ALL 10 CHAPTERS COVERED
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1 Digital Systems and Binary Numbers
2 Boolean Algebra and Logic Gates
3 GateLevel Minimization
4 Combinational Logic
5 Synchronous Sequential Logic
6 Registers and Counters
7 Memory and Programmable Logic
8 Design at the Register Transfer Level
9 Laboratory Experiments with Standard ICs and
FPGAs 10 Standard Graphic Symbols
,© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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CHAPTER 1
1.1 Base-10: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Octal: 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40
Hex: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20
Base-13 A B C 10 11 12 13 14 15 16 17 18 19 23 24 25 26
1.2 (a) 32,768 (b) 67,108,864 (c) 6,871,947,674
1.3 (4310)5 = 4 * 53 + 3 * 52 + 1 * 51 = 58010
(198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010
2 1 0
(735)8 = 7 * 8 + 3 * 8 + 5 * 8 = 477
10
2 1 0
(525)6 = 5 * 6 + 2 * 6 + 5 * 6 = 197 10
1.4 14-bit binary:
11_1111_1111_1111 Decimal:
214 -1 = 16,38310
Hexadecimal: 3FFF16
1.5 Let b = base
(a) 14/2 = (b + 4)/2 = 5, so b = 6
(b) 54/4 = (5*b + 4)/4 = b + 3, so 5 * b = 52 – 4, and b = 8
(c) (2 *b + 4) + (b + 7) = 4b, so b = 11
1.6 (x – 3)(x – 6) = x2 –(6 + 3)x + 6*3 = x2 -11x + 22
Therefore: 6 + 3 = b + 1m so b =
8 Also, 6*3 = (18)10 = (22)8
1.7 68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)8
1.8 (a) Results of repeated division by 2 (quotients are followed by remainders):
43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_10102 = FA16
(b) Results of repeated division by 16:
43110 = 26(15); 1(10) (Faster)
Answer: FA = 1111_1010
1.9 (a) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.516 = 16 + 6 + 5*(.0615) = 22.3125
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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,© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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(d) FAFA.B16 = 15*163 + 10*162 + 15*16 + 10 + 11/16 = 64,250.6875
(e) 1010.10102 = 8 + 2 + .5 + .125 = 10.625
1.10 (a) 1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310
(b) 110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510
Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.
1011.11
1.11 101 | 111011.0000
101
01001
101
1001
101
1000
101
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: = 1011.112 = 58.7510
1.12 (a) 10000 and 110111
1011 1011
+101
x101 10000 = 1610
1011
1011
110111 = 5510
(b) 62h and 958h
2Eh 0010_1110 2Eh
+34h 0011_0100 x34h
62h 0110_0010 = 9810 B38
2
8A
9 5 8h = 239210
1.13 (a) Convert 27.315 to binary:
Integer Remainder Coefficient
Quotient
27/2 = 13 + ½ a0 = 1
13/2 6 + ½ a1 = 1
6/2 3 + 0 a2 = 0
3/2 1 + ½ a3 = 1
½ 0 + ½ a4 = 1
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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2710 = 110112
Integer Fraction Coefficient
.315 x 2 = 0 + .630 a-1 = 0
.630 x 2 = 1 + .26 a-2 = 1
.26 x 2 = 0 + .52 a-3 = 0
.52 x 2 = 1 + .04 a-4 = 1
.31510 .01012 = .25 + .0625 = .3125
27.315 11011.01012
(b) 2/3
.6666666667 Integer Fraction Coefficient
.6666_6666_67 x = 1 + .3333_3333_34 a-1 = 1
2
.3333333334 x2 = 0 + .6666666668 a-2 = 0
.6666666668 x 2 = 1 + .3333333336 a-3 = 1
.3333333336 x 2 = 0 + .6666666672 a-4 = 0
.6666666672 x 2 = 1 + .3333333344 a-5 = 1
.3333333344 x 2 = 0 + .6666666688 a-6 = 0
.6666666688 x 2 = 1 + .3333333376 a-7 = 1
