Question 4 of 6: RS/T flip-flop circuit — analysis and machine classification

Question 4: RS/T flip-flop circuit — analysis and machine classification [3+4+3+2 = 12]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. RS flip-flop with state $A$ (output $A$, i.e. $Q_A$) and T flip-flop with state $B$ (output $B$, i.e. $Q_B$), both clocked on the same edge. From the figure: $S_A$ (RS flip-flop's Set input) is the AND of $X$ and $B$; $R_A$ (Reset input) is the OR of $X$ and $B$; $T_B$ (T flip-flop's toggle input) is driven directly by $A$; the circuit output $Y$ is the AND of $A$ and $\bar B$ (the T flip-flop's inverted output). RS characteristic behaviour: $S{=}1,R{=}0\Rightarrow Q^+{=}1$; $S{=}0,R{=}1\Rightarrow Q^+{=}0$; $S{=}0,R{=}0\Rightarrow Q^+{=}Q$ (hold); $S{=}1,R{=}1$ is the forbidden input combination. T flip-flop: $Q^+=T\oplus Q$.

Find. $R_A,S_A,T_B,Y$ as logic expressions; the full state transition table; the state diagram; whether the machine is Moore or Mealy.

Approach. Read the four expressions directly off the given circuit, then evaluate all 8 combinations of $A,B,X$: apply the RS rule to get $A^*$ (flagging the forbidden $S_A{=}R_A{=}1$ case explicitly rather than guessing a resolution) and the T rule to get $B^*$, then classify Moore/Mealy by checking whether $Y$ ever depends on $X$.

1. (a) Logic expressions. $$\boxed{S_A=XB,\quad R_A=X+B,\quad T_B=A,\quad Y=A\bar B}.$$
2. (b) State transition table. Applying $S_A,R_A,T_B,Y$ and then the RS/T update rules to all 8 combinations of $A,B,X$:

   A	B	X	$R_A$	$S_A$	$T_B$	$A^*$	$B^*$	Y
   0	0	0	0	0	0	0	0	0
   0	0	1	1	0	0	0	0	0
   0	1	0	1	0	0	0	1	0
   0	1	1	1	1	0	forbidden	1	0
   1	0	0	0	0	1	1	1	1
   1	0	1	1	0	1	0	1	1
   1	1	0	1	0	1	0	0	0
   1	1	1	1	1	1	forbidden	0	0

   Whenever $X{=}1$ and $B{=}1$ simultaneously, $S_A{=}XB{=}1$ and $R_A{=}X{+}B{=}1$ at the same time — the forbidden RS input. This is a genuine race/design flaw in the given circuit (not an authoring assumption): the two rows $(A,B,X){=}(0,1,1)$ and $(1,1,1)$ leave $A^*$ undefined by the RS flip-flop's own specification. $B^*$ and $Y$ remain well-defined in every row since they come from $T_B{=}A$ and $Y{=}A\bar B$, neither of which touches the forbidden condition.
3. (c) State transition diagram. Plotting only the well-defined transitions (the two forbidden rows are shown as dashed stubs rather than resolved arrows):
   Fig. 4(c) — state transition diagram for $AB$; solid arrows are well-defined RS/T transitions, dashed red stubs mark the $S_A{=}R_A{=}1$ forbidden condition at $AB{=}01,X{=}1$ and $AB{=}11,X{=}1$.
4. (d) Moore or Mealy? $Y=A\bar B$ depends only on the present state $(A,B)$ — it never appears as a function of $X$ in the table above (rows $(1,0,0)$ and $(1,0,1)$ both give $Y{=}1$; rows $(1,1,0)$ and $(1,1,1)$ both give $Y{=}0$; and so on for every state, regardless of $X$). $$\boxed{\text{This is a Moore machine}}\text{: the output is a function of state alone.}$$