Test: SOP Form - Electronics and Communication Engineering (ECE) MCQ

Canonical Form:  
Any Boolean function that expressed as a sum of min terms or as a product of max terms is said to be in its canonical form.
There are two types of canonical forms:
SOP:
Sum of products or sum of min terms
Example of SOP: XY + X’Y’
POS:
Product of sums or product of max terms
Example of POS: (X+Y) (X’+Y’)
Explanation:
SOP contains only sum of min terms hence option 1 is the correct answer.

Canonical Form:  Any Boolean function that is expressed as a sum of min terms or as a product of max terms is said to be in its canonical form.
There are two types of canonical forms:
SOP:

• Sum of products or sum of min terms

Example of SOP: XY + X’Y’
POS:

• Product of sums or product of max terms

Example of POS: (X+Y) (X’+Y’)

Canonical form:  Any Boolean function that expressed as a sum of minterms or as a product of max terms is said to be in its canonical form.
There are two types of canonical forms:
SOP:
Sum of products or sum of minterms
In SOP (sum of product) form, a minterm is represented by 1.
Example of SOP: XY + X’Y’
POS:
Product of sums or product of max terms
In POS (product of sum) form, a maxterm is represented by 0.
Example of POS: (X+Y) (X’+Y’)
Calculation:


F(A, B, C) = ∑ (2, 4, 5, 7)

Sum of Product:
SOP is an abbreviation for Sum of Product. Sum of product form is a Boolean algebra statement in which distinct product terms of inputs are added together. This product is not an arithmetic multiply, but rather a Boolean logical AND, with the Sum being a Boolean logical OR.
SOP gives a minterm of a logical expression. 
Example:
F= XY+YZ+ XZ
The above expression can be written like,
F= X.Y+Y.Z+ X.Z
F= (X) and (Y) or (Y) and (Z) or (X) and (Z)
In minterms X takes as '1' and X' takes 0 similarly same conventions follow for all variables.
XY= 110 (m6) and 111 (m7)
YZ= 011 (m3) and 111 (m7)
ZX= 101 (m5) and 111 (m7)
Min terms are, F= Σ(3,5,6,7)
Product of sum:
POS is an abbreviation for Product of Sum. Product of sum form is a Boolean algebra statement in which distinct product terms of inputs are multiplied together. This Sum is a Boolean logical OR, with the product being a Boolean logical AND.
POS gives a maxterm of a logical expression. 
Example:
F= (X+Y). (Y+Z). (X+Z)
The above expression can be written like,
F= X+Y). (Y+Z). (X+Z)
F= (X) or (Y) and (Y) or (Z) and (X) or (Z)
In max terms X takes as '0' and X' takes '1' similarly same conventions follow for all variables.
X+Y= 000 (M0), 001 (M1)
Y+Z=000 (M0), 100 (M4)
Z+X=000 (M0), 010 (M2)
Max terms are, F= Σ(0, 1, 2,4)
Option 1:(A + B)(C + D)
False, It is in product of sum.
Option 2:(A)-­B (CD) 
False, It is not in POS and SOP.
Option 3: AB (CD)
False, It is not in POS and SOP.
Option 4: AB + CD
True, It is a correct form Sum of Product. It is Boolean logical AND, with the Sum being a Boolean logical OR.
Hence the correct answer is AB+CD.

F(P, Q, R) = PQ + QR' + PR'
= PQ (R + R') + (P + P')QR' + P(Q + Q')R'
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
= m₇ + m₆ + m₂ + m₄
= m₂ + m₄ + m₆ + m₇

Some laws of Boolean Algebra:
Distributive Law:

• P +QR = (P + Q).(P + R)
• P(Q + R) = PQ + PR

Inverse Law:

• PP̅ =0 
• P + P̅ =1

F = (P + Q̅ + R̅)⋅(P + Q̅ + R)⋅(P + Q + R̅)
F= ((P + Q̅) + R̅.R)(P + Q + R̅)
F = (P + Q̅)(P + Q + R̅)
F = P + Q̅.(Q + R̅)
F = P + Q̅R̅

We can minimize the Boolean expression/truth table of ‘n’ variable using a K-map in which 2ⁿ cells are present.
Steps to solve expression using K-map:

• Select the K-map according to the number of variables (Cells = 2ⁿ)
• Identify maxterm or minterm as given as per the problem.
• For SOP, put 1’s in blocks of K-map respective to the minterms.
• For POS, put 0’s in blocks of K-map respective to the max terms.
• To minimize, make rectangular groups containing total terms in power of two (like 1, 2, 4, 8..).
• From the groups made in step-5, find the product terms & add them for SOP form.

Calculation:
The truth table given is:

The 3 – variable K-map corresponding to the given truth table is drawn with 2³ = 8 (cells) as shown:

By making a group of 1’s,
Output (X) = A̅ B̅ C + A B C̅ 

F = ∑m (0, 2, 3, 5)