chapter 3 boolean algebra and logic gate (part 2)
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Chapter 3
Boolean Algebra and Logic Gate (Part 2)
1. Identity Elements 2. Inverse Elements 1 . A = A A . A = 0 0 + A = A A + A = 1 3. Idempotent Laws 4. Boundess Laws A + A = A A + 1 = 1 A . A = A A . 0 = 0 5. Distributive Laws 6. Order Exchange Laws A . (B + C) = A.B + A.C A . B = B . A A + (B . C) = (A+B) . (A+C) A + B = B + A 7. Absorption Laws 8. Associative Laws A + (A . B) = A A + (B + C) = (A + B) + C A . (A + B) = A A . (B . C) = (A . B) . C 9. Elimination Laws 10. De Morgan Theorem A + (A . B) = A + B (A + B) = A . B A . (A + B) = A . B (A . B) = A + B
Basic Theorems of Boolean Algebra
Truth Table
A B C Q
0 0 1 0
1 0 1 0
0 1 1 0
1 1 0 1
A = 0, B = 0, C = 1, Q = 0
A = 1, B = 0, C = 1, Q = 0
A = 0, B = 1, C = 1, Q = 0
A = 1, B = 1, C = 0, Q = 1
Truth Table
A B Q
0 0 0
1 0 1
0 1 0
1 1 0
Truth Table
p q r p . qp . q qq q + rq + r
0 0 0 0 1 1
0 0 1 0 1 1
0 1 0 0 0 0
0 1 1 0 0 1
1 0 0 0 1 1
1 0 1 0 1 1
1 1 0 1 0 0
1 1 1 1 0 1
p q p+q p.(p+q) p.q p.p (p.p)+(p.q)
0 0 0 0 0 0 0
0 1 1 0 0 0 0
1 0 1 1 0 1 1
1 1 1 1 1 1 1
Logically equivalentp.(p+q) = (p.p)+(p.q)
Relationship Between Boolean Function and Logic Circuit
AB
F
A.B = AB
CD C + D
= AB + C + D
G = A . (B + C + D)
A
B
CD
G = A . (B + C + D)
C + D
B + C + D
A
B Q
AAB
B= AB + B
Produce a truth table from the logic circuit
A B A AB Q
0 0 1 0 0
0 1 1 1 1
1 0 0 0 0
1 1 0 0 1
Elimination Laws: A.(A + B) = A.B
Proof using truth table.
A B A A + B A.B A.(A + B)
0 0 1 1 0 0
0 1 1 1 0 0
1 0 0 0 0 0
1 1 0 1 1 1
Draw the logic circuit for output = A.(A + B)
A A
A.(A+B)
A
A + B
Draw the logic circuit for output = A.(A + B)
B
De Morgan Theorem : (A + B) = A . B
Proof using truth table.
A B A+B A B A.B (A + B)
0 0 0 1 1 1 1
0 1 1 1 0 0 0
1 0 1 0 1 0 0
1 1 1 0 0 0 0
Draw the logic circuit for output = (A + B)
Karnaugh Map
• A graphical way of depicting the content of a truth table where the adjacent expressions differ by only one variable
• For the purposes simplification, the Karnaugh map is a convenient way of representing a Boolean function of a small number (up to four) of variables
• The map is an array of 2n squares, representing all possible combination of values of n binary variables
• Example: 2 variables, A and B
A B A B
A B A B
00 01
10 11
BA B
A
B
A
BA
1 0
1
0
• The number of squares in Karnaugh map depends on the number of variables
• e.g., if 2 variables, A, and B, there are 22 = 4 squares in the Karnaugh map
• e.g., if 3 variables, A, B, and C, there are 23=8 squares
000 001
010 011
110 111
100 101
AB C
A B
C
A B
C
A B
A B
or
000
001
AB
C
A B
C
A BC A B A B
0000 0001
0100
1100
1000
AB C D
A B
C D
A B
CD
A B
A B
C DC D
4 variables, A, B, C, D 24 = 16 squares
000 010 110 100
001 011 111 101
AB
C
A B
C
A BC A B A B
000 001
010 011
110 111
100 101
AB C
A B
C
A B
A B
A B
00 01 11 10
0
1
00
01
11
10
0 1
• The adjacent differ by only one variable
• List combinations in the order 00, 01, 11, 10
C
A B C F
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0
Truth Table
Karnaugh Map
1 1
1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C
How to create Karnaugh Map
1. Place 1 in the corresponding square
1 1
0 0 0 1 1 1 1 0AB
F = AB + AB
A BA B A B A B
Karnaugh Maps to Represent Boolean Functions
1 1
1
0 0 0 1 1 1 1 0BCB CB C B C B CA
A
A
ABC
ABC
ABC
F=ABC + ABC + ABC
1
1
1
0 0 0 1 1 1 1 0CDC DC D C D C DAB
ABCD
ABCD
A B
A B
A B
A B
00
01
11
10
ABCD
F = + ABCD+ABCD ABCD
A B C F
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0
Truth Table
Karnaugh Map
1 1
1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C
Create Karnaugh Map
1. Place 1 in the corresponding square
2. Group the adjacent squares:Begin grouping square with 2n-1 for n variables• e.g. 3 variables, A, B, and C
23-1 = 22 = 4 = 21 = 2 = 20 = 1
1 1
1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C
ABBC ABC F = BC AB ABC+ +
Represent Boolean Functions
1 1
1 1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C3 variables: 23-1 = 22 = 4 22-1 = 21 = 2 21-1 = 20 = 1
C
AB
F = C + AB
1
1 1 1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C3 variables: 23-1 = 22 = 4 22-1 = 21 = 2 21-1 = 20 = 1
A
BC
F = A + BC
1 1
1
1
1 1 1
AB 01
01
00
00
CD
11
10
1011
4 variables, A, B, C, D 24-1 = 23 = 8 (maximum); 22 = 4;21 = 2; 20 = 1 (minimum);
CD + BD ABC+F =
1 1
AB 01
01
00
00
CD
11
10
1011
ABDF =
1
1
AB 01
01
00
00
CD
11
10
1011
F = BCD
1 1
1 1
AB 01
01
00
00
CD
11
10
1011
BCF =
1 1 1 1
1 1 1 1
AB 01
01
00
00
CD
11
10
1011
F = A
1 1
1 1
1 1
1 1
AB 01
01
00
00
CD
11
10
1011
F = D