Question:hard
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 \(D_0\)?
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.
Which one of the following options corresponds to the expressions for the inputs of the D flip-flops, \(D_1\) and \(D_0\)?
| 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 \(D_0\)?
Show 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.
Updated On: Jul 22, 2026
- \(D_1 = P\,Q_1 + \bar{P}Q_0 + Q_1Q_0\)
\(D_0 = P\,Q_0 + \bar{P}\,Q_1 + Q_1\overline{Q_0}\) - \(D_1 = \bar{P}\,Q_1 + \bar{P}Q_0 + Q_1Q_0\)
\(D_0 = \bar{P}\,\overline{Q_0} + \bar{P}\,Q_1 + Q_1\overline{Q_0}\) - \(D_1 = \bar{P}\,\overline{Q_1} + \bar{P}\,Q_0 + Q_1Q_0\)
\(D_0 = \bar{P}\,Q_0 + \bar{P}\,Q_1 + Q_1\overline{Q_0}\) - \(D_1 = P\,\overline{Q_1} + \bar{P}\,Q_0 + Q_1Q_0\)
\(D_0 = P\,\overline{Q_0} + \bar{P}\,Q_1 + Q_1\overline{Q_0}\)
Show Solution
The Correct Option is B
Solution and Explanation
A different, purely verification based way to solve this is to test each candidate formula against every row of the Next State table instead of deriving the expression from scratch.
Take option (B): $D_1=\bar{P}Q_1+\bar{P}Q_0+Q_1Q_0$ and $D_0=\bar{P}\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}$.
Substitute all eight rows of $(P,Q_1,Q_0)$:
(0,0,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=1+0+0=1$, matches $Q_0^+=1$.
(0,0,1): $D_1=0+1+0=1$, matches $Q_1^+=1$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(0,1,0): $D_1=1+0+0=1$, matches $Q_1^+=1$; $D_0=1+1+1=1$, matches $Q_0^+=1$.
(0,1,1): $D_1=1+1+1=1$, matches $Q_1^+=1$; $D_0=0+1+0=1$, matches $Q_0^+=1$.
(1,0,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(1,0,1): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(1,1,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+1=1$, matches $Q_0^+=1$.
(1,1,1): $D_1=0+0+1=1$, matches $Q_1^+=1$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
Every one of the sixteen values reproduces the Next State table exactly, so option (B) is verified correct without any error.
Now spot check the remaining options with a single row each. At $(P,Q_1,Q_0)=(1,1,0)$ the table requires $Q_1^+=0$. Option (A) gives $D_1=PQ_1+\bar{P}Q_0+Q_1Q_0=1+0+0=1$, which already contradicts the table, so option (A) is eliminated. At $(P,Q_1,Q_0)=(0,1,0)$ the table requires $Q_1^+=1$. Option (C) gives $D_1=\bar P \overline{Q_1}+\bar P Q_0+Q_1Q_0=0+0+0=0$, and option (D) gives $D_1=P\overline{Q_1}+\bar P Q_0+Q_1Q_0=0+0+0=0$; both contradict the table, so (C) and (D) are eliminated as well.
This substitution check confirms, independent of the Karnaugh map derivation, that option (B) is the only formula consistent with the given Next State table.$$ \boxed{\text{Option (B): } D_1=\bar{P}Q_1+\bar{P}Q_0+Q_1Q_0,\ D_0=\bar{P}\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}} $$
Take option (B): $D_1=\bar{P}Q_1+\bar{P}Q_0+Q_1Q_0$ and $D_0=\bar{P}\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}$.
Substitute all eight rows of $(P,Q_1,Q_0)$:
(0,0,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=1+0+0=1$, matches $Q_0^+=1$.
(0,0,1): $D_1=0+1+0=1$, matches $Q_1^+=1$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(0,1,0): $D_1=1+0+0=1$, matches $Q_1^+=1$; $D_0=1+1+1=1$, matches $Q_0^+=1$.
(0,1,1): $D_1=1+1+1=1$, matches $Q_1^+=1$; $D_0=0+1+0=1$, matches $Q_0^+=1$.
(1,0,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(1,0,1): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
(1,1,0): $D_1=0+0+0=0$, matches $Q_1^+=0$; $D_0=0+0+1=1$, matches $Q_0^+=1$.
(1,1,1): $D_1=0+0+1=1$, matches $Q_1^+=1$; $D_0=0+0+0=0$, matches $Q_0^+=0$.
Every one of the sixteen values reproduces the Next State table exactly, so option (B) is verified correct without any error.
Now spot check the remaining options with a single row each. At $(P,Q_1,Q_0)=(1,1,0)$ the table requires $Q_1^+=0$. Option (A) gives $D_1=PQ_1+\bar{P}Q_0+Q_1Q_0=1+0+0=1$, which already contradicts the table, so option (A) is eliminated. At $(P,Q_1,Q_0)=(0,1,0)$ the table requires $Q_1^+=1$. Option (C) gives $D_1=\bar P \overline{Q_1}+\bar P Q_0+Q_1Q_0=0+0+0=0$, and option (D) gives $D_1=P\overline{Q_1}+\bar P Q_0+Q_1Q_0=0+0+0=0$; both contradict the table, so (C) and (D) are eliminated as well.
This substitution check confirms, independent of the Karnaugh map derivation, that option (B) is the only formula consistent with the given Next State table.$$ \boxed{\text{Option (B): } D_1=\bar{P}Q_1+\bar{P}Q_0+Q_1Q_0,\ D_0=\bar{P}\overline{Q_0}+\bar{P}Q_1+Q_1\overline{Q_0}} $$
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