Logic Gates and Boolean Algebra MCQ Quiz - Objective Question with Answer for Logic Gates and Boolean Algebra - Download Free PDF

The correct option is 4

Concept:

From the truth table, it is clear that the output of an EX-OR gate is '0' when both inputs are '1'. This is because the EX-OR gate outputs '1' only when the inputs are different. When both inputs are the same (either both 0 or both 1), the output is '0'. Therefore, the correct answer is option 4.

The correct answer is: 4) An OR gate with an additional curve at the input end.

Explanation:

The XOR (Exclusive-OR) gate is represented by:

• An OR gate symbol (a curved shape)

• With an additional curved line at the input side (to distinguish it from a standard OR gate).

Additional Information

Diagram of XOR Gate:

The primary goal of simplifying a Boolean expression before implementing it with gates is: 3) To reduce the number of gates and interconnections

Explanation:

• Minimizing gates reduces the circuit's cost, complexity, and physical space required.
• Fewer interconnections improve reliability (less chance of wiring errors or signal interference).
• Optimized circuits consume less power and operate faster (fewer propagation delays).

Explanation:

Dual of a Boolean Expression

Definition: The dual of a Boolean expression is derived by replacing all AND (×) operations with OR (+) operations, all OR (+) operations with AND (×) operations, and swapping the constants 1 and 0 in the original expression. The variables in the Boolean expression remain unchanged.

To determine the dual of a given Boolean expression, follow these steps:

1. Identify all the AND (×) and OR (+) operations in the expression.
2. Replace AND (×) with OR (+) and vice versa.
3. Swap 1 with 0 wherever they appear in the expression.

Given Boolean Expression:

X + YZ = (X + Y) × (X + Z)

Let’s determine the dual of this expression:

• In the given expression, the OR (+) operation between X and YZ will be replaced with an AND (×) operation.
• The AND (×) operations within YZ and (X + Y) × (X + Z) will be replaced with OR (+) operations.
• Since there are no constants 1 or 0 in the expression, we do not need to swap them.

Replacing operations step by step:

• The left-hand side, X + YZ, becomes X × (Y + Z).
• The right-hand side, (X + Y) × (X + Z), becomes (X × Y) + (X × Z).

Thus, the dual of the given Boolean expression is:

X × (Y + Z) = (X × Y) + (X × Z)

Correct Option Analysis:

The correct option is:

Option 1: X × (Y + Z) = XY + YZ

This is the correct dual of the given Boolean expression. As derived above, the left-hand side X + YZ becomes X × (Y + Z), and the right-hand side (X + Y) × (X + Z) becomes (X × Y) + (X × Z). Hence, the correct dual is accurately represented in Option 1.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: X.(Y + Z) = X.Y + X.Z

This option is not the correct dual of the given Boolean expression. While it might seem similar, it actually represents the distributive property of Boolean algebra and does not match the dual derived from the original expression.

Option 3: X + (Y + Z) = X.Y + Z

This option introduces an additional OR (+) operation within (Y + Z) that was not present in the original expression. Moreover, the right-hand side, X.Y + Z, does not correspond to the structure of the dual derived from the given Boolean expression. Therefore, it is incorrect.

Option 4: X + (YZ) = X + Y + Z

This option does not align with the dual derived from the given Boolean expression. It incorrectly simplifies X + (YZ) to X + Y + Z, which is not mathematically valid based on Boolean algebra rules. Hence, this option is incorrect.

Conclusion:

By understanding the concept of duality in Boolean algebra, we can accurately derive the dual of any given Boolean expression. The dual is obtained by swapping AND (×) and OR (+) operations and interchanging the constants 1 and 0. For the given expression X + YZ = (X + Y) × (X + Z), the correct dual is X × (Y + Z) = (X × Y) + (X × Z), which is represented in Option 1.

Explanation:

Universal Gate

Definition: A universal gate is a type of logic gate that can be used to implement any Boolean function without needing to use any other type of gate. Universal gates are versatile and are fundamental in digital electronics because they can be used to construct other basic gates such as AND, OR, and NOT gates.

Correct Option: The correct answer is Option 3: NOR Gate.

Explanation:

The NOR gate is called a universal gate because it can be used to create all the other basic logic gates (AND, OR, NOT, NAND, and XOR). The NOR gate is a combination of the OR gate followed by a NOT gate. It outputs a logical "1" only when all inputs are logical "0". The truth table for a NOR gate is as follows:

Why NOR Gate is Universal:

• Implementation of NOT Gate: A single input NOR gate acts as a NOT gate. If the input is "A", the output will be "NOT A".
• Implementation of OR Gate: By combining two NOR gates, an OR gate can be constructed.
• Implementation of AND Gate: By using multiple NOR gates in a specific configuration, an AND gate can be created.
• Implementation of XOR Gate: XOR gates can also be constructed using NOR gates, but it requires a more complex arrangement.

Because of its ability to replicate all other logic gates, the NOR gate is classified as a universal gate. This feature is extremely useful in digital circuit design, as it allows for simplification and standardization in hardware implementations.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: AND Gate

The AND gate is a basic logic gate that outputs "1" only when all its inputs are "1". It is not a universal gate because it cannot be used to construct other gates by itself. While it is a fundamental gate in digital circuits, it lacks the versatility of universal gates like NOR and NAND.

