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SOLUTIONS MANUAL
DIGITAL DESIGN
FOURTH EDITION
M. MORRIS MANO
California State University, Los Angeles
MICHAEL D. CILETTI
University of Colorado, Colorado Springs
rev 01/21/
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1.8 (a) Results of repeated division by 2 (quotients are followed by remainders):
(b) Results of repeated division by 16:
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(b) 1800 o 01800 o 98199 (9s comp) o 98200 (10 comp) from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
- CHAPTER
- 1.1 Base-10:
- Octal:
- Hex: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
- Base-13 A B C
- 1.2 (a) 32,768 (b) 67,108,864 (c) 6,871,947,
- 1.3 (4310) 5 = 4 * 5 3 + 3 * 5 2 + 1 * 5 1 =
- (198) 12 = 1 * 12 2 + 9 * 12 1 + 8 * 12 0 =
- (735) 8 = 7 * 8 2 + 3 * 8 1 + 5 * 8 0 =
- (525) 6 = 5 * 6 2 + 2 * 6 1 + 5 * 6 0 =
- 1.4 14-bit binary: 11_1111_1111_
- Decimal: 2 14 -1 = 16,383
- Hexadecimal: 3FFF
- (a) 14/2 = (b + 4)/2 = 5, so b = 1.5 Let b = base
- (b) 54/4 = (5*b + 4)/4 = b + 3, so 5 * b = 52 – 4, and b =
- (c) (2 *b + 4) + (b + 7) = 4b, so b =
- 1.6 (x – 3)(x – 6) = x 2 –(6 + 3)x + 6*3 = x 2 -11x +
- Therefore: 6 + 3 = b + 1m so b =
- Also, 6*3 = (18) 10 = (22)
- 1.7 68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)
- Answer: 1111_1010 2 = FA 431 10 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
- Answer: FA = 1111_ 431 10 = 26(15); 1(10) (Faster)
- 1.9 (a) 10110.0101 2 = 16 + 4 + 2 + .25 + .0625 = 22.
- (b) 16.5 16 = 16 + 6 + 5*(.0615) = 22.
- (c) 26.24 8 = 2 * 8 + 6 + 2/8 + 4/64 = 22.
- 27 10 =
- .315 x 2 = 0 + .630 a -1 = Integer Fraction Coefficient
- .630 x 2 = 1 + .26 a -2 =
- .26 x 2 = 0 + .52 a -3 =
- .52 x 2 = 1 + .04 a -4 =
- .315 10 # .0101 2 = .25 + .0625 =.
- 27.315 # 11011.0101
- (b) 2/3 #.
- .6666_6666_67 x 2 = 1 + .3333_3333_34 a -1 = Integer Fraction Coefficient
- .3333333334 x 2 = 0 + .6666666668 a -2 =
- .6666666668 x 2 = 1 + .3333333336 a -3 =
- .3333333336 x 2 = 0 + .6666666672 a -4 =
- .6666666672 x 2 = 1 + .3333333344 a -5 =
- .3333333344 x 2 = 0 + .6666666688 a -6 =
- .6666666688 x 2 = 1 + .3333333376 a -7 =
- .3333333376 x 2 = 0 + .6666666752 a -8 =
- .6666666667 10 # .10101010 2 = .5 + .125 + .0313 + ..0078 = .6641
- 1.14 (a) 1000_0000 (b) 0000_0000 (c) 1101_ .101010102 = .1010_1010 2 = .AA 16 = 10/16 + 10/256 = .6641 10 (Same as (b)). - 1s comp: 0111_1111 1s comp: 1111_1111 1s comp: 0010_ - 2s comp: 1000_0000 2s comp: 0000_0000 2s comp: 0010_
- (d) 0111_0110 (e) 1000_0101 (f) 1111_
- 1s comp: 1000_1001 1s comp: 0111_1010 1s comp: 0000_
- 2s comp: 1000_1010 2s comp: 0111_1011 2s comp: 0000_
- 1.15 (a) 52,784,630 (b) 63,325, - 9s comp: 47,215,369 9s comp: 36,674, - 10s comp: 47,215,370 10s comp: 36,674,
- (c) 25,000,000 (d) 00,000,
- 9s comp: 74,999,999 9s comp: 99,999,
- 10s comp: 75,000,000 10s comp: 00,000,
- 1.16 B2FA B2FA: 1011_0010_1111_ - 15s comp: 4D05 1s comp: 0100_1101_0000_ - 16s comp: 4D06 2s comp: 0100_1101_0000_0110 = 4D - 0 6428 – 03409 = 06428 + 96591 = 1.17 (a) 3409 o 03409 o96590 (9s comp) o 96591 (10s comp) - Magnitude: 125 – 1800 = 00125 + 98200 = 98325 (negative) - Result: 125 – 1800 =
(c) 6152 o 06152 o 93847 (9s comp) o 93848 (10s comp) 2043 – 6152 = 02043 + 93848 = 95891 (Negative) Magnitude: 4109 Result: 2043 – 6152 = -
(d) 745 o 00745 o 99254 (9s comp) o 99255 (10s comp) 1631 -745 = 01631 + 99255 = 0886 (Positive) Result: 1631 – 745 = 886
