# Chapter 04 The sum of all min terms that are present in either

Type Homework Help
Pages 9
Words 3434
Textbook Fundamentals of Logic Design 7th Edition
Authors Jr.Charles H. Roth, Larry L Kinney

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31
Unit 4 Solutions
x y z0 z1 z2 z3 z4 z5 z6 z7 z8 z9 z10 z11 z12 z13 z14 z15
0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
2
2
n = 222 = 24 = 16
4.4 (a)
4.4 (b)
Alternate
Solutions
A B C D E F Z
0 0 0 1 1 X31
0 0 1 X2 X2 1 1
4.5
A B C D E F Z
1 These truth table entries were made don't cares because
ABC = 110 and ABC = 010 can never occur
2 These truth table entries were made don't cares because
when F is 1, the output Z of the OR gate will be 1 regardless
4.6 (b) By inspection, G = C when both don’t cares are set to 0.
Unit 4 Problem Solutions
See FLD p. 733 for solution.
A B C D E y z
0 0 0 0 0 (less than 10 gpm) +
4.2 Y = A'B'C'D'E' + AB'C'D'E' + ABC'D'E'
4.2 (a)
4.1
32
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Unit 4 Solutions
Exactly one variable not complemented: F = A'B'C
+ A'BC' + AB'C' = ∑ m(1, 2, 4)
A B C D F
0 0 0 0 0 × 0 = 0 ≤ 2 1
0 0 0 1 0 × 1 = 0 ≤ 2 1
0 0 1 0 0 × 2 = 0 ≤ 2 1
0 0 1 1 0 × 3 = 0 ≤ 2 1
0 1 0 0 1 × 0 = 0 ≤ 2 1
0 1 0 1 1 × 1 = 1 ≤ 2 1
4.8
F(A, B, C, D) = ∑ m(0, 1, 2, 3, 4, 5, 6, 8, 9, 12)
Refer to FLD p. 733 for full term expansion
4.8 (a)
F = abc' + b' (a + a') (c + c') = abc' + ab'c + ab'c' +
a'b'c + a'b'c'; F = ∑ m(0, 1, 4, 5, 6)
4.9 (a)
Remaining terms are maxterms: F = ∏ M(2, 3, 7)
4.9 (b)
Maxterms of F are minterms of F':
F' = ∑ m(2, 3, 7)
4.9 (c)
Minterms of F are maxterms of F':
F' = ∏ M(0, 1, 4, 5, 6)
4.9 (d)
4.11 (a) difference, di = xi yi bi; bi+1 = bi
xi' + xi'yi + bi
yi
xi yi bibi+1 di
0 0 0 0 0
0 0 1 1 1
4.11 (b) di = si; bi+1 is the same as ci+1 with xi replaced by xi'
See FLD p. 734 for solution.
4.10 (a) F = ∑ m(1, 4, 5, 6, 7, 10, 11)
4.10 (b) F = ∏ M(0, 2, 3, 8, 9, 12, 13, 14, 15)
F(a, b, c, d) = (a + b + d) (a' + c) (a' + b' + c') (a + b + c' + d')
4.10
4.7 (a)
4.12
33
Unit 4 Solutions
4.13 A B C D Z
0 0 0 0 1
0 0 0 1 1
0 0 1 0 0
4.14 A B C D Z
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
Z = A'B'C'D' + A'B'C'D + AB'C'D'
+ ABCD' + ABCD + A'BCD
= A'B'C' + ABC + AB'C'D' +
A'BCD
= A'B'C' + ABC + AB'C'D' +
Z = A'BC'D + A'BCD' + A'BCD +
AB'C'D' + AB'C'D + AB'CD'
+ AB'CD
= A'BD + AB'C' + AB'C + A'BCD'
= AB' + A'BD + A'BCD' + A'BC
4.16 Truth Table
x3 x2 x1 x0 z y1 y0
0 0 0 0 0 x x
0 1 1 1 1 1 0
1 0 0 0 1 1 1
1 0 0 1 1 1 1
1 0 1 0 1 1 1
(a) Prime digits are 1, 3, 5, and 7 represented as 0010, 0111, 1011 and 1110. The minterms are A'B'CD', A'BCD,
AB'CD and ABCD'. The don't care minterms are A'B'C'D', A'B'CD, A'BC'D, A'BCD', AB'C'D, AB'CD', ABC'D'
and ABCD.
4.15
(a)
minterms of z: 1, 2, 3, 4, 5, 6,
(b)
maxterms of z: 0
maxterms of y1: 1, 2, 3
don't care maxterm: 0
maxterms of y0: 1, 4, 5, 6, 7
Truth Table
A B C D W X Y Z
0 0 0 0 1 0 0 1
0 1 1 1 0 0 1 0
1 0 0 0 0 0 0 1
1 0 0 1 0 0 0 0
1 0 1 0 x x x x
4.17
(a)
minterms of w: 0, 1
7, 8, 9
maxterms of x: 0, 1, 6, 7, 8, 9
maxterms of y: 0, 1, 4, 5, 8, 9
maxterms of z: 1, 3, 5, 7, 9
don't care maxterms: 10, 11,
Unit 4 Solutions
The buzzer will sound if the key is in the ignition switch and the car door is open, or the seat belts are not fastened.
B K D S'
The two possible interpretations are: B = KD + S' and B = K(D + S')
4.19 (a)
The roads will be very slippery if it snows or it rains and there is oil on the road.
V S R O
The two possible interpretations are: V = (S + R) O and V = S + RO
4.19 (d)
Z = AB + AC + BC
4.20 Z = (ABCDE + A'B'C'D'E')'; Y = A'B'CD'E
4.21
Truth Table
A B C D W X Y Z
0 0 0 0 1 1 1 1
0 1 0 1 x x x x
0 1 1 0 1 0 0 1
0 1 1 1 x x x x
1 0 0 0 x x x x
1 0 0 1 0 1 1 0
4.18 (a),
(b)
(a)
minterms of W: 0, 1, 2, 3, 6
(b)
maxterms of W: 9, 12, 13,
14, 15
maxterms of X: 6, 12, 13, 14,
15
maxterms of Y: 2, 3, 6, 14,
Alternative Truth Table
A B C D W X Y Z
0 0 0 0 1 1 1 1
0 1 0 1 x x x x
0 1 1 0 x x x x
0 1 1 1 1 0 0 0
1 0 0 0 0 1 1 1
1 0 0 1 x x x x
4.18 (a),
(b)
(a)
minterms of W: 0, 1, 2, 3, 7
(b)
maxterms of W: 8, 12, 13,
14, 15
maxterms of X: 7, 12, 13,
14,15
maxterms of Y: 2, 3, 7, 14,
F1F2 = ∏ M(0, 4, 5, 6, 7). General rule: F1F2 is the product of all maxterms that are present in either F1 or F2.
Proof:
4.23
35
Unit 4 Solutions
A B C D F G H J
0 0 0 0 0 1 0 0
0 0 0 1 0 0 0 0
0 0 1 0 0 1 0 0
0 0 1 1 0 0 0 0
0 1 0 0 0 1 0 1
0 1 0 1 1 0 0 0
4.25 (a) F(A, B, C, D) =
m(5, 6, 7, 10, 11, 13, 14, 15)
= ∏ M(0, 1, 2, 3, 4, 8, 9, 12)
(b) G (A, B, C, D) =
m(0, 2, 4, 6)
= ∏ M(1, 3, 5, 7, 8, 9, 10, 11,
12, 13, 14, 15)
4.27 You can also work this problem using a truth table,
as in problem 4.28.
f(a, b, c) = a(b + c') = ab + ac' = ab (c + c') +
a(b + b') c' = abc + abc' + abc' + ab'c'
m7 m6 m6 m4
a b c d f
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
4.28 (a) f = ∑ m(1, 2, 4, 5, 6, 11, 12,
14, 15)
(b) f = ∏ M(0, 3, 7, 8, 9, 10, 13)
algebraically, as in problem 4.27.
