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CHAPTER 3

Boolean Algebra
Contents









Describing Logic Circuits Algebraically
rules of evaluating a Boolean expression
Boolean Theorems
DeMorgan's Theorem
Universality of NAND and NOR Gates
Alternate Logic Gate Representations
Minterms and Maxterms
STANDARD FORMS

2
Describing Logic Circuits Algebraically


OR gate, AND gate, and NOT circuit are
the basic building blocks of digital systems

3
Circuits containing Inverters

4
Evaluating Logic Circuit Outputs


Let A=0, B=0, C=1, D=1, E=1
X

= [D+ ((A+B)C)'] • E
= [1 + ((0+0)1 )'] • 1
= [1 + (0•1)'] • 1
= [1+ 0'] •1
= [1+ 1 ] • 1
=1

5
Rules of evaluating a Boolean expression
First, perform all inversions of single
terms; that is, 0 = 1 or 1 = 0.
 Then perform all operations within
parentheses.
 Perform an AND operation before an OR
operation unless parentheses indicate
otherwise.
 If an expression has a bar over it, perform
the operations of the expression first and
then invert the result.


6
Determining Output Level from a Diagram

7
Boolean
Theorems

8
Multivariable Theorems
(9)
(10)
(11)
(12)
(13.a)
(13.b)
(13.c)
(14)
(15)
(16)

x + y = y + x (Commutative law)
x • y = y • x (Commutative law)
x+ (y+z) = (x+y) +z = x+y+z (Associative law)
x (yz) = (xy) z = xyz (Associative law)
x (y+z) = xy + xz (Distributive law)
x + yz = (x + y) (x + z) (Distributive law)
(w+x)(y+z) = wy + xy + wz + xz
x + xy = x (Absorption) [proof]
x + x'y = x + y
(x +y)(x + z) = x +yz
9
Proof of (14, 15, 16)
x + xy

x + x’y

= x (1+y)
= x • 1 [using theorem (6)]
= x [using theorem (2)]
= ( x + x’) (x + y) [theorem 13b]
= 1 (x +y)
= (x + y)

(x +y)(x + z) =xx + xz + yx + yz
= x + xz + yx + yz
= x (1+z+y) +yz
= x . 1 + yz
= x + yz

10
DeMorgan's Theorem
(18) (x+y)' = x' • y'
 (19) (x•y)' = x' + y'
 Example


X

= [(A'+C) • (B+D')]'
= (A'+C)' + (B+D')'
= (AC') + (B'D)
= AC' + B'D

11
Three Variables DeMorgan's Theorem
(20) (x+y+z)' = x' • y' • z'
 (21) (xyz)' = x' + y' + z‘




EXAMPLE:
 Apply DeMorgan’s theorems to each of the
following expressions:
 (a)
( A + B + C) D


(b)

ABC + DEF



(c)

A B + CD + EF
12
Universality of NAND Gates

13
Universality of NOR Gates

14
Alternate Logic Gate
Representations

15
Minterms and Maxterms
x y

z

0 0

0

Minterms
Term Designation
x' y’ z' m0

Maxterms
Term
Designation
x+y+z
M0

0 0

1

x' y' z

M1

0 1

0

x' y z’

m2

0 1

1

x' y z

m3

1 0

0

x y' z’

m4

1 0

1

x y' z

m5

1 1

0

x y z’

m6

1 1

1

xyz

m7

x+y+z’
+y
x+y’+z
+y
-t- Z
x+y’+z’
+ y'
+2
x'+y+z
+y'
+ '+y+z’
xt
,
'+y
+ '+y’+z
x
z
4-'+y’+z’
x
2'

M1
M2
M3
M4
M5
M6
M7

2
16
Canonical FORMS


There are two types of canonical forms:



the sum of minterms
The product of maxterms

17
Sum of minterms


f1 = x'y'z + xy'z' + xyz = m1 + m4 +m7



f2 = x'yz + xy'z + xyz’ + xyz = m3 + m5+
m6 + m7
x
0
0
0
0
1
1
1
1

y
0
0
1
1
0
0
1
1

Z
0
1
0
1
0
1
0
1

f1
0
1
0
0
1
0
0
1

f2
0
0
0
1
0
1
1
1
18
Product of maxterms
The complement of f1 is read by forming a
minterm for each combination that
produces a 0 as:
 f1’=x’y’z’ + x’yz’ + x’yz + xy’z + xyz’




f1 = (x + y + z)(x + y' + z)(x + y' + z' )(x’+ y +
z)(x’ + y' + z)
= Mo M2 M3 M5 M6

Similarly
 f2 = ?