1.18 Note: Consider sign extension with 2s complement arithmetic.
(a) 10001 ( b) 100011 1s comp: 01110 1s comp: 1011100 with sign extension 2s comp: 01111 2s comp: 1011101 10011 0100010 Diff: 00010 1111111 sign bit indicates that the result is negative 0000001 2s complement -000001 result
(c) 101000 ( d ) 10101 1s comp: 1010111 1s comp: 1101010 with sign extension 2s comp: 1011000 2s comp: 1101011 001001 110000 Diff: 1100001 (negative) 0011011 sign bit indicates that the result is positive 0011111 (2s comp) Check: 48 -21 = 27 -011111 (diff is -31)
1.19 +9286 o 009286; +801 o 000801; -9286 o 990714; -801 o 999199
(a) (+9286) + (_801) = 009286 + 000801 = 010087
(b) (+9286) + (-801) = 009286 + 999199 = 008485
(c) (-9286) + (+801) = 990714 + 000801 = 991515
(d) (-9286) + (-801) = 990714 + 999199 = 989913
1.20 +49 o 0_110001 (Needs leading zero indicate + value); +29 o 0_011101 (Leading 0 indicates + value) -49 o 1_001111; -29 o 1_
(a) (+29) + (-49) = 0_011101 + 1_001111 = 1_101100 (1 indicates negative value.) Magnitude = 0_010100; Result (+29) + (-49) = -
(b) (-29) + (+49) = 1_100011 + 0_110001 = 0_010100 (0 indicates positive value) (-29) + (+49) = +
(c) Must increase word size by 1 (sign extension) to accomodate overflow of values: (-29) + (-49) = 11_100011 + 11_001111 = 10_110010 (1 indicates negative result) Magnitude: 1_001110 = 78 (^10) Result: (-29) + (-49) = -
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1.27 For a deck with 52 cards, we need 6 bits (32 < 52 < 64). Let the msb's select the suit (e.g., diamonds, hearts, clubs, spades are encoded respectively as 00, 01, 10, and 11. The remaining four bits select the "number" of the card. Example: 0001 (ace) through 1011 (9), plus 101 through 1100 (jack, queen, king). This a jack of spades might be coded as 11_1010. (Note: only 52 out of 64 patterns are used.)
1.28 G (dot) (space) B o o l e 01000111_11101111_01101000_01101110_00100000_11000100_11101111_
1.29 Bill Gates
1.30 73 F4 E5 76 E5 4A EF 62 73
73: 0_111_0011 s F4: 1_111_0100 t E5: 1_110_0101 e 76: 0_111_0110 v E5: 1_110_0101 e 4A: 0_100_1010 j EF: 1_110_1111 o 62: 0_110_0010 b 73: 0_111_0011 s
1.31 62 + 32 = 94 printing characters
1.32 bit 6 from the right
1.33 (a) 897 (b) 564 (c) 871 (d) 2,
1.34 ASCII for decimal digits with odd parity:
(0): 10110000 (1): 00110001 (2): 00110010 (3): 10110011 (4): 00110100 (5): 10110101 (6): 10110110 (7): 00110111 (8): 00111000 (9): 10111001
1.35 (a) a b c
f
g
a b c f g 1. a b
f
g
a b
f g
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CHAPTER 2
2.1 (a)
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
x + y + z
0 1 1 1 1 1 1 1
(x + y + z)'
1 0 0 0 0 0 0 0
x'
1 1 1 1 0 0 0 0
y'
1 1 0 0 1 1 0 0
z'
1 0 1 0 1 0 1 0
x' y' z'
1 0 0 0 0 0 0 0
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
(xyz)
0 0 0 0 0 0 0 1
(xyz)'
1 1 1 1 1 1 1 0
x'
1 1 1 1 0 0 0 0
y'
1 1 0 0 1 1 0 0
z'
