‘Sum of Product’ is commonly abbreviated as SOP form. When an expression is expressed in a sum of binary terms ( A term refers to a set of binary variables, where each binary variable is combined with an operation) called Minterms then it is said to be Sum of Products.
In other words, An expression consisting only of minterms is called Sum of the Product. For example, A.B+A’B is an SOP expression.
[other concept Maxterm, (A+B).(A’+B’)]
– Canonical or Standard SOP form: In this, each and every binary variable must have in each term. For example- A.B+A’B’
– Minimal SOP form: In this, the standard SOP expression is reduced up to the minimum possible expression.
1) In the form Truth table
2) In the form of non-canonical Expression
3) In the form of Boolean function
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 (m0) |
| 0 | 1 | 0 (m1) |
| 1 | 0 | 1 (m2) |
| 1 | 1 | 1 (m3) |
Points
Solution :
Step 1. Since there are 2 variables, so, therefore, there would be 2n combinations which are 22 =4.
We consider a high output as minterm. a minterm is denoted as m.
Y= m2+m3
Y = A.B’ +A.B :- It is in Canonical SOP form
In Sum of Product each term is combined with OR operation and within each term, every variable combined with AND operation.
Step 2. Now narrow the founded expression down to minimal SOP form. In this, you should know the rules of Boolean expression or K-map
Y= A.( B’ + B)
Y= A.1
Y= A
OR
Considering, A= 1, A’=0
Y= A
A is your answer.
Simplify the following Boolean function in SOP form
F(X,Y,Z) = Σm(1,4,5,6,7)
Points to Remember
Step 1. Make the truth table, if you want to see the actual picture or skip the step.
| X | Y | Z | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 (m0) |
| 0 | 0 | 1 | 1 (m1) |
| 0 | 1 | 0 | 0 (m2) |
| 0 | 1 | 1 | 0 (m3) |
| 1 | 0 | 0 | 1 (m4) |
| 1 | 0 | 1 | 1 (m5) |
| 1 | 1 | 0 | 1 (m6) |
| 1 | 1 | 1 | 1 (m7) |
In the above table, we put output as 1 or high at the mentioned position in the above Boolean function.
For example, in Boolean function first value is 1, so we put the output high to respective place in the truth table as output 1 at row 001 ( represents 1)
Similarly, we have 4 , so at row 100 (represents 4) in the truth table, we put high output.
Step 2. Now make the expression as per the above table,
F= m1 + m4 + m5 + m6 + m7
F= X’Y’Z + XY’Z’ + XY’Z+ XYZ’ + XYZ :- Canonical SOP form
Step 3. Now Solve the function to minimal SOP form using K map or Boolean rules
K-map is always considered the easiest and fastest way.
F= X + Y’Z :- Minimal SOP form
This your answer
Simplify the following expression in SOP form
F= x’z’+y’z’+yz’+xy
Simplify the following non-canonical expression in SOP form
F= x’z’+y’z’+yz’+xy
Point to remember
Step 1. First, check the given expression is in Canonical form or not, if yes simply make K-map (or jump to 4th step ) and simplify it, otherwise follow the steps.
Step 2. Now, the above expression isn’t in canonical form as there are three binary variables or inputs, x , y and z which can’t be together found in each and every term. So to do this, make it canonical form by doing the following:-
F = (x’.y’. 1) + (y’z’ .1) + (y z’ .1) + ( x. y.1 )
Since, in each term, a variable is missing, to bring it, in each term we have to put 1 to get the expression.
Step 3. Put relevant input in place of 1 on the basis of Boolean rules
There is a rule, where
x + x’ =1 in Boolean algebra, we will apply it.
In the first term, y is missing (x’y’), similarly in the second one, x is missing (y’z’) and so one.
F = x’z'( y +y’) + y’z'(x+x’) + yz'(x+x’)+ xy(z+z’)
F= x’z’y+x’z’y’+y’z’x+y’z’x’+yz’x+yz’x’+xyz+xyz’
Now, Arranging in it alphabetical order for ease of understanding
F=x’yz’+x’y’z’+xy’z’+x’y’z’+xyz’+x’yz’+xyz+xyz’
It is now in Canonical SOP or Sum of Product form
Step 3. This is the last step, shrink the expression down to minimal SOP form using K-map
or Boolean algebra
Which gives, F = z’ + xy
This post was last modified on December 27, 2020