.3333333376 x 2 = 0 + .6666666752 a-8 = 0
.666666666710 .101010102 = .5 + .125 + .0313 + ..0078 = .664110
.101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14 (a) 1000_0000 (b) 0000_0000 (c) 1101_1010
1s comp: 0111_1111 1s comp: 1111_1111 1s comp:
0010_0101 2s comp: 1000_0000 2s comp: 0000_0000
2s comp: 0010_0110
(d) 0111_0110 (e) 1000_0101 (f) 1111_1111
1s comp: 1000_1001 1s comp: 0111_1010 1s comp:
0000_0000 2s comp: 1000_1010 2s comp:
0111_1011 2s comp: 0000_0001
1.15 (a) 52,784,630 (b) 63,325,600
9s comp: 47,215,369 9s comp: 36,674,399
10s comp: 47,215,370 10s comp: 36,674,400
(c) 25,000,000 (d) 00,000,000
9s comp: 74,999,999 9s comp: 99,999,999
10s comp: 75,000,000 10s comp: 00,000,000
1.16 B2FA B2FA: 1011_0010_1111_1010
15s comp: 4D05 1s comp: 0100_1101_0000_0101
16s comp: 4D06 2s comp: 0100_1101_0000_0110 =
1.17 (a) 3409 03409 96590 (9s comp) 4D06 96591 (10s comp)
06428 – 03409 = 06428 + 96591 = 03019
(b) 1800 01800 98199 (9s comp) 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
Result: 125 – 1800 = 1675
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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(c) 6152 06152 93847 (9s comp) 93848 (10s comp)
2043 – 6152 = 02043 + 93848 = 95891 (Negative)
Magnitude: 4109
Result: 2043 – 6152 = -4109
(d) 745 00745 99254 (9s comp) 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 009286; +801 000801; -9286 990714; -801 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 0_110001 (Needs leading zero indicate + value); +29 0_011101 (Leading 0 indicates + valu
-49 1_001111; -29 1_100011
(a) (+29) + (-49) = 0_011101 + 1_001111 = 1_101100 (1 indicates negative value.)
Magnitude = 0_010100; Result (+29) + (-49) = -20
(b) (-29) + (+49) = 1_100011 + 0_110001 = 0_010100 (0 indicates positive
value) (-29) + (+49) = +20
(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 = 7810
Result: (-29) + (-49) = -78
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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,© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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1.21 +9742 009742 990257 (9's comp) 990258 (10s) comp
+641 000641 999358 (9's comp) 999359 (10s) comp
(a) (+9742) + (+641) 010383
(b) (+9742) + (-641) 009742 + 999359 = 009102
Result: (+9742) + (-641) = 9102
(c) -9742) + (+641) = 990258 + 000641 = 990899 (negative)
Magnitude: 009101
Result: (-9742) + (641) = -9101
(d) (-9742) + (-641) = 990258 + 999359 = 989617 (Negative)
Magnitude: 10383
Result: (-9742) + (-641) = -10383
1.22 8,723
BCD: 1000_0111_0010_0011
ASCII: 0_011_1000_011_0111_011_0010_011_0
001
1.23
1000 0100 0010 ( 842)
0101 0011 0111 (+537)
1101 0111 1001
0110
0001 0011 0111 0101
(1,379)
1.24 (a) (b)
6 3 1 1 Decimal 6 4 2 1 Decimal
0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 0 0 0 1 1
0 0 1 0 2 0 0 1 0 2
0 1 0 0 3 0 0 1 1 3
0 1 1 0 4 (or 0101) 0 1 0 0 4
0 1 1 1 5 0 1 0 1 5
1 0 0 0 6 1 0 0 0 6 (or 0110)
1 0 1 0 7 (or 1001) 1 0 0 1 7
1 0 1 1 8 1 0 1 0 8
1 1 0 0 9 1 0 1 1 9
1.25 BCD:
(a) 5,13710 0101_0011_0111
(b) Excess-3:
(c) 1000_0100_0110_1010
(d) 2421:
1011_0001_0011_01
1.26 5,137 9s Comp: 114,862
2421 code: 6311:
0100_1110_1100_1000
1s comp: 0111_0001_0100_10
1011_0001_0011_0111 same as (c) in
01
1.25
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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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
