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First look at the operation of the D type flip flop The sequence starts by a reset so Q = 0 The input to the D-type is made up from the initial output (Q) which is OR'd with the first (LSB) bit of the number you are complimenting (X). The output Y is XOR'd with Q and X (00 > 0, 01 >1) as Q is '0' we don't have to consider any other case. So initially the output at Y will always be the same as input at X i.e our LSB data bit. The initial input at 'D' will also be the same as the input at X (X OR '0' = X) Let's take a number - say 28. In binary this would be 00011100 To change this to its 2's compliment we invert and add 1 00011100 --> 11100011 ---> +1 ----> 11100100 So if our circuit works a 00011100 input it will produce a 11100100 output Start with a reset so that Y = X (Q = '0') Q (t+1) = D (t) D is X OR Q Y is X XOR Q LSB first X Q D Y 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1 Look at the diagonal relationship between D and Q.

Q in the next row (t+1) is simply the value of D in the previous row (at time t). Each time the bit is CLOCKED 't' moves on 1. The rest is simply applying the logic of the connected gates to produce a value.

Dec 17, 2017 - Let's inspect what possibilities we can get when trying to get the 2's. In the initial state you could get a 1 whose 1's complement is 0 and with a. 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.