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I: INTRODUCTION

The text, Fundamentals of Logic Design,7th edition, has been designed so that it

can be used either for a standard lecture course or for a self-paced course. The text is divided

into 20 study units in such a way that the average study time for each unit is about the same.

The units have undergone extensive class testing in a self-paced environment and have been

revised based on student feedback. The study guides and text material are sufficient to allow

almost all students to achieve mastery of all of the objectives. For example, the material on

Boolean algebra and algebraic simplification is 2½ units because students found this topic

difficult. There is a separate unit on going from problem statements to state graphs because

this topic is difficult for many students.

1.1 Using the Text in a Lecture Course

Even though the text was developed in a self-paced environment, the text is well

suited for use in a standard lecture course. Since the format of the text differs somewhat

from a conventional text, a few suggestions for using the text in a lecture course may be

appropriate. Except for the inclusion of objectives and study guides, the units in the text

differ very little from chapters in a standard textbook. The study guides contain very basic

questions, while the problems at the end of each unit are of a more comprehensive nature.

At the University of Texas a class composed largely of Electrical Engineering and

Computer Science sophomores and juniors covers 18 units (all units except 6 and 19) of

the text in one semester. Units 8, 10, 12, 16, 17, and 20 contain design problems that are

suitable for simulation and lab exercises. The design problems help tie together and review

the material from a number of preceding units. Units 10, 17, and 20 introduce the VHDL

2

1.2 Some Remarks About the Text

In this text, students are taught how to use Boolean algebra effectively, in contrast with

many texts that present Boolean algebra and a few examples of its application and then leave it to

the student to figure out how to use it effectively. For example, use of the theorem x + yz = (x + y)

(x + z) in factoring and multiplying out expressions is taught explicitly, and detailed guidelines are

given for algebraic simplification.

The most important and often most difficult part of sequential circuit design is formulating

the state table or graph from the problem statement, but most texts devote only a few paragraphs

to this subject because there is no algorithm. This text devotes a full unit to the subject, presents

guidelines for deriving state tables and graphs, and provides programmed exercises that help the

student learn this material. Most of the material in the text is treated in a fairly conventional

1.3 Using the Text in a Self-Paced Course

This section introduces the personalized system of self-paced instruction (PSI) and offers

suggestions for using the text in a self-paced course. PSI (Personalized System of Instruction)

is one of the most popular and successful systems used for self-paced instruction. The essential

features of the PSI method are

(a) Students are permitted to pace themselves through the course at a rate commensurate with

their ability and available time.

3

designed PSI course.

The PSI method of instruction and its implementation are described in detail in the following

references:

1. Keller, Fred S. and J. Gilmour Sherman, The Keller Plan Handbook, W. A. Benjamin, Inc.,

1974.

Results of applying PSI to a first course in logic design of digital systems are described in

Roth, C.H., Continuing Effectiveness of Personalized Self-Paced Instruction in Digital Systems

Engineering, Engineering Education, Vol. 63, No. 6, March 1973.

The instructor in charge of a self-paced course will serve as course manager in addition to

his role in the classroom. For a small class, he may spend a good part of his time acting as proctor

A progress chart showing the units completed by each student is very helpful in

monitoring student progress through the course. The instructor may wish to have individual

conferences with students who fall too far behind. The instructor needs to be available in the

classroom to answer individual student questions and to assist with grading of readiness tests

as needed. He should make a special point to speak with the weak or slow students and give

them a word of encouragement. From time to time he may need to settle differences which

arise between proctors and students.

4

period, most of the time was spent evaluating unit tests. We found that a standard 50 minute

class period was not long enough for a PSI session. We usually scheduled sessions of 1½ or

2 hours or longer depending on class size. This allowed adequate time for students to have

their questions answered, take a unit test, and have their tests graded. Interactive grading of

1.4. Use of Computer Software

Three software packages are included on the CD that accompanies the textbook. The first

is a logic simulator program called SimUaid, the second is a basic computer-aided logic design

program called LogicAid, and the third is a VHDL Simulator called DirectVHDL. In addition, we

use the Xilinx ISE software for synthesizing VHDL code and downloading to CPLD or FPGA circuit

boards. The Xilinx ISE software is available at nominal cost through the Xilinx University Program

(for information, go to www.xilinx.com/university/index.htm). A “Webpack” version of the Xilinx

software is also available for downloading from the Xilinx.com website.