Option 2: NOT Gate

The NOT gate is a unary operator that outputs the inverse of its input. While it is essential for negation in digital logic, it is not a universal gate. It cannot construct other gates without additional components.

Option 4: EX-OR Gate

The XOR gate outputs "1" when its inputs are different. Although it has unique properties and is widely used in applications like parity checking and error detection, it is not a universal gate. It cannot be used to construct other gates independently.

Option 5: NAND Gate

Although not listed as the correct answer in the question, the NAND gate is also a universal gate. Like the NOR gate, it can be used to construct all other basic logic gates, making it equally versatile. If the question had included NAND gate as an option, it would also have been correct.

Conclusion:

The NOR gate is a universal gate because it can be used to implement any Boolean function and construct all other basic logic gates. This versatility makes it essential in digital electronics and circuit design. While other gates like AND, NOT, and XOR serve specific purposes, they lack the ability to replicate all other gate functions, which is the defining characteristic of universal gates. Alongside NOR gates, NAND gates also share the universal gate classification.

Concept:

XNOR Gate:

Symbol:

Truth Table:

Output Equation: $Y=\overline{A\oplus B}$

1) If B is always Low, the output is the inverted value of the other input A, i.e. A̅.

2) The output is low when both the inputs are different.

3) The output is high when both the inputs are the same.

4) XNOR gate produces an output only when the two inputs are same.

Analysis:

$F=\overline{A+0}=\overline{A}$

F(P, Q, R) = PQ + QR' + PR'

= PQ (R + R') + (P + P')QR' + P(Q + Q')R'

= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'

= PQR + PQR' + P'QR' + PQ'R'

= m₇ + m₆ + m₂ + m₄

• AND, OR, NOT gates are the basic gates.
• The logic gates which are derived from the basic gates like AND, OR, NOT gates are known as derived gates. NAND, NOR, XOR, and XNOR are the derived gates.
• A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates.

(A.A̅) + A

= 0 + A = A

All Boolean algebra laws are shown below:

The number of 2-input NAND gates required to implement a 2-input XOR gate is 4.

Similarly, the number of 2-input NOR gates required to implement a 2-input XNOR gate is 4.

X = [(A + B̅) (B + C)] B

= (AB + AC + 0 +  B̅C)B

= AB + ABC

= AB(1 + C)

= AB

Concept:

K-map:

• K-map (Karnaugh Map) is a pictorial method used to minimize Boolean expression without having to use Boolean Algebra theorems and equation manipulation.
• K-map can be thought of as a special version of a truth table.
• Using K-map, expression with two to four variables are easily minimized.
• K-maps are also referred to as 2D truth tables as each K-map is nothing but a different format of representing the values present in a one-dimensional truth table.
• To simplify a logic expression with two inputs, we require a K-map with 4 cells (= 2²)
• Similarly, a logic expression with four inputs we require a K-map with 16 cells (= 2⁴)
• Each cell within K-map has a definite place value which is obtained by using on encoding technique known as Gray code.
• For n-variable K-map, with 2ⁿ cells, try to group 2ⁿ cells first, then for 2^n-1 cells, next for 2^n-2 cells, and so on until the group contains only 2° cells ie. Isolated bits (if any)
• Also remember, the number of cells in a group must be equal to an integer power to 2 i.e. 1, 2, 4, 8, ….

Calculation:

→ There are no 16 bits group, no 8-bits group, but there are 2-four bits group

→ Eliminate the variables for which the corresponding hit appears within the group as both 0 and 1.

• Group 1 → B̅₁ B₂
• Group 2 → B₁ B̅₂

→ Therefore in SOP form (sum of products) output $Y=\overline{B_1}{B}_2+\overline{B_2}{B}_1$

XOR GATE

Symbol:

Truth Table:

Output Equation: $Y=\mathbf{A}\oplus \mathbf{B}=\overline{A}B+A\overline{B}$

Key Points: 

1) If B is always High, the output is the inverted value of the other input A, i.e. A̅.

1) The output is low when both the inputs are the same. 

2) The output is high when both the inputs are different.

Explanation:

$Y=\overline{\mathbf{X}}\oplus \mathbf{X}=\overline{\overline{X}}X+\overline{X}\overline{X}$

$Y=XX+\overline{X}\overline{X}$

$Y=X+\overline{X}$

Y = 1

Concept:

Consensus Law is one of the most powerful theorems used in digital electronics for the minimization of Boolean function or equation either in the successive reduction method or in the K-Map method.

Statement:

• The consensus theorem states that the consensus term of a disjunction is defined when the terms in function are reciprocals to each other (such as A and A̅).
• The consensus theorem is defined in two statements (normal form and it's dual). They are
• AB + ĀC+BC = AB+ĀC
• (A+B)(Ā+C)(B+C) = (A+B)( Ā+C)

Calculation:

Y = P̅Q + QR + PR

Y = P̅Q + PR + QR (P̅ + P)

Y = P̅Q + PR + QRP̅ + QRP

Y = P̅Q(1 + R) + PR(1 + Q)

Y = P̅Q + PR where (1 + A = 1) according to Boolean algebra.