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91

Unit 8 Solutions

Unit 8 Problem Solutions

8.1

8.2 (a)

(cont.)

8.2 (a)

Y

X

WA B

C D 00 01 11 10

00

1 1

A B

C D 00 01 11 10

00

01

0

0

0

0

8.2 (b)

A B

C D 00 01 11 10

00

1 1

8.2 (c) A B

C D 00 01 11 10

00

01

0

0

0

0

8.3 (a)

F

E

C

A B

C D 00 01 11 10

00

01

1 1

Modified circuit (to avoid hazards)

8.3 (b) A = 1; B = Z; C = 1 ⋅ Z = X; D = 1 + Z = 1;

E = X' = X; F = 1' = 0; G = X ⋅ 0 = 0;

H = X + 0 = X

See FLD Table 8-1, p. 241.

8.4

A = B = 0, C = D = 1

8.5

92

Unit 8 Solutions

1

1

1

11

A B

C D 00 01 11 10

00

01

B

Hazard-free AND-OR circuit function:

f(A, B, C, D) = BD + A'C + AC'D + B'C'D' +

A'B'D' + AB'C'

8.6 (a)

G

F

C

8.6 (b)

A

8.7 (a)

0

0

0

0

a b

c d 00 01 11 10

00

01

8.7 (b)

a

d'

c'

The minimal POS expression for f is f(a,b,c,d)

= (a + d')(b' + d)(c' + d') but (a + b') and (b' + c')

must be added to eliminate the static-0 hazards.

8.8 (a)

1

1

1

11

A B

C D 00 01 11 10

00

01

B

D

A'

C

8.8 (a)

cont.

93

Unit 8 Solutions

8.8 (a)

cont.

1

1

1

11

A B

C D 00 01 11 10

00

01

B

D

A'

C

B

D

A'

A

C'

B'

8.8 (b) A B

C D 00 01 11 10

00

01

11

0

0

0

0

F

C

D

B'

B

C'

A'

B'

D

A'

8.8 (a)

cont.

8.8 (b)

cont.

A B

C D 00 01 11 10

00

01

11

10

0

0

0

0

0

0

F = (A + B + C + D')(B'+ C + D )

(A'+ B + C')(A'+ C'+ D )

Static-0 Hazard: 1100↔1110

A

D'

BC

C

B'

Hazard-free OR-AND circuit function:

f(A, B, C, D) = (A + B + C + D')(B'+ C + D)

(A'+ B + C') (A'+ B'+ D) (A'+ C'+ D)

A

BC

94

Unit 8 Solutions

8.10

X

W

1

1

1

1

A B

C D 00 01 11 10

00

01

A

B

8.11 (a)

cont.

f = (A + B)(B'C' + BD')

= AB'C' + ABD' + BB'C' + BD'

= (A + B)(B' + B)(B' + D')(B + C')(C' + D')

From the Karnaugh map and the BB'C' term

8.11 (a)

8.9 (a) A B

C D 00 01 11 10

00

01

1

1

1

1

1

f = (A'B'+AC')(A+D) = AA'B' + AC' + A'B'D +

AC'D) = AA'B' + AC' + A'B'D

static-1 hazard: 0001↔1001

static-0 hazard: 0010↔1010

potential dynamic hazards:

A B

C D 00 01 11 10

00

0

0

C

A

8.9 (b) Since a circuit with NOR gates is desired, start with

POS expressions for f that corresponds to a hazard-

free OR-AND (NOR-NOR) circuit. From the Kar-

naugh map, all prime implicants are required,

f = (A'+C')(A+B')(A+D)(C'+D)(B'+C').

f can be multiplied out as f = (A'B'D+C')(A+B'D).

When this expression is expanded to a POS, it does

not contain any sum of the form (X + X' + ß) so the

corresponding circuit is free of hazards. The three

level NOR circuit is.

95

Unit 8 Solutions

8.12

X

W

C D 00 01 11 10

00

01

11

1 1

1

1

A B

C D 00 01 11 10

00

01

0

0

0

0

D

C

B

f

From the Karnaugh map for f, it is seen that a hazard-free POS expression for f requires all prime implicants.

f = (A + B)(B' + D')(B + C')(C' + D')(A + D')

f can be multiplied out as f = (A + B)(B' + D')(B + C')(C' + D')(A + D') = (AC' + B)(AB'C' + D')

8.11 (a)

cont.

A = Z; B = 0; C = Z' = X; D = Z ⋅ 0 = 0;

E = Z; F = 0 + 0 + X = X; G = (0 ⋅ Z)' = 0' = 1;

8.14 A = B = C = 1, so F = (A + B' + C') (A' + B +

C') (A' + B' + C) = 1. But, in the figure, gate 4

outputs F = 0, indicating something is wrong. For

8.15

8.16 (a)

A B

C D 00 01 11 10

00

01

1

1

1

1

1

1

A B

C D 00 01 11 10

00

01

1

1

1

1

1

1

A B

C D 00 01 11 10

00

01

1

1

1

1

1

1

F (A, B, C, D) = ∑ m(0, 2, 5, 6, 7, 8, 9, 12, 13, 15)

There are 3 different minimum AND-OR solutions to this problem. The problem asks for any two of these.

96

Unit 8 Solutions

A B

C D 00 01 11 10

00

01

0

0

A B

C D 00 01 11 10

00

01

0

0

8.16 (b)

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