3 votes 3 votes The size of the $ROM$ required to build an 8-bit adder / subtractor with mode control, carry input, carry output and two's complement overflow output is given as $2^{16} \times 8$ $2^{18} \times 10$ $2^{16} \times 10$ $2^{18} \times 8$ Digital Logic ugcnetcse-dec2014-paper2 digital-logic rom + – makhdoom ghaya asked Jul 18, 2016 recategorized Jan 1, 2020 makhdoom ghaya 3.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Swati Tandel commented Jun 18, 2016 reply Follow Share result will be 8 bit so 8 vertical lines +( 1 for carry ) +1 ( for saying underflow) . ^^ i didnt get this 0 votes 0 votes Nikhil Patil commented Jul 23, 2018 reply Follow Share Underflow occurs when result produced is smallest no. which can be represented. 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes total input to the rom decoder will be (8+8 ( two 8 bit number ) +1( mode ) +1( carry in)) so total number of words out of decoder will be 2^18 . result will be 8 bit so 8 vertical lines +( 1 for carry ) +1 ( for saying underflow) . answer will be B. a little more here . https://www.google.co.in/search?q=8-bit+adder%2F+subtractor&oq=8-bit+adder%2F+subtractor+&aqs=chrome..69i57.4118j0j7&sourceid=chrome&es_sm=93&ie=UTF-8#q=rom+required+8-bit+adder%2F+subtractor Tendua answered Nov 17, 2015 Tendua comment Share Follow See all 3 Comments See all 3 3 Comments reply Chetana Tailor commented Nov 17, 2015 reply Follow Share Thank you sir. 0 votes 0 votes Tendua commented Nov 17, 2015 reply Follow Share thanks mat bolo . bas ek jodi kapda sil dena. ;) 7 votes 7 votes Ritwik commented Jun 15, 2017 reply Follow Share @Chetana Tailor : Tag the source of this question. Or else here is a source you could update it with : UGC-NET December 2014 Paper II Qn 10. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Refer this shekhar chauhan answered Jul 18, 2016 shekhar chauhan comment Share Follow See all 0 reply Please log in or register to add a comment.