6 votes 6 votes The following circuit compares two $2$-bit binary numbers, $X$ and $Y$ represented by $X_1X_0$ and $Y_1Y_0$ respectively. ($X_0$ and $Y_0$ represent Least Significant Bits) Under what conditions $Z$ will be $1$? $X>Y$ $X<Y$ $X=Y$ $X!=Y$ Digital Logic isro-2020 digital-logic digital-circuits circuit-output normal + – Satbir asked Jan 13, 2020 • retagged Dec 7, 2022 by Lakshman Bhaiya Satbir 4.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply Aman_777 commented Mar 7 reply Follow Share correct 0 votes 0 votes Please log in or register to add a comment.
10 votes 10 votes Ans: a) X > Y The function Z can be represented as: $Z = X_1.Y_1' + (X_1 \odot Y_1).(X_0.Y_0')$ The first term, $X_1.Y_1'$ would be 1 only when $X_1 > Y_1$ The second term would be 1 when $X_1 == Y_1$ and $X_0 > Y_0$ braindead answered Jan 13, 2020 • edited Feb 9, 2020 by Satbir braindead comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes The output equation will be $(X_1⊙Y_1).(X_0.\overline{Y}_0)+X_1.\overline{Y}_1$ Clearly, when $X=0$, ie, when $X_0$ and $X_1$ are both $0$, output can't be 1. So, Options B, C and D are incorrect, because in them, $X=0$ is valid. Option A JashanArora answered Feb 21, 2020 JashanArora comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes The The rhe the conditions ZZ will be 1 is X>Y THEREFORE: OPTION A) heisenberggg answered Apr 4, 2021 heisenberggg comment Share Follow See all 0 reply Please log in or register to add a comment.