CSOP and CPOS Functions | GCE A/L )/L ICT | Unit 4 | Boolean Logic and Digital Circuit | in English Tamil | தமிழில் Medium Notes Questions and Answers

[Boolean Logic] SOP/POS Terms Explained

GCE A/L ICT – Sri Lanka | Beginner Friendly Guide with Examples & Exercises

💡 Teacher's Tip: Min Terms and Max Terms are two sides of the same coin. Master them once — and K-Maps, SOP/POS forms will become easy!

9. CSOP and CPOS - The Canonical Forms

🔥 Key Distinction:
• 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"

Example 1: F(X,Y,Z) = X'Y + XY'Z + XYZ

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"

Example 2: F(X,Y,Z) = (X+Y)(X'+Z)(Y+Z)

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)

Example 3: F(X,Y,Z) = X'Y + XYZ

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)

Example 4: Convert F = Σm(1,3,5,6,7) to CPOS form

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

⚠️ Critical Note: CSOP and CPOS are NOT the same function! They represent the same logical function but in different canonical forms. CSOP focuses on when F=1, while CPOS focuses on when F=0.

10. Deriving Functions from Truth Tables

📝 Step-by-Step Method:
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

Example 5: Derive CSOP for F(X,Y,Z) from 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	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)

⚠️ Common Mistake: Students often confuse the row numbers. Remember:

• 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!

📚 A/L Exam Tip: When asked to "express in canonical form":

• 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!