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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??

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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

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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??

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