Chapter 2 : Digital Logic (Simplification of boolean equation)

2.2...COMBINATION CIRCUITS (Simplification of boolean equation)       

By: ABDULADHEM SAGHIR MESFER ALAZAM   B031210406         

Simplification of boolean equation.

There are two ways to simplify boolean equation ,laws of boolean algabra and

1.Laws of boolean algabra –rules to simplify boolean expression .

2.Karnaugh map –A grid –like representation of a truth table .

Simplification of boolean equation Rules

1.A  +  AB    =A

Examples
Here are some examples of Boolean algebra simplifications. Each line gives a form of the expression, and the rule or rules used to derive it from the previous one. Generally, there are several ways to reach the result. Here is the list of simplification rules.

• Simplify: C + BC:

Expression	Rule(s) Used
C + BC	Original Expression
C + (B + C)	DeMorgan's Law.
(C + C) + B	Commutative, Associative Laws.
T + B	Complement Law.
T	Identity Law.

• Simplify: AB(A + B)(B + B):

Expression	Rule(s) Used
AB(A + B)(B + B)	Original Expression
AB(A + B)	Complement law, Identity law.
(A + B)(A + B)	DeMorgan's Law
A + BB	Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication.
A	Complement, Identity.

• Simplify: (A + C)(AD + AD) + AC + C:

Expression	Rule(s) Used
(A + C)(AD + AD) + A + C	Original Expression
(A + C)A(D + D) + AC + C	Distributive.
(A + C)A + AC + C	Complement, Identity.
A((A + C) + C) + C	Commutative, Distributive.
A(A + C) + C	Associative, Idempotent.
AA + AC + C	Distributive.
A + (A + T)C	Idempotent, Identity, Distributive.
A + C	Identity, twice.

• You can also use distribution of or over and starting from A(A+C)+C to reach the same result by another route.
• Simplify: A(A + B) + (B + AA)(A + B):

Expression	Rule(s) Used
A(A + B) + (B + AA)(A + B)	Original Expression
AA + AB + (B + A)A + (B + A)B	Idempotent (AA to A), then Distributive, used twice.
AB + (B + A)A + (B + A)B	Complement, then Identity. (Strictly speaking, we also used the Commutative Law for each of these applications.)
AB + BA + AA + BB + AB	Distributive, two places.
AB + BA + A + AB	Idempotent (for the A's), then Complement and Identity to remove BB.
AB + AB + AT + AB	Commutative, Identity; setting up for the next step.
AB + A(B + T + B)	Distributive.
AB + A	Identity, twice (depending how you count it).
A + AB	Commutative.
(A + A)(A + B)	Distributive.
A + B	Complement, Identity.

Example

A	B	C	F
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	1
1	0	1	0
1	1	0	1
1	1	1	1

F=(A+B+C)(A+B+C’)(A+B’+C)(A+B+C)

Produc

A+B+C

A+B+C’

A+B’+C

A’+B+C’

POS expression

F=(A+B+C)(A+B+C’)(A+B’+C)(A’+B+C)

NOTE .this is not simplified version

Note that the method is reverible .You can find the POS

Expression from the truth table or build the truth table from the expression .