39 votes 39 votes Given the function $F = P' +QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P'=!P$, consider the following statements. $(S1) F = \sum(4, 5, 6)$ $(S2) F = \sum(0, 1, 2, 3, 7)$ $(S3) F = \Pi (4, 5, 6)$ $(S4) F = \Pi (0, 1, 2, 3, 7)$ Which of the following is true? (S1)-False, (S2)-True, (S3)-True, (S4)-False (S1)-True, (S2)-False, (S3)-False, (S4)-True (S1)-False, (S2)-False, (S3)-True, (S4)-True (S1)-True, (S2)-True, (S3)-False, (S4)-False Digital Logic gatecse-2015-set3 digital-logic canonical-normal-form normal + – go_editor asked Feb 15, 2015 • edited Feb 10, 2018 by go_editor go_editor 8.4k views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments Himanshu Kumar Gupta commented Aug 29, 2020 reply Follow Share A is right option 0 votes 0 votes palashbehra5 commented Aug 16, 2021 reply Follow Share It is one of the most basic questions I have come across but couldn't figure out what exactly it meant. Anyone else who found this ambiguous? 0 votes 0 votes Hira Thakur commented Nov 1, 2023 reply Follow Share $f=\sum(0,1,2,3,7)$ $f=\Pi(4.5.6)$ 0 votes 0 votes Please log in or register to add a comment.
Best answer 38 votes 38 votes $F=P'+QR$, draw the Kmap for this We can find the minterm $\sum (0,1,2,3,7)$ and maxterm $\Pi (4,5,6) $ So, option A is correct: $(S1)$-False, $(S2)$-True, $(S3)$-True, $(S4)$-False Anoop Sonkar answered Feb 16, 2015 • edited Jun 24, 2018 by Milicevic3306 Anoop Sonkar comment Share Follow See all 0 reply Please log in or register to add a comment.
17 votes 17 votes $F = P{}' + QR$for SOP we have :$F = P{}.1.1 + 1.QR'=P{}'(Q+Q{}')(R+R{}') + (P+P{}')QR$$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR+P{}'QR$$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR$$F=\sum(0, 1, 2, 3, 7)$ (considering barred terms as 0 and unbarred as 1 and converting them to binary and then to decimal).now for POS we have :$F = P{}' + QR = (P{}'+Q)(P{}'+R) = (P{}'+Q+0)(P{}'+R+0)$$(P{}'+Q+R.R{}')(P{}'+R+Q.Q{}') $$(P{}'+Q+R)(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$$(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$$F=\prod (4, 5, 6)$(considering barred terms as 1 and unbarred as 0 and converting them to binary and then to decimal).http://mcs.uwsuper.edu/sb/461/PDF/sop.html Tamojit Chatterjee answered Feb 27, 2015 • edited Feb 28, 2015 by Tamojit Chatterjee Tamojit Chatterjee comment Share Follow See all 4 Comments See all 4 4 Comments reply Arjun commented Feb 28, 2015 reply Follow Share But K-map is better time saving approach rt? 5 votes 5 votes Tamojit Chatterjee commented Feb 28, 2015 reply Follow Share may be but that also kind of varies from individual to individual don't you think ? 2 votes 2 votes Arjun commented Feb 28, 2015 reply Follow Share Yes. Sure :) But you needn't solve for both min terms and max terms rt? Solve one and take complement for other. 3 votes 3 votes Tamojit Chatterjee commented Feb 28, 2015 reply Follow Share i was thinking of doing that, but just did it to show the general method to solve it both ways( POS ans SOP). Feel free to edit. 3 votes 3 votes Please log in or register to add a comment.
15 votes 15 votes Answer = A amarVashishth answered Oct 8, 2015 amarVashishth comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes S2 and S3 are true abhishekmehta4u answered Mar 26, 2018 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.