1100101
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) 2,199
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
a
f
b
c
g
f
g
1.36
a b
a
f
b
g
f
g
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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CHAPTER 2
2.1 (a)
xyz x+y+z (x + y + z)' x' y' z' x' y' z' xyz (xyz) (xyz)' x' y' z' x' + y' + z'
000 0 1 1 1 1 1 000 0 1 1 1 1 1
001 1 0 1 1 0 0 001 0 1 1 1 0 1
010 1 0 1 0 1 0 010 0 1 1 0 1 1
011 1 0 1 0 0 0 011 0 1 1 0 0 1
100 1 0 0 1 1 0 100 0 1 0 1 1 1
101 1 0 0 1 0 0 101 0 1 0 1 0 1
110 1 0 0 0 1 0 110 0 1 0 0 1 1
111 1 0 0 0 0 0 111 1 0 0 0 0 0
(b) (c)
xyz x+ (x + y) (x + z) (x + y)(x + z) x y z x(y + z) x xz xy + xz
yz y
000 0 0 0 0 000 0 0 0 0
001 0 0 1 0 001 0 0 0 0
010 0 1 0 0 010 0 0 0 0
011 1 1 1 1 011 0 0 0 0
100 1 1 1 1 100 0 0 0 0
101 1 1 1 1 101 1 0 1 1
110 1 1 1 1 110 1 1 0 1
111 1 1 1 1 111 1 1 1 1
(c) (d)
xyz x y+z x + (y + z) (x + y) (x + y) + z xyz y x(yz) xy
z (xy)z
000 0 0 0 0 0 000 0 0 0 0
001 0 1 1 0 1 001 0 0 0 0
010 0 1 1 1 1 010 0 0 0 0
011 0 1 1 1 1 011 1 0 0 0
100 1 0 1 1 1 100 0 0 0 0
101 1 1 1 1 1 101 0 0 0 0
110 1 1 1 1 1 110 0 0 1 0
111 1 1 1 1 1 111 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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SOLUTIONS MANUAL
,TABLE OF CONTENTS
1 Digital Systems and Binary Numbers
2 Boolean Algebra and Logic Gates
3 GateLevel Minimization
4 Combinational Logic
5 Synchronous Sequential Logic
6 Registers and Counters
7 Memory and Programmable Logic
8 Design at the Register Transfer Level
9 Laboratory Experiments with Standard ICs and
FPGAs 10 Standard Graphic Symbols
,© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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CHAPTER 1
1.1 Base-10: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Octal: 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40
Hex: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20
Base-13 A B C 10 11 12 13 14 15 16 17 18 19 23 24 25 26
1.2 (a) 32,768 (b) 67,108,864 (c) 6,871,947,674
1.3 (4310)5 = 4 * 53 + 3 * 52 + 1 * 51 = 58010
(198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010
2 1 0
(735)8 = 7 * 8 + 3 * 8 + 5 * 8 = 477
10
2 1 0
(525)6 = 5 * 6 + 2 * 6 + 5 * 6 = 197 10
1.4 14-bit binary:
11_1111_1111_1111 Decimal:
214 -1 = 16,38310
Hexadecimal: 3FFF16
1.5 Let b = base
(a) 14/2 = (b + 4)/2 = 5, so b = 6
(b) 54/4 = (5*b + 4)/4 = b + 3, so 5 * b = 52 – 4, and b = 8
(c) (2 *b + 4) + (b + 7) = 4b, so b = 11
1.6 (x – 3)(x – 6) = x2 –(6 + 3)x + 6*3 = x2 -11x + 22
Therefore: 6 + 3 = b + 1m so b =
8 Also, 6*3 = (18)10 = (22)8
1.7 68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)8
1.8 (a) Results of repeated division by 2 (quotients are followed by remainders):
43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_10102 = FA16
(b) Results of repeated division by 16:
43110 = 26(15); 1(10) (Faster)
Answer: FA = 1111_1010
1.9 (a) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.516 = 16 + 6 + 5*(.0615) = 22.3125
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
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(d) FAFA.B16 = 15*163 + 10*162 + 15*16 + 10 + 11/16 = 64,250.6875
(e) 1010.10102 = 8 + 2 + .5 + .125 = 10.625
1.10 (a) 1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310
(b) 110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510
Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.
