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BasicEngineering

Boolean Algebra and Logic Gates

F Hamer, M Lavelle & D McMullanTheaim of this document is to provide a short, self assessment programme for students who wish to understand the basic techniques of logic gates.c ?2005Email: chamer,mlavelle,dmcmullan@plymouth.ac.uk

LastRevision Date: August 31, 2006Version 1.0

Table of Contents

1.Logic Gates (Introduction)

2.Truth Tables

3.Basic Rules of Boolean Algebra

4.Boolean Algebra

5.Final Quiz

Solutions to Exercises

Solutions to Quizzes

The full range of these packages and some instructions, should they be required, can be obtained from our web pageMathematics Support Materials.

Section 1: Logic Gates (Introduction) 3

1. Logic Gates (Introduction)

The packageTruth Tables and Boolean Algebraset out the basic principles of logic. Any Boolean algebra operation can be associated with an electronic circuit in which the inputs and outputs represent the statements of Boolean algebra. Although these circuits may be complex, they may all be constructed from three basic devices. These are theANDgate, theORgate and theNOTgate. x yx·y

AND gatex

yx+y

OR gatexx?

NOT gate

In the case of logic gates, a differentnotationis used:x?y, the logicalANDoperation, is replaced byx·y, orxy.

x?y, the logicalORoperation, is replaced byx+y. ¬x, the logicalNEGATIONoperation, is replaced byx ?orx. The truth valueTRUEis written as1(and corresponds to a high voltage), andFALSEis written as0(low voltage).

Section 2: Truth Tables 4

2. Truth Tables

x yx·yxyx·y000 010 100
111

Summary of AND gate

xyx+y000 011 101
111

Summary of OR gatex

yx+y xx?xx ?01 10

Summary of NOT gate

Section 3: Basic Rules of Boolean Algebra 5

3. Basic Rules of Boolean Algebra

The basic rules for simplifying and combining logic gates are called Boolean algebra in honour of George Boole (1815-1864) who was a self-educated English mathematician who developed many of the key ideas. The following set of exercises will allow you to rediscover the basic rules:Example 1x 1 Consider theANDgate where one of the inputs is1. By using the truth table, investigate the possible outputs and hence simplify the expressionx·1. SolutionFrom the truth table forAND, we see that ifxis1then

1·1 = 1, while ifxis0then0·1 = 0. This can be summarised in the

rule thatx·1 =x, i.e., x 1x

Section 3: Basic Rules of Boolean Algebra 6

Example 2

x 0 Consider theANDgate where one of the inputs is0. By using the truth table, investigate the possible outputs and hence simplify the expressionx·0. SolutionFrom the truth table forAND, we see that ifxis1then

1·0 = 0, while ifxis0then0·0 = 0. This can be summarised in the

rule thatx·0 = 0x 00

Section 3: Basic Rules of Boolean Algebra 7

Exercise 1.(Click on thegreenletters for the solutions.) Obtain the rules for simplifying the logical expressions(a)x+ 0which corresponds to the logic gatex 0 (b)x+ 1which corresponds to the logic gatex 1 Exercise 2.(Click on thegreenletters for the solutions.) Obtain the rules for simplifying the logical expressions:(a)x+xwhich corresponds to the logic gatex (b)x·xwhich corresponds to the logic gatex

Section 3: Basic Rules of Boolean Algebra 8

Exercise 3.(Click on thegreenletters for the solutions.) Obtain the rules for simplifying the logical expressions:(a)x+x?which corresponds to the logic gatex (b)x·x?which corresponds to the logic gatex QuizSimplify the logical expression(x?)?represented by the following circuit diagram.x (a)x(b)x ?(c)1(d)0

Section 3: Basic Rules of Boolean Algebra 9

Exercise 4.(Click on thegreenletters for the solutions.) Investi- gate the relationship between the following circuits. Summarise your conclusions using Boolean expressions for the circuits.(a)x yx y (b)x yx y The important relations developed in the above exercise are called De Morgan"s theorems and are widely used in simplifying circuits. These correspond to rules (8a) and (8b) in the table of Boolean identities on the next page.

Section 4: Boolean Algebra 10

4. Boolean Algebra

(1a)x·y=y·x(1b)x+y=y+x(2a)x·(y·z)=(x·y)·z(2b)x+ (y+z)=(x+y) +z(3a)x·(y+z)=(x·y) + (x·z)(3b)x+ (y·z)=(x+y)·(x+z)(4a)x·x=x

(4b)x+x=x (5a)x·(x+y)=x (5b)x+ (x·y)=x (6a)x·x?=0 (6b)x+x?=1 (7)(x?)?=x (8a)(x·y)?=x ?+y?(8b)(x+y)?=x ?·y?