F1 + F2 = ∏ M(0, 4). General rule: F1 + F2 is the product of all maxterms that are present in both F1 and F2.
Proof:
4.24
A B C D F G H J
0 0 0 0 0 1 0 0
0 0 0 1 0 1 0 0
0 0 1 0 0 1 0 0
0 0 1 1 0 0 0 0
0 1 0 0 0 1 0 1
0 1 0 1 1 0 0 0
4.26 (a) F(A, B, C, D) =
m(5, 7, 10, 11, 13, 14, 15)
= ∏ M(0, 1, 2, 3, 4, 6, 8, 9,
12)
(b) G (A, B, C, D) =
m(0, 1, 2, 4, 8)
= ∏ M(3, 5, 6, 7, 9, 10, 11, 12,
13, 14, 15)
36
Unit 4 Solutions
F(A, B, C, D) = ∑ m(3, 4, 5, 8, 9, 10, 11, 12, 14)
F = A'B'CD + A'BC'D' + A'BC'D + AB'C'D' +
AB'C'D + AB'CD' + AB'CD + ABC'D' + ABCD'
4.30 (a) 4.30 (b) F(A, B, C, D) = ∏ M(0, 1, 2, 6, 7, 13, 15)
F = (A + B + C + D)(A + B + C + D')
(A + B + C' + D)(A + B' + C' + D)
(A + B' + C' + D')(A' + B' + C + D')
(A' + B' + C' + D')
4.32 (a) If don't cares are changed to (1, 1), respectively,
A B C D E F Z
0 0 0 1 1 X20
4.33 1 These truth table entries
4.32 (b) If don't cares are changed to (1, 0), respectively
4.32 (c) If don't cares are changed to (1, 1), respectively
F3 = (A + B + C) (A + B + C') = A + B
4.32 (d) If don't cares are changed to (0, 1), respectively
F4 = A'B'C' + A'BC + AB'C' + ABC
= B'C' + BC
G2(A, B, C) = ∑ m(0, 1, 6, 7) = ∏ M(2, 3, 4, 5)
4.34 (a) G1(A, B, C) = ∑ m(0, 7) = ∏ M(1, 2, 3, 4, 5, 6) 4.34 (b)
f(A,B,C,D) = AB+ A'CD = ABC'D' + ABC'D
+ ABCD' + ABCD + A'B'CD + A'BCD
= (A+A'CD)(B+A'CD) = (A+C)(A+D)(A'+B)
(B+C)(B+D)
f(A,B,C,D) = (A+B'+C+D')(A+B'+C+D)
4.29 (a) f(A,B,C,D) = (A+B+D')(A'+C)(C+D)
= (A+B+D')(A'D+C) = AC+A'BD+BC+CD'
= AC(B+B')(D+D')+A'BD(C+C')
+BC(A+A')(D+D')+(A+A')(B+B')CD'
= ABCD+ABCD'+AB'CD+AB'CD'+A'BCD
4.29 (b)
37
Unit 4 Solutions
A B C D S T U V W X Y Z
0 0 0 0 0 × 5 = 00 0 0 0 0 0 0 0 0
0 0 0 1 1 × 5 = 05 0 0 0 0 0 1 0 1
0 0 1 0 2 × 5 = 10 0 0 0 1 0 0 0 0
0 0 1 1 3 × 5 = 15 0 0 0 1 0 1 0 1
4.37 A B C D S T U V W X Y Z
0 0 0 0 0 × 4 + 1 = 01 0 0 0 0 0 0 0 1
0 0 0 1 1 × 4 + 1 = 05 0 0 0 0 0 1 0 1
0 0 1 0 2 × 4 + 1 = 09 0 0 0 0 1 0 0 1
0 0 1 1 3 × 4 + 1 = 13 0 0 0 1 0 0 1 1
4.38
4.35 (a) A B C D 1's X Y Z
0 0 0 0 0 0 0 0
0 1 1 0 2 0 1 0
0 1 1 1 3 0 1 1
1 0 0 0 1 0 0 1
1 0 0 1 2 0 1 0
4.36 (a) A B C D W X Y Z
0 0 0 0 0 0 1 1
0 1 1 0 0 1 0 1
0 1 1 1 0 1 1 0
1 0 0 0 0 1 0 0
1 0 0 1 0 1 0 1
1 1 1 1 0 1 1 1
X = ABCD
Y = A'B'CD + A'BC'D +
A'BC'D' + A'BCD +
AB'C'D' + AB'CD +
ABC'D + ABCD'
Y = (A + B + C + D) (A + B + C + D')
(A + B + C' + D) (A + B' + C + D)
4.35 (b) Y = (A + B + C + D') (A + B + C' + D)
(A + B + C' + D') (A + B' + C + D)
4.36 (b)
X = A'B'C'D + A'B'CD'
+ A'B'CD + A'BC'D'
Y = A'B'C'D' + A'BCD +
ABC'D + ABCD' +
ABCD
Unit 4 Solutions
Notice that the sign bit X3 of the 4-bit number is
extended to the leftmost full adder as well.