LogicAid provides an easy way to introduce students to the use of the computer in the logic

design process. It enables them to solve larger, more practical design problems than they could by

hand. They can also use LogicAid to verify solutions that they have worked out by hand. Instructors

can use the program for grading homework and quizzes. We first introduce LogicAid in Unit 5. The

program has a Karnaugh Map Tutorial mode that is very useful in teaching students to solve Karnaugh

map problems. This tutorial mode helps students learn to derive minimum solutions from a Karnaugh

map by informing them at each step whether that step is correct or not. It also forces them to choose

essential prime implicants first. When in the KMap tutor mode, LogicAid prints “KMT” at the top of

each output page, so you can check to see if the problems were actually solved in the tutorial mode.

students construct state tables. It allows students to check their solutions without revealing the correct

5

1.5. Suggested Equipment for Laboratory Exercises

Many types of logic lab equipment are available that are adequate to perform the lab

exercises. Since most logic design is done today using programmable logic instead of individual

ICs, we now recommend use of CPLDs or FPGAs for hardware implementation of logic circuit

designs. At the University of Texas, we are presently using the XILINX Spartan-3 FPGA boards,

Unit 1 Solutions

7

II. SOLUTIONS TO HOMEWORK PROBLEMS

Unit 1 Problem Solutions

757.2510

16 | 757 0.25

1.1 (a) 123.1710

16 | 123 0.17

EB1.616 = E × 162 + B × 161 + 1 × 160 + 6 × 16–1

= 14 × 256 + 11 × 16 + 1 + 6/16 = 3761.37510

1110 1011 0001.011(0)2

E B 1 6

1.2 (a) 59D.C16 = 5 × 162 + 9 × 161 + D × 160 + C × 16–1

= 5 × 256 + 9 × 16 + 13 + 12/16 =

1437.7510

0101 1001 1101.110016

5 9 D C

1.1 (b)

1.2 (b)

3BA.2514 = 3 × 142 + 11 × 141 + 10 × 140 + 2 × 14–1

+ 5 ×14–2

= 588 + 154 + 10 + 0.1684 = 752.168410

1.3

1457.1110

16 | 1457 0.11

16 | 91 r1 16

1.4 (b)

Unit 1 Solutions

1.12 (c) A52.A411 = 10 × 121 + 5 × 11 + 2 + 10/11 + 4/121

= 1267.9410

∴ 1305.37510 = 519.60016

= 0101 0001 1001.0110 0000 00002

5 1 9 6 0 0

1644.87510

16 | 1644 0.875

16 | 102 r12 16

6 r6 (14).000

1.10 (c)

1.5 (a) 1 1 1

1111 (Add)

+1010

1111 (Multiply)

×1010

0000

See FLD p. 730 for solutions.

101 111 010 100.101 2 = 5724.58

= 5 × 83 + 7 × 82 + 2 × 81 + 4 × 80 + 5 × 8–1

= 5 × 512 + 7 × 64 + 2 × 8 + 4 + 5/8

= 3028.62510

1.11 (a)

375.548 = 3 × 64+ 7 × 8 + 5 + 5/8 + 4/64

= 253.687510

3 | 253 0.69

3 | 84 r1 3

3 | 28 r0 (2).07

301.1210

11.3310

16 | 111 0.33

6 r15 = F16 16

(5).28

100 001 101 111.0102 = 4157.28

= 4 × 83 + 1 × 82 5 × 81 + 7 × 80 + 2 × 8–1

= 4 × 512 + 1 × 64 + 5 × 8 + 7 + 2/8

= 2159.2510

384.7410

4 | 384 0.74

4 | 96 r0 4

1.12 (a)

1.6, 1.7,

1.8, 1.9

1.5 (b, c) See FLD p. 730 for solutions.

