Digital Logic Design Fundamentals: A Comprehensive Guide

Posted on Jul 23, 2024 in Computers

Unit I: Number System, Binary Codes, and Data Representation

1. Explain the process of converting a decimal number to its binary, octal, and hexadecimal equivalents. Provide an example for each conversion.

Binary Conversion:

To convert a decimal number to binary, repeatedly divide the number by 2 and record the remainder. Read the remainders in reverse order to get the binary equivalent.

Example:

Convert 45 to binary.

1. 45 ÷ 2 = 22 remainder 1
2. 22 ÷ 2 = 11 remainder 0
3. 11 ÷ 2 = 5 remainder 1
4. 5 ÷ 2 = 2 remainder 1
5. 2 ÷ 2 = 1 remainder 0
6. 1 ÷ 2 = 0 remainder 1

Reading the remainders in reverse order: 45 in binary is 1011012101101_21011012​.

Octal Conversion:

To convert a decimal number to octal, repeatedly divide the number by 8 and record the remainder. Read the remainders in reverse order to get the octal equivalent.

Example:

Convert 45 to octal.

1. 45 ÷ 8 = 5 remainder 5
2. 5 ÷ 8 = 0 remainder 5

Reading the remainders in reverse order: 45 in octal is 55855_8558​.

Hexadecimal Conversion:

To convert a decimal number to hexadecimal, repeatedly divide the number by 16 and record the remainder. Read the remainders in reverse order to get the hexadecimal equivalent.

Example:

Convert 45 to hexadecimal.

1. 45 ÷ 16 = 2 remainder 13 (D in hexadecimal)
2. 2 ÷ 16 = 0 remainder 2

Reading the remainders in reverse order: 45 in hexadecimal is 2D162D_{16}2D16​.

2. What are BCD, Excess-3, and Gray codes? Discuss their advantages and disadvantages.

BCD (Binary Coded Decimal):

BCD is a binary-encoded representation of integer values that uses a 4-bit nibble to represent each digit of a decimal number.

Advantages:

• Easier to convert to and from human-readable decimal form.
• Useful in financial and business applications where decimal accuracy is crucial.

Disadvantages:

• Less efficient in terms of storage compared to pure binary representation.
• Arithmetic operations are more complex and slower.

Excess-3 Code:

Excess-3 is a non-weighted code used to express decimal numbers. Each decimal digit is represented by its corresponding 4-bit binary number plus 3.

Advantages:

• Self-complementary, which simplifies the design of digital systems.

Disadvantages:

• More complex to convert from/to decimal compared to BCD.

Gray Code:

Gray code is a binary numeral system where two successive values differ in only one bit.

Advantages:

• Reduces the chance of errors in digital systems, particularly in analog-to-digital converters.

Disadvantages:

• Not suitable for arithmetic operations.
• Conversion to and from binary can be complex.

Unit II: Binary Arithmetic and Logic Gates

3. Demonstrate binary addition and binary subtraction using 1’s and 2’s complement with examples.

Binary Addition:

Add the binary numbers 1011 and 1101.

1011 + 1101 ------ 11000

The sum is 11000.

Binary Subtraction Using 1’s Complement:

Subtract 1101 from 1011 using 1’s complement.

1. Find the 1’s complement of 1101: 0010

2. Add this to 1011:

   1011 + 0010 ------ 1101

3. Since there is no carry, take the 1’s complement of the result: 0010. Therefore, the answer is -0010 in 1’s complement form.

Binary Subtraction Using 2’s Complement:

Subtract 1101 from 1011 using 2’s complement.

1. Find the 2’s complement of 1101: 0010 + 1 = 0011

2. Add this to 1011:

   1011 + 0011 ------ 1110

3. Since there is a carry, the answer is positive. Therefore, the result is 1110.

4. Explain the function and properties of NAND and NOR gates. Why are they considered universal gates?

NAND Gate:

A NAND gate is a digital logic gate that outputs false only when all its inputs are true. Its truth table is as follows:

NOR Gate:

A NOR gate is a digital logic gate that outputs true only when all its inputs are false. Its truth table is as follows:

Properties:

• Both gates can be used to implement any other gate (AND, OR, NOT) and thus any digital circuit.
• They are called universal gates because any logical function can be implemented using only NAND or NOR gates.

