Digital Design With an Introduction to the Verilog HDL VHDL and System Verilog 6th Edition - Mano - Solutions Manual

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

(735)8 = 7 * 82 + 3 * 81 + 5 * 80 = 47710

(525)6 = 5 * 62 + 2 * 61 + 5 * 60 = 19710

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.


www.Mohandesyar.com

, © 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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(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
+34 h 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.


www.Mohandesyar.com