when an input is given to system S,1^{st} stage gives 1's complement of input, at 2^{nd} stage gives 2's complement of 1^{st }stage inputs. correct??

The Gateway to Computer Science Excellence

0 votes

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??

0 votes

Best answer

no....

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**

52,345 questions

60,504 answers

201,906 comments

95,336 users