Related Concepts: 11 SOP, POS, Canonical & Standard Forms | 07 Boolean Algebra Foundations & Duality Principle | 08 Boolean Algebra Significance & Circuit Minimization

12 Mathematical Conversions & Expansion of SOP and POS

Concept Overview

Converting standard Boolean expressions into canonical {fully expanded} Sum of Products (SOP) or Product of Sums (POS) form is a core skill in digital design.

Canonical expansion can be accomplished either visually via a Truth Table or mathematically using Algebraic Missing-Variable Expansion.

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1. Missing Variable Algebraic Expansion Rules

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]

A. Rule for Sum of Products (SOP) Expansion

• Algebraic Rule: Multiply each non-canonical product term by $(x+x^{\prime })=1$ for every missing variable $x$. Since $x+x^{\prime }=1$, multiplying by $1$ preserves identity while introducing the missing literal.
• Class Example: Convert $F(A,B)=\bar{A}+B$ to Canonical SOP. $F=\bar{A}(B+\bar{B})+B(A+\bar{A})$ $F=\bar{A}B+\bar{A}\bar{B}+AB+\bar{A}B$ Remove duplicate terms ($\bar{A}B$): $\mathbf{F}=\bar{\mathbf{A}}\bar{\mathbf{B}}+\bar{\mathbf{A}}\mathbf{B}+\mathbf{AB}=\sum \mathbf{m}(0,1,3)$

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B. Rule for Product of Sums (POS) Expansion

• Algebraic Rule: Add $(x\cdot x^{\prime })=0$ to each non-canonical sum term for every missing variable $x$. Then apply the distributive law $A+BC=(A+B)(A+C)$ to expand into linear sum terms.
• Class Example: Convert $F(A,B,C)=\bar{A}(B+C)$ to Canonical POS.
  1. Term 1 ($\bar{A}$) is missing variables $B$ and $C$. Add $B\bar{B}$ and $C\bar{C}$: $\bar{A}+B\bar{B}+C\bar{C}=(\bar{A}+B+C)(\bar{A}+B+\bar{C})(\bar{A}+\bar{B}+C)(\bar{A}+\bar{B}+\bar{C})$
  2. Term 2 ($B+C$) is missing variable $A$. Add $A\bar{A}$: $B+C+A\bar{A}=(A+B+C)(\bar{A}+B+C)$
  3. Combine all terms and eliminate duplicate maxterms: $\mathbf{F}=(\mathbf{A}+\mathbf{B}+\mathbf{C})(\bar{\mathbf{A}}+\mathbf{B}+\mathbf{C})(\bar{\mathbf{A}}+\mathbf{B}+\bar{\mathbf{C}})(\bar{\mathbf{A}}+\bar{\mathbf{B}}+\mathbf{C})(\bar{\mathbf{A}}+\bar{\mathbf{B}}+\bar{\mathbf{C}})=\prod \mathbf{M}(0,4,5,6,7)$

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2. Deriving Canonical Forms Using Truth Tables

Any Boolean function can be converted directly into Canonical SOP or Canonical POS by inspecting its truth table:

• Canonical SOP (Sum of Minterms $\Sigma m$): Formed by selecting every input combination row where the function output is $F=1$. Write the minterm for each such row and OR them together.
• Canonical POS (Product of Maxterms $\Pi M$): Formed by selecting every input combination row where the function output is $F=0$. Write the maxterm for each such row and AND them together.

De Morgan's Expansion Problem

Question: Use De Morgan’s theorem to express $(A+B+C{)}^{\prime }$ in both Canonical SOP and Canonical POS forms.

Solution:

1. Apply De Morgan’s Law: $(A+B+C{)}^{\prime }=A^{\prime }\cdot B^{\prime }\cdot C^{\prime }$
2. This expression is a single canonical product term (minterm $m_0$ where $A=0,B=0,C=0$).
   • Canonical SOP: $\mathbf{F}=\sum \mathbf{m}(0)={\mathbf{A}}^{\prime }{\mathbf{B}}^{\prime }{\mathbf{C}}^{\prime }$
3. Since minterm 0 is the only state yielding $F=1$, all remaining 7 input combinations ($1$ through $7$) yield $F=0$:
   • Canonical POS: $\mathbf{F}=\prod \mathbf{M}(1,2,3,4,5,6,7)$
4. Expanding the maxterm product: $F=(A+B+C^{\prime })(A+B^{\prime }+C)(A+B^{\prime }+C^{\prime })(A^{\prime }+B+C)(A^{\prime }+B+C^{\prime })(A^{\prime }+B^{\prime }+C)(A^{\prime }+B^{\prime }+C^{\prime })$

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3. Complex Exam Conversion Numerical Problems

Major PYQ Problem 1 (2022 - 10 Marks)

Question: Convert the Boolean expression $(AB+C)(B+C^{\prime }D)$ into both SOP and POS forms.