1 0 1 0 1 0 1 0
x' + y' + z'
1 1 1 1 1 1 1 0
(b) (c)
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
x + yz
0 0 0 1 1 1 1 1
(x + y)
0 0 1 1 1 1 1 1
(x + z)
0 1 0 1 1 1 1 1
(x + y)(x + z)
0 0 0 1 1 1 1 1
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
x(y + z)
0 0 0 0 0 1 1 1
xy
0 0 0 0 0 0 1 1
xz
0 0 0 0 0 1 0 1
xy + xz
0 0 0 0 0 1 1 1
(c) (d)
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
x 0 0 0 0 1 1 1 1 y + z
0 1 1 1 0 1 1 1
x + (y + z)
0 1 1 1 1 1 1 1
(x + y)
0 0 1 1 1 1 1 1
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
yz
0 0 0 1 0 0 0 1
x(yz)
0 0 0 0 0 0 0 1
xy
0 0 0 0 0 0 1 1
(xy)z
0 0 0 0 0 0 0 1
(x + y) + z
0 1 1 1 1 1 1 1
2.2 (a) xy + xy' = x(y + y') = x
(b) (x + y)(x + y') = x + yy' = x(x +y') + y(x + y') = xx + xy' + xy + yy' = x
(c) xyz + x'y + xyz' = xy(z + z') + x'y = xy + x'y = y
(d) (A + B)'(A' + B') = (A'B')(A B) = (A'B')(BA) = A'(B'BA) = 0
(e) xyz' + x'yz + xyz + x'yz' = xy(z + z') + x'y(z + z') = xy + x'y = y
(f) (x + y + z')(x' + y' + z) = xx' + xy' + xz + x'y + yy' + yz + x'z' + y'z' + zz' =
= xy' + xz + x'y + yz + x'z' + y'z' = x y + (x z)' + (y z)'
2.3 (a) ABC + A'B + ABC' = AB + A'B = B
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x y z F (^) simplified
F
(d) B
F
Fsimplified
A 0
(e)
x y z F (^) simplified
F
(f)
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x y z
F
F (^) simplified
2.6 (a) A B C
F
F (^) simplified
(b) x y z
F
Fsimplified
(c) x y
F
Fsimplified
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(b) w x y z
Fsimplified
F
(c) A B C D
Fsimplified
F
(d) A B C D
Fsimplified
F
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(e) A B C D
F
Fsimplified
2.8 F' = (wx + yz)' = (wx)'(yz)' = (w' + x')(y' + z')
FF' = wx(w' + x')(y' + z') + yz(w' + x')(y' + z') = 0 F + F' = wx + yz + (wx + yz)' = A + A' = 1 with A = wx + yz
2.9 (a) F' = (xy' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') = xy + x'y'
(b) F' = [(A'B + CD)E' + E]' = [(A'B + CD) + E]' = (A'B + CD)'E' = (A'B)'(CD)'E' F' = (A + B')(C' + D')E' = AC'E' + A D'E' + B'C'E' + B'D'E'
(c) F' = [(x' + y + z')(x + y')(x + z)]' = (x' + y + z')' + (x + y')' + (x + z)' = F' = xy'z + x'y + x'z'
2.10 (a) F 1 + F 2 = 6 m (^) 1i + 6 m (^) 2i = 6 (m (^) 1i + m (^) 2i)
(b) F1 F2 = 6 m i 6 m j where m i m j = 0 if i z j and m i m j = 1 if i = j
2.11 (a) F(x, y, z) = 6 (1, 4, 5, 6, 7)
(b) F(x, y, z) = 6 (0, 2, 3, 7)
x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
F 0 1 0 0 1 1 1 1
F = xy + xy' + y'z x y z
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
F 1 0 1 1 0 0 0 1
F = x'z' + yz
2.12 A = 1011_
B = 1010_
(a) A AND B = 1010_ (b ) A OR B = 1011_ (c) A XOR B = 0001_
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2.14 (a)
x y
F =xy + x'y' + y'z
z
(b)
x y
F = xy + x'y' + y'z
= ( x' + y' )' + ( x + y )' + ( y + z' )'
z
(c) x y
F = xy + x'y' + y'z
= [(xy)' (x'y')' (y'z)']'
z
(d) x y
F = xy + x'y' + y'z
