Conversion from Minterm Expression to Maxterm Expression
- Find the minterm designations: For each min term, find its designation.
- Identify the missing minterms: A boolean function with n variables will have 2n possible minterms. Identify the minterms that are not present in the given expression. These will be the maxterms in the maxterm expression.
- Find the maxterm designations: For each missing minterm, find its maxterm designation. The maxterm designation is the complement of the minterm designation.
- Write the maxterms from their designations: Convert each maxterm designation into a term by replacing 0 with the uncomplemented variable and 1 with the complemented variable.
- Form the max termexpression: Finally, form the maxterm expression by taking the product of all the maxterms.
Solved Examples:
1. Convert the Minterm Expression to Maxterm Expression f(x, y, z) = x’yz + xy’z + x’y’z’
Minterm designation of x’yz = 011 = 3 = m3
Minterm designation of xy’z = 101 = 5 = m5
Minterm designation of x’y’z’ = 000 = 0 = m0
The function mentioned above has 3 variables. There can be 23 = 8 possible terms, out of which only three terms are used to form the expression. It means, the maxterm expression should have remaining 5 terms. Which are M1 , M2 , M4 , M6 , M7
Maxterm for M1 = 001 = (x + y + z’)
Maxterm for M2 = 010 = (x + y’ + z)
Maxterm for M4 = 100 = (x’ + y + z)
Maxterm for M6 = 110 = (x’ + y’ + z)
Maxterm for M7 = 111 = (x’ + y’ + z’)
The boolean function using maxterms will be: f(x, y, z) = (x + y + z’)(x + y’ + z)(x’ + y + z)(x’ + y’ + z)(x’ + y’ + z’)
2. Convert the Minterm Expression to Maxterm Expression f(x, y, z) = xyz’ + xy’z’ + x’yz’ + xyz
Minterm designation of xyz’ = 110 = 6 = m6
Minterm designation of xy’z’ = 100 = 4 = m4
Minterm designation of x’yz’ = 010 = 2 = m2
Minterm designation of xyz = 111 = 7 = m7
The function has 3 variables. There can be 23 = 8 possible terms, out of which only four terms are used to form the expression. It means, the maxterm expression should have the remaining 4 terms, which are M0, M1, M3, M5.
Maxterm for M0 = 000 = (x + y + z)
Maxterm for M1 = 001 = (x + y + z’)
Maxterm for M3 = 011 = (x + y’ + z’)
Maxterm for M5 = 101 = (x’ + y + z’)
The boolean function using maxterms will be: f(x, y, z) = (x + y + z)(x + y + z’)(x + y’ + z’)(x’ + y + z’)
3. Convert the Minterm Expression to Maxterm Expression f(x, y) = xy + xy’
Minterm designation of xy = 11 = 3 = m3
Minterm designation of xy’ = 10 = 2 = m2
The function has 2 variables. There can be 22 = 4 possible terms, out of which only two terms are used to form the expression. It means, the maxterm expression should have the remaining 2 terms, which are M0, M1.
Maxterm for M0 = 00 = (x + y)
Maxterm for M1 = 01 = (x + y’)
The boolean function using maxterms will be: f(x, y) = (x + y)(x + y’)
Conversion From Minterm Expression to Maxterm Expression
Minterm is the product of N distinct literals where each literal occurs exactly once. The output of the minterm functions is 1. Maxterm is the sum of N distinct literals where each literals occurs exactly once. The output of the maxterm functions is 0. The conversion from minterm to maxterm involves changing the representation of the function from a Sum of Products (SOP) to a Product of Sums (POS).
In this article, we will cover prerequisites like minterm, maxterm, minterm designation, maxterm designation, conversion from Cardinal form to Minterm Expression, and conversion from Cardinal form to Maxterm Expression with a detailed explanation of conversion from minterm expression to maxterm expression with solved examples and FAQs.