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Class 9 Computer Science Notes Chapter 3

Chapter 3: Digital Systems and Logic Design

Exercise Solution

PART I: MULTIPLE-CHOICE QUESTIONS (MCQs)

1. Which of the following Boolean expressions represents the OR operation?

a) (A \cdot B)
b) (A+B)
c) (\overline{A})
d) (A \oplus B)

Correct Answer: (b) (A+B)

2. What is the dual of the Boolean expression (A \cdot 0=0)?

a) (A+1=1)
b) (A+0=A)
c) (A \cdot 1=A)
d) (A \cdot 0=0)

Correct Answer: (a) (A+1=1)

3. Which logic gate outputs true only if both inputs are true?

a) OR gate
b) AND gate
c) XOR gate
d) NOT gate

Correct Answer: (b) AND gate

4. In a half-adder circuit, the carry is generated by which operation?

a) XOR operation
b) AND operation
c) OR operation
d) NOT operation

Correct Answer: (b) AND operation

5. What is the decimal equivalent of the binary number 1101?

a) 11
b) 12
c) 13
d) 14

Correct Answer: (c) 13

PART II: SHORT QUESTIONS

1. Define a Boolean function and give an example.

Answer: A Boolean function is an algebraic expression formed using binary variables, binary operators (AND (\cdot), OR (+), NOT), and logical constants (0 and 1). It outputs a single binary value (0 or 1) based on its input variables.

Example:

[
F(A,B)=A\cdot B+\overline{A}
]

2. What is the significance of the truth table in digital logic?

Answer: A truth table provides a complete tabular representation listing all possible combinations of input binary values alongside their corresponding output value. It helps logic designers analyze, define, verify, and simplify digital circuits and Boolean functions.

3. Explain the difference between analog and digital signals.

Answer:

Analog Signals: Continuous signals that vary smoothly over time and can take an infinite number of intermediate voltage or amplitude values. Examples include human voice and temperature sensor output.
Digital Signals: Discrete signals that represent data in distinct binary states (0 and 1, or LOW and HIGH voltages), offering higher noise immunity and precision. An example is computer clock signals.

4. Describe the function of a NOT gate with its truth table.

Answer: A NOT gate, also called an inverter, is a single-input, single-output logic gate that outputs the logical complement or inverse of its input. If the input is 0, the output is 1; if the input is 1, the output is 0.

Truth Table:

Input ((A))	Output ((\overline{A}))
0	1
1	0

5. What is the purpose of a Karnaugh map in simplifying Boolean expressions?

Answer: A Karnaugh map (K-map) provides a visual geometric method to simplify complex Boolean expressions. By grouping adjacent cells containing 1s according to specific powers of 2, K-maps eliminate redundant variables and yield minimal Sum-of-Products (SOP) or Product-of-Sums (POS) expressions without requiring tedious algebraic manipulation.

PART III: LONG QUESTIONS

1. Explain the usage of Boolean functions in computers.

Answer:

Digital Circuit Design: Computer hardware operates exclusively on digital logic levels. Boolean functions define how physical logic gates (AND, OR, NOT, NAND, NOR, XOR) are interconnected inside microprocessors, memory controllers, and field-programmable gate arrays (FPGAs).
Arithmetic and Logic Unit (ALU): The ALU performs arithmetic operations, such as addition and subtraction, and logical evaluations, such as AND, OR, and NOT, using hardware implementations of Boolean expressions, including adders, comparators, and multipliers.
Control Units and Multiplexers: Directing data flow along system buses, decoding instruction opcodes, and routing control signals depend on complex Boolean functions designed to select specific hardware pathways.
Software Conditionals and Algorithms: Programming statements such as if, while, AND, OR, and NOT evaluate logical flags and conditional branches using Boolean logic.

2. Describe how to construct a truth table for a Boolean expression with an example.

Answer:

Steps to Construct a Truth Table:

Determine the number of input variables ((n)). The total number of rows will be (2^n).
Write down all binary combinations for the inputs in systematic order.
Break down the expression into smaller sub-expressions or operations.
Evaluate each sub-expression column by column.
Compute the final output value for each row.

Example: Construct a truth table for:

[
F(A,B)=A\cdot\overline{B}+B
]

There are two input variables, so (2^2=4) rows are required.

