EE208Chap2-2.0

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Chapter 2 : Boolean Algebra and Logic Gates

Chapter 2 page: 1EE208: Logic Design 1431-1432Dr. Ridha Jemal

By Dr. Ridha Jemal

Electrical Engineering Department

College of Engineering

King Saud University

1431-1432

2.1. Basic Definitions

2.2. Basic Theorems and Properties of Boolean Algebra

2.3. Boolean Function Examples

2.4. Canonical and Standard Forms

2.5. Example of three variables functions

2.6. SOP and POS Conversion

2.7. SOP and POS Forms

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Basic Definitions


The Boolean Algebra may be defined with a set of elements, a set of operators and a

number ofpostulates to deduce rules, theorems andproperties of the system. Themost common postulates used to formulate various algebraic structure are:

o Closure: Given a binary operators * and a set of elements S. S is closed tothe binary operator (*) if for any a, b S we obtain a unique c by theoperation a*b =c

o Commutative law: A binary operator * on a set S is said to be

commutative whenever : x*y = y*z for all x,y S

o Associative law: A binary operator * on a set S is said to be associative

whenever : (x*y)*z =x*(y*z) for all x,y,z S

o Distributive law: If * and : are two binary operators on a set S, * is said to

be distributive over whenever : x*(y z) = (x*y) (x*z)

o Identity Element: A set S is said to have an identity element with respectto a binary operation * if there exists an element e S with the property:

e*x = x*e = x for every So Inverse: A set S having the identity element e with respect to a binary

operator * is said to have an inverse, for every S, there exists an element

y S such that x*y = e

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Boolean Algebra is an algebraic structure defined by a set of elements B, with two

operators + and

, provided that the following postulates are satisfied:

o Closure with respect to the operator +

Closure with respect to the operator

o Commutative with respect to + : x+y=y+x

Commutative with respect to : x y=y x

o An identity element with respect to + designated by 0

An identity element with respect to designated by 1

o Distributive over + : x (y+z) = (x y)+(x z)

o + Distributive over : x + (y z) = (x + y) (x + z)

o For every element x B, there exists an element x B called the

complement of x such that x+x=1 and x x=0

Basic Definitions

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o Identity Law 1(a): X+X =X

1(b): X.X = X

o Rule 4(a): (X)=X

Basic Theorems and Properties of Boolean Algebra

Note that every law has two expressions, (a) and (b). This is known as duality.

These are obtained by changing every AND(.) to OR(+), every OR(+) to AND(.) andall 1's to 0's and vice-versa.

It has become conventional to drop the . (AND symbol) i.e. A.B is written as AB.

o Associative Law 3(a): X+(Y+Z) = (X+Y)+Z

3(b): X.(Y.Z) =(X.Y).Z

o Distributive Law 4(a): X.(Y+Z) = XY +XZ

4(b): X+(Y.Z) =(X+Y).(X+Z)

o Commutative Law 2(a): X+Y= Y+X

2(b): X.Y =Y.X

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o

Theorem 2(a): (X+Y) = XY DeMorgan2(b): (X.Y) =X+Y

o Theorem 3(a): X+XY = X Absorption

3(b): X.(X+Y) =X

o Theorem 4(a): X+XY = X+Y

4(b): X.(X+Y) = XY

Basic Theorems and Properties of Boolean Algebra

o Theorem 1(a): X+1=1

1(b): X.0=0 By duality

Theorems

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A+ AB = A.1 + AB= A.(1 + B) +AB= A .1+ AB + AB

= A. + B(A + A)= A + B

2. Simplify the function: F(A,B,C) = (A + B + C)(A + BC) using only rules and Theorems

1. Prove the Rule Rx Algebraically: A + AB = A+B

F(A,B,C) = (A + B + C)(A + BC)= AA + ABC + BA + BBC + CA + CBC

= A(1+ BC + B + C) + BC + BCC

= A + BC

Boolean Functions Examples

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A Boolean function described by an algebraic expression consists of

binary variables, the constants 0 and 1, and the logic operation symbols.

F2=xyz + xyz + xy

Simplify the Boolean functions:

1. x(x+y)

2. x+xy

3. (x+y)(x+y)

4. xy+ xz +yz

Find the complement of the functions F1 and F2:

1. F1= (xyz +xyz)

2. F2= x(yz+yz)

Boolean Functions Examples

F1=x+xy

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Minterms and Maxterms

Definition: a minterm of n variables is a product of the variables in

which each appears exactly once in true or complemented form.

e.g.: minterms of 3 variables:

Each minterm = 1 for only one combination ofvalues of the variables,= 0 otherwise

Canonical and Standard Forms

Definition: a maxterm of n variables is a sum of the variables

in which each appears exactlyonce in true or complemented form.

e.g.: maxterm of 3 variables:

Each maxterm = 0for only one combination of values of the variables,

= 1 otherwise

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Minterms and Maxterms

Consider 2 binary variables, there are 4 possible configurations:

xy, xy, xy and xy

Each of these four AND terms is called minterm or standard product.

