RS + T flip-flop state machine: 22-Elec-A4 Digital Systems and Computers, May 2018, Q3

Question 3: RS + T flip-flop state machine (12 marks)

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. Two negative-edge-triggered flip-flops: flip-flop A is an RS type with outputs $Q_A,\overline{Q}_A$; flip-flop B is a T type with outputs $Q_B,\overline{Q}_B$. Reading the gate network from the figure (three 2-input ANDs, two 2-input ORs, and the clock inverter), the input equations are $R_A=X\,Q_A$, $S_A=X\,\overline{Q}_A$, $T_B=\overline{Q}_A\,(X+Q_B)$ and the output $Y=Q_A+Q_B$.

Find. (a) those four expressions; (b) the state-transition table over present state $(Q_A,Q_B)$ and input $X$; (c) the state diagram; (d) the Moore/Mealy classification with justification.

Approach. Apply each flip-flop's characteristic equation — RS gives $Q_A^+=S_A+\overline{R}_A Q_A$, T gives $Q_B^+=T_B\oplus Q_B$ — to all four present states under $X=0$ and $X=1$, then read the output and inspect its dependence on $X$.

(a) Input/output equations (from the schematic): $$R_A=X\cdot Q_A,\quad S_A=X\cdot\overline{Q}_A,\quad T_B=\overline{Q}_A\,(X+Q_B),\quad Y=Q_A+Q_B.$$ Note $R_A S_A=X Q_A\cdot X\overline{Q}_A=0$ always, so the RS forbidden state never occurs.
Simplify the next-state of A. Substituting into $Q_A^+=S_A+\overline{R}_A Q_A$: for $X=0$, $R_A=S_A=0\Rightarrow Q_A^+=Q_A$ (hold); for $X=1$, $Q_A^+=\overline{Q}_A+\overline{Q_A}\,Q_A=\overline{Q}_A$ (toggle). So $X$ acts as a "toggle $Q_A$" command.
Next-state of B. $Q_B^+=T_B\oplus Q_B=\big[\overline{Q}_A(X+Q_B)\big]\oplus Q_B$. For $X=1$, $T_B=\overline{Q}_A$; for $X=0$, $T_B=\overline{Q}_A Q_B$.
(b) Evaluate all eight (state, input) combinations and the Moore output $Y=Q_A+Q_B$.

(b) State transition table — next state $(Q_A^{+}Q_B^{+})$ and output
$Q_A Q_B$	$X=0\;\to\;Q_A^{+}Q_B^{+}$	$X=1\;\to\;Q_A^{+}Q_B^{+}$	$Y$
0 0	0 0	1 1	0
0 1	0 0	1 0	1
1 0	1 0	0 0	1
1 1	1 1	0 1	1

(c) State diagram. Each state is labelled with its output $Y$ (Moore); arcs are labelled with the input $X$ that causes the transition.

(d) Moore or Mealy? The output is $Y=Q_A+Q_B$, a function of the present state only — $X$ does not appear. Confirming from the table, both entries of every row share the same $Y$. Therefore this is a Moore machine: the output is associated with the state, and it changes only after a clock edge moves the machine to a new state, never combinationally with $X$.

Question 3 — summary
Item	Result
$Q_A^{+}$	hold on $X=0$, toggle on $X=1$ ($=\,X\oplus Q_A$)
$Q_B^{+}$	$\big[\overline{Q}_A(X+Q_B)\big]\oplus Q_B$
Output	$Y=Q_A+Q_B$ (state-only)
Machine type	Moore

Topic: Analysis of a clocked sequential circuit; Moore vs. Mealy.

Key relations: RS characteristic $Q^+=S+\overline{R}Q$ with constraint $RS=0$; T characteristic $Q^+=T\oplus Q$; Moore output $=f(\text{state})$, Mealy output $=f(\text{state},\text{input})$.

Why this works: Each flip-flop's next output depends only on its own excitation inputs and current output, so substituting the driving-network equations into the characteristic equations yields the next-state logic directly. Because the excitation network here makes $R_A S_A=0$ identically, the RS device behaves safely as a general set/reset element, and $X=1$ turns it into a toggle. The output is wired straight from the state variables with no path from $X$, which is the defining signature of a Moore machine.

Common pitfall: reading a Moore machine as Mealy because an input feeds a next-state gate — what matters for the classification is only the output function. Also, applying the RS characteristic without checking $RS=0$ can hide a design error; here the check passes.

Source: Mano & Ciletti, Digital Design 6th ed., §5.4–5.7 (analysis of clocked sequential circuits, state tables/diagrams, Moore/Mealy).

Question 3 of 6: RS + T flip-flop state machine

Question 3: RS + T flip-flop state machine (12 marks)