Boolean Algebra
In this tutorial we will learning about Sum of Products and Product of Sums.
A boolean expression consisting entirely either of minterm or maxterm is called canonical expression.
Example
if we have two variables X and Y then,
Following is a canonical expression consisting of minterms XY + X’Y’
and
Following is a canonical expression consisting of maxterm (X+Y) . (X’ + Y’)
There are two forms of canonical expression.
A boolean expression consisting purely of Minterms (product terms) is said to be in canonical sum of products form.
Example
lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when any one of the input is true or 1. Now we draw the truth table for F.
| A | B | F |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Now we will create a column for the minterm using the variables A and B. If input is 0 we take the complement of the variable and if input is 1 we take the variable as is.
| A | B | F | Minterm |
|---|---|---|---|
| 0 | 0 | 0 | A'B' |
| 0 | 1 | 1 | A'B |
| 1 | 0 | 1 | AB' |
| 1 | 1 | 1 | AB |
To get the desired canonical SOP expression we will add the minterms (product terms) for which the output is 1.
F = A’B + AB’ + AB
From the previous example we have
F = A’B + AB’ + AB
Now, lets say we want to express the SOP using shorthand notation.
we have F = A’B + AB’ + AB
First we need to denote the minterms in shorthand notation.
A’B = (01)2 = m1
AB’ = (10)2 = m2
AB = (11)2 = m3
We saw the conversion of SOP to shorthand notation. Lets check the conversion of shorthand notation to SOP.
Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, the minterms are expressed as shorthand notation given below.
F = ∑(1, 2, 3)
our task is to get the SOP.
F has two input variables A and B and output of F = 1 for m1, m2 and m3 i.e., 2nd, 3rd and 4th combination.
we have,
F = ∑(1, 2, 3)
= m1 + m2 + m3
= 01 + 10 + 11
To convert from shorthand notation to SOP we follow the given rules. If the variable is 1 then it is taken "as is" and if the variable is 0 then we take its "complement".
F = ∑(1, 2, 3)
= A’B + AB’ + AB
And we have the required SOP
A boolean expression consisting purely of Maxterms (sum terms) is said to be in canonical product of sums form.
Example
Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when only one of the input is true or 1.
now we draw the truth table for F
| A | B | F |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Now we will create a column for the maxterm using the variables A and B. If input is 1 we take the complement of the variable and if input is 0 we take the variable as is.
| A | B | F | Maxterm |
|---|---|---|---|
| 0 | 0 | 0 | A + B |
| 0 | 1 | 1 | A + B' |
| 1 | 0 | 1 | A' + B |
| 1 | 1 | 0 | A' + B' |
To get the desired canonical POS expression we will multiply the maxterms (sum terms) for which the output is 0.
F = (A+B) . (A’+B’)
From the previous example we have
F = (A+B) . (A’+B’)
Now, lets say we want to express the POS using shorthand notation.
we have F = (A+B) . (A’+B’)
First we need to denote the maxterms in shorthand notation.
A+B = (00)2 = M0
A’+B’ = (11)2 = M3
Now we express F using shorthand notation.
F = M0 . M3
This can also be written as F = ∏(0, 3)
We saw the conversion of POS to shorthand notation. Lets check the conversion of shorthand notation to POS.
Lets say, we have a boolean function F defined on two variables A and B so, A and B are the inputs for F and lets say, the maxterm are expressed as shorthand notation given below.
F = ∏(1, 2, 3)
Our task is to get the POS.
F has two input variables A and B and output of F = 0 for M1, M2 and M3 i.e., 2nd, 3rd and 4th combination.
we have, F = ∏(1, 2, 3)
= M1 . M2 . M3
= 01 . 10 . 11
To convert from shorthand notation to POS we follow the given rules. If the variable is 0 then it is taken as is and if the variable is 1 then we take its complement.
we have, F = ∏(1, 2, 3)
= (A+B’) . (A’+B) . (A’+B’)
And we have the required POS.
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