In a similar manner, n variables can be combined to form 2n

minterms represented by a symbol mj

Similarly, n variables forming 2n possible combinations of an ORterms called Maxterms or standard sums represented by Mj


F(a,b,c) = miWhere mi represent all minterms for which F(a,bc,)=1

F(a,b,c) = MjWhere Mj represent all minterms for which F(a,bc,)=0

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SOP and POS Forms

A Boolean function can be expressed algebraically from a given truth

table by forming a minterm for each combination of the variables that

produces 1 in the function, and then taking the OR of all those terms.


All possible minterms and maxterms are obtained from the truth table:

Every function can be written as a sum ofminterms, which is a special

kind ofSum Of Products form (SOP).

The sum ofminterms form for any function is unique

Every function can be written as a unique product ofmaxterms (POS).

If you have a truth table for a function, you can write a product of

maxterms (POS) expression by picking out the rows of the table where the

function output is 0. (Be careful ifyoure writing the actual literals!)

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Minterms Maxterms

x y z Term Design. Term Desig.

0 0 0 xyz m0 x+y+z M0

0 0 1 xyz m1 x+y+z M1

0 1 0 xyz m2 x+y+z M2

0 1 1 xyz m3 x+y+z M3

1 0 0 xyz m4 x+y+z M4

1 0 1 xyz m5 x+y+z M5

1 1 0 xyz m6 x+y+z M6

1 1 1 xyz m7 x+y+z M7


Minterms and Maxterms for Three variables

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x y z F1 F2

0 0 0 0 0

0 0 1 1 0

0 1 0 0 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

F1=m1+m4+m7

= m(1,4,7)F2=m3+m5+m6+m7

= m(3,5,6,7)

F1=M0.M2.M3.M5.M6

F2=M0.M1.M2.M4


If you have a truth table for a function, you can write a sum of minterms

expression just by picking out the rows of the table where the function

output is 1.

Express F1 and F2 as a SOP form and its equivalent form using Maxterms.

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x y z F F

0 0 0 1 0

0 0 1 1 0

0 1 0 1 0

0 1 1 1 0

1 0 0 0 1

1 0 1 0 1

1 1 0 1 0

1 1 1 0 1

F=m0+m1+m2+m3+m6

=m(0,1,2,3,6)

F=m4+m5+m7

=m(4,5,7)

Example of three variables functions

F contains all the minterms not in F.

Express F and F as a SOP form.

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x y z F F

0 0 0 1 0

0 0 1 1 0

0 1 0 1 0

0 1 1 1 0

1 0 0 0 1

1 0 1 0 1

1 1 0 1 0

1 1 1 0 1

F = M4+M5+M7

=M(4,5,7)F = M0+M1+M2+M3+M6

=M (0,1,2,3,6)

Express F and F as a POS form (using Maxterms).

If you have a truth table for a function, you can write a product ofmaxterms (POS) expression by picking out the rows of the table where the

function output is 0. (Be careful ifyoure writing the actual literals!)

F contains all the maxterms not in F.

Example of three variables functions

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We can convert a sum ofminterms to a product ofmaxterms

Consider F(A,B,C)=m(1,3,5,7)

And F(A,B,C) = m(0,2,4,6)

= m0 + m2+m4+m6Complementing (F) = (m0 + m2+m4+m6)

So F = m0. m2. m4 . m6

= M0.M2.M4.M6=M (0,2,4,6)

SOP and POS Conversion

In general, just replace the minterms with maxterms, using maxterm numbers

that dont appear in the sum ofminterms:

F(A,B,C)= m(1,3,5,7)=M (0,2,4,6)

The same thing works for converting from a product of maxterms to a sum of

minterms.

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Sum of Minterms

Sometimes, its convenient to express a Boolean function in its sum-of-mintermsform. If the function is not in this form, it can be made by first expanding the

expression into a sum ofAND terms. Each term is then inspected to see if it contains

all the variables. If it misses one or more, it is ANDed with an expression such as

x+x, where x is the missing variable.

Example: Express F=A+BCas a sum of minterms or a SOP.

Then first term A is missing 2 variables B and C

A=A(B+B) = A B +AB

AB =AB(C+C)

= ABC +ABC

AB =AB(C+C)=ABC+ABC

Then second term BCis missing one variables A

BC = BC(A+A) =BCA+BCA

SOP and POS Forms

F(A,B,C)=ABC + ABC+ABC+ABC+ABC= m(1,4,5,6,7)

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EE208Chap2-2.0