Electronic – Building a 2’s complementer (need help visualising how to make a state diagram and table)

digital-logic

Just to be fair, this is a previous exam question and was an assignment question. I've already handed in the assignment and now I'm trying to see why I got it wrong — to study for my final.

I fully understand how to find the 2's complement of a binary number, however, implementing a circuit to find the complement is driving me crazy. I'm given a shift register (that stores my binary number) and a flip flop (using D for simplicity). The process I follow for building sequential circuits is this:

  • I look at the question and derive a state diagram.
  • From the state diagram, I draw a state table and find the input of my flip flops
  • I use maps and stuff to finally build the circuit.

I'm trying to visualize how to actually draw the state diagram for this complementer. I know it depends on the previous value and stuff, but if someone could take the time to explain in plain english how he/she would go about building the state diagram, I would be vey grateful.

Best Answer

That link you gave says it all:

Hint: - 2’s complement of a number can be obtained by keeping the least significant bits as such until the first 1, and then complementing all bits

"example 001010 → 110110" that is to say

   001010 -> flip bits -> 110101 -> add 1 -> 110110
msb     lsb             msb    lsb         msb    lsb


      msb    lsb
input   001010
output  110110
         <--- time

So there are two states - "I haven't seen a one yet" and "I have seen a one previously". Given my state and my current input, my output and next state are fully determined.

If my current state is "I haven't seen a one yet" and my input is a 0, my output is a 0 and my next state is (still) "I haven't seen a one yet"

If my current state is "I haven't seen a one yet" and my input is a 1, my output is a 1 and my next state is (changes to) "I have seen a one previously"

If my current state is "I have seen a one previously" and my input is a 0, my output is a 1 and my next state is (still) "I have seen a one previously"

If my current state is "I have seen a one previously" and my input is a 1, my output is a 0 and my next state is (still) "I have seen a one previously"

Write that out as a truth table encoding "I haven't seen a one yet" as 0 and "I have seen a one previously" as 1, and you should be home free.

To "wire it up" the current data input bit and the current state go into logic that feeds the input to the D-flip flop and separate logic that feeds the input to the shift register, the D-flip-flop holds the "state variable", and both are clocked by the data clock. And to be complete you need some kind of reset logic - left as an exercise to the reader. The "current state" is the output of the D-flip-flop, and next state is the input to it...

       +----------------------------+
       |                            |
       |  +---------+     +-----+   |
       +--+ Logic A +-----+ DFF +---+
Data-+-|--+         |     |     |
     | |  +---------+     | CLK |
     | |                  +--+--+
     | |                     |
     | |                     +--------Clock 
     | |                  +--+--+
     | |  +---------+     | CLK |
     | +--+ Logic B +-----+     |
     +----+         |     | S R |
          +---------+     +-----+
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