Solution:

1. Converting to SOP Form:

Multiply out the factors directly using distribution: $F=(AB+C)(B+C^{\prime }D)=ABB+AB{C}^{\prime }D+BC+C{C}^{\prime }D$

Apply Boolean laws ($B\cdot B=B$ and $C\cdot C^{\prime }=0$): $F=AB+AB{C}^{\prime }D+BC+0$

Factor $AB$ from the first two terms: $F=AB(1+C^{\prime }D)+BC$

Since $(1+\text{anything})=1$: $\text{Final Simplified SOP: }\mathbf{F}=\mathbf{AB}+\mathbf{BC}$

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2. Converting to POS Form:

First, apply distributive law $X+YZ=(X+Y)(X+Z)$ to expand $(AB+C)$: $(AB+C)=(A+C)(B+C)$ Thus, the unexpanded POS form is: $F=(A+C)(B+C)(B+C^{\prime })(B+D)$

Next, insert missing variables as $(x\cdot x^{\prime })=0$:

• Term 1 $(A+C)$: Add $B{B}^{\prime }$ and $D{D}^{\prime }$.
• Term 2 $(B+C)$: Add $A{A}^{\prime }$ and $D{D}^{\prime }$.
• Term 3 $(B+C^{\prime })$: Add $A{A}^{\prime }$ and $D{D}^{\prime }$.
• Term 4 $(B+D)$: Add $A{A}^{\prime }$ and $C{C}^{\prime }$.

Expanding all four terms yields the complete canonical product of maxterms: $\text{Final POS: }\prod \mathbf{M}(0,1,2,3,4,5,8,9,10,11)$ (Note: Its equivalent canonical SOP minterms would be $\Sigma m(6,7,12,13,14,15)$).

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Major PYQ Problem 2 (2022 - 10 Marks)

Question: Convert the Boolean expression $x^{\prime }+x(x+y^{\prime })(y+z^{\prime })$ into both SOP and POS forms.

Solution:

1. Converting to SOP Form:

Expand the product terms: $F=x^{\prime }+x(x+y^{\prime })(y+z^{\prime })$ $F=x^{\prime }+(xx+x{y}^{\prime })(y+z^{\prime })$ $F=x^{\prime }+(x+x{y}^{\prime })(y+z^{\prime })\text{(since }xx=x\text{)}$ $F=x^{\prime }+x(1+y^{\prime })(y+z^{\prime })=x^{\prime }+x(y+z^{\prime })$ $F=x^{\prime }+xy+x{z}^{\prime }$

Apply absorption law ($x^{\prime }+xy=x^{\prime }+y$): $F=(x^{\prime }+y)+x{z}^{\prime }$ Apply absorption law again ($x^{\prime }+x{z}^{\prime }=x^{\prime }+z^{\prime }$): $\text{Final Simplified SOP: }\mathbf{F}={\mathbf{x}}^{\prime }+\mathbf{y}+{\mathbf{z}}^{\prime }$

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2. Converting to POS Form:

Notice that the fully simplified SOP expression derived above is: $F=x^{\prime }+y+z^{\prime }$ Because this simplified expression consists of a single OR sum term containing no product multiplications, it is intrinsically already in POS form.

$\text{Final POS: }\mathbf{F}=({\mathbf{x}}^{\prime }+\mathbf{y}+{\mathbf{z}}^{\prime })$

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Major PYQ Problem 3 (4-Variable Canonical Expansion - 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:

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

• Term 1 $(A+B^{\prime }+C)$ is missing $D$: $(A+B^{\prime }+C+D{D}^{\prime })=(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })$ In binary ($A=0,B=1,C=0$):

  • $0100_2=4⟹M_4$
  • $0101_2=5⟹M_5$

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

  • $(A+B+C+D^{\prime })\to 0001_2=1⟹M_1$
  • $(A+B+C^{\prime }+D^{\prime })\to 0011_2=3⟹M_3$
  • $(A^{\prime }+B+C+D^{\prime })\to 1001_2=9⟹M_9$
  • $(A^{\prime }+B+C^{\prime }+D^{\prime })\to 1011_2=11⟹M_{11}$

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

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

Since a 4-variable function has $2^4=16$ total minterms/maxterms ($0$ through $15$), the minterms are all the remaining index numbers not present in the maxterm list:

$\text{Remaining Minterm Indices: }\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}\setminus \{1,3,4,5,9,11\}$

$\text{Sum of Minterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\sum \mathbf{m}(0,2,6,7,8,10,12,13,14,15)■$

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Past Year Questions (PYQs)

• 2022 (10 Marks): Convert $(AB+C)(B+C^{\prime }D)$ and $x^{\prime }+x(x+y^{\prime })(y+z^{\prime })$ into SOP and POS forms.
  • Solution: See Problem 1 and Problem 2 above.
• 2015, 2017, 2021 (12-10 Marks): Express the function in a Sum of Minterms and a Product of Maxterms.
  • Solution: Follow the 4-variable algebraic missing variable expansion method demonstrated in Problem 3.