1.10 (b)

1.10 (d)

1.11 (b)

1.12 (b)

Unit 1 Solutions

9

3 | 97 .7

3 |32 r1 3

3 |10 r2 (2).1

1110212.202113

01 11 02 12.20 21 10 = 1425.6739

Base 3 Base 9

12 5

20 6

21 7

22 8

544.19 = 5 × 92 + 4 × 91 + 4 × 90 + 1 × 9–1

= 5 × 81 + 4 × 9 + 4 + 1/9

= 445 1/910

16 | 445 1/9

1.13

1.15

(c) 16 | 97 .7

16 | 6 r1 16

0 r6 (11).2

16

1.14 (a),

(b), (c)

1.14 (d)

5 | 97 .7

5 |19 r2 5

5 |3 r4 (3).5

0 r3 5

(2).5

∴ 97.710 = 342.322..5

1.14 (e)

2983 63/6410 =

8 | 2983 0.984

1.16 (a) 93.7010

8 | 93 0.70

1.16 (b)

1900 31/3210

8 | 1900 0.969

8 | 273 r4 8

109.3010

8 | 109 0.30

8 | 13 r5 8

1.16 (c) 1.16 (d)

Unit 1 Solutions

11011 Quotient

1110 )110000001

1110

10100

1 1 1 1 1

110010 (Add) 110010 (Sub)

11101 11101

1001111 10101

110010 (Mult)

11101

110010

1 1 1 1 1

(a) 10100100 (b) 10010011

01110011 01011001

0110001 00111010

1 1

(c) 11110011

10111 Quotient

110 )10001101

110

1.17(c)

101110 Quotient

101 )11101001

101

1001

1100 Quotient

1001 )1110010

1001

1 1 1 1 1 1

1111 (Add) 1111 (Subtract)

1001 1001

11000 0110

1 1 1 1 1

1101001(Add) 1101001 (Sub)

110110 110110

10011111 110011

1011000100110

1.17 (a) 1.17 (b)

1.18

1.19(a) 1.19(b)

1.19(c) 1.20(a)

Unit 1 Solutions

11

(a) 4 + 3 is 10 in base 7, i.e., the sum digit is

0 with a carry of 1 to the next column. 1 + 5 +

4 is 10 in base 7. 1 + 6 + 0 is 10 in base 7. This

overows since the correct sum is 10007.

1.21 If the binary number has n bits (to the right of the

radix point), then its precision is (1/2n+1). So to

have the same precision, n must satisfy

1.22

8(10/11) = 7 + 3/11

8(3/11) = 2 + 2/11

8(2/11) = 1 + 5/11

8(5/11) =3 + 7/11

8(7/11) = 5+ 1/11

8(1/11) = 0 + 8/11

8(8/11) = 5 + 9/11

8(9/11) = 6 + 6/11

8(6/11) = 4 + 4/11

Expand the base b number into a power series

N = d3k-1b3k-1 + d3k-2b3k-2 + d3k-3b3k-3 + …. +

d5b5 + d4b4 + d3b3 + d2b2 + d1b1 + d0b0 + d-1b-1 +

d-2b-2 + d-3b-3 + …. + d-3m+2b-3m+2 + d-3m+1b-3m+1

+ d-3mb-3m where each di has a value from 0 to

(b-1). (Note that 0’s can be appended to the number

1.24 (a)

1011 Quotient

1010 )1110100

100011 Quotient

1011 )110000011

1.20(b) 1.20(c)

(5 - 1) = 45, (52 - 1) = 445 and (53 - 1) = 4445

(bn-1) = (b - 1)(bn-1) + bn-1

1.25(a)

1.25(b)

1.26(a) (b + 1)2 = b2 + 2b + 1 so (11b)2 = 121b if b > 2.

(b2 + b + 1)2 = b4 + 2b3 + 3b2 + 2b + 1 so

1.26(b)

Unit 1 Solutions

12

20 0 1 0

30 0 1 1

40 1 1 0

50 1 1 1

61 0 0 0

71 0 0 1

81 0 1 0

91 0 1 1

(0100)

(0101)

(1100)

(1101)

20 0 1 0

30 0 1 1

40 1 1 0

51 0 0 0

61 0 0 1

71 0 1 0

81 0 1 1

91 1 1 0

(0100)

(0101)

(1100)

(1101)

Alternate

Solutions:

7 3 2 1

00 0 0 0

10 0 0 1

20 0 1 0

30 1 0 0

(0011)

1.33

71 0 0 1

81 0 0 0

91 1 1 1

and 1's with

0's (bit-by-bit

complement).

