《逻辑与计算机设计基础》（第五版）课后习题答案pdf

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《逻辑与计算机设计基础》（原书第五版）课后习题答案.zip （13个子文件）

《逻辑与计算机设计基础》（原书第五版）课后习题答案

chapter10_solutions-5th.pdf 1.76MB

chapter04_solutions-5th.pdf 4.33MB

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chapter11_solutions-5th.pdf 1.61MB

chapter03_solutions-5th.pdf 4.83MB

chapter05_solutions-5th.pdf 1.48MB

chapter12_solutions-5th.pdf 1.36MB

chapter08_solutions-5th.pdf 1.95MB

chapter06_solutions-5th.pdf 2.11MB

chapter02_solutions-5th.pdf 2.18MB

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chapter07_solutions-5th.pdf 1.22MB

新建文本文档.txt 0B

Problem Solutions – Chapter 3

CHAPTER 3

3-1.

Place a 1 in each K-map cell where 2 or more inputs are equal to 1.

1 1

F = XZ + XY + YZ

This is the same function as the

carry f or the full adder.

Place a 1 in each K-map cell where 2 or more inputs are equal to 1.

F = XZ + XY + YZ

This is the same function as the carry

for the full adder.

3-2.*

1 1 1

F = AB + AC

3-3.

Assuming inputs G3, G2, G1, G0 and outputs B3, B2, B1, B0, with G3 and B3 being the most significant bits, and treating the invalid

input combinations as don’t cares:

BG

2 3 2

B G G

1 2 1 3 2 1

B G G G G G

0 3 0 2 1 0 2 1 0

2 1 0 3 2 1 0

  



B G G G G G G G G

G GG G G G G

3-4. a) For the 3 x 3 pattern, there are exactly three row, three column and two diagonal combinations that represent a win for the X

player: W = X1 X2 X3 + X4 X5 X6 + X7 X8 X9 + X1 X4 X7 + X2 X5 X8 + X3 X6 X9 + X1 X5 X9 + X3 X5 X7 Gate Input cost = 32

b) W = X5 (X1 X9 + X2 X8 + X3 X7 + X4 X6) + X1 X2 X3 + X1 X4 X7 + X7 X8 X9 + X3 X6 X9 Gate Input Cost = 30

3-5. a) For the 4 x 4 pattern, there are exactly four row, four column and two diagonal combinations that represent a win for the X

player: W = X1 X2 X3 X4 + X5 X6 X7 X8 + X9 X10 X11 X12 + X13 X14 X15 X16 + X1 X5 X9 X13 X2 X6 X10 X14 + X3 X7 X11

X15 + X4 X8 X12 X16 + X1 X6 X11 X16 + X4 X7 X10 X13 Gate Input cost = 50

b) W = X1(X2 X3 X4 + X5 X9 X13 + X6 X11 X15) + X7(X5 X6 X8 + X3 X11 X15 + X4 X10 X13) + X9 X10 X 11 X12

+ X13 X14 X15 X16 + X2 X6 X10 X14 + X4 X8 X12 X16 Gate Input Cost = 48

This work is protected by United States copyright laws

and is provided solely for the use of instructors in teaching

their courses and assessing student learning. Dissemination

or sale of any part of this work (including on the World Wide Web)

will destroy the integrity of the work and is not permitted.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem Solutions – Chapter 3

3-6.

a) Detecting a change in one-out-of-

three inputs can be done using a parity

function as Z. The truth table shown is

for even parity. For this case,

1 2 3  Z X X X

If odd parity is chosen, then an

alternative result for Z is:

1 2 3  Z X X X

3-7.

