Boolean Logic 1

Technician Series

Boolean 1.1

©Paul GodinCreated Jan 2015

prgodin @ gmail.com

Boolean Simplification

◊ Boolean equations are used to describe a logic circuit’s function.

◊ Equations can become complex and require simplification.

◊ There are laws and theorems to help simplify complex Boolean problems.

◊ Manipulating Boolean equations follows many of the rules of standard algebra.

Boolean 1.2

The 3 Boolean Laws

◊ Commutative:◊ Addition: A + B = B + A◊ Multiplication: AB = BA

◊ Associative:◊ Addition: A + (B + C) = (A + B) + C◊ Multiplication: A(BC) = (AB)C

◊ Distributive:◊ A(B + C) = AB + AC◊ (A + B)(C + D) = AC + AD + BC + BD

Boolean 1.3

The 10 Basic Rules (part 1)

1. Anything ANDed with a 0 is equal to 0: A ● 0 = 0

2. Anything ANDed with a 1 is equal to itself: A ● 1 = A

3. Anything ORed with a 0 is equal to itself: A + 0 = A

4. Anything ORed with a 1 is equal to 1: A + 1 = 1

5. Anything ANDed with itself is equal to itself: A ● A = A

Boolean 1.4

The 10 Basic Rules (part 2)

6. Anything ORed with itself is equal to itself: A + A = A

7. Anything ANDed with its own complement equals 0: A ● A = 0

8. Anything ORed with its own complement equals 1: A + A = 1

9. Anything complemented twice is equal to the original: A = A

10. The two variable rules:

a) A + AB = A + Bb) A + AB = A + Bc) A + AB = A

Boolean 1.5

De Morgan’s Theorem Review

◊ De Morgan’s Theorem allows the inversion of an expression to be broken up into inversions of individual variables.

◊ Inversion of an expression: A + B

◊ Inversion of individual variables: A ● B

“Break the bar and change the sign”

Boolean 1.6

7. A ● 0 = ___

8. A + A = ____

9. A + A = ____

10. A + AB = ____

11. A ● 1 = ____

12. A = ____

Basic Boolean Rules Exercise 1

1. A + 0 = ____

2. A + AB = ____

3. A + 1 = ____

4. A + AB = ____

5. A ● A = ____

6. A ● A = ____

Determine the outcome of the following:

Boolean 1.7

Basic Boolean Rules Exercise 2

Determine the output of the following gates

?0

A

?1

?0

?1

AA

A’A

A’A

Boolean 1.8

Boolean Simplification

◊ Boolean equations can be simplified using algebraic methods, using the Boolean rules and laws to reduce the equation.

Boolean 1.9

Examples of Boolean Reduction 1

◊ Consider CD(D+DF)◊ CD(D) Rule 10a where D+DF=D◊ CDD Associative Law◊ CD Rule 5 where DD=D

◊ Consider C’D’(C+D)’◊ C’D’(C’D’) DeMorgan where (C+D)=(CD)◊ C’C’D’D’ Associative where brackets removed◊ C’D’ Rule 5 where C’C’=C’ and D’D’=D’

Boolean 1.10

Examples of Boolean Reduction 2◊ Consider (C+D)(C+D’)

◊ CC+CD’+DC+DD’ Distributive◊ C+CD’+CD+0 Rule 5: CC=C; Rule

7:DD’=0, ◊ C+CD’+CD Rule 3: (A+0=A)◊ (C+CD’)+CD Associative◊ C+CD Rule 10c: C+CD’=C◊ C+C Rule 10c: C+CD=C◊ C Rule 6: C+C=C

◊ Consider C’+CDE+E◊ (C’+CDE)+E Associative◊ C’+DE+E Rule 10b: C’+CDE=C’+DE◊ C’+(E+DE) Associative, Commutative◊ C’+E Rule 10c: E+DE=E

Boolean 1.11

Exercise 3

◊ Simplify the following:◊ A’+AB’+B

◊ A+A’B+B’C+AC

◊ AB’+A’CD+B+C’+D’

Other examples may be given in class

Boolean 1.12

Pushing a Signal

◊ Signal Pushing is a technique of applying an input value and following its progression through a circuit.

◊ This method is used extensively when analysing and troubleshooting circuits.

Boolean 1.13

Truth Tables and Signal Pushing

◊ A truth table can be derived from a circuit by signal pushing.

◊ All possible input combinations are applied to determine the output of the circuit.

Boolean 1.14

1

Example 1: Signal Pushing

INPUT OUTPUT

A B W

0 0

0 1

1 0

1 1

Apply all input combinations, follow the logic through the circuit and complete the truth table.

0 1

0

1

011

10 0

1

0

11

01

0

1

1

1

1

11

0

0

0

Animated

Boolean 1.15

Example 2: Signal Pushing

INPUT OUTPUT

A B C Y

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

Boolean 1.16

Exercise 1: Signal Pushing

Apply all input combinations and complete the truth table.

INPUT OUTPUT

A B C W

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Boolean 1.17

Exercise 2: Signal Pushing

INPUT OUTPUT

A B C D Z

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

Boolean 1.18

Sum-of-Products (SOP)

Boolean 1.19

Boolean from the Truth Table

◊ A Boolean equation derived from the truth table takes on a “Sum-of-Products” form.

◊ Sum-of-Products are ANDed statements (products) that are ORed together (sum). Example: (A●B)+(C●D)

◊ From the Truth Table, for all output values that equal “1”, the ANDed input values are written. ◊ If the input value is “0”, the complimented input is

indicated.

Boolean 1.20

SOP from the Truth Table

This example demonstrates how the S.O.P. equation is determined.

INPUT OUTPUT

A B W

0 0 1

0 1 0

1 0 1

1 1 0

(A●B)+(A●B)=W

Boolean 1.21

S.O.P. Simplification

◊ Once the Boolean equation in S.O.P. form is determined, standard simplification rules are applied.

◊ Example: (A●B)+(A●B)=W

B(A+A)=W

B(1)=W

B=W

Boolean 1.22

Example 2: Sum of Products

(ABC)+(ABC)+(ABC)+(ABC) = WINPUT OUTPUT

A B C W

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

Boolean 1.23

3-Boolean Expression Simplified

(ABC)+(ABC)+(ABC)+(ABC) = W

(ABC)+(ABC)+(ABC)+(ABC)+(ABC)+(ABC) = W

BC(A+A)+AB(C+C)+AC(B+B) = W

BC+AB+AC = W

Boolean 1.24

©Paul R. Godinprgodin°@ gmail.com

END

Boolean 1.25