Fundamentals of Digital, Analog Electronics, Measurement and Electrical Safety (10hrs) · Basics of Electrical & Electronics Engineering

Boolean algebra

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Boolean algebra

1. What Boolean Algebra Is (for Digital)

Boolean algebra = algebra for binary variables (0/10/1 or false/true) with operations AND (·), OR (+), NOT (¯), obeying postulates parallel to ordinary algebra but with idempotence and complementation. Basis for minimizing logic circuits.

2. Axioms / Laws (memorize 8 with dual)

Law Form (OR) Form (AND) Dual
Commutative A+B=B+AA+B=B+A A⋅B=B⋅AA·B=B·A swap +↔·
Associative (A+B)+C=A+(B+C)(A+B)+C=A+(B+C) (A⋅B)⋅C=A⋅(B⋅C)(A·B)·C=A·(B·C)
Distributive A⋅(B+C)=A⋅B+A⋅CA·(B+C)=A·B+A·C A+B⋅C=(A+B)(A+C)A+B·C=(A+B)(A+C) (unlike ordinary!)
Identity A+0=AA+0=A, A+1=1A+1=1 A⋅1=AA·1=A, A⋅0=0A·0=0
Idempotent A+A=AA+A=A A⋅A=AA·A=A
Complement A+Aˉ=1A+\bar{A}=1 A⋅Aˉ=0A·\bar{A}=0
Involution Aˉˉ=A\bar{\bar{A}}=A
Absorption A+A⋅B=AA+A·B=A, A⋅(A+B)=AA·(A+B)=A
De Morgan's A+B‾=Aˉ⋅Bˉ\overline{A+B}= \bar{A}·\bar{B} A⋅B‾=Aˉ+Bˉ\overline{A·B}= \bar{A}+\bar{B} key for universal

3. Simplification Example

Y = A·B + A·\bar{B} = A(B+\bar{B})=A·1=A ⇒ needs one wire, not two ANDs + OR.

Y = (A+B)(A+C)=A+BC (distributive variant)

Use Karnaugh map for 3+ variables, but Boolean laws suffice for exam.

Prove: \overline{A·B+C}=? Use De Morgan stepwise.

4. Duality Principle

Dual of expression: swap +↔·, 0↔1, keep variables — if law holds, dual holds (e.g., A+1=1A+1=1 dual is A⋅0=0A·0=0).

5. Exam Points

  • List laws (commutative, associative, distributive, De Morgan) — 5-7 marks, very frequent (almost certain to ask laws).
  • State De Morgan's A+B‾=AˉBˉ\overline{A+B}=\bar{A}\bar{B}, AB‾=Aˉ+Bˉ\overline{AB}=\bar{A}+\bar{B} — 5 marks, very frequent.
  • Simplify A+AB=AA+AB=A, A(A+B)=AA(A+B)=A, AB+ABˉ=AAB+A\bar{B}=A — 5-7 marks, very frequent (common simplification numerical: minimize Y=AˉB+ABY=\bar{A}B+AB).
  • Show A+AˉB=A+BA+\bar{A}B=A+B — 5 marks, frequent.
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