1011.11
1.11 101 | 111011.0000
101
01001
101
1001
101
1000
101
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: = 1011.112 = 58.7510
1.12 (a) 10000 and 110111
1011 1011
+101
x101 10000 = 1610
1011
1011
110111 = 5510
(b) 62h and 958h
2Eh 0010_1110 2Eh
+34h 0011_0100 x34h
62h 0110_0010 = 9810 B38
2
8A
9 5 8h = 239210
1.13 (a) Convert 27.315 to binary:
Integer Remainder Coefficient
Quotient
27/2 = 13 + ½ a0 = 1
13/2 6 + ½ a1 = 1
6/2 3 + 0 a2 = 0
3/2 1 + ½ a3 = 1
½ 0 + ½ a4 = 1
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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2710 = 110112
Integer Fraction Coefficient
.315 x 2 = 0 + .630 a-1 = 0
.630 x 2 = 1 + .26 a-2 = 1
.26 x 2 = 0 + .52 a-3 = 0
.52 x 2 = 1 + .04 a-4 = 1
.31510 .01012 = .25 + .0625 = .3125
27.315 11011.01012
(b) 2/3
.6666666667 Integer Fraction Coefficient
.6666_6666_67 x = 1 + .3333_3333_34 a-1 = 1
2
.3333333334 x2 = 0 + .6666666668 a-2 = 0
.6666666668 x 2 = 1 + .3333333336 a-3 = 1
.3333333336 x 2 = 0 + .6666666672 a-4 = 0
.6666666672 x 2 = 1 + .3333333344 a-5 = 1
.3333333344 x 2 = 0 + .6666666688 a-6 = 0
.6666666688 x 2 = 1 + .3333333376 a-7 = 1
.3333333376 x 2 = 0 + .6666666752 a-8 = 0
.666666666710 .101010102 = .5 + .125 + .0313 + ..0078 = .664110
.101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14 (a) 1000_0000 (b) 0000_0000 (c) 1101_1010
1s comp: 0111_1111 1s comp: 1111_1111 1s comp:
0010_0101 2s comp: 1000_0000 2s comp: 0000_0000
2s comp: 0010_0110
(d) 0111_0110 (e) 1000_0101 (f) 1111_1111
1s comp: 1000_1001 1s comp: 0111_1010 1s comp:
0000_0000 2s comp: 1000_1010 2s comp:
0111_1011 2s comp: 0000_0001
1.15 (a) 52,784,630 (b) 63,325,600
9s comp: 47,215,369 9s comp: 36,674,399
10s comp: 47,215,370 10s comp: 36,674,400
(c) 25,000,000 (d) 00,000,000
9s comp: 74,999,999 9s comp: 99,999,999
10s comp: 75,000,000 10s comp: 00,000,000
1.16 B2FA B2FA: 1011_0010_1111_1010
15s comp: 4D05 1s comp: 0100_1101_0000_0101
16s comp: 4D06 2s comp: 0100_1101_0000_0110 =
1.17 (a) 3409 03409 96590 (9s comp) 4D06 96591 (10s comp)
06428 – 03409 = 06428 + 96591 = 03019
(b) 1800 01800 98199 (9s comp) 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
Result: 125 – 1800 = 1675
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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(c) 6152 06152 93847 (9s comp) 93848 (10s comp)
2043 – 6152 = 02043 + 93848 = 95891 (Negative)
Magnitude: 4109
Result: 2043 – 6152 = -4109
(d) 745 00745 99254 (9s comp) 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 009286; +801 000801; -9286 990714; -801 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 0_110001 (Needs leading zero indicate + value); +29 0_011101 (Leading 0 indicates + valu
-49 1_001111; -29 1_100011
(a) (+29) + (-49) = 0_011101 + 1_001111 = 1_101100 (1 indicates negative value.)
