1 votes 1 votes Digital Logic test-series minimum-number-of-gates digital-logic + – sidlewis asked Sep 2, 2018 sidlewis 323 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments MiNiPanda commented Sep 2, 2018 reply Follow Share We should know that an expression in POS form is always easier to be realized by NOR gates. Given is the SOP ∑m(0,2,6,10,14) so we should convert it to POS. Hence piM(1,3,4,5,7,8,9,11,12,13,15) [I couldn't find the "pi" symbol :P] So the expression is D'.(B'+C).(A'+C) Now we have to convert this expression in such a way that we can realize it with NOR ( ( D'.(B'+C).(A'+C) )' )' = ( (D')' + (B'+C)' + (A'+C)' )' = (D + BC' + AC' )' = (D + C'(A+B) )' = ( D + ( ( C'(A+B) )' )' )' = (D + ( (C')' + (A+B)' )' )' = (D + ( C + (A+B)' )' )' X=(A+B)' --> 1 NOR Y=(C+ X)' --> 1 NOR (D+Y)' --> 1 NOR 0 votes 0 votes Shaik Masthan commented Sep 2, 2018 reply Follow Share By k-map, SOP format ===> CD' + A' B' D' ===> D' (C+A'B') ==> D' . ( C + (A+B)' ) ===> ( D + ( C + (A+B)' )' ) ' TOTAL = 3 NOR GATES 2 votes 2 votes MiNiPanda commented Sep 2, 2018 reply Follow Share Yeah.. Shaik's one is faster.. usually converting to POS works better..but in this case it's not 0 votes 0 votes Please log in or register to add a comment.