DeMorgan's theorems state that the complement of a product of variables is equal to the sum of the complemented variables and the complement of a sum of variables is equal to the product of the complemented variables. Boolean expressions can be written in sum-of-products (SOP) form or product-of-sums (POS) form, where in SOP two or more product terms are summed and in POS two or more product terms are multiplied.
4. DeMorgan's 1st Theorem :
The complement of a product of variables is
equal to the sum of the complemented
variables.
AB = A + B
Applying DeMorgan's 1st Theorem to gates:
A B AB A+B
0 0 1 1
0 1 1 1
1 0 1 1
1 1 0 0
5. DeMorgan's 2st Theorem :
The complement of a sum of variables is equal to
the product of the complemented variables.
A + B = A B
Applying DeMorgan's 2st Theorem to gates:
A B AB A+B
0 0 1 1
0 1 0 0
1 0 0 0
1 1 0 0
A B
6. Example:
Apply DeMorgan's theorem to remove the over bar
covering both terms from the expression X= C + D.
X= C . D
X= C . D
X= C . D
8. SOP and POS forms:
Boolean expressions can be written in the sum-of-product form (SOP) or
in the product-of-sum (POS). These forms can simplify the
implementation of combinational logic. In both forms, an over-bar cannot
extend over more than one variable.
An expression is in SOP form when two or more product terms are summed as in
the following examples:
A B C + A B A B C + C D C D + E
An expression is in POS form when two or more product terms are multiplied as
in the following examples:
(A + B) (A + C) (A + B + C) (B + D) (A + B) C
9. Sum of Products (SOP)
ABCCBACBACBAF
AC
BBAC
)(
CB
AACB
)(
BA
BA
CCBA
)1(
)(
)()()( BBACCCBAAACBF
ACBACBF
13. Question
One of De Morgan's theorems states that . Simply
stated, this means that logically there is no
difference between:
A. a NOR and an AND gate with inverted inputs
B. a NAND and an OR gate with inverted inputs
C. an AND and a NOR gate with inverted inputs
D .a NOR and a NAND gate with inverted inputs
14. Question
Which of the following expressions is in the sum-
of-products (SOP) form?
A. (A + B)(C + D)
B. (A)B(CD)
C. AB(CD)
D. AB + CD
15. Question
Use Boolean algebra to find the most simplified SOP
expression for F = ABD + CD + ACD +ABC + ABCD.
A. F = ABD + ABC + CD
B. F = CD + AD
C. F = BC + AB
D. F = AC + AD