

0 1 0 1

A 00 01 10 0 0

B 01 00 00 0 1

C 10 11 00 0 1

D 11 00 11 0 1

(a) The excitation table is the same as the transition

table where Di=Q+i.

- 7.23 -

7.19. (continued)

(b) Excitation table:

Present state(Q1Q2)


Output(z)

Input (x) Input (x)

0 1 0 1

00 0-,1- 1-,0- 0 0

01 0-,-1 0-,-1 0 1

10 -0,1- -1,0- 0 1

11 -1,-1 -0,-0 0 1

The output map and expression are the same as part (a).

- 7.24 -

7.19. (continued)

(c) Excitation table:

Present state(Q1Q2)

Excitation(T1T2)

Output(z)

Input (x) Input (x)

0 1 0 1

00 01 10 0 0

01 01 01 0 1

10 01 10 0 1

11 11 00 0 1


- 7.25 -

7.19. (continued)

(d) Excitation table:

Present state(Q1Q2)

Excitation(S1R1,S2R2)

Output(z)

Input (x) Input (x)

0 1 0 1

00 0-,10 10,0- 0 0

01 0-,01 0-,01 0 1

10 -0,10 01,0- 0 1

11 01,01 -0,-0 0 1


- 7.26 -

7.20. Adjacency conditions to be satisfied:

Rule I: (B,D), (B,C)

Rule II: (B,C), (A,D)(2×)

Rule III: (B,C,D)

State assignment map:

Transition table:

Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)

Input (x) Input (x)

0 1 0 1

A 00 11 10 0 0

B 11 00 00 0 1

C 10 01 00 0 1

D 01 00 01 0 1

- 7.27 -

7.20. (continued)


table where Di=Q+i.

- 7.28 -

7.20. (continued)


Present state(Q1Q2)


Output(z)

Input (x) Input (x)

0 1 0 1

00 1-,1- 1-,0- 0 0

11 -1,-1 -1,-1 0 1

10 -1,1- -1,0- 0 1

01 0-,-1 0-,-0 0 1


- 7.29 -

7.20. (continued)


Present state(Q1Q2)

Excitation(T1T2)

Output(z)

Input (x) Input (x)

0 1 0 1

00 11 10 0 0

11 11 11 0 1

10 11 10 0 1

01 01 00 0 1


- 7.30 -

7.20. (continued)


Present state(Q1Q2)


Output(z)

Input (x) Input (x)

0 1 0 1

00 10,10 10,0- 0 0

11 01,01 01,01 0 1

10 01,10 01,0- 0 1

01 0-,01 0-,-0 0 1


- 7.31 -


Present state(Q1Q2)

Next state(Q+1Q

+2)

Output(z)

Input (x)

0 1

A 00 00 01 1

B 01 10 00 0

C 10 00 11 0

D 11 10 10 1


table where Di=Q+i.

- 7.32 -

7.21. (continued)


Present state(Q1Q2)


Output(z)

Input (x)

0 1

00 0-,0- 0-,1- 1

01 1-,-1 0-,-1 0

10 -1,0- -0,1- 0

11 -0,-1 -0,-1 1


- 7.33 -

7.21. (continued)


Present state(Q1Q2)

Excitation(T1T2)

Output(z)

Input (x)

0 1

00 00 01 1

01 11 01 0

10 10 01 0

11 01 01 1


- 7.34 -

7.21. (continued)


Present state(Q1Q2)


Output(z)

Input (x)

0 1

00 0-,0- 0-,10 1

01 10,01 0-,01 0

10 01,0- -0,10 0

11 -0,01 -0,01 1


- 7.35 -


Present State(Q1Q2Q3)

Next State(Q+1Q

+2Q

+3)

Output(z)

Input (x) Input (x)

0 1 0 1

A 000 001 000 0 0

B 001 000 010 1 0

C 010 011 000 0 0

D 011 011 100 1 0

E 100 010 011 1 1


table where Di=Q+i.

- 7.36 -

7.22. (continued)


Present state(Q1Q2Q3)

Excitation(J1K1,J2K2,J3K3)

Output(z)

Input (x) Input (x)

0 1 0 1

000 0-,0-,1- 0-,0-,0- 0 0

001 0-,0-,-1 0-,1-,-1 1 0

010 0-,-0,1- 0-,-1,0- 0 0

011 0-,-0,-0 1-,-1,-1 1 0

100 -1,1-,0- -1,1-,1- 1 1

- 7.37 -

7.22. (continued)

- 7.38 -

7.22. (continued)




Excitation(T1T2T3)

Output(z)

Input (x) Input (x)

0 1 0 1

000 001 000 0 0

001 001 011 1 0

010 001 010 0 0

011 000 111 1 0

100 110 111 1 1

- 7.39 -

7.22. (continued)


