Minimization of Boolean Expression MCQ Quiz - Objective Question with Answer for Minimization of Boolean Expression - Download Free PDF

F (P, Q, R, S) = ∑ 0, 2, 5, 7, 8, 10, 13, 15

Don’t care min terms are 2, 7, 8, 13

By plotting the K-map, the minimal SOP (sum of products) can be found.

Explanation –

While putting the terms to k-map following things happen,

• 3^rd and 4^th columns are swapped
• 3^rd and 4^th rows.
• term 2 is going to (0, 3) column instead of (0, 2)
• 8 is going to (3, 0) instead of (2,0)

Solving, the above K-map, we get Q̅S̅ + QS

Explanation:

To understand why the complement law holds, we need to analyze the truth table for the expression A + A'. A truth table lists all possible values of the variables involved and the resulting value of the expression for each combination of variable values. In this case, we are dealing with a single variable A, which can either be 0 or 1.

Concept:

We are given a Boolean function in the nested form: \( f(f(xy, y), z) \). To simplify this, we need to understand how the function \( f \) behaves and use Boolean algebra rules.

Step 1: Assume definition of function

Let us assume \( f(a, b) = a + b \), a common definition used in Boolean simplification unless otherwise specified.

So,

\( f(xy, y) = xy + y \)

Using the identity: \( xy + y = y \) (Since \( y(x + 1) = y \))

Step 2: Substitute into outer function

Now substitute into the outer function:

\( f(y, z) = y + z \)

Hence, the simplified expression is:

\( y + z \)

Compare with Options:

• Option 1: \( \bar{x} + z \) ❌
• Option 2: \( xyz \) ❌
• Option 3: \( xy̅ + z \) ❌
• Option 4: None of the options ✅

Final Answer: ✅ None of the options

The correct answer is Remains unchanged when n is even.

Key Points

• When a bit string is XORed (exclusive OR) with itself, the result is always 0.
• For example, if you have a bit string B, XORing B with B will result in 0 (B ⊕ B = 0).
• XOR operation is both associative and commutative, meaning the order of operations does not change the result.
• When the bit string is XORed with itself an even number of times, the result remains unchanged.
• Mathematically, B ⊕ B ⊕ B ⊕ B (n times, where n is even) simplifies to 0 ⊕ 0 = 0.

Additional Information

• The XOR operation is widely used in various fields such as cryptography, error detection, and correction algorithms.
• In computer science, XOR is used in binary addition without carrying.
• When n is odd, XORing the bit string with itself results in the bit string itself (B ⊕ B ⊕ B = B).
• XOR can be used to swap values of two variables without using a temporary variable.

The correct answer is Option 3) (A + B')(C' + D).

Key Points

Given Boolean function: A'B + CD' + A'B + CD'

• First simplify the given expression:
  • A'B + A'B = A'B (redundant term)
  • CD' + CD' = CD' (redundant term)
• So, the simplified function = A'B + CD'

Now, we are asked to find the complement of this function:

Let F = A'B + CD'

Then the complement is: F' = (A'B + CD')'

Apply De Morgan’s Law:

• (A'B + CD')' = (A'B)' · (CD')'
• (A'B)' = A + B'
• (CD')' = C' + D

Therefore, F' = (A + B')(C' + D)

Hence, the correct answer is: Option 3) (A + B')(C' + D)

Concept

1.) De-morgan's law: \(\overline{A+B}=\overline{A}\space\overline{B}\)

2.) \(\overline{A}{A}=0\)

3.) \(\overline{A}+{A}=1\)

Explanation

\(Y=\rm\overline{(A+\bar{B}+C)+(B+\bar{C})}\)

Using De-morgan's Law:

\(Y=\overline{(A+\overline{B}+C)}\space \overline{(B+\overline{C})}\)

\(Y=(\overline{A}B\overline{C})\space (\overline{B}C)\)

\(Y=\overline{A}(B\overline {B})(C\overline {C})\)

Y = 0

FROM laws of Boolean algebra

1 + any variable = 1

Concept:

All Boolean algebra laws are shown below

Calculation:

F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZ

= X̅ Z̅ + Z̅ (Y̅ + Y) + XYZ

= X̅ Z̅ + Z̅ + XYZ

= Z̅ (1 + X̅) + XYZ

= Z̅ + XYZ 

Now using Distributive Law

= (Z̅ + Z)(Z̅ + XY)

= Z̅ + XY 

The correct answer is option 1.