ABCD

GNS

YNS

RNS

GEW

YEW

REW

0000

0001

0011

0010

0110

0111

0101

0100

1100

1101

1111

1110

1010

1011

1001

1000

ABCD GNS YNS RNS GEW YEW REW

0000 1 0 0 0 0 1

0001 1 0 0 0 0 1

0011 1 0 0 0 0 1

0010 1 0 0 0 0 1

0110 1 0 0 0 0 1

0111 1 0 0 0 0 1

0101 0 1 0 0 0 1

0100 0 0 1 0 0 1

1100 0 0 1 1 0 0

1101 0 0 1 1 0 0

1111 0 0 1 1 0 0

1110 0 0 1 1 0 0

1010 0 0 1 1 0 0

1011 0 0 1 1 0 0

1001 0 0 1 0 1 0

1000 0 0 1 0 0 1

GNS AC AB+=

YNS ABCD=

GEW AB AC+=

YEW ABCD=

REW A BCD+=

YNS

GNS

GEW

YEW

RNS A BCD+=

RNS

GNS AC AB

GEW AB AC

ABCD GNS YNS RNS GEW YEW REW

0000 1 0 0 0 0 1

0001 1 0 0 0 0 1

0011 1 0 0 0 0 1

0010 1 0 0 0 0 1

0110 1 0 0 0 0 1

0111 1 0 0 0 0 1

0101 0 1 0 0 0 1

0100 0 0 1 0 0 1

1100 0 0 1 1 0 0

1101 0 0 1 1 0 0

1111 0 0 1 1 0 0

1110 0 0 1 1 0 0

1010 0 0 1 1 0 0

1011 0 0 1 1 0 0

1001 0 0 1 0 1 0

1000 0 0 1 0 0 1

GNS AC AB+=

YNS ABCD=

GEW AB AC+=

YEW ABCD=

REW A BCD+=

YNS

GNS

GEW

YEW

RNS A BCD+=

RNS

YNS ABCD

YEW ABCD

ABCD GNS YNS RNS GEW YEW REW

0000 1 0 0 0 0 1

0001 1 0 0 0 0 1

0011 1 0 0 0 0 1

0010 1 0 0 0 0 1

0110 1 0 0 0 0 1

0111 1 0 0 0 0 1

0101 0 1 0 0 0 1

0100 0 0 1 0 0 1

1100 0 0 1 1 0 0

1101 0 0 1 1 0 0

1111 0 0 1 1 0 0

1110 0 0 1 1 0 0

1010 0 0 1 1 0 0

1011 0 0 1 1 0 0

1001 0 0 1 0 1 0

1000 0 0 1 0 0 1

GNS AC AB+=

YNS ABCD=

GEW AB AC+=

YEW ABCD=

REW A BCD+=

YNS

GNS

GEW

YEW

RNS A BCD+=

RNS

RNS A BCD

REW A BCD

3-8.

S0 C

S1 0

S2 ABC ABC

S3 ABC ABC

S4 AB AC

S5 AB







This work is protected by United States copyright laws

and is provided solely for the use of instructors in teaching

their courses and assessing student learning. Dissemination

or sale of any part of this work (including on the World Wide Web)

will destroy the integrity of the work and is not permitted.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem Solutions – Chapter 3

3-9.

S0 BCD BCD AB ACD ABCD

S1 AB AB ACD BCD

S2 ABC ABD

    

   



3-10.

1010 to

XXXX

1111

W AC BD BD

X BCD BC+BD

Y CD +CD

  





This work is protected by United States copyright laws

and is provided solely for the use of instructors in teaching

their courses and assessing student learning. Dissemination

or sale of any part of this work (including on the World Wide Web)

will destroy the integrity of the work and is not permitted.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem Solutions – Chapter 3

3-11.

PL PS

LL PS LS RS PS LS RR

RL PS LS RS PS RS RR





PS LS RS RR PL LL RL

0 0 0 0 0 0 0

0 0 0 1 0 0 0

0 0 1 0 0 0 1

0 0 1 1 0 0 1

0 1 0 0 0 1 0

0 1 0 1 0 1 0

0 1 1 0 0 0 1

0 1 1 1 0 1 0

1 0 0 0 1 0 0

1 0 0 1 1 0 0

1 0 1 0 1 0 0

1 0 1 1 1 0 0

1 1 0 0 1 0 0

1 1 0 1 1 0 0

1 1 1 0 1 0 0

1 1 1 1 1 0 0

PL = PS

LL = PS LS RS + PS LS RR

RL = PSLS RS + PS RS RR

3-12.

a = AC + ABD+ ABD + ABC

b = AB +BC + ACD+ ACD

c = AB+BC + AD

d = ABCD + ABC + ABD+ ABC + ACD

e = BCD+ ACD

f = ABC + ABD+ ABC + ACD

g = ABC + ABC + ABC + ACD

c) The following gate input counts include input inverters and share AND gates.

Total gate inputs for this solutions = 74. Total gate inputs for book solution is 70. The book solution is better by 4 gate inputs.

1 1

a=AC+ A B D + ABD + ABC

b= AB + B C + A C D + ACD

c= AB + B C + AD

d= ABCD + AB C + A B D + A BC + ACD

e=BC D + ACD

f= ABC + ABD + ABC + A C D

g= ABC + ABC + A BC + ACD

Total gate inputs for this solutions = 74. Total gate inputs for book solution is 70. The book solution is better by 4 gate inputs.

c) The following gate input counts include input inverters and share AND gates.

This work is protected by United States copyright laws

and is provided solely for the use of instructors in teaching

their courses and assessing student learning. Dissemination

or sale of any part of this work (including on the World Wide Web)

will destroy the integrity of the work and is not permitted.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem Solutions – Chapter 3

3-13.

W=XZ + YZ

Hierarchy

G = B(CE + DE) + BC

F = A(CE + DE) + AD

3-14.

Hierarchy

BC+BD

G = A(BC + BD) + A(BC + BD)

= ABC + ABD + ABC + ABD

G= A(BC+BD)+ A(BC+BD)

= ABC+ ABD+ ABC+ ABD

3-15.

Part b requires 6 f ewer gates.

Part b requires 6 fewer gates.

This work is protected by United States copyright laws

and is provided solely for the use of instructors in teaching

their courses and assessing student learning. Dissemination

or sale of any part of this work (including on the World Wide Web)

will destroy the integrity of the work and is not permitted.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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