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
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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.
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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
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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.
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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.
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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)