SOLUTIONS MANUAL DIGITAL DESIGN FOURTH EDITION BY M. MORRIS MANO MICHAEL D. CILETTI

SOLUTIONS MANUAL
DIGITAL DESIGN

FOURTH EDITION

M. MORRIS MANO

MICHAEL D. CILETTI

,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

(735)8 = 7 * 82 + 3 * 81 + 5 * 80 = 477 10

(525)6 = 5 * 62 + 2 * 61 + 5 * 60 = 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

, (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

, 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 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

.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 = 4D06
1.17 (a) 3409 → 03409 →96590 (9s comp) → 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