Boolean Algebra•Purpose of BA is to facilitates design and analysis of digital circuits.

•For example the value of boolean function F=A + BC’ with the

following gate implementation can be shown by this truth table:

ABC

A+BC

A B C BC A+BC00001111

00110011

01010101

00100010

00101111

Boolean Algebra• Basic identities of boolean algebra

1. X 0 X

3. X 1 1

5. X X X 7. X X ’ 1excluded

middle

9. (X ’)’ X involution

10. X Y Y X

12. XYZ ) (XY )Z

14. XYZ ) XY XZ

16. X Y ) X Y 18. X + XY = X

2. X1 X identity

4. X0 0 base

6. XX X idempotence

8. XX’ 0 non contradiction

11. XY YX commutative

13. XYZ ) (XY )Z associative

15. XYZ ) XY )XZ ) distributive

17. XY)’ = XY demorgan

19. X.(X+Y) = x absorbtion

Basic identities of B.A. can be proven by truth table:

XYZ ) XY )XZ )YZ

00010001

XYZ000001010011100101110111

XYZ )00011111

XY00111111

XZ01011111

XY )XZ )00011111

X Y ) X Y XY00011011

X Y 1000

X Y )1000

demorgan

distributive

Boolean Algebra

• For each algebraic expression the dual of of algebraic expression achieved by interchanging AND and OR operators and replacing 0’s and 1’s.

• Parallel columns illustrate duality principle. The duality principle states that if E1 and E2 are Boolean expressions then

E1= E2 dual (E1)=dual (E2)where dual(E) is the dual of E

• Note: 15-17 have no counterpart in ordinary algebra.

• Other handy identity.X+X’Y=X+Y (15, 7 and 2)

Boolean Algebra

• By using boolean algebra rules, a simpler expression may be obtained

• Operator Precedence: when evaluating boolean expression , order of precedence is:

1. Parentheses

2. NOT

3. AND

4. OR

For example :Look at DeMorgan truth table first (x+Y) is computed then complement of (X+Y).

But for x’.y’ first the complement of x and complement of y is computed and then the result is ANDed

Boolean Function Simplification

Boolean function simplification

• It means by manipulation of B.A. reducing the number of terms and literals in the function. For example:

f= x’y’z + x’yz + xy’

= x’(y’z + yz) + xy’

= x’ (z(y’+y)) + xy’

= x’(z.1) +xy’

= x’z + xy’

The Consensus Theorem

Theorem. XY +YZ + X Z = XY + X ZProof. XY +YZ + X Z = XY + (X + X )YZ + X Z

2,7

= XY + XYZ + X YZ + X Z 14

= XY(1 + Z ) + X Z(Y + 1) 2,11,14

= XY + X Z 3,2

Dual. X + Y )Y + Z )X + Z ) = X + Y )X + Z )

Complement of a Function

There are two ways for doing that:1) Using DeMorgan’s theorem2) Taking the dual of the function and

complement each literalFor example complements of x’yz’ + x’y’z = (x+y’+ z)(x+ y+z’) x(y’z’ + yz) = x’ + (y+z)(y’ + z’)

Canonical and standard Forms• The sum of products is one of two standard forms for Boolean

expressions.sum-of-products-expression = p-term + p-term ... + p-termp-term = literal literal literal– example. X Y Z + X Z + XY + XYZ

• A minterm is a term that contains every variable, in either complemented or un-complemented form.– example. in expression above, X Y Z is minterm, but X Z is

not

• A sum of minterms expression is a sum of products expression in which every term is a minterm.– example: X Y Z + X YZ + XYZ + XYZ is sum of minterms

expression that is equivalent to expression above.– shorthand : list minterms numerically, so X Y Z + X YZ +

XYZ + XYZ becomes 001+011+110+111 or m (1,3,6,7)

Canonical and standard Forms• The product of sums is the second standard form for Boolean

expressions.product-of-sums-expression = s-term s-term ... s-terms-term = literal literal literal– example. (X +Y +Z )(X +Z )(X +Y )(X +Y +Z )