Concept:

The given K-Map is,

F = a'b'c+a'bc'+ab'c'+abc

F= a'(b'c+bc')+a (b'c'+bc)

F=a'(b⊕c)+a(b⊙c)

F=a'(b⊕c)+a(b⊕c)'

F=a⊕b⊕c

Hence the correct answer is Exclusive OR.

The correct answer is option 4

K-maps

F(A, B, C, D) = ∑ (0, 1, 2, 3, 6, 12, 13, 14, 15)

Two K-Maps can be constructed from the given boolean function

The expression for K-Map 1 is AB + A'B' + A'CD'

The expression for K-Map 2 is AB +A'B' + BCD'

• A variable is a symbol that may take on the value 0 or 1.
• A literal is the use of a variable or its complement in an expression.
• A term is an expression formed by literals and operations at one level.

For example, the following function:

F₁ = xy + xy'z + x'yz

Has 3 variables (x,y,z),

8 literals (x,y,x,y',z,x',y,z), and

4 terms (xy, xy'z, x'yz, and the OR term that combines the first level AND terms).

Concept:

Implicants:  Every min-term in SOP form or max-term in POS form in a Boolean function is termed as an implicant. 
For example,
F = AB + AC 
AB and AC are called implicants.

Prime Implicants:  All pairs that cannot be a part of any quad or all quads that cannot be a part of any octet in a K-map are termed as prime implicants.

Essential Prime Implicants:  Those prime implicants that cover at least one min-term that can’t be covered by any other prime implicant are called essential prime implicants.

Calculation:

Given the Boolean function,

F (A, B, C, D) = A'B'C'D + A'BCD' + ABC'D'

For the above Boolean function, the K – map representation is:

Hence we can see there three minterms in the given function.

Essential prime implicants are also three as they are not covered by any other prime implicant.

Hence option (3) is the correct answer.

Concept:

Draw the K- map, convert the K-map into a SOP (sum of product) or POS (product of sum) form. While reducing the K-map in these forms, a don’t care will be needed only when with the use of don’t cares we can reduce the term size.

Diagram: K – Map

From the K-map simplification:

F(a, b, c, d) = a̅.c

F(a, b, c, d) = \(\overline {\left( {a + \bar c} \right)}\)

Diagram:

Therefore, only one NOR gate is needed to implement the minimized function

Concept:

We can simplify the given boolean function with the help of K-Map.

The K-map is a systematic way of simplifying Boolean expressions. With the help of the K-map method, we can find the simplest POS and SOP expression, which is known as the minimum expression.

Just like the truth table, a K-map contains all the possible values of input variables and their corresponding output values.

The K-map method is used for expressions containing 2, 3, 4, and 5 variables.

Calculation:

Given ;  F (x, y, z) = Σ(0, 2, 4, 5, 6) 

3 variable K-map:

The grouping of cells has shown below

The expression obtained from the K-Map → F = z' + xy'

Additional Information Note 1 − If outputs are not defined for some combination of inputs, then those output values will be represented with the don’t care symbol ‘x’. That means, we can consider them as either ‘0’ or ‘1’.

Note 2 − If don’t care terms are also present, then place doesn’t care ‘x’ in the respective cells of the K-map. Consider only the don’t care ‘x’ that are helpful for grouping the maximum number of adjacent ones. In those cases, treat the don’t care value as ‘1’.