1. Simplify the Boolean expressions:
a) Y = (AB + C)(AB + D)
b) Y = ABCD + ABbarCD
Answers
BOOLEAN THEOREMS:
Boolean algebra helps to analyze a logic circuit and express its operation mathematically. We have several Boolean Theorems that helps us to simplify logic expressions and logic circuits.
Single Variable Theorems:
AND Operation OR Operation
A . 0 = 0 A + 0 = A
A . 1 = A A + 1 = 1
A . A = A A + A = A
A . A’ = 0 A + A’ = 1
Multivariable Theorems:
The multivariable theorems involves more than one variable.
Commutative laws:
A + B = B + A
A . B = B . A
Associative laws:
A + (B + C) = (A + B) + C = A + B + C
A (BC) = (AB) C = ABC
Distributive laws:
A (B + C) = AB + AC
A + BC = (A+B) (A+C)
(A + B) (C + D) = AC + AD + BC + BD
Absorption laws:
A + AB = A
A (A + B) = A
Redundant Literal Rule:
A + A’B = A + B
A’ + AB = A’ + B
A(A’ + B) = AB
Demorgan’s Theorem:
(A+B)’ = A’ . B’
(AB)’ = A’ + B’
This law can be extended to any number of variables or combinations of variables.
How to apply Demorgan’s theorem to an expression?
Complement the entire given expression.
Change all the AND’s to OR’s and all the OR’s to AND’s.
Complement each of the individual variables.
Change all 0s to 1s and all 1s to 0s.
This procedure is called as Demorganization or Complementation of switching expressions. In simple words, we can say it as ‘Break the line, change the sign’.
f(A, B, ………, 0, 1, . , +) = f(A’, B’, ………., 1, 0, +, .)
Consensus Theorem:
AB + A’C + BC = AB + A’C
(A + B)(A’ + C)(B + C) = (A + B)(A’ + C)
If one term is containing A and the other term is containing A’ and the third term containing the left-out literals of the first two terms then the third term is redundant. It means the function remains same with and without the third term.
This theorem can be extended to any number of variables.
AB + A’C + BCD = AB + A’C
(A + B)(A’ + C)(B + C + D) = (A + B)(A’ + C)
Transposition Theorem:
AB + A’C = (A + C)(A’ + B)
DUAL OF A BOOLEAN FUNCTION:
The DUAL of a boolean function is obtained by interchanging OR and AND operations and replacing 1’s by 0’s and 0’s by 1’s.
For example, DUAL of (ABCD…F)’ = A’ + B’ + C’ + D’ + … + F’
COMPLEMENT OF A BOOLEAN FUNCTION:
To compute the complement of a boolean function, we use Demorgan’s theorems.
How to complement a boolean function?
Parenthesize product terms.
Take the dual of a function.
Complement each literal.
For example, F(A,B,C) = AB’ + BC’ + A’C’
F'(A,B,C) = (AB’ + BC’ + A’C’)’
= (AB’)’ . (BC’)’ . (A’C’)’
= A’B . B’C . AC
REDUCING BOOLEAN EXPRESSIONS:
Realization of a digital circuit with the minimal expression results in reduction of cost and complexity and the corresponding increase in reliability.
How to reduce Boolean Expressions?
Multiply all variables necessary to remove parentheses.
Look for identical terms. Only one of those terms be retained and all others are dropped. For example, AB+AB+AB+AB=AB
Look for a variable and its negation in the same term. This term can be dropped. For example, A + BB’C = A
Look for pair of terms that are identical except for one variable which may be missing in one of the terms. The larger term can be dropped. For example, ABCD + ABC = ABC
Look for pair of terms which have the same variables with one or more variables complemented. If a variable in one term of such a pair is complemented while in the second term it is not, then such terms can be combined into a single term with that variable dropped. For example, ABC’D’ + ABC’D = ABC’