4.39
X Y Sum Cout
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
4.40
S3
S2
S1
S0
S4
S3
S1
S0
3
S2
X
Y
Sum
Cout
4.41 (a),
(b), (c)
(a)
f = x(y+y')+y(x+x') = xy+xy'+x'y
(sum-of-minterms)
f = x+y already in product-of-maxterms form
(b)
f = ax+by = ax(y+y')+by(x+x')
= axy+axy'+bxy+bx'y = (a+b)xy+axy'+bx'y
= xy+axy'+bx'y
Alternatively,
f = ax+by = (a+by)(x+by) = (a+b)(a+y)(x+b)(x+y)
= (a+xx'+y)(b+yy'+x)(x+y)
= (a+x+y)(a+x'+y)(b+x+y)(b+x+y')(x+y)
= [(a+x+y)(b+x+y)(x+y)](a+x'+y)(b+x+y')
= (ab+x+y)(a+x'+y)(b+x+y')
= (x+y)(a+x'+y)(b+x+y')
(d)
x y f
4.41 (d),
(e)
(e)
f(x,y) is completely
(a) m1 + m2 = m1(m2' + m2) + (m1' + m1)m2
= m1m2' + m1m2 + m1'm2
4.42
x y f
39
Unit 4 Solutions
(a) Sum = X Y Cin and
Cout = X′YCin + XY′Cin + XYC′in + XYCin
= (X′Y + XY′)Cin + XY(C′in + Cin)
= (X Y)Cin + XY
4.43
bi+1 = (xi + yi + bi)(xi′ + yi + bi)
(xi′ + yi + bi′)(xi′ + yi′ + bi)
= [(xi + yi)(xi′ + yi′) + bi](xi′ + yi)
= [(xi yi) + bi](xi′ + yi)
4.44
(a) si = ai bi E1′ci E1E0aibi and
ci+1 = (E0 ai)bi + (E0 ai)ci + bici
E1E0sici+1
00si = ai bi cici+1 = aibi + aici
+ bici
01si = ai bi cici+1 = ai′bi +
ai′ci + bici
(b)
E1E0Function
4.45
the outputs of the NANDs can be moved forward
through the XOR’s where they cancel.
(a) si = ai bi E1ci E1′E0′ai′bi′ and
ci+1 = (E0 ai)bi + (E0 ai)ci + bici
E1E0sici+1
00si = ai′ + bici+1 = aibi + aici
1 1 si = ai bi cici+1 = ai′bi +
ai′ci + bici
(b)
1 0 Add (A + B)
1 1 Subt (A - B)
4.47
(a) si = ai bi (E1′ + ci′) (E1 + E0 + ai + bi)
and ci+1 = (E0 ai)bi + (E0 ai)ci + bici
E1E0sici+1
00si = ai′ + bici+1 = aibi + aici
+ bici
01si = ai bici+1 = ai′bi +
ai′ci + bici
4.48
40
Unit 4 Solutions

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