21 votes 21 votes Let $f=(\bar{w} + y)(\bar{x} +y)(w+\bar{x}+z)(\bar{w}+z)(\bar{x}+z)$ Express $f$ as the minimal sum of products. Write only the answer. If the output line is stuck at $0$, for how many input combinations will the value of $f$ be correct? Digital Logic gate1997 digital-logic min-sum-of-products-form numerical-answers + – go_editor asked Oct 14, 2015 • edited Jun 19, 2018 by Pooja Khatri go_editor 3.2k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply set2018 commented Sep 2, 2017 reply Follow Share @bikram sir what is the meaning of second point .m not getting this k map 0 votes 0 votes Bikram commented Sep 3, 2017 reply Follow Share @ set2018 second point want to say , first find the f as the minimal sum of products , now consider that the output is 0 , now find for how many input combinations you are getting output as 0. output line is stuck at 0 means output is 0 , i.e f = 0. 1 votes 1 votes Please log in or register to add a comment.
Best answer 35 votes 35 votes Answer of question A: $w'x'+yz$ Answer of question B: Stuck at $0$, means output is fixed at $0$ (No matter what the input is). We got $0$ for $9$ input combinations (Check K-Map). So, answer is 9. Akash Kanase answered Nov 18, 2015 • edited Apr 27, 2019 by ajaysoni1924 Akash Kanase comment Share Follow See all 6 Comments See all 6 6 Comments reply set2018 commented Nov 25, 2017 reply Follow Share how 9 combinations? 0 votes 0 votes air1ankit commented Dec 31, 2017 reply Follow Share how to make k-map for the given function , please explain it is in sop form so ??? 0 votes 0 votes Mk Utkarsh commented Aug 5, 2018 reply Follow Share set2018 The POS expression is the equation of the logic function as read off the truth table to specify the input combinations when the output is a logical 0. So we have 9 Max-terms in K-map. 0 votes 0 votes VIDYADHAR SHELKE 1 commented Nov 19, 2019 reply Follow Share take two times complement and reduced one complete using demorgans law then u will get sum of product(min term) after that take complements of min terms 1 votes 1 votes Gupta731 commented Aug 11, 2020 reply Follow Share In case anyone wondering how to solve without K-map $f=(w'+y)(x'+y)(w+x'+z)(w'+z)(x'+z)$ $f'=wy'+xy'+w'xz'+wz'+xz'$ $=wy'+xy'+wz'+xz'(w'+1)$ $=wy'+xy'+wz'+xz'$ $=y'(w+x)+z'(w+x)$ $=(w+x)(y'+z')$ $(f')'=x'w'+yz$ 8 votes 8 votes neel19 commented Jul 3, 2022 reply Follow Share Our SOP is: $w’x’ + yz$. To get output 0, both the first term and second term should be zero $w’x’ $ can be zero in three ways. Similarily, $yz$ can be zero in three ways. Total: $3 * 3 = 9 $ ways. 6 votes 6 votes Please log in or register to add a comment.