Magnitude = 0_010100; Result (+29) + (-49) = -20
(b) (-29) + (+49) = 1_100011 + 0_110001 = 0_010100 (0 indicates positive
value) (-29) + (+49) = +20
(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 = 7810
Result: (-29) + (-49) = -78
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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1.21 +9742 009742 990257 (9's comp) 990258 (10s) comp
+641 000641 999358 (9's comp) 999359 (10s) comp
(a) (+9742) + (+641) 010383
(b) (+9742) + (-641) 009742 + 999359 = 009102
Result: (+9742) + (-641) = 9102
(c) -9742) + (+641) = 990258 + 000641 = 990899 (negative)
Magnitude: 009101
Result: (-9742) + (641) = -9101
(d) (-9742) + (-641) = 990258 + 999359 = 989617 (Negative)
Magnitude: 10383
Result: (-9742) + (-641) = -10383
1.22 8,723
BCD: 1000_0111_0010_0011
ASCII: 0_011_1000_011_0111_011_0010_011_0
001
1.23
1000 0100 0010 ( 842)
0101 0011 0111 (+537)
1101 0111 1001
0110
0001 0011 0111 0101
(1,379)
1.24 (a) (b)
6 3 1 1 Decimal 6 4 2 1 Decimal
0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 0 0 0 1 1
0 0 1 0 2 0 0 1 0 2
0 1 0 0 3 0 0 1 1 3
0 1 1 0 4 (or 0101) 0 1 0 0 4
0 1 1 1 5 0 1 0 1 5
1 0 0 0 6 1 0 0 0 6 (or 0110)
1 0 1 0 7 (or 1001) 1 0 0 1 7
1 0 1 1 8 1 0 1 0 8
1 1 0 0 9 1 0 1 1 9
1.25 BCD:
(a) 5,13710 0101_0011_0111
(b) Excess-3:
(c) 1000_0100_0110_1010
(d) 2421:
1011_0001_0011_01
1.26 5,137 9s Comp: 114,862
2421 code: 6311:
0100_1110_1100_1000
1s comp: 0111_0001_0100_10
1011_0001_0011_0111 same as (c) in
01
1.25
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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
1100101
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) 2,199
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
a
f
b
c
g
f
g
1.36
a b
a
f
b
g
f
g
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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CHAPTER 2
2.1 (a)
xyz x+y+z (x + y + z)' x' y' z' x' y' z' xyz (xyz) (xyz)' x' y' z' x' + y' + z'
000 0 1 1 1 1 1 000 0 1 1 1 1 1
001 1 0 1 1 0 0 001 0 1 1 1 0 1
010 1 0 1 0 1 0 010 0 1 1 0 1 1
011 1 0 1 0 0 0 011 0 1 1 0 0 1
100 1 0 0 1 1 0 100 0 1 0 1 1 1
101 1 0 0 1 0 0 101 0 1 0 1 0 1
110 1 0 0 0 1 0 110 0 1 0 0 1 1
111 1 0 0 0 0 0 111 1 0 0 0 0 0
(b) (c)
xyz x+ (x + y) (x + z) (x + y)(x + z) x y z x(y + z) x xz xy + xz
yz y
000 0 0 0 0 000 0 0 0 0
001 0 0 1 0 001 0 0 0 0
010 0 1 0 0 010 0 0 0 0
011 1 1 1 1 011 0 0 0 0
100 1 1 1 1 100 0 0 0 0
101 1 1 1 1 101 1 0 1 1
110 1 1 1 1 110 1 1 0 1
111 1 1 1 1 111 1 1 1 1
(c) (d)
xyz x y+z x + (y + z) (x + y) (x + y) + z xyz y x(yz) xy
z (xy)z
000 0 0 0 0 0 000 0 0 0 0
001 0 1 1 0 1 001 0 0 0 0
010 0 1 1 1 1 010 0 0 0 0
011 0 1 1 1 1 011 1 0 0 0
100 1 0 1 1 1 100 0 0 0 0
101 1 1 1 1 1 101 0 0 0 0
110 1 1 1 1 1 110 0 0 1 0
111 1 1 1 1 1 111 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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