(A)	(B)	(\overline{B})	(A\cdot\overline{B})	(F=A\cdot\overline{B}+B)
0	0	1	0	0
0	1	0	0	1
1	0	1	1	1
1	1	0	0	1

3. Describe the concept of duality in Boolean algebra and provide an example to illustrate it.

Answer:

Concept: The Duality Principle states that any true Boolean identity or algebraic expression remains true if all AND ((\cdot)) operators are swapped with OR ((+)) operators, all OR operators are swapped with AND operators, and all constants 0 and 1 are swapped ((0\leftrightarrow1)), while variables remain unchanged.

Example:

Original identity: (A+1=1)
Dual identity: Swap (+) with (\cdot) and (1) with (0).

[
A\cdot0=0
]

Both statements hold universally true in Boolean algebra.

4. Compare and contrast half-adders and full-adders, including their truth tables, Boolean expressions, and circuit diagrams.

Answer:

Comparison:

Half-Adder: Adds two single binary bits ((A) and (B)). It produces a Sum ((S)) and a Carry ((C)). It cannot take an incoming carry from a previous stage into account.
Full-Adder: Adds three single binary bits ((A), (B), and an incoming carry (C_{in})). It produces a Sum ((S)) and an outgoing Carry ((C_{out})).

Half-Adder Truth Table:

(A)	(B)	Sum ((S))	Carry ((C))
0	0	0	0
0	1	1	0
1	0	1	0
1	1	0	1

Full-Adder Truth Table:

(A)	(B)	(C_{in})	Sum ((S))	Carry ((C_{out}))
0	0	0	0	0
0	0	1	1	0
0	1	0	1	0
0	1	1	0	1
1	0	0	1	0
1	0	1	0	1
1	1	0	0	1
1	1	1	1	1

Boolean Expressions:

Half-Adder:

[
S=A\oplus B
]

[
C=A\cdot B
]

Full-Adder:

[
S=A\oplus B\oplus C_{in}
]

[
C_{out}=(A\cdot B)+[C_{in}\cdot(A\oplus B)]
]

Circuit Diagrams:

Half-Adder Circuit: Constructed using an XOR gate for the Sum ((S)) and an AND gate for the Carry ((C)), with inputs (A) and (B) connected to both gates.
Full-Adder Circuit: Constructed using two half-adders and an OR gate. The first half-adder adds (A) and (B). Its Sum output and (C_{in}) feed the second half-adder to produce (S). The carry outputs from both half-adders enter the OR gate to form (C_{out}).

5. How do Karnaugh maps simplify Boolean expressions? Provide a detailed example with steps.

Answer:

Simplification Mechanism: K-maps arrange minterms in a two-dimensional grid based on Gray Code sequence, where adjacent cells differ by only one bit. This arrangement allows terms in which a variable appears in both complemented and uncomplemented forms to be grouped and simplified using the identity:

[
X+\overline{X}=1
]

Detailed Example: Simplify:

[
F(A,B)=\overline{A},\overline{B}+\overline{A}B+A\overline{B}
]

Step 1: Construct a two-variable K-map grid, with (A) along the rows and (B) along the columns.

Step 2: Fill 1s into the cells corresponding to the minterms:

(\overline{A},\overline{B}): cell ((0,0)=1)
(\overline{A}B): cell ((0,1)=1)
(A\overline{B}): cell ((1,0)=1)

Step 3: Form adjacent groups of 1s in sizes that are powers of 2, such as 2 and 4.

Group 1: The top row contains (\overline{A},\overline{B}) and (\overline{A}B). Variable (B) changes, so (B) and (\overline{B}) cancel, leaving (\overline{A}).
Group 2: The left column contains (\overline{A},\overline{B}) and (A\overline{B}). Variable (A) changes, so (A) and (\overline{A}) cancel, leaving (\overline{B}).

Step 4: Combine the minimized group terms.

[
\boxed{F(A,B)=\overline{A}+\overline{B}}
]

6. Design a 4-bit binary adder using both half-adders and full-adders. Explain each step with truth tables, Boolean expressions, and circuit diagrams.

