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In the sum of the product form of representation, The product num is logical and operation of the different input variables where the variables could be in the true form or in the complemented form.
The Sum of Product (SOP) form refers to a Boolean expression where several product terms (involving AND operations) are added up. This is perhaps the most popular form used in the design and optimization of digital systems. In SOP format, each of those elements is a combination (ANDing) of some variables (or their negations); all of them are combined by means of disjunction (ORing). This is very helpful in, for instance, the creation of logic circuits from truth tables or achieving algorithmic simplifications of logic functions.
Example:
A.B, A.B̅.C (example of product term)
In SOP sum refers to logical OR Operation. Therefore, in this sum of product form of expression, we perform logical or operations on different product terms. Therefore it is known as the Sum of Product form.
Example:
A+ B̅.C + A.C
as we can see in above example the product terms(A , B̅.C , A.C) which are created by multiplying input variables are are summed with each other .
The SOP form can be in either canonical form or non-canonical form.
In this form each product term between may or may not contain all the variables of the function.
Example:
F(A,B,C) = A+ B̅.C + A.C
as we can see in above example the function have variables A, B, C but we are not including each variable in each product term. in first product term (A) we have not included B & C. In second product term we (B̅.C) have not included A. While in last product term we have not included B.
In canonical SOP form each product term contains all the variables of the function, where variables in each product term can be in true form or complemented form.
Example :
F(A,B) = Ā.B + A.B̅
As we can see in above example each product term contain all the variables which are present in function. In first one(Ā.B) A is present in complementary form while B is in true form. In second one A is present in true form while B is in complementary form.
Suppose following is the truth table for an electronic circuit:
A | B | C | F |
|---|---|---|---|
0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 0 | 1 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
The SOP form is written when the function if high. i.e we will take conditions only in which the output is one. In given truth table the columns with F = 1 are the rows with high output. Now we will write function in the form of given variables which are A, B and C. If we have 1 in the cell of given variable we will write it as it is , and if we have 0 in the given cell then we will write it's complement.
F = Ā.B.C̅ + A.B̅.C̅ + A.B̅.C + A.B.C
This is the POS in canonical form for the given truth table. Each expressions which lies in between the + are known as Minterms.
Example :
A.B.C, A.B̅.C
Therefore we can also write it as
F(A, B, C) = M2 + M4 + M5 + M6 + M7
F(A,B,C) = ∑m(2, 4, 5, 6, 7)
we can further reduce it to non-canonical SOP form by using boolean algebra.
F = Ā.B.C̅ + A.B̅.C̅ + A.B̅.C + A.B.C
F = Ā.B.C̅ + A.B̅.[C̅ + C] + A.B.[C̅ + C]
F = Ā.B.C̅ + A.B̅ + A.B
F = Ā.B.C̅ + A.[B̅ + B]
F = A + B.C̅
This is how we can get non-canonical SOP form of Boolean expression.
Suppose following is the SOP expression for which we will make truth table now
F(A, B, C) = A'B' + AB + AC
1. As we have three input values (A, B, C) we will create 2^3 = 8 rows in truth table to include all input combinations.
A | B | C |
|---|---|---|
0 | 0 | 0 |
0 | 0 | 1 |
0 | 1 | 0 |
0 | 1 | 1 |
1 | 0 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
1 | 1 | 1 |
2. for each input combination we will put values of input in expression and calculate the output:
F(A, B, C) = A'B' + AB + AC
F(0, 0, 0) = 0'0' + 00 + 00 = 0 + 0 + 0 = 0
F(0, 0, 1) = 0'0' + 00 + 01 = 0 + 0 + 0 = 0
F(0, 1, 0) = 0'1' + 01 + 00 = 0 + 0 + 0 = 0
F(0, 1, 1) = 0'1' + 01 + 01 = 0 + 0 + 0 = 0
F(1, 0, 0) = 1'0' + 10 + 10 = 0 + 0 + 0 = 0
F(1, 0, 1) = 1'0' + 10 + 11 = 0 + 0 + 1 = 1
F(1, 1, 0) = 1'1' + 11 + 10 = 0 + 1 + 0 = 1
F(1, 1, 1) = 1'1' + 11 + 11 = 0 + 1 + 1 = 1
3. Create truth table form given input combinations and calculated output.
A | B | C | F |
|---|---|---|---|
0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 |
1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |