Boolean Equation Forms

BOOLEAN EQUATION FORMS

      ·         Can be represented in two forms:

                                 i.     Sum-of-Products (SOP)

Ø  Combination of input values that produces one (1)

Ø  Easier to derive from truth table

A	B	C	X	Product term
0	0	0	0
0	0	1	1	A’B’C
0	1	0	0
0	1	1	0
1	0	0	1	AB’C’
1	0	1	0
1	1	0	0
1	1	1	1	ABC

SOP Expression:

X = A’B’C + AB’C’ + ABC

                               ii.     Product-of-Sums (POS)

Ø  Input combinations that produces zero (0) in sum terms

Ø  Usually use if more one (1) produce in output function

A	B	C	X	Product term
0	0	0	0	A+B+C
0	0	1	0	A+B+C’
0	1	0	0	A+B’+C
0	1	1	1
1	0	0	1
1	0	1	0	A’+B+C’
1	1	0	1
1	1	1	1

POS Expression:

                X = (A+B+C)(A+B+C’)(A+B’+C)(A’+B+C’)

(Note that the method is reversible. We can find SOP/POS expression from truth table or build the truth table from the expression.)

      ·         Two ways to simplify Boolean equation:

1.       Laws of Boolean Algebra

	AND Form	OR Form	Notes
Identity Law	A•1 = A	A+0 = A
Zero and One Law	A•0 = 0	A+1 = 1
Inverse Law	A•Ā = 0	A+Ā = 1
Idempotent Law	A•A = A	A+A = A
Communication Law	A•B = B•A	A+B = B+A
Association Law	A•(B•C) = (A•B)•C	A+(B+C) = (A+B)+C
Distribution Law	A+(B•C)=(A+B)•(A+C)	A•(B+C) = (A•B)+(A•C)	Reverse Derivation (A+B)•(A+C)=AA+AC+AB+BC→AA=A                        =A+AC+AB+BC                        =A+AB+BC                        =A(1+B)+BC                        =A•1+BC                        =A+BC
Absorption Law	A(A+B) = A	A+A•B = A A+A’B = A+B	A(A+B)=A(1+B)           →1+B=1               =A(1)              →A•1=A               =A A(A+B)=AA+AB           →A•A=A               =A+AB            →A(1+B)=A(1)               =A A+A’B=(A+AB)A’B       →A=A•A             =(AA+AB)+A’B             =AA+AB+A’B             =(A+A’)(A+B)             =1•(A+B)           →A+A’=1             =(A+B)
DeMorgan’s Law	(A•B)’ = A’+B’	(A+B)’ = A’•B’	If you break the line, change the sign.
Double Complement Law	X” = X

2.       Karnaugh Map