- 7.40 -

7.22. (continued)



Excitation(S1R1,S2R2,S3R3)

Output(z)

Input (x) Input (x)

0 1 0 1

000 0-,0-,10 0-,0-,0- 0 0

001 0-,0-,01 0-,10,01 1 0

010 0-,-0,10 0-,01,0- 0 0

011 0-,-0,-0 10,01,01 1 0

100 01,10,0- 01,10,10 1 1

- 7.41 -

7.22. (continued)


- 7.42 -

7.23. Adjacency conditions to be satisfied:

Rule I: (C,D), (A,C)

Rule II: (A,B), (A,C), (A,D), (D,E), (C,D)

Rule III: (B,D,E)


Transition table:


Next State(Q+1Q

+2Q

+3)

Output(z)

Input (x) Input (x)

0 1 0 1

A 000 001 000 0 0

B 001 000 100 1 0

C 100 101 000 0 0

D 101 101 111 1 0

E 111 100 101 1 1

- 7.43 -

7.23. (continued)


table where Di=Q+i.

- 7.44 -

7.23. (continued)




Output(z)

Input (x) Input (x)

0 1 0 1

000 0-,0-,1- 0-,0-,0- 0 0

001 0-,0-,-1 1-,0-,-1 1 0

100 -0,0-,1- -1,0-,0- 0 0

101 -0,0-,-0 -0,1-,-0 1 0

111 -0,-1,-1 -0,-1,-0 1 1

- 7.45 -

7.23. (continued)


- 7.46 -

7.23. (continued)



Excitation(T1T2T3)

Output(z)

Input (x) Input (x)

0 1 0 1

000 001 000 0 0

001 001 101 1 0

100 001 100 0 0

101 000 010 1 0

111 011 010 1 1

- 7.47 -

7.23. (continued)




Excitation(S1R1,S2R2,S3R3)

Output(z)

Input (x) Input (x)

0 1 0 1

000 0-,0-,10 0-,0-,0- 0 0

001 0-,0-,01 10,0-,01 1 0

100 -0,0-,10 01,0-,0- 0 0

101 -0,0-,-0 -0,10,-0 1 0

111 -0,01,01 -0,01,-0 1 1

- 7.48 -

7.23. (continued)

- 7.49 -

7.23. (continued)


7.24. (a) Transition table:


Next State(Q+1Q

+2Q

+3)

Output(z)

Input (x)

0 1

A 000 000 001 0

B 001 010 101 0

C 010 011 000 1

D 011 100 010 0

E 100 011 101 0

F 101 100 011 1

- 7.50 -

7.24. (continued)

The excitation table is the same as the transition

table where Di=Q+i.

- 7.51 -

7.24. (continued)

(b) Adjacency conditions to be satisfied:

Rule I: (C,E), (D,F), (B,E)

Rule II: (A,B), (C,F), (A,D), (C,E), (D,F), (D,E)

Rule III: (C,F)


Transition table:


Next State(Q+1Q

+2Q

+3)

Output(z)

Input (x)

0 1

A 000 000 001 0

B 001 111 110 0

C 111 010 000 1

D 010 011 111 0

E 011 010 110 0

F 110 011 010 1

- 7.52 -

7.24. (continued)

The excitation table is the same as the transition

table where Di=Q+i.

- 7.53 -

7.25. (a) Transition table:


Next State(Q+1Q

+2Q

+3)

Output(z)

Input (x)

0 1

A 000 000 001 0

B 001 010 101 0

C 010 011 000 1

D 011 100 010 0

E 100 011 101 0

F 101 100 011 1

Excitation table:



Output(z)

Input (x)

0 1

000 0-,0-,0- 0-,0-,1- 0

001 0-,1-,-1 1-,0-,-0 0

010 0-,-0,1- 0-,-1,0- 1

011 1-,-1,-1 0-,-0,-1 0

100 -1,1-,1- -0,0-,1- 0

101 -0,0-,-1 -1,1-,-0 1

- 7.54 -

7.25. (continued)

- 7.55 -

7.25. (continued)

- 7.56 -

7.25. (continued)

(b) Use the state assignment established in Problem

7.24(b).

Excitation table:



Output(z)

Input (x)

0 1

000 0-,0-,0- 0-,0-,1- 0

001 1-,1-,-0 1-,1-,-1 0

111 -1,-0,-1 -1,-1,-1 1

010 0-,-0,1- 1-,-0,1- 0

011 0-,-0,-1 1-,-0,-1 0

110 -1,-0,1- -1,-0,0- 1

- 7.57 -

7.25. (continued)

Documents

Chapter 7