Solution Digital Design 4th Ed- Morris Mano.pdf-资料库

© 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. 1 SOLUTIONS MANUAL DIGITAL DESIGN FOURTH EDITION M. MORRIS MANO California State University, Los Angeles MICHAEL D. CILETTI University of Colorado, Colorado Springs rev 01/21/2007 Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.

© 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. 2 CHAPTER 1 1.1 1.2 1.3 1.4 Base-10: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40 Octal: 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 (a) 32,768 (b) 67,108,864 (c) 6,871,947,674 (4310)5 = 4 * 53 + 3 * 52 + 1 * 51 = 58010 (198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010 (735)8 = 7 * 82 + 3 * 81 + 5 * 80 = 47710 (525)6 = 5 * 62 + 2 * 61 + 5 * 60 = 19710 14-bit binary: 11_1111_1111_1111 Decimal: Hexadecimal: 3FFF16 214 -1 = 16,38310 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 68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)8 1.7 1.8 (a) Results of repeated division by 2 (quotients are followed by remainders): 1.9 43110 = 215(1); 107(1); Answer: 1111_10102 = FA16 53(1); 26(1); 13(0); 6(1) 3(0) 1(1) (b) Results of repeated division by 16: Answer: FA = 1111_1010 43110 = 26(15); 1(10) (Faster) (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.

© 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. 3 (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. 1.11 1.12 1.13 1011.11 101 | 111011.0000 101 01001 101 1001 101 0110 1000 101 The quotient is carried to two decimal places, giving 1011.11 Checking: 1110112 / 1012 = 5910 / 510 1011.112 = 58.7510 (a) 10000 and 110111 1011 +101 (b) 62h and 958h 10000 = 1610 2Eh 0010_1110 +34 h 0011_0100 0110_0010 = 9810 62h 1011 x101 1011 1011 110111 = 5510 2Eh x34h B38 82A 9 5 8h = 239210 (a) Convert 27.315 to binary: 27/2 = 13/2 6/2 3/2 ½ Integer Quotient 13 6 3 1 0 Remainder Coefficient + + + + + ½ ½ ½ ½ 0 a0 = 1 a1 = 1 a2 = 0 a3 = 1 a4 = 1 Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.

© 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. 4 1.14 1.15 1.16 1.17 2710 = 110112 .315 x 2 = .630 x 2 = .26 x 2 .52 x 2 = = Integer 0 1 0 1 Fraction Coefficient + .630 + .26 + .52 + .04 a-1 = 0 a-2 = 1 a-3 = 0 a-4 = 1 .31510 .01012 = .25 + .0625 = .3125 27.315 11011.01012 (b) 2/3 .6666666667 .6666_6666_67 x 2 = 1 = 0 .3333333334 x 2 = 1 .6666666668 x 2 = 0 .3333333336 x 2 .6666666672 x 2 = 1 = 0 .3333333344 x 2 = 1 .6666666688 x 2 .3333333376 x 2 = 0 Integer Fraction + .3333_3333_34 + .6666666668 + .3333333336 + .6666666672 + .3333333344 .6666666688 + .3333333376 + + .6666666752 Coefficient a-1 = 1 a-2 = 0 a-3 = 1 a-4 = 0 a-5 = 1 a-6 = 0 a-7 = 1 a-8 = 0 .666666666710 .101010102 = .5 + .125 + .0313 + ..0078 = .664110 .101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)). (a) 1000_0000 1s comp: 0111_1111 2s comp: 1000_0000 (d) 0111_0110 1s comp: 1000_1001 2s comp: 1000_1010 (a) 52,784,630 9s comp: 47,215,369 10s comp: 47,215,370 (c) 25,000,000 9s comp: 74,999,999 10s comp: 75,000,000 (b) 0000_0000 1s comp: 1111_1111 2s comp: 0000_0000 (e) 1000_0101 1s comp: 0111_1010 2s comp: 0111_1011 (c) 1101_1010 1s comp: 0010_0101 2s comp: 0010_0110 (f) 1111_1111 1s comp: 0000_0000 2s comp: 0000_0001 (b) 63,325,600 9s comp: 36,674,399 10s comp: 36,674,400 (d) 00,000,000 9s comp: 99,999,999 10s comp: 00,000,000 B2FA 4D05 4D06 15s comp: 16s comp: (a) 3409 03409 96590 (9s comp) 96591 (10s comp) B2FA: 1011_0010_1111_1010 1s comp: 0100_1101_0000_0101 2s comp: 0100_1101_0000_0110 = 4D06 06428 – 03409 = 06428 + 96591 = 03019 125 – 1800 = 00125 + 98200 = 98325 (negative) (b) 1800 01800 98199 (9s comp) 98200 (10 comp) Magnitude: 1675 Result: 125 – 1800 = 1675 Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.