Answer:

Design Steps:

A 4-bit binary adder adds two 4-bit numbers:
[
A=A_3A_2A_1A_0,\quad B=B_3B_2B_1B_0
]
Stage 0 (LSB): Use a half-adder, or a full-adder with (C_{in}=0), to add bits (A_0) and (B_0). The outputs are Sum bit (S_0) and Carry (C_1).
[
S_0=A_0\oplus B_0,\quad C_1=A_0\cdot B_0
]
Stage 1: Use a full-adder to add (A_1), (B_1), and incoming carry (C_1). The outputs are (S_1) and (C_2).
Stage 2: Use a full-adder to add (A_2), (B_2), and incoming carry (C_2). The outputs are (S_2) and (C_3).
Stage 3 (MSB): Use a full-adder to add (A_3), (B_3), and incoming carry (C_3). The outputs are (S_3) and the final carry (C_4).

Circuit Diagram:

A 4-bit ripple-carry adder consists of four stages connected sequentially. The carry output ((C_{out})) of each stage is connected directly to the carry input ((C_{in})) of the next stage. The first stage can use a half-adder, while the remaining three stages use full-adders.

7. Simplify the following Boolean function using Boolean algebra rules.

[
F(A,B)=A\cdot B+A\cdot\overline{B}
]

Solution:

Factoring out (A) using the Distributive Law:

[
F(A,B)=A\cdot(B+\overline{B})
]

Since (B+\overline{B}=1), by the Complement Law:

[
F(A,B)=A\cdot1
]

Since (A\cdot1=A), by the Identity Law:

[
\boxed{F(A,B)=A}
]

8. Use De Morgan’s laws to simplify the following function.

[
F(A,B,C)=\overline{A+B+AC}
]

Solution:

Applying De Morgan’s First Law:

[
F(A,B,C)=\overline{A}\cdot\overline{B}\cdot\overline{AC}
]

Applying De Morgan’s Second Law to (\overline{AC}):

[
F(A,B,C)=\overline{A}\cdot\overline{B}\cdot(\overline{A}+\overline{C})
]

Distributing (\overline{A}\cdot\overline{B}) inside the bracket:

[
F(A,B,C)=(\overline{A}\cdot\overline{B}\cdot\overline{A})+(\overline{A}\cdot\overline{B}\cdot\overline{C})
]

Since (\overline{A}\cdot\overline{A}=\overline{A}), by the Idempotent Law:

[
F(A,B,C)=(\overline{A}\cdot\overline{B})+(\overline{A}\cdot\overline{B}\cdot\overline{C})
]

Factoring out (\overline{A}\cdot\overline{B}):

[
F(A,B,C)=\overline{A}\cdot\overline{B}\cdot(1+\overline{C})
]

Since (1+\overline{C}=1):

[
\boxed{F(A,B,C)=\overline{A}\cdot\overline{B}}
]

9. Simplify the following expressions.

(a) ((A+\overline{B})\cdot(A+B))

Applying the Distributive Law:

[
=A+(\overline{B}\cdot B)
]

Since (\overline{B}\cdot B=0):

[
=A+0
]

[
\boxed{=A}
]

(b) ((A+\overline{B})\cdot(\overline{A}+B))

Expanding using the FOIL method:

[
=(A\cdot\overline{A})+(A\cdot B)+(\overline{B}\cdot\overline{A})+(\overline{B}\cdot B)
]

Since (A\cdot\overline{A}=0) and (\overline{B}\cdot B=0):

[
=0+A\cdot B+\overline{A}\cdot\overline{B}+0
]

[
\boxed{=A\cdot B+\overline{A}\cdot\overline{B}=A\odot B}
]

This is equivalent to the XNOR operation.

(c) (A+\overline{A}\cdot(\overline{B}+C))

Applying the Distributive Law:

[
=A+(\overline{B}+C)
]

[
\boxed{=A+\overline{B}+C}
]

(d) (A\cdot B+A\cdot B)

Applying the Idempotent Law, (X+X=X):

[
\boxed{=A\cdot B}
]

(e) ((A\cdot B)+(\overline{A}\cdot\overline{B}))

This expression represents the XNOR function and is already in its minimal Sum-of-Products (SOP) form.

[
\boxed{=A\odot B=A\cdot B+\overline{A}\cdot\overline{B}}
]

Notes by Sir Dilawar

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