Q. Illustration to demonstrate design of sequential circuits?

Let us take an illustration to demonstrate above process. Suppose we want to design 2-bit binary counter employing D flip-flop. Circuit goes through repeated binary states 00, 01, 10 and 11 whenever external input X = 1 is applied. State of circuit won't change when X = 0. State table and state diagram for this is displayed in figure. Although how do we make the state diagram? Please note number of flip-flops- 2 in our illustration as we are designing 2 bits counter. Different states of two bit input will be 00 01 10 and 11.  These are displayed in circle. The arrow denotes transitions on an input value X.  For illustration when counter is in state 00 and input value X=0 happens counter remains in 00 state.  Therefore loop back on X= 0.  But on encountering X=1 counter moves to state 01.  Similarly in all other states similar transition happen.  For making state table consider excitation table of D flip-flop given in figure (c).

Present state of two flip-flops and subsequent states of flip-flops are put into the table with any input value.  For illustration if present state of flip-flops is 01 and input value is 1 then counter will move to state 10.  Notice these values in fourth row of values in state table (figure (a) Or we can write as below:

This denotes flip-flop A has moved from state clear to set. As we are constructing the counter using D flip-flop the main question is what would be input D_A value of A flip-flop which allows this transition which  is Q(t) = 0 to Q(t+1) =1 possible for A flip flop.  On checking excitation table for D Flip-flop we find value of D input of a flip-flop (known as D_A in this illustration) would be 1.  Similarly B flip-flop have a transition Q(t) = 1 to Q(t+1)=0 so D_B, would be 0.  Therefore notice the values of flip-flop inputs D_A and D_B. (Row 3).