[Boolean Logic] SOP/POS Terms Explained
GCE A/L ICT – Sri Lanka | Beginner Friendly Guide with Examples & Exercises
9. CSOP and CPOS - The Canonical Forms
• CSOP = Canonical Sum of Products = Sum (OR) of ALL Min Terms where F=1
• CPOS = Canonical Product of Sums = Product (AND) of ALL Max Terms where F=0
What is CSOP?
- CSOP stands for Canonical Sum of Products
- It's the standard form where the function is expressed as the sum (OR) of ALL minterms where F=1
- Each minterm contains ALL variables in either true or complemented form
- Also called "Full SOP" or "Standard SOP"
Step 1: Expand each term to include ALL variables
• X'Y = X'Y(Z + Z') = X'YZ + X'YZ'
• XY'Z = XY'Z (already has all 3 variables)
• XYZ = XYZ (already has all 3 variables)
Step 2: Combine all minterms
F = X'YZ + X'YZ' + XY'Z + XYZ
Step 3: Convert to minterm notation
X'YZ = 011 → m₃
X'YZ' = 010 → m₂
XY'Z = 101 → m₅
XYZ = 111 → m₇
CSOP: F = Σm(2,3,5,7)
What is CPOS?
- CPOS stands for Canonical Product of Sums
- It's the standard form where the function is expressed as the product (AND) of ALL maxterms where F=0
- Each maxterm contains ALL variables in either true or complemented form
- Also called "Full POS" or "Standard POS"
Step 1: Expand each term to include ALL variables
• (X+Y) = (X+Y+Z)(X+Y+Z')
• (X'+Z) = (X'+Y+Z)(X'+Y'+Z)
• (Y+Z) = (X+Y+Z)(X'+Y+Z)
Step 2: Combine all maxterms (remove duplicates)
F = (X+Y+Z)(X+Y+Z')(X'+Y+Z)(X'+Y'+Z)
Step 3: Convert to maxterm notation
X+Y+Z = 000 → M₀
X+Y+Z' = 001 → M₁
X'+Y+Z = 010 → M₂
X'+Y'+Z = 011 → M₃
CPOS: F = ΠM(0,1,2,3)
CSOP vs SSOP (Simplified SOP)
- CSOP: All terms have ALL variables (canonical form)
- SSOP: Some terms may have fewer variables (simplified form)
CSOP: F = X'Y(Z+Z') + XYZ = X'YZ + X'YZ' + XYZ = Σm(2,3,7)
SSOP: F = X'Y + XYZ (not all terms have all 3 variables)
Key Difference: SSOP is simplified and shorter, but CSOP is complete and systematic
Converting CSOP to CPOS (Step-by-Step)
Step 1: Identify total number of possible terms
3 variables → 2³ = 8 terms (0 to 7)
Step 2: Find which terms are NOT in CSOP
CSOP has m₁, m₃, m₅, m₆, m₇
Missing terms: m₀, m₂, m₄
Step 3: These missing terms become the maxterms in CPOS
Missing minterms → m₀, m₂, m₄
Corresponding maxterms → M₀, M₂, M₄
Step 4: Write CPOS
F = ΠM(0,2,4)
Why? Because CSOP = 1 for m₁,m₃,m₅,m₆,m₇ → CPOS = 0 for m₀,m₂,m₄
10. Deriving Functions from Truth Tables
1. Create truth table with all variables
2. For CSOP: Find rows where F=1
3. For each such row, write the minterm
4. OR all minterms together
5. For CPOS: Find rows where F=0
6. For each such row, write the maxterm
7. AND all maxterms together
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Step 1: Find rows where F=1 → Rows 1, 3, 4, 6, 7
Step 2: Write minterms for these rows
Row 1 (0,0,1): X'Y'Z = m₁
Row 3 (0,1,1): X'YZ = m₃
Row 4 (1,0,0): XY'Z' = m₄
Row 6 (1,1,0): XYZ' = m₆
Row 7 (1,1,1): XYZ = m₇
Step 3: Write CSOP
F = m₁ + m₃ + m₄ + m₆ + m₇ = Σm(1,3,4,6,7)
Step 4: Write CPOS (find rows where F=0)
F=0 for rows 0, 2, 5 → M₀, M₂, M₅
F = ΠM(0,2,5)
- For minterms: Row 0 = 000 = m₀, Row 1 = 001 = m₁, etc.
- For maxterms: Row 0 = 000 = M₀, Row 1 = 001 = M₁, etc.