19
Example: Sum of Minterms


Express the Boolean function F = A + B'C
in a sum of minterms.



F=A+B'C
= ABC + ABC' + AB'C + AB'C' + AB'C + A'B'C

20
Example: Product of Maxterms


Express the Boolean function F =xy' + yz in
a product of maxterm form.



F = xy' + yz = (xy' + y)(xy' + z) = (x + y)(y' + y)(x + z)(y' + z)
= (x + y)(x + z)(y' + z) = (x + y + zz')(x + yy' + z)(xx' + y' + z)
= (x + y + z)(x + y + z')(x+y + z)(x+y’+ z)(x + y' + z)(x'+y'+z)
= (x + y + z)(x + y + z') (x + y' + z) (x'+y'+z)
= M0 M1 M2 M6
= Π (0,1,2,6)

21
STANDARD FORMS


There are two types of standard forms:



the sum of products (SOP)
The product of sums (POS).

22
Sum of Products
The sum of products is a Boolean
expression containing AND terms, called
product terms, of one or more literals
each. The sum denotes the ORing of these
terms.
 F = xy + z +xy'z'. (SOP)


23
Product of Sums
A product of sums is a Boolean expression
containing OR terms, called sum terms.
Each term may have any number of
literals. The product denotes the ANDing
of these terms.
 F = z(x+y)(x+y+z) (POS)
 F = x (xy' + zy)
(nonstandard form)


24

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Boolean Algebra Fundamentals

