Test: Boolean Functions - Electronics and Communication Engineering (ECE) MCQ

• we are given the boolean function
• we need to simplify the function in the sum of minterms
• we will use the K-Map to simplify the function 
• the K-Map is a very useful tool to simplify long boolean expressions into simplified form in a very systematic approach

From the K-Map we get 

• F(A, B, C, D) = Σ(0, 6, 8, 13, 14)
• d (A, B, C, D) = Σ(2, 4, 10)

A = a₁ a₀  B = b₁ b₀
The Boolean function S satisfies the condition,
If A> B then  = 1.
We can represent the above condition through a truth table.



By using k-maps

Minimized form:

Logic: From the option we will find the question.
(A+B) (A+C) = A.A + A.C + B.A + B.C
Since A.A = A, the above expression becomes:
= A + A.C + A.B + B.C
= A(1 + C + B) + B.C
Since 1 + any variable = 1, the above expression becomes:
= A + B.C
(A+B) (A+C) = A + B.C

The above equation can also be written as :

Hence option 3 is correct
Formulae:

De Morgan’s law states that:

Applying the De-Morgans theorem in the above function F

Hence option (4) is correct.

Any variable 'a' can have 2 value i.e, 0 or 1.
For 'n' variables there are 2ⁿ entries in the truth table.
And each output of any particular row in the truth table can be 0 or 1.
Hence, we have 2²ⁿ  different Boolean functions with n variables.
Let's understand with the help of an example - 
Let's say there are 2 variables a, and b: n = 2
There are  2² entries in the truth table, and each entry can be 0 or 1
So, we have  2²² = 16  different Boolean functions with 2 variables.

All Boolean algebra laws are shown below


The given Boolean function can be simplified as :
Z =  X + XY
Z = X (1 + Y)
Z = X 

Let,

To solve X+X'Y
Use X.1 = X, 
= X.1 + X'Y 
Use 1 + Y = 1
= X(1 + Y) + X'Y 
= X + XY + X'Y
= X + Y(X + X')
Use X + X' = 1 
= X + Y.1
Use Y.1 = Y
= X + Y
Hence answer is X+Y.

Given logical expression can be written as:
(A+B) (A+C) = A.A + A.C + B.A + B.C
Since A.A = A, the above expression becomes:
= A + A.C + A.B + B.C
= A(1 + C + B) + B.C
Since 1 + any variable = 1, the above expression becomes:
= A + B.C
(A+B) (A+C) = A + B.C