2's complement (2's complement) = original number,
in above fig in the 2nd stage we get the original number after that we again take 2's compliment??
3 times 2's complement of input??
if i/p is A then 1st stage ( 2's com( 1's compl ( A ) ) )
at 2nd stage o/p of 1st stage is i/p
( 2's com( 1's compl ( ( 2's com( 1's compl ( A ) ) ) ) ) ) ----> you can't get original no...
let take i/p = y ===> 1's complement(y) = X
2's complement (X) = 1's complement(X) +1 = y+1
therefore your System S giving output as increment by 1 the input
O/P of given circuit = i/p + 3
The system s "1's comp+ 2's" comp is increment the value .
when an input is given to system S,1st stage gives 1's complement of input, at 2nd stage gives 2's complement of 1st stage inputs. correct??