Boolean Algebra Revision Notes
Subject: Computer Science | Level: A-Level | Exam Board: Edexcel
Master the core of digital logic with this guide to Boolean Algebra for Edexcel A-Level Computer Science. We break down complex simplification, Karnaugh maps, and logic circuits into exam-focused, easy-to-revise chunks, helping you secure top marks.
Revision Notes & Key Concepts
Key Terms & Definitions
- Boolean Algebra
- A branch of algebra in which the variables' values are the truth values true (1) and false (0), and the primary operations are the logical operations AND, OR, and NOT.
- Logic Gate
- A basic building block of a digital circuit that performs a logical operation on one or more binary inputs to produce a single binary output.
- Truth Table
- A table that shows the output of a logic circuit for all possible combinations of its inputs.
- Karnaugh Map (K-map)
- A graphical method used to simplify Boolean algebra expressions, where cells are ordered using Gray code.
- Sum of Products (SOP)
- A Boolean expression containing ORed AND terms.
- D-type Flip-Flop
- A sequential logic circuit that can store one bit of information. The output (Q) takes the value of the data input (D) on the rising edge of a clock signal.
Worked Examples
Worked Example
Question: Simplify the following Boolean expression using algebraic laws: F = (A + B) · (A + ¬B)
Solution: Step 1: Apply the distributive law: F = A·A + A·¬B + B·A + B·¬B Step 2: Apply the idempotent law (A·A = A) and complement law (B·¬B = 0): F = A + A·¬B + A·B + 0 Step 3: Factor out A: F = A(1 + ¬B + B) Step 4: Apply the complement law (¬B + B = 1) and then the null law (1 + 1 = 1): F = A(1 + 1) = A(1) Step 5: Apply the identity law (A·1 = A): F = A Final answer: F = A
Worked Example
Question: Use De Morgan's laws to simplify the expression: Q = ¬(¬A + B) · ¬(A · ¬B)
Solution: Step 1: Apply De Morgan's law to the first term ¬(¬A + B) ≡ ¬(¬A) · ¬B. Then apply double negation ¬(¬A) = A. So the first term becomes A · ¬B. Step 2: Apply De Morgan's law to the second term ¬(A · ¬B) ≡ ¬A + ¬(¬B). Then apply double negation ¬(¬B) = B. So the second term becomes ¬A + B. Step 3: Combine the simplified terms: Q = (A · ¬B) · (¬A + B) Step 4: Apply the distributive law: Q = (A · ¬B · ¬A) + (A · ¬B · B) Step 5: Apply the complement law (A · ¬A = 0 and B · ¬B = 0): Q = 0 + 0 Final answer: Q = 0
Worked Example
Question: A logic circuit is represented by the Boolean expression Z = A·B + ¬A·C + B·C. Simplify this expression using a Karnaugh map.
Solution: Step 1: Draw a 3-variable K-map with columns for A and rows for BC (labelled 00, 01, 11, 10). Step 2: Populate the map. For A·B, place 1s in cells (A=1, B=1). For ¬A·C, place 1s in cells (A=0, C=1). For B·C, place 1s in cells (B=1, C=1). Step 3: The map will have 1s in cells 3, 5, 6, and 7 (if numbered 0-7). Step 4: Group the 1s. Create a group of two for the 1s in cells (A=1, B=1, C=0) and (A=1, B=1, C=1). This simplifies to A·B. Create another group of two for the 1s in cells (A=0, B=1, C=1) and (A=1, B=1, C=1). This simplifies to B·C. Create a third group of two for the 1s in cells (A=0, B=0, C=1) and (A=0, B=1, C=1). This simplifies to ¬A·C. Step 5: Identify the redundant group. The term B·C is redundant because both of its 1s are already covered by the other two groups. This is a key simplification step. Step 6: Write the final expression from the essential prime implicants. Z = A·B + ¬A·C. Final answer: Z = A·B + ¬A·C
Practice Questions
Question: Draw a logic gate circuit to represent the un-simplified Boolean expression F = ¬(A + B) + (A · B).
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Question: Simplify the expression Q = (X + Y) · (¬X + Y) using Boolean algebra.
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Question: A system has three inputs, A, B, and C. The output Z is 1 only if a majority of the inputs are 1. Derive the simplified sum-of-products expression for Z using a Karnaugh map.
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Question: Construct a NOT gate, an AND gate, and an OR gate using only two-input NAND gates.
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Question: Explain the purpose of the Gray code ordering in a Karnaugh map.
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