• A maxterm is a sum term that contains every variable, in complemented or uncomplemented form.– example. in exp. above, X +Y +Z is a maxterm, but X +Z is

not

• A product of maxterms expression is a product of sums expression in which every term is a maxterm.– example. (X +Y +Z )(X +Y+Z )(X+Y+Z )(X+Y+Z ) is product of

maxterms expression that is equivalent to expression above.– shorthand : list maxterms numerically: so, (X +Y +Z )(X

+Y+Z) (X+Y+Z )(X+Y+Z ) becomes 110+100+001+000 or– M(6,4,1,0)

How to build the boolean function from truth table

• One way is to find a minterms or standard products by ANDing the terms of the n variable, each being primed if it is 0 and unprimed if it is 1. A boolean function can be formed by forming a minterm for each combination of variables that produce 1 in the function and then taking OR of all those forms.

• Another way is by finding maxterms or standard sums by OR term of the n variables, with each variable being unprimed if corresponding bit is 0 and primed if it is 1. A boolean function can be formed as a product of maxterms for each combination of variables that produce 0 in the function and then form And of all those forms

For example:

x y z function f1 function f2 minterms maxterms 0 0 0 0 0 m0 M0 0 0 1 1 0 m1 M1 0 1 0 0 0 m2 M2 0 1 1 0 1 m3 M3 1 0 0 1 0 m4 M4 1 0 1 0 1 m5 M5 1 1 0 0 1 m6 M6 1 1 1 1 1 m7 M7 Sum of mintermsf1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7f2= x’yz + xy’z + xyz’ + xyx = m3 + m5 + m6 + m7

Product of maxtermsf1= (x+y+ z)(x+y’+z)(x’+ y + z’)(x’ + y + z) = M0M2M3M5M6f2= (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z) = M0M1M2M4

Conversion between Canonical Forms

• By reading from a truth table the two canonical forms ( sum of minterms and product of maxterms) can be obtained.

• A boolean function can be converted to the canonical form. For example:

F= xy + x’y (in form of sum of the products S.O.P)

= (z+ z’)xy + x’z(y+y) = xyz + xyz’ + x’zy + x’zy’ can be converted to sum of minterms that can be shown by m1+m3+m6+m7 or

∑(1,3,6,7)

Conversion between Canonical Forms

• To convert it to the product of maxterms

F= xy + x’y = (xy + x’)(xy +z) =

(x’ + x) (x’+ y)(x+z)(y+z) = (x’+y)(x+z)(y+z)

it is in the form of products of sums (P.O.S) but we want the product of maxterms. So

= (x’+y+(z.z’)) (x + z + (y.y’))(y +z + (x.x’))

= (x’ + y + z)(x + z + y)(x+ z + y’)(x’ + y +z’)

= M0M2M4M5=∏(0,2,4,5)

Conversion between Canonical Forms• To convert from one canonical form to another,

interchange the symbol ∑ and ∏ and list those numbers missing from the total number of minterms or maxterms which is 2n,where n is number of variables.

• To verify that we convert sum of minterms to the product of maxterms by finding the complement of a function presented as sum of minterms. From the result the product of maxterms can be easily obtained

For example: F(x,y,z) = m1+m3+m6+m7= ∑(1,3,6,7) the complement of F (presented in the form of sum of the minterms) are the minterms that makes F to be zero thus

F’(x,y,z) = (∑(1,3,6,7))’= (m0 + m2 + m4 + m5) F = (F’(x,y,z))’ = (m0 + m2 + m4 + m5)’ Using Demorgan’s =m0’m2’m4’m5’= since each m’j = Mj it is M0M2M4M5 = ∏(0,2,4,5)

Standard forms• Sometimes boolean functions are shown as standard forms.

For example: F1 =y’ + xy + x’y’z’ ( sum of products) F2 = x(y’ + z) (x’ + y + z’) (product of sums) the product and sum can be used to make the gate structure

consist of AND and OR gates

• Sometimes boolean function can be shown in non standard forms:

F3= AB + C(D + E) can be changed to AB + CD + CE

• Different forms results different level of implementation of logical gates (see the next slide)