# Chapter 05 Indicates a min term that makes the corresponding

Type Homework Help
Pages 10
Words 3630
Authors Jr.Charles H. Roth, Larry L Kinney

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Unit 5 Solutions
41
Unit 5 Problem Solutions
5.3 (a) a
b c 0 1
00
1
5.3 (b) 5.3 (c) r
s t 0 1
00
1
1
5.3 (d) x
y z 0 1
00
0
1
5.4 (a) A B
C D 00 01 11 10
00
01
1
0
1
0
1
0
1
1
5.4 (b) 5.4 (c)
d
e f 0 1
00
1
1
A B
C D 00 01 11 10
00
01
1
1
1
1
1
A B
C D 00 01 11 10
00
01
1
1
1
1
1
See FLD p. 735
for solution.
C C
X X 00 01 11 10
00
01
0
1
0
1
0
0
1
0
1 2
1 2
2 1 1 212 11 11
Z = C 'X 'X + C 'X X ' + C C X 'X ' + C X X + C 'C 'X
2 2 1 221
5.5 (b)
5.6 (a)
*
a b
c d 00 01 11 10
00
01
1
1
1
1
*
5.6 (b) a b
c d 00 01 11 10
00
01
1
0
1
1
1
0
0
1
*
5.5 (a)
Unit 5 Solutions
42
5.7 (b) a b
c d 00 01 11 10
00
01
X
1
1
5.7 (c) a b
c d 00 01 11 10
00
01
1
0
0
1
1
0
1
1
5.7 (d) A B
C D 00 01 11 10
00
01
0
X
0
1
0
X
X
1
a b
c d 00 01 11 10
00
01
0
0
1
1
0
1
0
1
5.8 (a)
5.8 (b) a b
c d 00 01 11 10
00
01
0
1
1
0
X
0
X
0
a b
c d 00 01 11 10
00
01
0
0
1
1
0
1
0
1
a b
c d 00 01 11 10
00
01
0
1
1
0
X
0
X
0
5.6 (c) a b
c d 00 01 11 10
00
01
1
X
1
0
1
X
1
0
*
a b
c d 00 01 11 10
00
01
1
1
1
5.7 (a)
Unit 5 Solutions
43
5.9 (a) B C
D E
A
00 01 11 10
00
01
1
1
X
0
0
1
X
0
1
0
1
1
0
0
0
0
5.9 (b) B C
D E
A
00 01 11 10
00
01
0
X
1
1
1
0
X
X
1
1
0
1
0
0
0
0
B C
D E
A
00 01 11 10
00
01
1
1
X
0
0
1
X
0
1
0
1
1
0
0
0
0
B C
D E
A
00 01 11 10
00
01
0
X
1
1
1
0
X
X
1
1
0
1
0
0
0
0
5.10 (a) b c
d e
a
00 01 11 10
00
01
1
1
1
11
1
1
1
1
1
5.10 (b) b c
d e
a
00 01 11 10
00
01
1
1
1
11
1
1
1
1
1
5.14 (a) A
B C 0 1
00
d
e f 0 1
00
5.14 (b) r
s t 0 1
00
1
1
5.14 (c) a
b c 0 1
00
1
5.14 (d)
A B
C D 00 01 11 10
00
01
11
0
0
1
1
1
1
1
0
1
0
0
1
5.12 (a)
5.11
0
b c
d e
a
00 01 11 10
00
01
0
X
1
1
1
0
1
0
1
0
0
0
0
1
1
A B
C D 00 01 11 10
00
01
1
1
0
0
0
1
1
1
F' = A'B'C' + A B D' + A C'D
5.12 (b)
A B
C D 00 01 11 10
00
01
0
0
1
1
1
0
0
0
5.12 (c)
5.13 A B
C D 00 01 11 10
00
01
1
1
1
1
1
F = A C D' + B C'D + B'C + A'C'
Minterms m0, m1, m2, m3, m4, m10, and m11 can be
made don’t cares, individually, without changing
the given expression. However, if m13 or m14 is
made a don’t care, the term BC'D or the term ACD'
(respectively) is not needed in the expression.
