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Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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
© 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.
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
2
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 * 5
3
+ 3 * 5
2
+ 1 * 5
1
= 580
10
(198)
12
= 1 * 12
2
+ 9 * 12
1
+ 8 * 12
0
= 260
10
(735)
8
= 7 * 8
2
+ 3 * 8
1
+ 5 * 8
0
= 477
10
(525)
6
= 5 * 6
2
+ 2 * 6
1
+ 5 * 6
0
= 197
10
1.4 14-bit binary: 11_1111_1111_1111
Decimal: 2
14
-1 = 16,383
10
Hexadecimal: 3FFF
16
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) = x
2
–(6 + 3)x + 6*3 = x
2
-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):
431
10
= 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_1010
2
= FA
16
(b) Results of repeated division by 16:
431
10
= 26(15); 1(10) (Faster)
Answer: FA = 1111_1010
1.9 (a) 10110.0101
2
= 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.5
16
= 16 + 6 + 5*(.0615) = 22.3125
(c) 26.24
8
= 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
© 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.
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
3
(d) FAFA.B
16
= 15*16
3
+ 10*16
2
+ 15*16 + 10 + 11/16 = 64,250.6875
(e) 1010.1010
2
= 8 + 2 + .5 + .125 = 10.625
1.10 (a) 1.10010
2
= 0001.1001
2
= 1.9
16
= 1 + 9/16 = 1.563
10
(b) 110.010
2
= 0110.0100
2
= 6.4
16
= 6 + 4/16 = 6.25
10
Reason: 110.010
2
is the same as 1.10010
2
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: 111011
2
/ 101
2
= 59
10
/ 5
10
# 1011.11
2
= 58.75
10
1.12 (a) 10000 and 110111
1011 1011
+101
x101
10000 = 16
10
1011
1011
110111 = 55
10
(b) 62
h
and 958
h
2E
h
0010_1110 2E
h
+34
h
0011_0100 x34
h
62
h
0110_0010 = 98
10
B
3
8
8
2
A
9 5 8
h
= 2392
10
1.13 (a) Convert 27.315 to binary:
Integer Remainder Coefficient
Quotient
27/2 = 13 + ½ a
0
= 1
13/2 6 + ½ a
1
= 1
6/2 3 + 0 a
2
= 0
3/2 1 + ½ a
3
= 1
½ 0 + ½ a
4
= 1
© 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.
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
4
27
10
= 11011
2
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
.315
10
# .0101
2
= .25 + .0625 = .3125
27.315 # 11011.0101
2
(b) 2/3 # .6666666667
Integer Fraction Coefficient
.6666_6666_67 x 2 = 1 + .3333_3333_34 a
-1
= 1
.3333333334 x 2 = 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
.6666666667
10
# .10101010
2
= .5 + .125 + .0313 + ..0078 = .6641
10
.101010102 = .1010_1010
2
= .AA
16
= 10/16 + 10/256 = .6641
10
(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 = 4D06
1.17 (a) 3409 o 03409 o96590 (9s comp) o 96591 (10s comp)
06428 – 03409 = 06428 + 96591 = 03019
(b) 1800 o 01800 o 98199 (9s comp) o 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
Result: 125 – 1800 = 1675
© 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.
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
5
(c) 6152 o 06152 o 93847 (9s comp) o 93848 (10s comp)
2043 – 6152 = 02043 + 93848 = 95891 (Negative)
Magnitude: 4109
Result: 2043 – 6152 = -4109
(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_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 = 78
10
Result: (-29) + (-49) = -78
© 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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