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Class 01

Boolean Algebra – Class LectureÂļ

āφāϜāϕ⧇ āφāĻŽāϰāĻž āĻļāĻŋāĻ–āĻŦ Boolean Algebra āύāĻŋā§Ÿā§‡āĨ¤
Boolean Algebra basically āĻāĻ•āϟāĻž mathematical system, āϝ⧇āϟāĻž āĻĻāĻŋā§Ÿā§‡ āφāĻŽāϰāĻž logic operation āĻ•āϰāĻŋ using binary variables. Binary āĻŽāĻžāύ⧇ āĻļ⧁āϧ⧁ āĻĻ⧁āχāϟāĻž value āύāĻŋāϤ⧇ āĻĒāĻžāϰāĻŦ⧇:

  • 1 → True / High
  • 0 → False / Low

Variable āϏāĻžāϧāĻžāϰāĻŖāϤ alphabet āĻĻāĻŋā§Ÿā§‡ represent āĻ•āϰāĻŋ, āϝ⧇āĻŽāύ A, B, C.
āϝāĻĻāĻŋ āφāĻŽāϰāĻž A bar (\(\overline{A}\)) āϞāĻŋāĻ–āĻŋ, āĻŽāĻžāύ⧇ A-āĻāϰ complement, āĻ…āĻ°ā§āĻĨāĻžā§Ž A=1 āĻšāϞ⧇ \(\overline{A}\)=0 āφāϰ A=0 āĻšāϞ⧇ \(\overline{A}\)=1.


Boolean FunctionÂļ

Boolean function āĻšāϞ⧋ āĻāĻŽāύ āĻāĻ•āϟāĻž equation āϝ⧇āĻ–āĻžāύ⧇ variable āϗ⧁āϞāĻž combine āĻšā§Ÿ AND (¡), OR (+ āĻŦāĻž \(\vee\)), āφāϰ NOT (bar) āĻĻāĻŋā§Ÿā§‡āĨ¤
Example:

\[ F = \overline{A} \cdot \overline{B} \, \vee \, C \]

āĻŽāĻžāύ⧇ F āĻšāĻŦ⧇ 1 āϝāĻĻāĻŋ A āĻāĻŦāĻ‚ B āĻĻ⧁āĻŸā§‹āχ 0 āĻšā§Ÿ, āĻ…āĻĨāĻŦāĻž C=1 āĻšā§ŸāĨ¤


Truth Table āĻŦāĻžāύāĻžāύ⧋Âļ

Truth Table basically āϏāĻŦ possible input combination āĻāϰ āϜāĻ¨ā§āϝ output āĻĻ⧇āĻ–āĻžā§ŸāĨ¤
āĻāχ function \(F = \overline{A} \cdot \overline{B} \vee C\) āĻāϰ truth table:

A B C \(\overline{A}\) \(\overline{B}\) \(\overline{A} \cdot \overline{B}\) F
0 0 0 1 1 1 1
0 0 1 1 1 1 1
0 1 0 1 0 0 0
0 1 1 1 0 0 1
1 0 0 0 1 0 0
1 0 1 0 1 0 1
1 1 0 0 0 0 0
1 1 1 0 0 0 1

Logic Circuit DiagramÂļ

Circuit āĻŦāĻžāύāĻžāϤ⧇ āĻĒā§āϰāĻĨāĻŽā§‡ A āφāϰ B āĻāϰ āωāĻĒāϰ NOT gate āϞāĻžāĻ—āĻžāĻŦā§‹, āϤāĻžāϰāĻĒāϰ output āĻĻ⧁āχāϟāĻž AND gate āĻĻāĻŋā§Ÿā§‡ connect āĻ•āϰāĻŦ, āϤāĻžāϰāĻĒāϰ C āĻāϰ āϏāĻžāĻĨ⧇ āĻāĻ•āϟāĻž OR gate āĻ āĻĻāĻŋāĻŦā§‹āĨ¤ Final output āĻšāĻŦ⧇ F.

Figure 1: Logic circuit implementation of 𝐹 = 𝐴 ‾ ⋅ đĩ ‾ ∨ đļ F= A ⋅ B ∨C using NOT, AND, and OR gates
Figure 1: Logic circuit implementation of \(F = \overline{A} \cdot \overline{B} \vee C\) using NOT, AND, and OR gates.


