16 votes 16 votes The output $F$ of the below multiplexer circuit can be represented by $AB+B\bar{C}+\bar{C}A+\bar{B}\bar{C}$ $A\oplus B\oplus C$ $A \oplus B$ $\bar{A} \bar{B} C+ \bar{A} B \bar{C}+A \bar{B} \bar{C}$ Digital Logic gate1987 digital-logic combinational-circuit multiplexer circuit-output + – makhdoom ghaya asked Nov 8, 2016 • recategorized Apr 22, 2021 by Lakshman Bhaiya makhdoom ghaya 4.5k views answer comment Share Follow See 1 comment See all 1 1 comment reply krish__ commented Dec 8, 2017 reply Follow Share An easy would be to convert the expression obtained to 0s and 1s. And then, you can notice the fact that there are all possible combinations of odd $\#1$s in each of the terms. Thus it would be the XOR of the inputs. 3 votes 3 votes Please log in or register to add a comment.
Best answer 21 votes 21 votes Answer is B) $A'B'C+AB'C'+A'BC'+ABC$ $=(A+B')(A'+B)C+A'BC'+AB'C'$ $=$$A\oplus B\oplus C$ srestha answered Nov 8, 2016 • edited Jun 26, 2018 by Milicevic3306 srestha comment Share Follow See all 12 Comments See all 12 12 Comments reply Show 9 previous comments srestha commented Nov 16, 2019 reply Follow Share @air1ankit it is simple $A'B'C+ABC=\left ( A\odot B \right )C$ Now $\left ( A\odot B \right )=\left ( A '+B\right )\left ( A+B' \right )$ 2 votes 2 votes Abhineet Singh commented Nov 10, 2020 reply Follow Share how have you assumed msb and lsb? 0 votes 0 votes raja11sep commented Sep 13, 2022 reply Follow Share @Abhineet Singh If nothing mentioned then by default rightmost is LSB and leftmost is MSB. 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes Option B Put c=1 Ram Swaroop answered Jan 6, 2020 Ram Swaroop comment Share Follow See 1 comment See all 1 1 comment reply shashankrustagi commented Dec 18, 2020 reply Follow Share cool thanks 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes (B) You wont find any explanation like this shashankrustagi answered Dec 18, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.