11. More Practice Exercises (With Answers)
Exercise 4
Find CSOP for F(x,y,z) = (x'y) + (xy')
✅ Click to see answer
Solution:
Step 1: Expand each term to include all 3 variables
x'y = x'y(z + z') = x'yz + x'yz'
xy' = xy'(z + z') = xy'z + xy'z'
Step 2: Combine all minterms
F = x'yz + x'yz' + xy'z + xy'z'
Step 3: Convert to minterm numbers
x'yz = 011 → m₃
x'yz' = 010 → m₂
xy'z = 101 → m₅
xy'z' = 100 → m₄
CSOP: F = Σm(2,3,4,5)
Exercise 5
Convert F = Σm(2,4,5,6,7) to CPOS form
✅ Click to see answer
Solution:
Step 1: Total terms for 3 variables = 8 (0-7)
Step 2: Find missing minterms
Present: m₂, m₄, m₅, m₆, m₇
Missing: m₀, m₁, m₃
Step 3: Convert missing minterms to maxterms
Missing minterms → m₀, m₁, m₃
Corresponding maxterms → M₀, M₁, M₃
CPOS: F = ΠM(0,1,3)
Exercise 6
Derive CSOP and CPOS from the following truth table:
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
✅ Click to see answer
Solution:
CSOP: Find rows where F=1 → Rows 0, 2, 3, 5, 7
Row 0: X'Y'Z' = m₀
Row 2: X'YZ' = m₂
Row 3: X'YZ = m₃
Row 5: XY'Z = m₅
Row 7: XYZ = m₇
CSOP: F = Σm(0,2,3,5,7)
CPOS: Find rows where F=0 → Rows 1, 4, 6
Row 1: X'Y'Z = M₁
Row 4: XY'Z' = M₄
Row 6: XYZ' = M₆
CPOS: F = ΠM(1,4,6)
Exercise 7
Given F(x,y,z) = Σm(1,3,5,6,7), find:
(a) The truth table
(b) The CPOS form
✅ Click to see answer
Solution:
(a) Truth table:
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
(b) CPOS form:
F=0 for rows 0, 2, 4 → M₀, M₂, M₄
CPOS: F = ΠM(0,2,4)
Exercise 8
Given F(x,y,z) = ΠM(0,4,6,8,10,12,14,15), find CSOP
✅ Click to see answer
Solution:
Step 1: Total terms for 4 variables = 16 (0-15)
Step 2: Find missing maxterms
Present: M₀, M₄, M₆, M₈, M₁₀, M₁₂, M₁₄, M₁₅
Missing: M₁, M₂, M₃, M₅, M₇, M₉, M₁₁, M₁₃
Step 3: Convert missing maxterms to minterms
Missing maxterms → M₁, M₂, M₃, M₅, M₇, M₉, M₁₁, M₁₃
Corresponding minterms → m₁, m₂, m₃, m₅, m₇, m₉, m₁₁, m₁₃
CSOP: F = Σm(1,2,3,5,7,9,11,13)
Exercise from your Image
F(x,y,z) = Σm(1,3,5,6,7)
F(x,y,z) = Σm(2,4,5,6,7)
F(x,y,z) = ΠM(0,4,6,8,10,12,14,15)
Solutions:
1. F = Σm(1,3,5,6,7) → CPOS = ΠM(0,2,4)
2. F = Σm(2,4,5,6,7) → CPOS = ΠM(0,1,3)
3. F = ΠM(0,4,6,8,10,12,14,15) → CSOP = Σm(1,2,3,5,7,9,11,13)
9. Quick Summary Cheat Sheet
- ✓ Min Term = AND of all variables → Output = 1 for ONE row → Notation: m₀, m₁...
- ✓ Max Term = OR of all variables → Output = 0 for ONE row → Notation: M₀, M₁...
- ✓ CSOP = Sum (OR) of ALL Min Terms where F=1 → Σm(...)
- ✓ CPOS = Product (AND) of ALL Max Terms where F=0 → ΠM(...)
- ✓ SSOP = Simplified SOP (not all terms have all variables)
- ✓ SPOS = Simplified POS (not all terms have all variables)
- ✓ For n variables: Total terms = 2ⁿ
- ✓ Min/Max conversion: Missing terms swap between Σm and ΠM
- ✓ mᵢ = (Mᵢ)' → They are complements!
- If given SOP → Expand to Min Terms → Write Σm(...)
- If given POS → Expand to Max Terms → Write ΠM(...)
Prepared with care for Sri Lankan A/L ICT Students | Practice Truth Tables Daily!