38 votes 38 votes What is the minimum number of $2$-input NOR gates required to implement a $4$ -variable function expressed in sum-of-minterms form as $f=\Sigma(0,2,5,7, 8, 10, 13, 15)?$ Assume that all the inputs and their complements are available. Answer: _______ Digital Logic gatecse-2019 numerical-answers digital-logic canonical-normal-form 2-marks + – Arjun asked Feb 7, 2019 retagged Nov 30, 2022 by Lakshman Bhaiya Arjun 30.1k views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments vishnu_m7 commented Nov 25, 2020 reply Follow Share @mrinmoyh 4 NOR gates are required when complemented inputs are not available. 4 votes 4 votes Shiva Sagar Rao commented Dec 10, 2020 reply Follow Share Here important statement given in question is: Assume that all the inputs and their complements are available. If complements are NOT available then minimum $4$ NOR gates($2$ – input) are required to implement Ex – NOR. 0 votes 0 votes Vishal_kumar98 commented Feb 11, 2021 reply Follow Share Fascinated 0 votes 0 votes Please log in or register to add a comment.
Best answer 46 votes 46 votes $f = ( B' + D ) . ( B + D' )$ It is mentioned that both Complimentary as well as Uncomplimentary forms are available. $B'\text{ NOR }D = (B' + D)'$ $B \text{ NOR } D' = (B + D')'$ $(B' \text{ NOR } D) \text{ NOR } (B \text{ NOR } D')$ $= ( (B' + D)' + (B + D')' )'$ $= ( (B' + D)'' . (B + D')'' )$ $= ((B' + D).(B + D'))$ $= f$ Thus, $3 \text{ NOR }$ Gates are required. Balaji Jegan answered Feb 7, 2019 edited May 13, 2019 by Krithiga2101 Balaji Jegan comment Share Follow See all 11 Comments See all 11 11 Comments reply Show 8 previous comments Asim Siddiqui 4 commented Nov 5, 2020 reply Follow Share would you please provide the source of this statement? 0 votes 0 votes pankya commented Sep 5, 2021 reply Follow Share Can anyone please explain me why in this question POS(Product Of Sum) is calculated? In the question they have given the minterms,why are we solving the POS form and not SOP? 2 votes 2 votes rupesh17 commented Nov 15, 2021 reply Follow Share pankya because they are asking minimum nor gates… OR AND(POS) realization is implemented by NOR gates AND OR(SOP) is implemented by NAND 3 votes 3 votes Please log in or register to add a comment.
55 votes 55 votes $3$ NOR GATE is required. Lakshman Bhaiya answered Feb 7, 2019 edited Feb 14, 2019 by Lakshman Bhaiya Lakshman Bhaiya comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments ayushsomani commented Nov 15, 2019 reply Follow Share @Lakshman Patel RJIT "NOR-NOR Realization is equal to OR-AND Realization. " Please, Correct this. Thanks. I got it. 1 votes 1 votes ayushsomani commented Nov 15, 2019 reply Follow Share @Lakshman Patel RJIT NAND-NAND realisation is AND-OR Realisation, therefore, we need SOP form. X-OR = $A\bar{B}+\bar{A}B$. The given Boolean expression is already in SOP form. Complements are not directly supplied. Stuck!! 1 votes 1 votes Lakshman Bhaiya commented Nov 15, 2019 reply Follow Share We can design using a $5$ NAND gate but we want the minimum number of NAND gates. Try to minimize the NAND gates. These three NAND gates we can't reduce but we can reduce the NAND gate, where complement input needs individual gates. 0 votes 0 votes Please log in or register to add a comment.
20 votes 20 votes Using K-map method we can reduce this to xnor and to implement xnor we will only need 3 nor gates if both inputs and their complements are available. Pankaj Joshi 3 answered Feb 5, 2019 edited Feb 5, 2019 by Pankaj Joshi 3 Pankaj Joshi 3 comment Share Follow See 1 comment See all 1 1 comment reply Smarty19 commented Sep 5, 2022 reply Follow Share Best explanation 0 votes 0 votes Please log in or register to add a comment.
15 votes 15 votes I THINK ANSWER IS 4 ! PLEASE CHECK THIS PIC teja2608 answered Feb 20, 2019 teja2608 comment Share Follow See all 4 Comments See all 4 4 Comments reply Ashok commented Feb 23, 2019 reply Follow Share Minimum number of NOR Gates required are 3. Note:- Minterm and Product term are different. Every Minterm can be Product term but not vice versa. For example, in a 4 variable function, term "BD" is product term but not minterm. Because, Minterms are terms that should contain every variable either in true or complemented form. The term "BD" doesn't contain both A and C either in true or complemented term. Whereas the term "ABCD" is a minterm for a 4 variable function and it is equal to 15. Let's come to the question now. The function is expressed in sum of minterms. Now we have to minimize it. We can minimize it into either SOP or POS as there are no restrictions on the questions. So, using 3 NOR gate, we can minimize. 1 votes 1 votes srilekha panda commented Mar 16, 2019 reply Follow Share Dear how you directly took { complement B and complement D} without using NOT gate. Because for Complement B and Complement D, we need 2 NOT gates. So total 5 nor gates needed. 0 votes 0 votes jlimbasiya commented Nov 4, 2019 reply Follow Share All variable available in it's complemented form also ,that is written in question. 2 votes 2 votes anurags commented Nov 14, 2019 reply Follow Share I think you have made a mistake. Saying F is "sum of products" means nothing in this case. F can be represented as a sum of minterms or as a product of maxterms. This will not change the number of NOR gates needed to realize the logic. 0 votes 0 votes Please log in or register to add a comment.