Question

The characteristic equation of SR flip-flop is

• A. \( Q(t+1) = SQ(t) + R \)
• B. \( Q(t+1) = S + RQ(t) \)
• C. \( Q(t+1) = \overline{S}Q(t) + \overline{R} Q(t) \)
• D. \( Q(t+1) = S + \overline{R} Q(t) \)

Hint:

The characteristic equation defines the next state \(Q(t+1)\) of a flip-flop in terms of its present state \(Q(t)\) and inputs.
For an SR flip-flop, assuming the condition \(SR=0\) (S and R are not simultaneously 1), the characteristic equation is \(Q(t+1) = S + \overline{R}Q(t)\).

Answer 1

The Correct Option is D

The SR flip-flop has inputs S (Set) and R (Reset), and output Q. The truth table is: \begin{tabular}{|c|c|c|p{3cm}|} \hline S & R & $Q(t+1)$ & Comment
\hline 0 & 0 & $Q(t)$ & No change (Hold)
0 & 1 & 0 & Reset
1 & 0 & 1 & Set
1 & 1 & - & Forbidden
\hline \end{tabular} \medskip Characteristic equation derivation:
For $S=1,R=0 \Rightarrow Q(t+1)=1$
For $S=0,R=1 \Rightarrow Q(t+1)=0$
For $S=0,R=0 \Rightarrow Q(t+1)=Q(t)$ Using K-map with $SR=0$ constraint: \[ Q(t+1) = S + \overline{R}Q(t) \] Let's test this: If S=0, R=0: \(Q(t+1) = 0 + \overline{0}Q(t) = 1 \cdot Q(t) = Q(t)\). (Hold - Correct) If S=0, R=1: \(Q(t+1) = 0 + \overline{1}Q(t) = 0 \cdot Q(t) = 0\). (Reset - Correct) If S=1, R=0: \(Q(t+1) = 1 + \overline{0}Q(t) = 1 + 1 \cdot Q(t) = 1\). (Set - Correct) If S=1, R=1 (forbidden): \(Q(t+1) = 1 + \overline{1}Q(t) = 1 + 0 \cdot Q(t) = 1\). This equation sets Q to 1 for the forbidden state, which is one possible outcome if S dominates R, or it's simplified under \(SR=0\). Option (d) is \( Q(t+1) = S + \overline{R} Q(t) \). This matches the commonly used characteristic equation under the constraint \(SR=0\). \[ \boxed{Q(t+1) = S + \overline{R} Q(t)} \]