Upcoming SlideShare
×

# Logic Design - Chapter 3: Boolean Algebra

700
-1

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
4 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
700
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
78
0
Likes
4
Embeds 0
No embeds

No notes for slide
• We can say that X = A’BC (A’D’).
• X = [D+ ((A+B)C)&apos;]•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. 
• X in the first diagram can = A’BC (A’D’)
• Commutative = Eltabdel - Associative = El2edmag - Distributive = Eltawze3 - Absorption = El2emtsas (mass)
Proof:- x+xy = x.1+x.y = x(1+y) = x.1 = x 
• AND will be OR and nwza3 el bar 3lehom el 2, w el3ks sa7e7 
• Universal because it do the 3 operation INVERTER, AND &amp; OR
• Universal because it do the 3 operation INVERTER, OR &amp; AND.
• ### 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
1. #### A particular slide catching your eye?

Clipping is a handy way to collect important slides you want to go back to later.