Question

Minimise the following function in POS minimal form using K-Maps: F(A, B, C, D) = M(0, 1, 2, 3, 4, 5) + d(10, 11, 12, 13, 14, 15)​

Answer 1

Explanation:

Minimize the following Boolean function using sum of products (SOP):. • f(a,b,c,d) = ∑m(3,7, 11,12,13,14,15) abcd. 3 0011. 7 0111. 11 1011

Answer 2

Given is a boolean function, find its minimal POS form using K-maps.

Explanation:

• Given is the boolean expression in four variables :  $F(A, B, C, D)=M(0,1,2,3,4,5)+d(10,11,12,13,14,15)$    
• In the above equation the terms represented in 'M()' are the max-terms of the function and are represented by zero in the K-map.
• In the above equation the terms represented in 'd()' are the don't-care terms and are represented by a cross in the K-map.  
• The K-map of the given function is ,  
• $AB/CD\ \ \ 00\ \ \ 01\ \ \ 11\ \ \ 10 \\00\ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ 0\ \ \ \ \ 0\ \ \ \ \ 0\ \ \ \ \ \ \ \\01\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ 0 \\11\ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ x\ \ \ \ \ x\ \ \ \ \ x\\10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ x$      
• Now apply the rules for k-maps minimization and form the maximum possible groups of multiple of two terms $2,4,8...$.
• Hence, we get the minimized POS form as,  $F=(A+B)(A+C)(B'+C)$