SOLUTIONS MANUAL
DIGITAL DESIGN
WITH AN INTRODUCTION TO THE VERILOG
HDL
Fifth Edition
N
U
M. MORRIS MANO
Professor Emeritus
California State University, Los Angeles
R
SE
MICHAEL D. CILETTI
Professor Emeritus
University of Colorado, Colorado Springs
TE
rev 02/14/2012
ST
jhgfdsa
,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-12 14 15 16 17 18 19 1A 1B 20 21 22 23 24 25 26 27 28
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
(435)8 = 4 * 82 + 3 * 81 + 5 * 80 = 28510
(345)6 = 3 * 62 + 4 * 61 + 5 * 60 = 13710
N
1.4 16-bit binary: 1111_1111_1111_1111
Decimal equivalent: 216 -1 = 65,53510
Hexadecimal equivalent: FFFF16
1.5 Let b = base
U
(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
R
(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
SE
Also, 6*3 = (18)10 = (22)8
1.7 64CD16 = 0110_0100_1100_11012 = 110_010_011_001 _101 = (62315 )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
TE
(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
ST
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
(d) DADA.B16 = 14*163 + 10*162 + 14*16 + 10 + 11/16 = 60,138.6875
jhgfdsa
, (e) 1010.11012 = 8 + 2 + .5 + .25 + .0625 = 10.8125
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
N
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: = 1011.112 = 58.7510
U
1.12 (a) 10000 and 110111
1011 1011
+101 x101
10000 = 1610 1011
R
1011
110111 = 5510
(b) 62h and 958h
2Eh 0010_1110 2Eh
SE
+34 h 0011_0100 x34h
62h 0110_0010 = 9810 B38
82A
9 5 8h = 239210
1.13 (a) Convert 27.315 to binary:
TE
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
ST
jhgfdsa
, 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
N
.6666666688 x 2 = 1 + .3333333376 a-7 = 1
.3333333376 x 2 = 0 + .6666666752 a-8 = 0
.666666666710 .101010102 = .5 + .125 + .0313 + ..0078 = .664110
U
.101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14 (a) 0001_0000 (b) 0000_0000 (c) 1101_1010
1s comp: 1110_1111 1s comp: 1111_1111 1s comp: 0010_0101
2s comp: 1111_0000 2s comp: 0000_0000 2s comp: 0010_0110
R
(d) 1010_1010 (e) 1000_0101 (f) 1111_1111
1s comp: 0101_0101 1s comp: 0111_1010 1s comp: 0000_0000
2s comp: 0101_0110 2s comp: 0111_1011 2s comp: 0000_0001
`
SE
1.15 (a) 25,478,036 (b) 63,325,600
9s comp: 74,521,963 9s comp: 36,674,399
10s comp: 74,521,964 10s comp: 36,674,400
(c) 25,000,000 (d) 00000000
9s comp: 74,999,999 9s comp: 99999999
10s comp: 75,000,000 10s comp: 100000000
1.16 C3DF C3DF: 1100_0011_1101_1111
TE
15s comp: 3C20 1s comp: 0011_1100_0010_0000
16s comp: 3C21 2s comp: 0011_1100_0010_0001 = 3C21
1.17 (a) 2,579 → 02,579 →97,420 (9s comp) → 97,421 (10s comp)
4637 – 2,579 = 2,579 + 97,421 = 205810
(b) 1800 → 01800 → 98199 (9s comp) → 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
ST
Result: 125 – 1800 = 1675
jhgfdsa
DIGITAL DESIGN
WITH AN INTRODUCTION TO THE VERILOG
HDL
Fifth Edition
N
U
M. MORRIS MANO
Professor Emeritus
California State University, Los Angeles
R
SE
MICHAEL D. CILETTI
Professor Emeritus
University of Colorado, Colorado Springs
TE
rev 02/14/2012
ST
jhgfdsa
,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-12 14 15 16 17 18 19 1A 1B 20 21 22 23 24 25 26 27 28
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
(435)8 = 4 * 82 + 3 * 81 + 5 * 80 = 28510
(345)6 = 3 * 62 + 4 * 61 + 5 * 60 = 13710
N
1.4 16-bit binary: 1111_1111_1111_1111
Decimal equivalent: 216 -1 = 65,53510
Hexadecimal equivalent: FFFF16
1.5 Let b = base
U
(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
R
(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
SE
Also, 6*3 = (18)10 = (22)8
1.7 64CD16 = 0110_0100_1100_11012 = 110_010_011_001 _101 = (62315 )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
TE
(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
ST
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
(d) DADA.B16 = 14*163 + 10*162 + 14*16 + 10 + 11/16 = 60,138.6875
jhgfdsa
, (e) 1010.11012 = 8 + 2 + .5 + .25 + .0625 = 10.8125
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
N
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: = 1011.112 = 58.7510
U
1.12 (a) 10000 and 110111
1011 1011
+101 x101
10000 = 1610 1011
R
1011
110111 = 5510
(b) 62h and 958h
2Eh 0010_1110 2Eh
SE
+34 h 0011_0100 x34h
62h 0110_0010 = 9810 B38
82A
9 5 8h = 239210
1.13 (a) Convert 27.315 to binary:
TE
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
ST
jhgfdsa
, 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
N
.6666666688 x 2 = 1 + .3333333376 a-7 = 1
.3333333376 x 2 = 0 + .6666666752 a-8 = 0
.666666666710 .101010102 = .5 + .125 + .0313 + ..0078 = .664110
U
.101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14 (a) 0001_0000 (b) 0000_0000 (c) 1101_1010
1s comp: 1110_1111 1s comp: 1111_1111 1s comp: 0010_0101
2s comp: 1111_0000 2s comp: 0000_0000 2s comp: 0010_0110
R
(d) 1010_1010 (e) 1000_0101 (f) 1111_1111
1s comp: 0101_0101 1s comp: 0111_1010 1s comp: 0000_0000
2s comp: 0101_0110 2s comp: 0111_1011 2s comp: 0000_0001
`
SE
1.15 (a) 25,478,036 (b) 63,325,600
9s comp: 74,521,963 9s comp: 36,674,399
10s comp: 74,521,964 10s comp: 36,674,400
(c) 25,000,000 (d) 00000000
9s comp: 74,999,999 9s comp: 99999999
10s comp: 75,000,000 10s comp: 100000000
1.16 C3DF C3DF: 1100_0011_1101_1111
TE
15s comp: 3C20 1s comp: 0011_1100_0010_0000
16s comp: 3C21 2s comp: 0011_1100_0010_0001 = 3C21
1.17 (a) 2,579 → 02,579 →97,420 (9s comp) → 97,421 (10s comp)
4637 – 2,579 = 2,579 + 97,421 = 205810
(b) 1800 → 01800 → 98199 (9s comp) → 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
ST
Result: 125 – 1800 = 1675
jhgfdsa
