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

The correct answer is C' + D'

Key PointsTo simplify the Boolean function F(A, B, C, D) using the Karnaugh Map (K-map) method, we first need to construct a 4-variable K-map and then fill in the values based on the given minterms (0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14). The minterms can be represented in binary form to determine their positions on the K-map.

The K-map for F(A, B, C, D) would look like this:

simplified expression =C’ + D’

Hence the correct answer is C’ + D’

Explanation:

Digital Circuit Implementation Using Multiplexers

Problem Statement: Given the function B̅C̅ + AC̅, the task is to implement this function using only 2:1 multiplexers. The inputs available are signals A, B, C, and their complements (A̅, B̅, C̅) are not directly available. Additionally, +5V supply and ground can be used as logic '1' and '0', respectively.

Objective: Determine the minimum number of 2:1 multiplexers required to implement the given function.

Solution:

To implement the given Boolean function B̅C̅ + AC̅ using 2:1 multiplexers, we must break down the function and map it into a structure suitable for multiplexer-based design. Here's the step-by-step solution:

1. Analyze the Boolean Function:

• The function is B̅C̅ + AC̅.
• This is a Sum of Products (SOP) expression.
• It consists of two terms: B̅C̅ and AC̅.
• The complement signals (B̅ and C̅) are not directly available, so they must be generated using multiplexers.

2. Generate Complement Signals:

• To generate B̅, a 2:1 multiplexer can be used:
  • Connect B to one input of the multiplexer.
  • Connect ground ('0') to the other input.
  • Use B as the select line.
  • Output will be B̅.
• Similarly, to generate C̅, another 2:1 multiplexer can be used:
  • Connect C to one input of the multiplexer.
  • Connect ground ('0') to the other input.
  • Use C as the select line.
  • Output will be C̅.

3. Implement the SOP Expression:

• The terms B̅C̅ and AC̅ need to be constructed:
  • For B̅C̅, use a 2:1 multiplexer:
    • Connect B̅ to one input of the multiplexer.
    • Connect ground ('0') to the other input.
    • Use C as the select line.
    • Output will be B̅C̅.
  • For AC̅, use another 2:1 multiplexer:
    • Connect A to one input of the multiplexer.
    • Connect ground ('0') to the other input.
    • Use C as the select line.
    • Output will be AC̅.
• Finally, combine B̅C̅ and AC̅ using another 2:1 multiplexer:
  • Connect B̅C̅ to one input.
  • Connect AC̅ to the other input.
  • Use ground ('0') as the select line (since no additional select signal is required).
  • Output will be B̅C̅ + AC̅.

4. Count the Number of Multiplexers:

• To generate B̅: 1 multiplexer.
• To generate C̅: 1 multiplexer.
• To implement B̅C̅: 1 multiplexer.
• To implement AC̅: 1 multiplexer.
• To combine B̅C̅ and AC̅: 1 multiplexer.

Total multiplexers required = 5.

Correct Option Analysis:

The correct option is:

Option 2: Minimum number of multiplexers required is 3.

However, this is incorrect because the analysis above shows that a minimum of 5 multiplexers is required to implement the given function

Simplified Boolean Function in POS Form

Problem Statement: The problem requires simplifying the given Boolean function f(A, B, C, D), which is expressed in terms of minterms (Σm) as:

f(A, B, C, D) = Σ(0, 2, 4, 5, 6, 7, 8, 10, 13, 15)

The task is to express the given Boolean function in its Product of Sums (POS) form and determine the correct simplified expression among the given options. The correct answer is Option 4.

Steps to Solve:

To simplify the Boolean function in POS form, follow these steps:

Step 1: Identify the Minterms and Maxterms

The given function is expressed in minterms as Σ(0, 2, 4, 5, 6, 7, 8, 10, 13, 15). To convert this into POS form, we first determine the maxterms, which are the complements of the minterms. The maxterms are the indices not included in the minterm list:

Minterms: 0, 2, 4, 5, 6, 7, 8, 10, 13, 15

Maxterms: 1, 3, 9, 11, 12, 14

Step 2: Write the Maxterms in Standard POS Form

Each maxterm corresponds to the complement of the minterm. For a 4-variable function, the maxterms are expressed using the variables A, B, C, D in their complemented or uncomplemented forms:

• Minterm 1 → Maxterm: A + B' + C' + D'
• Minterm 3 → Maxterm: A + B' + C + D'
• Minterm 9 → Maxterm: A' + B + C' + D
• Minterm 11 → Maxterm: A + B' + C + D
• Minterm 12 → Maxterm: A' + B' + C + D'
• Minterm 14 → Maxterm: A + B' + C' + D

The POS form is the product (AND operation) of these maxterms.

Step 3: Simplify the POS Expression Using Boolean Algebra

Using Boolean algebra, the given function can be simplified step by step. After simplification, the Boolean function in POS form is:

f(A, B, C, D) = (B + D')(A' + B' + D)

Step 4: Match the Simplified Expression with the Given Options

The simplified POS expression matches Option 4. Hence, the correct answer is:

Option 4: (B + D')(A' + B' + D)

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

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’.