Unit III: Logic Family

5. Describe the construction and working of TTL NAND and NOR gates.

TTL NAND Gate:

• TTL (Transistor-Transistor Logic) NAND gates are constructed using multiple bipolar junction transistors (BJTs) arranged in a totem-pole output stage.
• The inputs are connected to the base of transistors, and the output is taken from the collector.
• When all inputs are high, the output is low due to saturation of transistors. If any input is low, the output is high due to cutoff of the transistors.

TTL NOR Gate:

• Similar to TTL NAND, the NOR gate is constructed using BJTs, but the configuration ensures that when all inputs are low, the output is high.
• If any input is high, the output transistor is driven into saturation, resulting in a low output.

6. Compare TTL and CMOS logic families with respect to propagation delay time, power consumption, noise immunity, noise margin, fan-in, and fan-out.

TTL vs. CMOS:

Unit IV: Boolean Algebra

7. State and prove De Morgan’s Theorems. Provide examples of how they are used to simplify Boolean expressions.

De Morgan’s Theorems:

1. A⋅B‾=A‾+B‾\overline{A \cdot B} = \overline{A} + \overline{B}A⋅B=A+B
2. A+B‾=A‾⋅B‾\overline{A + B} = \overline{A} \cdot \overline{B}A+B​=A⋅B

Proof:

1. A⋅B‾=A‾+B‾\overline{A \cdot B} = \overline{A} + \overline{B}A⋅B=A+B

2. A+B‾=A‾⋅B‾\overline{A + B} = \overline{A} \cdot \overline{B}A+B​=A⋅B

Example:

Simplify A⋅B‾+A⋅B\overline{A \cdot B} + A \cdot BA⋅B+A⋅B.

Using De Morgan’s theorem: A⋅B‾+A⋅B=(A‾+B‾)+A⋅B\overline{A \cdot B} + A \cdot B = (\overline{A} + \overline{B}) + A \cdot BA⋅B+A⋅B=(A+B)+A⋅B

8. Simplify the following Boolean expression using Karnaugh maps: A′B+AB′+ABA’B + AB’ + ABA′B+AB′+AB.

Construct a Karnaugh map (K-map) for the given expression.

Step 1:

Plot the values.

Step 2:

Group the 1s in pairs.

From the K-map, group 1s:

• Group 1: A′BA’BA′B
• Group 2: AB′AB’AB′
• Group 3: ABABAB

Simplified Expression:

A′B+AB′+AB=B′+B=1A’B + AB’ + AB = B’ + B = 1A′B+AB′+AB=B′+B=1

Unit V: Combinational Circuits

9. Design a 4-bit binary adder/subtractor circuit and explain its working.

A 4-bit binary adder/subtractor circuit can be built using full adders and XOR gates.

Components:

• 4 Full Adders
• 4 XOR Gates
• Mode Control Bit (M)

Operation:

• When M = 0, the circuit performs addition.
• When M = 1, the circuit performs subtraction.

Steps:

1. Connect the inputs A and B to the XOR gates.
2. The XOR gates invert the B inputs when M = 1.
3. Connect the outputs of the XOR gates and A to the full adders.
4. The carry-in for the first full adder is M.
5. The carry-out from each full adder is connected to the next full adder’s carry-in.

Working:

• Addition: The full adders add A and B normally.
• Subtraction: B is inverted, and the circuit performs A + (~B + 1) using 2’s complement

10. What is a multiplexer and a demultiplexer? Explain their construction and working principles with examples.

Multiplexer (MUX):

A multiplexer is a combinational circuit that selects one of many inputs and forwards the selected input to a single output line.

Example:

A 4-to-1 MUX has 4 input lines (I0, I1, I2, I3), 2 select lines (S0, S1), and 1 output (Y).

Working:

• The select lines determine which input is connected to the output.
• If S1S0 = 00, Y = I0.
• If S1S0 = 01, Y = I1.
• If S1S0 = 10, Y = I2.
• If S1S0 = 11, Y = I3.

Demultiplexer (DEMUX):

A demultiplexer is a combinational circuit that takes a single input and routes it to one of many output lines.

Example:

A 1-to-4 DEMUX has 1 input line (I), 2 select lines (S0, S1), and 4 output lines (Y0, Y1, Y2, Y3).

Working:

• The select lines determine which output line the input is routed to.
• If S1S0 = 00, I is routed to Y0.
• If S1S0 = 01, I is routed to Y1.
• If S1S0 = 10, I is routed to Y2.
• If S1S0 = 11, I is routed to Y3.

These detailed answers provide a comprehensive understanding of each topic within the units, suitable for an 8-mark long answer question.