3 3 votes The output $Y$ of a $2$ -bit comparator is logic $1$ whenever the $2$ -bit input $A$ is greater than the $2$ -bit input $B$ . The number of combinations for which the output is logic $1$ , is$4$$6$$8$$10$ Digital Logic goclasses digital-logic goclasses-cs-dpp goclasses-cs-dpp-day-43 goclasses-digital-logics-practice-questions + – GO Classes 440 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote \begin{array}{ccc} A & B & \text { No.ofcases } \\ 01 & 00 & 1 \\ 10 & 00,01 & 2 \\ 11 & 00,01,10 & 3 \\ & & \text { total }=6 \end{array} GO Classes answered Jul 12, 2025 • reshown Jul 14, 2025 by GO Classes Support GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Answer- shashank sharma_1 answered Jul 12, 2025 shashank sharma_1 comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes For 2 bit we can find by finding exact combination but for n bits if we comparing n bits then for A>B or A<B or A=B first find equal condition and total possible cases for A=a0a1......an and B=b0b1b2......bn so truth table creating implies 2n variables implies total 2"2n rows implies total possible cases are 2'2n for A=B if we select some value for a0a1.......an then B values must be same so for n variable we have 2'n possiblities to put values hence for equality we have 2'n cases for unequal it is 2'2n-2'n and since A>B and B<A has same number of cases so A>B=A<B=2'2n-2'n/2 jacknroll answered Aug 27, 2025 jacknroll comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Option B is correct Jay Thakar answered Oct 10, 2025 Jay Thakar comment Share Follow 0 reply Please log in or register to add a comment.