© 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. 5 1.18 2043 – 6152 = 02043 + 93848 = 95891 (Negative) (c) 6152 06152 93847 (9s comp) 93848 (10s comp) Magnitude: 4109 Result: 2043 – 6152 = -4109 (d) 745 00745 99254 (9s comp) 99255 (10s comp) Result: 1631 – 745 = 886 1631 -745 = 01631 + 99255 = 0886 (Positive) Note: Consider sign extension with 2s complement arithmetic. (a) Diff: 10001 1s comp: 01110 2s comp: 01111 10011 00010 100011 1s comp: 1011100 with sign extension 2s comp: 1011101 0100010 1111111 sign bit indicates that the result is negative 0000001 2s complement -000001 result (b) (d) 10101 101000 1s comp: 1010111 2s comp: 1011000 001001 (c) Diff: 1s comp: 1101010 with sign extension 2s comp: 1101011 110000 1100001 (negative) 0011111 (2s comp) -011111 (diff is -31) 0011011 sign bit indicates that the result is positive Check: 48 -21 = 27 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 + value) -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.

© 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. 6 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: ASCII: 1.23 1000_0111_0010_0011 0_011_1000_011_0111_011_0010_011_0001 1000 0100 0010 ( 842) 0101 0011 0111 (+537) 1101 0111 0110 0001 0011 0111 0101 (1,379) 1001 1.24 (a) (b) 6 3 1 1 Decimal 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 1 0 0 3 0 1 1 0 4 (or 0101) 0 1 1 1 5 1 0 0 0 6 1 0 1 0 7 (or 1001) 1 0 1 1 8 1 1 0 0 9 6 4 2 1 Decimal 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 0 1 1 3 0 1 0 0 4 0 1 0 1 5 1 0 0 0 6 (or 0110) 1 0 0 1 7 1 0 1 0 8 1 0 1 1 9 0101_0011_0111 (a) 5,13710 BCD: (b) (c) (d) 2421: 6311: Excess-3: 1000_0100_0110_1010 1011_0001_0011_0111 0111_0001_0100_1001 5,137 9s Comp: 2421 code: 1s comp: 4,862 0100_1110_1100_1000 1011_0001_0011_0111 same as (c) in 1.25 Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved. 1.25 1.26

© 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. 7 1.27 1.28 1.29 1.30 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.) G o o l e 01000111_11101111_01101000_01101110_00100000_11000100_11101111_11100101 (space) B (dot) Bill Gates 73 F4 E5 76 E5 4A EF 62 73 1.31 1.32 1.33 73: F4: E5: 76: E5: 4A: EF: 62: 73: 0_111_0011 s 1_111_0100 t 1_110_0101 e 0_111_0110 v 1_110_0101 e 0_100_1010 j 1_110_1111 o 0_110_0010 b 0_111_0011 s 62 + 32 = 94 printing characters bit 6 from the right (a) 897 (b) 564 (c) 871 (d) 2,199 1.34 ASCII for decimal digits with odd parity: 1.35 1.36 (0): 10110000 (4): 00110100 (8): 00111000 (1): 00110001 (5): 10110101 (9): 10111001 (2): 00110010 (6): 10110110 (3): 10110011 (7): 00110111 (a) a b c f g a b c f g a b f g a b f g Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.

© 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. 8 CHAPTER 2 2.1 (a) x y z x + y + z (x + y + z)' x' y' z' x' y' z' x y z (xyz) (xyz)' x' 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 (b) (c) 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 (c) 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 0 x y z x + yz (x + y) (x + z) (x + y)(x + z) x y z x(y + z) xy xz xy + xz 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 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1 1 1 x 0 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 (d) y + z x + (y + z) (x + y) (x + y) + z x y z yz x(yz) xy (xy)z 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 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 Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.

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