The next state table of a 2−bit saturating up-counter is given below.…
2017
The next state table of a 2−bit saturating up-counter is given below.
\(\begin{array}{cc|cc} Q_1 & Q_0 & Q_1^+ & Q_0^+ \\ \hline 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}\)
The counter is built as a synchronous sequential circuit using \(T\) flip-flops. The expressions for \(T_1\) and \(T_0\) are
- A.
\(T_1 = Q_1Q_0, \quad T_0= \bar{Q_1} \bar{Q_0}\) - B.
\(T_1 = \bar{Q_1}Q_0, \quad T_0= \bar{Q_1} + \bar{Q_0}\) - C.
\(T_1 = Q_1+Q_0, \quad T_0= \bar{Q_1} \bar{Q_0}\) - D.
\(T_1 = \bar{Q_1}Q_0, \quad T_0= Q_1 + Q_0\)
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Correct answer: B
Derivation of T inputs for the T flip-flops
Find the next-state expressions from the table.
Q1+ is 1 for states 01, 10, 11, so Q1+ = Q1 + Q0.
Q0+ is 0 only for state 01, and 1 otherwise, so Q0+ = not(Q1=0 and Q0=1) = Q1 + ~Q0.
Use T = Q xor Q+ to get T inputs.
T1 = Q1 xor Q1+ = Q1 xor (Q1 + Q0). Simplify: when Q1=0, T1 = Q0; when Q1=1, T1 = 0. So T1 = ~Q1·Q0.
T0 = Q0 xor Q0+ = Q0 xor (Q1 + ~Q0). Evaluating the truth table gives T0 = 1 for states 00, 01, 10 and 0 for 11, so T0 = ~(Q1·Q0) = ~Q1 + ~Q0.
Final expressions: T1 = ~Q1·Q0, T0 = ~Q1 + ~Q0
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