Question

Consider a 2-bit saturating up/down counter that performs the saturating up count when the input \(P\) is 0, and the saturating down count when \(P\) is 1. The Next State table of the counter is as shown. The counter is built as a synchronous sequential circuit using D flip-flops.

Input P	Current State Q1	Current State Q0	Next State Q1+	Next State Q0+
0	0	0	0	1
0	0	1	1	0
0	1	0	1	1
0	1	1	1	1
1	0	0	0	0
1	0	1	0	0
1	1	0	0	1
1	1	1	1	0

Which one of the following options corresponds to the expressions for the inputs of the D flip-flops, \(D_1\) and \(D_0\)?

• A. \(D_1 = P\,Q_1 + \bar{P}Q_0 + Q_1Q_0\)
  \(D_0 = P\,Q_0 + \bar{P}\,Q_1 + Q_1\overline{Q_0}\)
• B. \(D_1 = \bar{P}\,Q_1 + \bar{P}Q_0 + Q_1Q_0\)
  \(D_0 = \bar{P}\,\overline{Q_0} + \bar{P}\,Q_1 + Q_1\overline{Q_0}\)
• C. \(D_1 = \bar{P}\,\overline{Q_1} + \bar{P}\,Q_0 + Q_1Q_0\)
  \(D_0 = \bar{P}\,Q_0 + \bar{P}\,Q_1 + Q_1\overline{Q_0}\)
• D. \(D_1 = P\,\overline{Q_1} + \bar{P}\,Q_0 + Q_1Q_0\)
  \(D_0 = P\,\overline{Q_0} + \bar{P}\,Q_1 + Q_1\overline{Q_0}\)

Hint:

For a D flip-flop, the flip-flop input equals the next state value, so write \(D_1=Q_1^+\) and \(D_0=Q_0^+\) directly from the table and simplify each using a Karnaugh map.

Answer 1

The Correct Option is B

Step 1: Recall the D flip-flop property.
For a D flip-flop, whatever value is applied at the D input appears at the output after the next clock edge. So the required excitation equations are simply the next state functions read directly off the table:\[ D_1 = Q_1^{+}, \qquad D_0 = Q_0^{+} \]Step 2: List the minterms of \(D_1\) from the table.
Reading the \(Q_1^+\) column, \(D_1=1\) for the rows \((P,Q_1,Q_0)=(0,0,1),(0,1,0),(0,1,1),(1,1,1)\).
Step 3: Group the minterms on a Karnaugh map for \(D_1\).
The pair \((0,1,0)\) and \((0,1,1)\) (Q0 varies) gives the term \(\bar{P}Q_1\).
The pair \((0,0,1)\) and \((0,1,1)\) (Q1 varies) gives the term \(\bar{P}Q_0\).
The pair \((0,1,1)\) and \((1,1,1)\) (P varies) gives the term \(Q_1Q_0\).
These three groups cover all four minterms exactly, so\[ D_1 = \bar{P}\,Q_1 + \bar{P}\,Q_0 + Q_1Q_0 \]Step 4: List the minterms of \(D_0\) from the table.
Reading the \(Q_0^+\) column, \(D_0=1\) for the rows \((0,0,0),(0,1,0),(0,1,1),(1,1,0)\).
Step 5: Group the minterms for \(D_0\).
The pair \((0,0,0)\) and \((0,1,0)\) (Q1 varies) gives \(\bar{P}\,\overline{Q_0}\).
The pair \((0,1,0)\) and \((0,1,1)\) (Q0 varies) gives \(\bar{P}\,Q_1\).
The pair \((0,1,0)\) and \((1,1,0)\) (P varies) gives \(Q_1\overline{Q_0}\).
These groups cover all four minterms exactly, so\[ D_0 = \bar{P}\,\overline{Q_0} + \bar{P}\,Q_1 + Q_1\overline{Q_0} \]Step 6: Compare with the given options.
The pair of expressions obtained in Steps 3 and 5 is exactly option (B).
Step 7: Eliminate the remaining options with direct counterexamples from the table.
Option (A) uses \(P\,Q_1\) in place of \(\bar{P}\,Q_1\) in \(D_1\). At row \(P=1,Q_1=1,Q_0=0\) the table gives \(Q_1^+=0\), but option (A) evaluates to \(P\,Q_1+\bar{P}Q_0+Q_1Q_0 = 1+0+0=1\), which is wrong.
Option (C) uses \(\bar{P}\,\overline{Q_1}\) in place of \(\bar{P}\,Q_1\) in \(D_1\). At row \(P=0,Q_1=1,Q_0=0\) the table gives \(Q_1^+=1\), but option (C) evaluates to \(\bar{P}\overline{Q_1}+\bar{P}Q_0+Q_1Q_0=0+0+0=0\), which is wrong.
Option (D) uses \(P\,\overline{Q_1}\) in place of \(\bar{P}\,Q_1\) in \(D_1\). At the same row \(P=0,Q_1=1,Q_0=0\) option (D) evaluates to \(P\overline{Q_1}+\bar{P}Q_0+Q_1Q_0=0+0+0=0\), again wrong, and its \(D_0\) uses \(P\overline{Q_0}\) instead of \(\bar{P}\,\overline{Q_0}\), which fails at row \(P=0,Q_1=0,Q_0=0\) where \(Q_0^+=1\) but \(P\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}=0+0+0=0\).
So only option (B) satisfies every row of the Next State table.\[ \boxed{D_1=\bar{P}Q_1+\bar{P}Q_0+Q_1Q_0,\ \ D_0=\bar{P}\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}\ \text{(Option B)}} \]