Given \(f(w, x, y, z) = \Sigma_m(0,1, 2, 3, 7, 8, 10) + \Sigma_d(5, 6, 11,…
2017
Given \(f(w, x, y, z) = \Sigma_m(0,1, 2, 3, 7, 8, 10) + \Sigma_d(5, 6, 11, 15)\); where \(d\) represents the 'don't-care' condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of \(f(w, x, y, z)\)?
- A.
\(f=(\bar{w}+\bar{z}) (\bar{x}+z)\) - B.
\(f=(\bar{w}+z) (x+z)\) - C.
\(f=(w+z) (\bar{x}+z)\) - D.
\(f=(w+\bar{z}) (\bar{x}+z)\)
Attempted by 127 students.
Show answer & explanation
Correct answer: A
Given f = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15).
Step 1: Find the minterms where f = 0 (maxterms).
The indices not listed as 1 or don't-care are 4, 9, 12, 13, 14, so these are the required zeros.
Step 2: Use don't-cares to form larger groups of zeros (POS grouping).
Group 1: Cover x = 1 and z = 0. This group can include indices 4, 6 (don’t-care), 12, 14. For a zero-group with x = 1 and z = 0 the corresponding sum term is (not x + z).
Group 2: Cover w = 1 and z = 1. This group can include indices 9, 11 (don’t-care), 13, 15 (don’t-care). For a zero-group with w = 1 and z = 1 the corresponding sum term is (not w + not z).
Final POS: (not w + not z)(not x + z)
Verification: The derived product of sums yields zeros for all required indices {4,9,12,13,14} and only extends zeros into don’t-care indices (6,11,15) where values are unspecified. It does not force any minterm that must be 1 to become 0, so it is the minimal POS.
A video solution is available for this question — log in and enroll to watch it.