81 1 0 1

91 1 1 0

9154 =

1110 0001 1001 1000

1.28

5-3-1-1 is possible, but

30 1 0 0

40 1 0 1

51 0 0 0

61 0 0 1

71 0 1 1

5-4-1-1 is not possible, because there is no way to

represent 3 or 8. 6-3-2-1 is possible:

6 3 2 1

00 0 0 0

10 0 0 1

1.29

1.30

(0110)

(1010)

1.27(a) (0.12)3 = (1/3 + 2/9)10 = (2/6 + 8/36)10

= (3/6 + 2/36)10 = (0.32)6

Unit 1 Solutions

13

In 2’s complement In 1’s complement

(–8) + (–11) (–8) + (–11)

In 2’s complement In 1’s complement

11 + 9 11 + 9

1.36 (c) 1.36(d)

1.35 (a) 222.2210

16 | 222 0.22

1.35 (b) 183.8110

16 | 183 0.81

In 2’s complement In 1’s complement

(–10) + (–6) (–10) + (–6)

1.36 (a) In 2’s complement In 1’s complement

(–10) + (–11) (–10) + (–11)

1.36 (b)

In 2’s complement In 1’s complement

10110 10110

+ 10011 + 10010

(1)01001 (1)01000

overow 1

01001

overow

1.37 (a)

1.37 (c)

In 2’s complement In 1’s complement

(–11) + (–4) (–11) + (–4)

01001-11010

In 2’s complement In 1’s complement

1.37 (b) In 2’s complement In 1’s complement

11010 11010

+ 00111 + 00110

(1)00001 (1)00000

1

00001

1.36 (e)

In 2’s complement In 1’s complement

11011 11011

In 2’s complement In 1’s complement

11100 11100

1.37 (e)1.37 (d)

Unit 1 Solutions

In 2’s complement In 1’s complement

10001 10001

In 2’s complement In 1’s complement

10101 10101

1.38 (c) 1.38 (d)

(a) add subt

101010 101010

overow

1.39 (a) complement

i) 00000000 (0) 11111111 (-0)

1.40

(a) (16)(4) = 64, add 2 0’s to get 6400

(10)(2) = 20, add 1 0 to get 200

(7)(1) = 7

6400 + 200 + 7 = 6607

(b) (dn-1dn-2 … d1d0)20 = dn-1(20)n-1 +

1.41 (a) If A + B < 2n-1 - 1, then the sign bit of A + B is

0 indicating a positive number with magnitude

A + B. If A + B > 2n-1 - 1, then the sign bit

of A + B is 1 indicating a negative number with

magnitude 2n - (A + B).

1.42

A and B positive: Overow does not occur if

A + B < 2n - 1 - 1 in which case A + B has a sign

bit of 0 and has a magnitude of A + B.

A positive and B negative and A > |B|: A + 2n -

1.43 1.43

cont.

A and B negative: (2n - 1 - A) + (2n - 1 - B) =

2n - 1 - (A + B) after the end-around carry. There

is no overow if A + B < 2n - 1 - 1 in which case

Unit 1 Solutions

15

1.44 Two positive numbers

No overow: 0x…x + 0x…x = 0x…x

carry in = 0 = carry out

Overow: 0x…x + 0x…x = 1x…x

carry in = 1, carry out = 0

1.45 There is no overow if the carry into the sign

position equals the carry out of the sign position.

There is overow if the carry into the sign position

does not equal the carry out of the sign position

unless an end around carry causes a carry into the

sign position.

1.46 If bn-1 = 0, then bn-22n-2 + bn-32n-3 + … + b12

+ b0 is the value of the positive number. If bn-1 =

1, then

- 2n-1 + bn-22n-2 + bn-32n-3 + … + b12 + b0

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