Related Concepts: 03 SOP, POS, Canonical & Standard Forms | 01 Boolean Algebra Foundations & Duality Principle | 05 Tabular Method (Quine-McCluskey) & Prime Implicants

3.04 Mathematical Conversions & Expansion of SOP and POS

Concept Overview

Converting standard Boolean expressions into canonical Sum of Products (SOP) or Product of Sums (POS) form is accomplished algebraically via Missing-Variable Expansion.

graph TD
    Expression[Non-Canonical Expression] --> Type{Expansion Target}
    
    Type -->|Expand to SOP| SOPRule[Multiply term by x + x' = 1]
    SOPRule --> ExpandSOP[Apply Distributive Law: AB x+x' = ABx + ABx']
    ExpandSOP --> ListMinterms[List Sum of Minterms Σm]
    
    Type -->|Expand to POS| POSRule[Add xx' = 0 to sum term]
    POSRule --> ExpandPOS[Apply Distributive Law: X + YZ = X+Y X+Z]
    ExpandPOS --> ListMaxterms[List Product of Maxterms ΠM]

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1. Algebraic Expansion Rules

• SOP Expansion Rule: Multiply each non-canonical AND product term by $(x+x^{\prime })=1$ for every missing variable $x$.
• POS Expansion Rule: Add $(x\cdot x^{\prime })=0$ to each non-canonical OR sum term for every missing variable $x$, then expand via distribution: $A+BC=(A+B)(A+C)$.

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2. 4-Variable Canonical Expansion (Major PYQ)

Major PYQ Problem (2015, 2017, 2021 - 12 Marks)

Question: Express $F(A,B,C,D)=(A+B^{\prime }+C)(B+D^{\prime })$ in Product of Maxterms ($\Pi M$) and Sum of Minterms ($\Sigma m$) notation.

Solution:

Part 1: Product of Maxterms ($\Pi M$)

• Term 1 $(A+B^{\prime }+C)$ is missing $D$: $(A+B^{\prime }+C+D{D}^{\prime })=(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })$ $\text{Binary: }0100_2=M_4,0101_2=M_5$

• Term 2 $(B+D^{\prime })$ is missing $A$ and $C$: $B+D^{\prime }+A{A}^{\prime }+C{C}^{\prime }$ Expanding via distribution yields 4 maxterms:

  • $(A+B+C+D^{\prime })\to 0001_2=M_1$
  • $(A+B+C^{\prime }+D^{\prime })\to 0011_2=M_3$
  • $(A^{\prime }+B+C+D^{\prime })\to 1001_2=M_9$
  • $(A^{\prime }+B+C^{\prime }+D^{\prime })\to 1011_2=M_{11}$

$\text{Product of Maxterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\prod \mathbf{M}(1,3,4,5,9,11)$

Part 2: Sum of Minterms ($\Sigma m$)

The minterm indices are the remaining numbers out of $0\dots 15$: $\text{Sum of Minterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\sum \mathbf{m}(0,2,6,7,8,10,12,13,14,15)■$

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3. Class Test 01 Master Problem: SOP vs POS Gate Optimization

Class Test 01: SOP vs POS Gate Optimization

Question: Given $F=\Pi M(0,2,5,7,8,10,13,15)$: a) Convert into canonical SOP. b) Which implementation (SOP or POS) requires fewer logic gates?

Part A: Canonical SOP The minterm list is the exact inverse of the given maxterm list for a 4-variable system (0-15):

• Minterms = $1,3,4,6,9,11,12,14$
• Canonical SOP: $\mathbf{F}=\mathbf{\Sigma m}(1,3,4,6,9,11,12,14)$

Part B: Gate Optimization Justification To determine which implementation requires fewer gates, simplify both using K-maps:

• SOP Simplification: Plotting 1s yields a quad $(1,3,9,11)\to B^{\prime }D$ and a quad $(4,6,12,14)\to B{D}^{\prime }$. $F_{\text{SOP}}=B^{\prime }D+B{D}^{\prime }=\mathbf{B}\oplus \mathbf{D}$
• POS Simplification: Plotting 0s yields $F^{\prime }=B^{\prime }{D}^{\prime }+BD=(B\oplus D{)}^{\prime }$. Complementing gives: $F_{\text{POS}}=(B+D)(B^{\prime }+D^{\prime })=\mathbf{B}\oplus \mathbf{D}$
• Conclusion: Both simplified expressions mathematically reduce to the exact same logic (an XOR gate, or 2 ANDs + 1 OR). Therefore, both SOP and POS require the exact same number of logic gates. $■$

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Past Year Questions (PYQs)

• [PYQ 2015, 2017, 2021, 2024]: Express equations into full SOP (Sum of Minterms) and POS (Product of Maxterms) forms. (10 to 12 Marks)
• [Class Test 01]: Canonical SOP conversion and SOP vs POS gate optimization analysis ($F=\Pi M(0,2,5,7,8,10,13,15)$). (10 Marks)