Section 4: Boolean Algebra 11

These rules are a direct translation into the notation of logic gates of the rules derived in the packageTruth Tables and Boolean Algebra. We have seen that they can all be checked by investigating the corresponding truth tables. Alternatively, some of these rules can

be derived from simpler identities derived in this package.Example 3Show how rule (5a) can be derived from the basic iden-

tities derived earlier.Solutionx·(x+y)=x·x+x·yusing (3a) =x+x·yusing (4a) =x·(1 +y)using (3a) =x·1using Exercise 1 =xas required.Exercise 5.(Click on thegreenletter for the solution.) (a)Show how rule (5b) can be derived in a similar fashion.

Section 4: Boolean Algebra 12

The examples above have all involved at most two inputs. However, logic gates can be put together to join an arbitrary number of inputs. The Boolean algebra rules of the table are essential to understand

when these circuits are equivalent and how they may be simplified.Example 4Let us consider the circuits which combine three inputs

viaANDgates. Two different ways of combining them are x y z(x·y)·z and x y zx·(y·z)

Section 4: Boolean Algebra 13

However, rule (2a) states that these gates are equivalent. The order of takingANDgates is not important. This is sometimes drawn as a three (or more!) inputANDgate xyzx·y·z but really this just means repeated use ofANDgates as shown above. Exercise 6.(Click on thegreenletter for the solution.) (a)Show two different ways of combining three inputs viaORgates and explain why they are equivalent. This equivalence is summarised as a three (or more!) inputORgate xyzx+y+z this just means repeated use ofORgates as shown in the exercise.

Section 5: Final Quiz 14

5. Final Quiz

Begin Quiz

1.Select the Boolean expression that isnotequivalent tox·x+x·x?(a)x·(x+x?)(b)(x+x?)·x(c)x

?(d)x

2.Select the expression which is equivalent tox·y+x·y·z(a)x·y(b)x·z(c)y·z(d)x·y·z3.Select the expression which is equivalent to(x+y)·(x+y?)(a)y(b)y

?(c)x(d)x

?4.Select the expression that isnotequivalent tox·(x?+y) +y(a)x·x?+y·(1 +x)(b)0 +x·y+y(c)x·y(d)y

End Quiz

Solutions to Exercises 15

Solutions to Exercises

Exercise 1(a)From the truth table forOR, we see that ifxis1then

1 + 0 = 1, while ifxis0then0 + 0 = 0. This can be summarised in

the rule thatx+ 0 =xx 0x

Click on the green square to return?

Solutions to Exercises 16

Exercise 1(b)From the truth table forORwe see that ifxis1then

1 + 1 = 1, while ifxis0then0 + 1 = 1. This can be summarised in

the rule thatx+ 1 = 1x 11

Click on the green square to return?

Solutions to Exercises 17

Exercise 2(a)From the truth table forOR, we see that ifxis1then x+x= 1+1 = 1, while ifxis0thenx+x= 0+0 = 0. This can be summarised in the rule thatx+x=xxx

Click on the green square to return?

Solutions to Exercises 18

Exercise 2(b)From the truth table forAND, we see that ifxis1 thenx·x= 1·1 = 1, while ifxis0thenx·x= 0·0 = 0. This can be summarised in the rule thatx·x=xxx

Click on the green square to return?

Solutions to Exercises 19

Exercise 3(a)From the truth table forOR, we see that ifxis1then x+x?= 1 + 0 = 1, while ifxis0thenx+x?= 0 + 1 = 1. This can be summarised in the rule thatx+x?= 1x1

Click on the green square to return?

Solutions to Exercises 20

Exercise 3(b)From the truth table forAND, we see that ifxis1 thenx·x?= 1·0 = 0, while ifxis0thenx·x?= 0·1 = 0. This can be summarised in the rule thatx·x?= 0x0

Click on the green square to return?

Solutions to Exercises 21

Exercise 4(a)The truth tables are:x

yxyx+y(x+y)?0001 0110
1010
1110
x yxyx ?y ?x ?·y?00111 01100
10010
11000

From these we deduce the identity

x y(x+y)?=x yx?·y?

Click on the green square to return?

Solutions to Exercises 22

Exercise 4(b)The truth tables are:x

yxyx·y(x·y)?0001 0101
1001
1110
x yxyx ?y ?x ?+y?00111 01101
10011
11000

From these we deduce the identity

x y(x·y)?=x yx?+y?

Click on the green square to return?

Solutions to Exercises 23

Exercise 5(a)

x+x·y=x·(1 +y)using (3a) =x·1using Exercise 1 =xas required.?

Solutions to Exercises 24

Exercise 6(a)Two different ways of combining them arex y z(x+y)+z and x y zx+(y+z) However, rule (2b) states that these gates are equivalent. The order of takingORgates is not important.?

Solutions to Quizzes 25

Solutions to Quizzes

Solution to Quiz:From the truth table forNOTwe see that ifx is1then(x?)?= (1?)?= (0)?= 1, while ifxis0then(x?)?= (0?)?= (1) ?= 0. This can be summarised in the rule that(x?)?=xxx

End Quiz

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