  • 2. Contents         Describing Logic Circuits Algebraically rules of evaluating a Boolean expression Boolean Theorems DeMorgan's Theorem Universality of NAND and NOR Gates Alternate Logic Gate Representations Minterms and Maxterms STANDARD FORMS 2
  • 3. Describing Logic Circuits Algebraically  OR gate, AND gate, and NOT circuit are the basic building blocks of digital systems 3
  • 5. Evaluating Logic Circuit Outputs  Let A=0, B=0, C=1, D=1, E=1 X = [D+ ((A+B)C)'] • E = [1 + ((0+0)1 )'] • 1 = [1 + (0•1)'] • 1 = [1+ 0'] •1 = [1+ 1 ] • 1 =1 5
  • 6. Rules of evaluating a Boolean expression First, perform all inversions of single terms; that is, 0 = 1 or 1 = 0.  Then perform all operations within parentheses.  Perform an AND operation before an OR operation unless parentheses indicate otherwise.  If an expression has a bar over it, perform the operations of the expression first and then invert the result.  6
  • 7. Determining Output Level from a Diagram 7
  • 9. Multivariable Theorems (9) (10) (11) (12) (13.a) (13.b) (13.c) (14) (15) (16) x + y = y + x (Commutative law) x • y = y • x (Commutative law) x+ (y+z) = (x+y) +z = x+y+z (Associative law) x (yz) = (xy) z = xyz (Associative law) x (y+z) = xy + xz (Distributive law) x + yz = (x + y) (x + z) (Distributive law) (w+x)(y+z) = wy + xy + wz + xz x + xy = x (Absorption) [proof] x + x'y = x + y (x +y)(x + z) = x +yz 9
  • 10. Proof of (14, 15, 16) x + xy x + x’y = x (1+y) = x • 1 [using theorem (6)] = x [using theorem (2)] = ( x + x’) (x + y) [theorem 13b] = 1 (x +y) = (x + y) (x +y)(x + z) =xx + xz + yx + yz = x + xz + yx + yz = x (1+z+y) +yz = x . 1 + yz = x + yz 10
  • 11. DeMorgan's Theorem (18) (x+y)' = x' • y'  (19) (x•y)' = x' + y'  Example  X = [(A'+C) • (B+D')]' = (A'+C)' + (B+D')' = (AC') + (B'D) = AC' + B'D 11
  • 12. Three Variables DeMorgan's Theorem (20) (x+y+z)' = x' • y' • z'  (21) (xyz)' = x' + y' + z‘   EXAMPLE:  Apply DeMorgan’s theorems to each of the following expressions:  (a) ( A + B + C) D  (b) ABC + DEF  (c) A B + CD + EF 12
  • 16. Minterms and Maxterms x y z 0 0 0 Minterms Term Designation x' y’ z' m0 Maxterms Term Designation x+y+z M0 0 0 1 x' y' z M1 0 1 0 x' y z’ m2 0 1 1 x' y z m3 1 0 0 x y' z’ m4 1 0 1 x y' z m5 1 1 0 x y z’ m6 1 1 1 xyz m7 x+y+z’ +y x+y’+z +y -t- Z x+y’+z’ + y' +2 x'+y+z +y' + '+y+z’ xt , '+y + '+y’+z x z 4-'+y’+z’ x 2' M1 M2 M3 M4 M5 M6 M7 2 16
  • 17. Canonical FORMS  There are two types of canonical forms:   the sum of minterms The product of maxterms 17
  • 18. Sum of minterms  f1 = x'y'z + xy'z' + xyz = m1 + m4 +m7  f2 = x'yz + xy'z + xyz’ + xyz = m3 + m5+ m6 + m7 x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 f1 0 1 0 0 1 0 0 1 f2 0 0 0 1 0 1 1 1 18
  • 19. Product of maxterms The complement of f1 is read by forming a minterm for each combination that produces a 0 as:  f1’=x’y’z’ + x’yz’ + x’yz + xy’z + xyz’   f1 = (x + y + z)(x + y' + z)(x + y' + z' )(x’+ y + z)(x’ + y' + z) = Mo M2 M3 M5 M6 Similarly  f2 = ?  19
  • 20. Example: Sum of Minterms  Express the Boolean function F = A + B'C in a sum of minterms.  F=A+B'C = ABC + ABC' + AB'C + AB'C' + AB'C + A'B'C 20
  • 21. Example: Product of Maxterms  Express the Boolean function F =xy' + yz in a product of maxterm form.  F = xy' + yz = (xy' + y)(xy' + z) = (x + y)(y' + y)(x + z)(y' + z) = (x + y)(x + z)(y' + z) = (x + y + zz')(x + yy' + z)(xx' + y' + z) = (x + y + z)(x + y + z')(x+y + z)(x+y’+ z)(x + y' + z)(x'+y'+z) = (x + y + z)(x + y + z') (x + y' + z) (x'+y'+z) = M0 M1 M2 M6 = Π (0,1,2,6) 21
  • 22. STANDARD FORMS  There are two types of standard forms:   the sum of products (SOP) The product of sums (POS). 22
  • 23. Sum of Products The sum of products is a Boolean expression containing AND terms, called product terms, of one or more literals each. The sum denotes the ORing of these terms.  F = xy + z +xy'z'. (SOP)  23
  • 24. Product of Sums A product of sums is a Boolean expression containing OR terms, called sum terms. Each term may have any number of literals. The product denotes the ANDing of these terms.  F = z(x+y)(x+y+z) (POS)  F = x (xy' + zy) (nonstandard form)  24

Notas del editor

  1. We can say that X = A’BC (A’D’).
  2. X = [D+ ((A+B)C)']•E = [D+((A+B)’+C’)]•E = [D+(A’B’+C’)]•E = [1+(1•1+0)]•1 = [1+(1+0)]•1 = [1+1] •1 = 1•1 = 1. 
  3. X in the first diagram can = A’BC (A’D’)
  4. Commutative = Eltabdel - Associative = El2edmag - Distributive = Eltawze3 - Absorption = El2emtsas (mass) Proof:- x+xy = x.1+x.y = x(1+y) = x.1 = x 
  5. AND will be OR and nwza3 el bar 3lehom el 2, w el3ks sa7e7 
  6. Universal because it do the 3 operation INVERTER, AND & OR
  7. Universal because it do the 3 operation INVERTER, OR & AND.