= [(xy)' (x'y')' (y'z)']'
z
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(e) x y
F = xy + x'y' + y'z
= ( x' + y' )' + ( x + y )' + ( y + z' )'
z
2.15 (a) T 1 = A'B'C' + A'B'C + A'BC' = A'B'(C' + C) +A'C'(B' + B) = A'B' +A'C' = A'(B' + C')
(b) T 2 =T 1 ' = A'BC + AB'C' + AB'C + ABC' + ABC = BC(A' + A) + AB'(C' + C) + AB(C' + C) = BC + AB' + AB = BC + A(B' + B) = A + BC
¦ (3, 5, 6, 7) 3 (0, 1, 2, 4) T 1 = A'B'C' + A'B'C + A'BC'
A'B' A'C' T 1 = A'B' A'C' = A'(B' + C')
T 2 = A'BC + AB'C' + AB'C + ABC' + ABC
AC
T 2 =AC' + BC + AC = A+ BC
BC
AC'
2.16 (a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC = A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC) = (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC = B'(C' + C) + B(C' + C) = B' + B = 1
(b) F(x 1 , x 2 , x 3 , ..., xn ) = 6 m i has 2 n/2 minterms with x 1 and 2 n^ /2 minterms with x' 1 , which can be factored and removed as in (a). The remaining 2n-1^ product terms will have 2 n-1/2 minterms with x 2 and 2 n-1/ minterms with x ' 2 , which and be factored to remove x 2 and x' 2. continue this process until the last term is left and xn + x' (^) n = 1. Alternatively, by induction, F can be written as F = xn G + x' (^) nG with G = 1. So F = (xn + x' (^) n)G = 1.
2.17 (a) (xy + z)(y + xz) = xy + yz + xyz + xz = 6 (3, 5, 6, 7) = 3 (0, 1, 2, 4)
(b) (A' + B)(B' + C) = A'B' + A'C + BC = 6 (0, 1, 3, 7) = 3 (2, 4, 5, 6)
(c) y'z + wxy' + wxz' + w'x'z = 6 (1, 3, 5, 9, 12, 13, 14) = 3 (0, 2, 4, 6, 7, 8, 10, 11, 15)
(d) (xy + yz' + x'z)(x + z) = xy + xyz' + xyz + x'z = 6 (1, 3, 9, 11, 14, 15) = 3 (0, 2, 4, 5, 6, 7, 8, 10, 12, 13)
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2.19 F = B'D + A'D + BD
ABCD
-B'-D 0001 = 1 0011 = 3 1001 = 9 1011 = 11
A'--D 0001 = 1 0011 = 3 0101 = 5 0111 = 7
-B-D 0101 = 5 0111 = 7 1101 = 13 1111 = 15
ABCD ABCD
F = 6 (1, 3, 5, 7, 9, 11,13, 15) = 3 (0, 2, 4, 6, 8, 10, 12, 14)
2.20 (a) F(A, B, C, D) = 6 (3, 5, 9, 11, 15) F'(A, B, C, D) = 6 (0, 1, 2, 4, 6, 7, 8, 10, 12, 13, 14)
(b) F(x, y, z) = 3 (2, 4, 5, 7) F' = 6 (2, 4, 5, 7)
2.21 (a) F(x, y, z) = 6 (2, 5, 6) = 3 (0, 1, 3, 4, 7)
(b) F(A, B, C, D) = 3 (0, 1, 2, 4, 7, 9, 12) = 6 (3, 5, 6, 8, 10, 11, 13, 14, 15)
2.22 (a) ( AB + C)(B + C'D) = AB + BC + ABC'D + CC'D = AB(1 + C'D) + BC = AB + BC (SOP form) = B(A + C) (POS form)
(b) x' + x(x + y')(y + z') = (x' + x)[x' + (x + y')(y + z')] = = (x' + x + y')(x' + y + z') = x c + y + z c
2.23 (a) B'C +AB + ACD A B C
F
D
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(b) (A + B)(C + D)(A' + B + D)
A B C
F
D
(c) (AB + A'B')(CD' + C'D)
B C D
F
A
(d) A + CD + (A + D')(C' + D)
B C D
F
A
2.24 x y = x'y + xy' and (x y)' = (x + y')(x' + y)
Dual of x'y + xy' = (x' + y)(x + y') = (x y)'
2.25 (a) x| y = xy' z y | x = x'y Not commutative
(x | y) | z = xy'z' z x | (y | z) = x(yz')' = xy' + xz Not associative
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