Show understanding of Boolean algebra
15.2 Boolean Algebra and Logic Circuits
🔍 Boolean algebra is the mathematical foundation of digital logic. It lets us simplify complex circuits and prove that two different-looking expressions are actually the same. Think of it as the “grammar” of electronic logic, where the only words are TRUE (1) and FALSE (0).
Key Concepts
- Variables – Symbols like
A,B,Cthat can be 0 or 1. - Operators –
AND (∧),OR (∨),NOT (¬),XOR (⊕),NAND,NOR,XNOR. - Expressions – Combinations of variables and operators, e.g.
(A ∧ B) ∨ ¬C. - Truth Table – A table that lists all possible input combinations and the resulting output.
- Logic Gate – Physical device that implements an operator (e.g., a NAND gate implements the NAND operator).
Boolean Laws & Identities
| Law | Expression | Result |
|---|---|---|
| Identity | A ∧ 1 = A | A |
| Null | A ∨ 0 = A | A |
| Complement | A ∨ ¬A = 1 | 1 |
| De Morgan | ¬(A ∧ B) = ¬A ∨ ¬B | ¬A ∨ ¬B |
| Distributive | A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) | (A ∧ B) ∨ (A ∧ C) |
Truth Table Example
Expression: (A ∧ B) ∨ ¬C
| A | B | C | Result |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Logic Gates & Circuits
🎛️ Each logic gate is like a tiny decision-maker. For example:
- AND gate – Only outputs 1 if both inputs are 1. Think of it as a “yes‑yes” rule.
- OR gate – Outputs 1 if at least one input is 1. Think of it as a “yes‑or‑yes” rule.
- NOT gate – Inverts the input. 0 becomes 1 and 1 becomes 0. It’s like a mirror.
- XOR gate – Outputs 1 only when inputs differ. It’s the “different‑only” rule.
Exam Tips
📌 Tip 1: Always start by writing a truth table if the problem asks for simplification.
📌 Tip 2: Use De Morgan’s laws to convert between NAND/NOR and the basic operators.
📌 Tip 3: Remember that ¬(A ∧ B) = ¬A ∨ ¬B and ¬(A ∨ B) = ¬A ∧ ¬B – they’re your best friends for negations.
📌 Tip 4: When simplifying, look for common factors and apply the distributive law.
📌 Tip 5: Check your final expression against the original truth table to avoid mistakes.
Practice Problems
- Simplify the expression
(A ∧ ¬B) ∨ (¬A ∧ B) ∨ (A ∧ B)using Boolean algebra. - Draw a circuit that implements the function
F = (A ∨ B) ∧ (¬A ∨ C)using only NAND gates. - Given the truth table below, find a minimal sum‑of‑products expression.
| A | B | C | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Quick Reference Cheat Sheet
| Operator | Symbol | Truth Table (A,B) |
|---|---|---|
| AND | ∧ | 0∧0=0, 0∧1=0, 1∧0=0, 1∧1=1 |
| OR | ∨ | 0∨0=0, 0∨1=1, 1∨0=1, 1∨1=1 |
| NOT | ¬ | ¬0=1, ¬1=0 |
| XOR | ⊕ | 0⊕0=0, 0⊕1=1, 1⊕0=1, 1⊕1=0 |
Revision
Log in to practice.