 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing by the publisher.For exclusive use of adopters of the book Digital Design Principles and Practices, Fourth Edition, by John F. Wakerly, ISBN 0-13-186389-4.2.1 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)2.3 (a)(b)(c) (d) (e) (f) 2.5 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)2.6 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)2.20Suppose a 3n-bit number B is represented by an n-digit octal number Q. Then the two’s-complement of B isrepresented by the 8’s-complement of Q.2.24 Starting with the arrow pointing at any number, adding a positive number causes overflow if the arrow isadvanced through the +7 to –8 transition. Adding a negative number to any number causes overflow if thearrow is not advanced through the +7 to –8 transition.3e2.1110101126B16= 174003811111000000000112=101101112B716=67.248110111.01012=10100.1101214.D16=F3A51611110011101001012=1101100123318=AB3D1610101011001111012=101111.0111257.348=15C.3816101011100.001112=3e2.310231610000001000112100438==7E6A161111110011010102771528==ABCD16101010111100110121257158==C35016110000110101000021415208==9E36.7A161001111000110110.01111012117066.3648==DEAD.BEEF161101111010101101.10111110111011112157255.5756748==3e2.51101011210710= 17400386349110=10110111218310=67.24855.312510=10100.1101220.812510=F3A5166237310=12010313810=AB3D164383710=71568369410=15C.3816348.2187510=3e2.61251011111012=34891066418=20910110100012=971410227628=1321010001002= 23851105D2B16=72710104025= 5719010DF6616=14351026338= 6511310FE5916=3e2.18hjb4ji+2j⋅i 0=∑=Therefore,Bbi2i⋅i 0–4n 1–∑hi16i⋅i 0=n 1–∑==B–24nbii 0=4n 1–∑2i⋅–16nhi16i⋅i 0=n 1–∑–==3e2.22 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under allcopyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing by the publisher.For exclusive use of adopters of the book Digital Design Principles and Practices, Fourth Edition, by John F. Wakerly, ISBN 0-13-186389-4.2.26 Let the binary representation of be . Then we can write the binary representation of as, where . Note that is the sign bit of . The value of isCase 1 In this case, if and only if , which is true if andonly if all of the discarded bits are 0, the same as .Case 2 In this case, if and only if , whichis true if and only if all of the discarded bits are 1, the same as .2.27 If the radix point is considered to be just to the right of the leftmost bit, then the largest number is andthe 2’s complement of is obtained by subtracting it from 2 (singular possessive). Regardless of the positionof the radix point, the 1s’ complement is obtained by subtracting from the largest number, which has all 1s(plural).2.30Case 1 First term is 0, summation terms have shifted coefficients as specified. Overflow if.Case 2 Split first term into two halves; one half is cancelled by summation term if. Remaining half and remaining summation terms have shifted coefficients as specified. Overflow if.2.35 001–010, 011–100, 101–110, 111–000. 2.38 The manufacturer’s code fails every 4th time, for a total of for an n-bit encoding disc. A standard binarycode fails when the LSB changes from 1 to 0, which changes the next bit (and possibly others), guaranteeing aproblem. So, half the boundaries in a standard binary code are bad, a total of for an n-bit encoding disc.The manufacturer’s code is only half as bad as a standard binary code.2.41 In the string representation, each position may have a 0, a 1, or an , a total of three possibilities per position,and combinations in all. 3e2.24Xxn 1–xn 2–…x1x0Yxmxm 1–…x1x0mnd–= xm 1–YYY 2m 1–– xm 1–xi2i⋅i 0=∑+⋅=The value of X isX 2n 1–– xn 1–xi2i⋅i 0=n 2–∑+⋅=2n 1–– xn 1–Y 2m 1–xm 1–⋅ xi2i⋅im1–=n 2–∑++ +⋅=2n 1–– xn 1–Y 22m 1–⋅ xi2i⋅n 2–∑++ +⋅=xm 1–0=() XY=2n 1–– xn 1–⋅ xi2i⋅im=n 2–∑+0=xm…xn 1–() xm 1–xm 1–1=() XY=2n 1–– xn 1–⋅ 22m 1–⋅ xi2i⋅im=n 2–∑++ 0=xm…xn 1–() xm 1–3e2.251.11…1DD3e2.28Bbn 1––2n 1–⋅ bi2i⋅i 0=n 2∑+=2Bbn 1––2nbi2i 1+⋅i 0=n 2–∑+⋅=bn 1–0=()bn 2–1=bn 1–1=() bn 2–2n 1–⋅bn 2–1=bn 2–0=4e2.334e2.372n 2–2n