Number System Conversions and Boolean Simplification

Posted on Oct 30, 2024 in Design and Engineering

Number System Conversions

A. Binary to Hexadecimal

Convert (10000101)₂ to hexadecimal:

1. Group the binary digits into sets of four, starting from the right: 1000 0101

2. Convert each group to its hexadecimal equivalent:

• 1000₂ = 8₁₆
• 0101₂ = 5₁₆

Therefore, (10000101)₂ = (85)₁₆

B. Hexadecimal to Decimal

Convert (AA)₁₆ to decimal:

(AA)₁₆ = (10 × 16¹) + (10 × 16⁰) = 160 + 10 = (170)₁₀

C. Decimal to Binary

Convert (5.15)₁₀ to binary:

Integer Part

Convert 5₁₀ to binary:

• 5 ÷ 2 = 2 remainder 1
• 2 ÷ 2 = 1 remainder 0
• 1 ÷ 2 = 0 remainder 1

Reading the remainders from bottom to top: (101)₂

Fractional Part

Convert 0.15₁₀ to binary:

• 0.15 × 2 = 0.30 → 0
• 0.30 × 2 = 0.60 → 0
• 0.60 × 2 = 1.20 → 1
• 0.20 × 2 = 0.40 → 0
• 0.40 × 2 = 0.80 → 0
• 0.80 × 2 = 1.60 → 1
• 0.60 × 2 = 1.20 → 1

(0.15)₁₀ ≈ (0.0010011)₂

Combining both parts: (5.15)₁₀ ≈ (101.0010011)₂

D. Two’s Complement

Find the two’s complement of (01001)₂:

1. Invert all bits (one’s complement): 10110
2. Add 1: 10110 + 1 = (10111)₂

Boolean Simplification using Karnaugh Map

Simplify F(A, B, C, D) = Σm(0, 2, 3, 8, 10, 11, 12, 14)

K-map Setup and Grouping

The K-map is arranged as follows, with minterms grouped for simplification:

Group 1: m(0, 2, 8, 10) = B̅D̅

Group 2: m(3, 11) = A̅C

Group 3: m(12, 14) = AB

Simplified Expression and Logic Gate Implementation

The simplified expression is:

F(A, B, C, D) = B̅D̅ + A̅C + AB

This can be implemented using AND gates for each product term and an OR gate to sum the outputs of the AND gates.