I know this question has been asked and there already are answers to this question on the Internet. I just want to ask you where I am wrong in my way of thinking. When I first saw this question, the shortcut -keep copying until you see a 1 and then complement the rest- hadn't occurred to me. So I thought I would just complement the bit, add 1 to it and if I get a carry then I would keep it in the flip flop to add it to the next bit coming from the input. Meaning that, In State-0 if the input is 0 then I should output '0' (after complementing) and keep that carry '1' in the flip-flop so flip-flop goes from state-0 to state-1. If the input is 1 then I should output '1' but the flip-flop remains in the same state which is state-0.

In State-1 if the input is 0 then I should output '1' -since 2's complement would be 10 and I have a carry '1' in the flip-flop- I would get another carry so the state remains the same. If the input is 1 then I should output '0' and the flip-flop remains in the same state as well. Here is my state diagram: +---------+-------+-------+--------+ Present Next State Input State Output A x A y 0 0 1 0 0 1 0 1 1 0 1 1 1 1 1 0 +---------+-------+-------+--------+ After getting the equations from the table above.

I get a circuit like so: – Schematic created using Can you please help me to figure out what's wrong with my way of thinking? Thank you in advance.:). I just figured out the problem and here I am explaining.

If you were trying to solve this problem the same way I was trying (without thinking about the shortcut like 'output the same value till you see a bit of value 1 and then complement others in the string.' ) then we will have 2 states. State-1 (Initial State - Carry-1 State) Which we have a carry-1 to add to the least significant bit of the string after complementing. Because that's what you would normally do when you are trying to get the 2's complement of a bit string. You just complement the entire string and add 1 to the least significant bit.

You could name this state, State-0 as well but I am just naming it State-1 not to get confused. So that I know I have a carry 1 in the flip-flop. State-0 (Which signifies that you don't have any carry left) This state means that you have no carry left in hand. So you just complement the bit at the input.

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2's complementer with shift register Design a 2's complementer with a shift register and a flip-flop. The binary # is shifted out from one side and it's 2's complement shifted into the other side of the shift register. 62 include a synchronous clear input to the register. 6.2 Include a synchronous clear input to the register of Fig. (See Problem 4.12). (HDL—see Problem 6.35(i).) 6.10 Design a serial 2’s complementer with a shift register and a flip‐flop.

Let's inspect what possibilities we can get when trying to get the 2's complement of a bit string. • In the initial state you could get a 1 whose 1's complement is 0 and with a carry in hand equals to 1. Since that sum doesn't generate a carry.

You have no carry left in hand and you go to State-0. • Again in the initial state you could get a 0 whose 1's complement is 1 and with a carry in hand this sum equals to 10, which means you will again have a carry in hand and therefore remain in the same state which is State-1. But the output will be 0.

• In State-0 you could get a 0 again and since you don't have any carry in hand you just output the 1's complement of 0 which is equal to 1. • Again in State-0 you could have 1 at the input and since you don't have any carry your output would be equal to 0. Having that said, we draw a state table as follows: +---------+-------+-------+--------+ Present Next State Input State Output A x A y 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 0 +---------+-------+-------+--------+ Looking at the table, we get a state diagram like so: Then we get this circuit: – Schematic created using.