Unit 5 Solutions
A
B C 0 1
00
01
1
1
5.16 (a) d
e f 0 1
00
01
1
1
1
5.16 (b) 5.16 (c) r
s t 0 1
00
01
1
1
1
5.16 (d) a
b c 0 1
00
01
1
x
y z 0 1
00
1
1
5.14 (f)
n
p q 0 1
00
1
5.14 (e) 5.15 (a) A
B C 0 1
00
0
0
d
e f 0 1
00
0
0
r
s t 0 1
00
01
5.15 (c) a
b c 0 1
00
01
0
0
0
5.15 (b)
5.15 (d) n
p q 0 1
00
01
0
5.15 (e) x
y z 0 1
00
01
0
5.15 (f)
5.16 (e) n
p q 0 1
00
x
y z 0 1
00
1
1
5.16 (f) 5.17 (a)
& (b)
A B
C D 00 01 11 10
00
1
Unit 5 Solutions
5.20 (e) a
b c 0 1
00
01
1
1 1
5.20 (f) D
E F 0 1
00
01
X
1
X
5.20 (d) s
t u 0 1
00
01
X
1
X
5.18 (a) & (b)
A B
C D 00 01 11 10
00
01
1
1
1
1
1
1
0 0
A B
C D 00 01 11 10
00
01
1
1
1
1
0
1
0
1
F = (A'+ C + D ) (A'+ C'+ D') (A'+ B'+ C')
Alt:
5.18 (c)
5.17 (c)
A B
C D 00 01 11 10
00
01
1
1
0
0
0
0
0
0
5.20 (a)
a
5.20 (b)
d
5.20 (c) p
q r 0 1
00
01
1
1
1
5.19 (a) 5.19 (b)
C1 C2 X1 X2 Z
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
1 0 1 1 1
1 1 0 0 1
1 1 0 1 0
1 1 1 0 0
C C
X X 00 01 11 10
00
01
11
0
0
1
0
1
0
1
1
1
1
0
1
1 2
1 2
Unit 5 Solutions
47
a b
c d 00 01 11 10
00
01
1
1 1
1
1
5.21
A B
C D 00 01 11 10
00
01
1
X
X
X
1
1
5.22 (a)
A B
C D 00 01 11 10
00
01
X
X
X
1
1
5.22 (b)
A B
C D 00 01 11 10
00
01
1
1
X
X
X
1
5.22 (c)
A B
C D 00 01 11 10
00
01
1
X
X
X
5.22 (d)
A B
C D 00 01 11 10
00
01
1
X
X
X
5.22 (e)
A B
C D 00 01 11 10
00
01
1
X
X
X
X
X
5.22 (f)
A B
C D 00 01 11 10
00
01
X
X
1
X
X
X
5.22 (g)
A B
C D 00 01 11 10
00
01
0
0
X
X
X
5.23 (a)
A B
C D 00 01 11 10
00
01
0
0
0
X
X
X
5.23 (b)
A B
C D 00 01 11 10
00
01
0
X
X
X
0
5.23 (c)
A B
C D 00 01 11 10
00
01
0
0
X
X
X
0
0
5.23 (d)
Unit 5 Solutions
48
A B
C D 00 01 11 10
00
01
1
1
1
1
5.24 (a) 5.24 (b)
A B
C D 00 01 11 10
00
01
X
1
1
5.24 (c)
A B
C D 00 01 11 10
00
01
1
X
1
X
1
5.24 (d)
w x
y z 00 01 11 10
00
01
1
X
1
1
1
1
1
A B
C D 00 01 11 10
00
01
0
0
X
0
1
0
1
X
5.24 (e)
A B
C D 00 01 11 10
00
01
0
0
X
X
X
X
X
5.23 (f)
A B
C D 00 01 11 10
00
01
X
X
X
X
X
0
0
5.23 (g)
A B
C D 00 01 11 10
00
01
0
0
X
X
X
0
0
5.23 (e)
5.25 (b)
a b
c d 00 01 11 10
00
01
0
1
1
0
0
1
1
1
5.25 (c)
a b
c d 00 01 11 10
00
01
1
X
1
1
5.25 (a)
a b
c d 00 01 11 10
00
01
1
1
1
11
5.25 (d)
a b
c d 00 01 11 10
00
01
0
X
1
X
1
0
0
0
Unit 5 Solutions
49
w x
y z 00 01 11 10
00
01
1
X
0
1
1
1
1
1
5.27 (b) w x
y z 00 01 11 10
00
01
1
X
0
1
1
1
1
1
5.26 (a) A B
C D 00 01 11 10
00
01
0
0
0
0
A B
C D 00 01 11 10
00
01
0
1
0
0
1
1
X
0
5.26 (b)
A B
C D 00 01 11 10
00
01
0
1
1
X
1
1
0
1
A B
C D 00 01 11 10
00
01
0
1
1
X
1
1
0
1
5.27 (a)
a b
c d 00 01 11 10
00
01
1
1
X
1
1
1
5.28
5.29 (a)
A B
C D 00 01 11 10
00
01
0
0
1
1
1
0
0
0
A B
C D 00 01 11 10
00
01
1
1
0
0
0
1
1
1
5.29 (b)
Unit 5 Solutions
50
5.29 (c) A B
C D 00 01 11 10
00
01
0
0
1
1
1
0
0
0
5.30 A B
C D 00 01 11 10
00
01
1
X
X
1
Prime implicants for f ': abc'e, ac'd', ab'e', a'ce,
b'c'de', c'd'e, a'd'e
5.31 5.32 For F: b'c'de', a'ce, ab'e', ac'd', abc'e, c'd'e, a'd'e
5.33
a b c
