Consider 2-bit saturating up/down counter for p = 0 and p = 1 respectively. Find excitation D₁ and D₀ for the given table.

P	Q1	Q0	Q1+	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

Find D₁ and D₀

The correct option is:
D₁ = P̅Q₁ + P̅Q₀ + Q₁Q₀ , D₀ = P̅Q̅₀ + P̅Q₁Q₀ (represented in the options as D1 = PQ1 + PQ0 + Q1Q0 , D0 = PQ0 + PQ1Q0, where spaces/formatting indicate complemented variables).

Step-by-step Explanation:

1. Understanding the State Table and Flip-Flop Excitation
For D flip-flops, the excitation input equation is simply equal to the next state:
$$D_1=Q_1^{+}$$
$$D_0=Q_0^{+}$$

From the provided state table, we can list the minterms for $Q_1^{+}$ and $Q_0^{+}$ based on the inputs $P$, $Q_1$, and $Q_0$:
• For $D_1$, the output is $1$ at rows:
- $(P,Q_1,Q_0)=(0,0,1)\Rightarrow m_1$
- $(P,Q_1,Q_0)=(0,1,0)\Rightarrow m_2$
- $(P,Q_1,Q_0)=(0,1,1)\Rightarrow m_3$
- $(P,Q_1,Q_0)=(1,1,1)\Rightarrow m_7$

• For $D_0$, the output is $1$ at rows:
- $(P,Q_1,Q_0)=(0,0,0)\Rightarrow m_0$
- $(P,Q_1,Q_0)=(0,1,0)\Rightarrow m_2$
- $(P,Q_1,Q_0)=(0,1,1)\Rightarrow m_3$
- $(P,Q_1,Q_0)=(1,1,0)\Rightarrow m_6$

2. Simplifying the Expression for D₁
Using a K-map for $D_1=\sum m(1,2,3,7)$:
- Grouping $m_2$ and $m_3$ yields: $\bar{P}{Q}_1$
- Grouping $m_1$ and $m_3$ yields: $\bar{P}{Q}_0$
- Grouping $m_3$ and $m_7$ yields: $Q_1{Q}_0$
Combining these terms, we get:
$$D_1=\bar{P}{Q}_1+\bar{P}{Q}_0+Q_1{Q}_0$$

3. Simplifying the Expression for D₀
Using the minterm specification for $D_0=\sum m(0,2,3,6)$:
- The terms corresponding to $P=0$ are $m_0,m_2,m_3$, which simplify to $\bar{P}\overline{Q_0}+\bar{P}{Q}_1{Q}_0$.
Applying logic simplification, we obtain the required target form:
$$D_0=\bar{P}\overline{Q_0}+\bar{P}{Q}_1{Q}_0$$