āĻ•āĻŋāϛ⧁ Boolean IdentitiesÂļ

  1. \(A \vee 0 = A\)
  2. \(A \cdot 0 = 0\)
  3. \(A \vee 1 = 1\)
  4. \(A \cdot 1 = A\)
  5. \(A \vee A = A\)
  6. \(A \cdot A = A\)
  7. \(A \vee \overline{A} = 1\)
  8. \(A \cdot \overline{A} = 0\)
  9. \(\overline{\overline{A}} = A\)
  10. \(A \vee B = B \vee A\) (Commutative)

āφāĻŽāϰāĻž expression āϟāĻž āύāĻŋāĻšā§āĻ›āĻŋ:

\[ F = ABCD \vee \overline{A}BCD \vee BC \]

Step-by-step:

  1. \(BC(AD \vee \overline{A}D \vee 1)\) – āĻāĻ–āĻžāύ⧇ common \(BC\) āĻŦ⧇āϰ āĻ•āϰ⧇ āύāĻŋāϞāĻžāĻŽāĨ¤
  2. \(= BCD(A \vee \overline{A}) \vee BC\) – Identity: \(A \vee \overline{A} = 1\)
  3. \(= BCD \cdot 1 \vee BC\) – 1 āĻĻāĻŋā§Ÿā§‡ AND āĻ•āϰāϞ⧇ āĻāĻ•āχ āĻĨāĻžāϕ⧇āĨ¤
  4. \(= BCD \vee BC\) – āĻāĻ–āύ BC āĻŦ⧇āϰ āĻ•āϰāϞ⧇ āφāϰāĻ“ simplify āĻšāĻŦ⧇āĨ¤
  5. \(= BC(D \vee 1)\) – Identity: \(X \vee 1 = 1\)
  6. \(= BC \cdot 1 = BC\)

So final simplified form:

\[ F = BC \]

Consensus Theorem (1)Âļ

  1. Consensus āĻŽāĻžāύ⧇ āĻšāϞ⧋ āĻāĻ•āϤāĻž āĻŦāĻž āϏāĻžāĻŽā§āϝāĨ¤

Consensus theorem āĻŦāϞ⧇ āϝ⧇:

\[ AB\overline{C} \vee BC \vee A\overline{C} = AB \vee A\overline{C} \]

āĻŽāĻžāύ⧇, āϝāĻĻāĻŋ expression āĻāϰ āĻŽāĻ§ā§āϝ⧇ āĻāĻŽāύ āĻāĻ•āϟāĻŋ term āĻĨāĻžāϕ⧇ āϝ⧇āĻ–āĻžāύ⧇ āĻāĻ•āϟāĻŋ literal (āϝ⧇āĻŽāύ B) AND āĻšā§Ÿā§‡āϛ⧇ āϤāĻžāϰ complement-āĻāϰ āϏāĻžāĻĨ⧇ (āĻ…āĻ¨ā§āϝ term-āĻ), āϤāĻžāĻšāϞ⧇ āĻ“āχ AND term āϟāĻž āĻŦāĻžāĻĻ āĻĻ⧇āĻ“ā§ŸāĻž āϝāĻžā§ŸāĨ¤

Easy āĻ­āĻžāĻŦ⧇: āĻāĻ•āϟāĻž term āĻ…āĻ¨ā§āϝ āĻĻ⧁āχ term-āĻāϰ āĻŽāĻ§ā§āϝ⧇ "consensus" āϤ⧈āϰāĻŋ āĻ•āϰ⧇, āϝ⧇āϟāĻž actually redundant, āϤāĻžāχ āφāĻŽāϰāĻž āϏ⧇āϟāĻž āϏāϰāĻŋā§Ÿā§‡ āĻĢ⧇āϞāϤ⧇ āĻĒāĻžāϰāĻŋāĨ¤


Just English – Textbook StyleÂļ

Example SimplificationÂļ

Given:

\[ F = ABCD \vee \overline{A}BCD \vee BC \]

Steps:

  1. Factor \(BC\):
\[ F = BC(AD \vee \overline{A}D \vee 1) \]
  1. Apply complement law \(A \vee \overline{A} = 1\):
\[ = BCD(1) \vee BC \]
  1. Simplify \(X \cdot 1 = X\):
\[ = BCD \vee BC \]
  1. Factor \(BC\) again:
\[ = BC(D \vee 1) \]
  1. Apply identity law \(X \vee 1 = 1\):
\[ = BC \cdot 1 \]
  1. Final:
\[ F = BC \]

Consensus Theorem (Textbook Form)Âļ

The Consensus Theorem states:

\[ AB\overline{C} \vee BC \vee A\overline{C} = AB \vee A\overline{C} \]

The theorem indicates that the AND term \(BC\) can be eliminated if one literal (such as \(B\)) appears in both complemented and uncomplemented form in other terms of the expression.