d e 000 001 011 010 110 111 101 100
00
01
1
1
1
1
1
1
1
1
1
b c
d e
0 0
0 1
1 1
1 0
0 0
1
1
1
1
5-variable mirror image map 5-variable diagonal map
5.34 (b)
& (c)
5.34 (d)
& (e)
5.34 (a)
a b
c d 00 01 11 10
00
01
X
1
X
X
X
a b
c d 00 01 11 10
00
01
X
1
X
X
X
a b
c d 00 01 11 10
00
01
X
0
X
0
X
X
0
Unit 5 Solutions
51
5.35 (a),
(b) &
(c)
A B C
D E 000 001 011 010 110 111 101 100
00
01
X
1
X
1
X
X
1
1
X
X
X
X
X
X
X
5-variable mirror image map
B C
D E
0 0
0 1
1 1
1 0
0 0
X
1
X
1
X
X
X
5-variable diagonal map
5.35 (d),
& (e)
A B C
D E 000 001 011 010 110 111 101 100
00
01
X
0
X
X
X
X
X
X
X
X
X
X
5.36 (a),
(b) &
(c)
A B C
D E 000 001 011 010 110 111 101 100
00
01
X
X
X
X
X
X
1
X
X
X
1
X
X
B C
D E
0 0
0 1
1 1
1 0
0 0
X
X
X
X
X
X
5-variable mirror image map 5-variable diagonal map
B C
D E
0 0
0 1
1 1
1 0
0 0
X
0
X
X
X
X
5-variable mirror image map 5-variable diagonal map
5.36 (d),
(e)
A B C
D E 000 001 011 010 110 111 101 100
00
01
X
X
X
X
0
X
X
0
X
X
0
X
X
X
B C
D E
0 0
0 1
1 1
1 0
0 0
X
X
0
X
0
X
X
X
5-variable mirror image map 5-variable diagonal map
Unit 5 Solutions
52
b c
d e
a
00 01 11 10
00
01
1
1
1
X
1
X
1
1
1
1
1
5.39 (a)
5.38 (a) b c
d e
a
00 01 11 10
00
01
1
1
1
1 1
1
1
1
1
1
*
*
*
*
5.38 (b) b c
d e
a
00 01 11 10
00
01
1
1
1
1 1
1
1
1
1
1
5.37 (a),
(b) &
(c)
B C
D E
A
00 01 11 10
00
01
X
1
X
1
X
X
1
1
X
X
X
X
= A'C E + B'C'E + C'D'E'
5.37 (d),
& (e)
B C
D E
A
00 01 11 10
00
01
X
0
X
X
X
X
X
0
X
0
0
X
5.39 (b) b c
d e
a
00 01 11 10
00
01
1
1
1
1
1 1
1
1
Unit 5 Solutions
53
5.42 (a) W X
Y Z
V
00 01 11 10
00
01
0
X
1
0
0
1
1
0
X
X
0
1
0
1
1
1
W X
Y Z
V
00 01 11 10
00
01
0
X
1
0
0
1
1
0
X
X
0
1
0
1
1
1
5.42 (b)
5.40
1
B C
D E
A
00 01 11 10
00
01
1
00
0
1 1
1
1
1 1
1
1
00
0
1
5.41 B C
D E
A
00 01 11 10
00
01
1
1
1
X
1
1 X
1
b c
d e
a
00 01 11 10
00
01
0
1
1
1
0
0
0
0
0
0
0
0
0
1
1
1
5.43 (a) 5.43 (b) b c
d e
a
00 01 11 10
00
01
X
1
X
1
0
0
1
1
1
1
0
1
1
0
0
0
Unit 5 Solutions
5.44 (a) w x
y z
v
00 01 11 10
00
01
0
0
1
1
1
1
1
1
0
0
1
1
0
1
1
0
5.45 (b)
A B
C D 00 01 11 10
00
01
1 1
1
1
1
5.46 (a)
W X
Y Z
V
00 01 11 10
00
01
X
1
1
1
X
X
1
1
1
V'WZ'm8; XYZm31; V'XY'm4
5.46 (b)
5.44 (b) b c
d e
a
00 01 11 10
00
01
X
0
0
0
0
0
0
0
0
5.45 (a)
A B
C D 00 01 11 10
00
01
1
1
1
1
1
5.47 (a) f
a b
c d 00 01 11 10
00
1
1
1
1
Nonessential: ac′, ad, bc′, bd
f = ab′ + ab + cd′ + cd + ad
f = ab′ + ab + cd′ + cd + bd
5.47 (c)
Unit 5 Solutions
55
5.48 (a) f
a b
c d 00 01 11 10
00
01
1
1
1
1
1
1
1
1
Essential: c'
Nonessential: a'b', a'd, b'd', bd, ab, ad'
5.48 (b)
5.49 (a) f
a b
c d 00 01 11 10
00
01
1
1
1
1
1
1
1
1
13, 14)
Essential: b', c', a'd, ad'
f = b' + c' + a'd + ad'
5.49 (b)
5.49 (c)
5.49 (a)
f
a b
c d 00 01 11 10
00
01
1
1
1
1
1
1
1
5.50 (a)
f
a b
c d 00 01 11 10
00
01
1
1
1
1
X
X
1
0
Unit 5 Solutions
56
5.50 (d)
f
a b
c d 00 01 11 10
00
01
1
1
1
1
X
X
1
0

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