Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Upcoming SlideShare
Loading in …5
×

# Logic Design - Chapter 3: Boolean Algebra

822 views

Published on

- 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.

Published in: Education, Technology
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

### Logic Design - Chapter 3: Boolean Algebra

1. 1. CHAPTER 3 Boolean Algebra
2. 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. 3. Describing Logic Circuits Algebraically  OR gate, AND gate, and NOT circuit are the basic building blocks of digital systems 3
4. 4. Circuits containing Inverters 4
5. 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. 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. 7. Determining Output Level from a Diagram 7
8. 8. Boolean Theorems 8
9. 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. 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. 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. 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
13. 13. Universality of NAND Gates 13
14. 14. Universality of NOR Gates 14
15. 15. Alternate Logic Gate Representations 15
16. 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. 17. Canonical FORMS  There are two types of canonical forms:   the sum of minterms The product of maxterms 17
18. 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. 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. 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. 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. 22. STANDARD FORMS  There are two types of standard forms:   the sum of products (SOP) The product of sums (POS). 22
23. 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. 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