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
Current
State
Next
State
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
Which one of the following options corresponds to the expressions for the inputs of
the D flip-flops, 𝐷1 and 𝐷0?

• A. 𝐷1 = 𝑃 𝑄1 + 𝑃̅𝑄0 + 𝑄1𝑄0 𝐷0 = 𝑃 𝑄0 + 𝑃̅ 𝑄1 + 𝑄1 𝑄0 ̅̅̅
• B. 𝐷1 = 𝑃̅ 𝑄1 + 𝑃̅𝑄0 + 𝑄1𝑄0 𝐷0 = 𝑃̅ 𝑄0 ̅̅̅ + 𝑃 ̅𝑄1 + 𝑄1 𝑄0 ̅̅̅
• C. 𝐷1 = 𝑃̅ 𝑄1 ̅̅̅ + 𝑃 ̅𝑄0 + 𝑄1 𝑄0 𝐷0 = 𝑃̅ 𝑄0 + 𝑃̅ 𝑄1 + 𝑄1 𝑄0 ̅̅̅
• D. 𝐷1 = 𝑃 𝑄1 ̅̅̅ + 𝑃̅ 𝑄0 + 𝑄1 𝑄0 𝐷0 = 𝑃 𝑄0 ̅̅̅ + 𝑃̅ 𝑄1 + 𝑄1 𝑄0 ̅̅̅

Hint:

Since D flip-flops directly store the next state, write $D_1$ and $D_0$ as Boolean functions of $P, Q_1, Q_0$ from the next-state columns, then simplify each using a K-map.

Answer 1

The Correct Option is B

Step 1: List the next-state table. From the given table, this is a saturating up/down counter with input P, current state \(Q_1Q_0\), and next state \(Q_1^+Q_0^+\), which equal the flip-flop inputs \(D_1, D_0\) since D flip-flops are used. For P=0: 00->01, 01->10, 10->11, 11->11 (saturating up count). For P=1: 00->00, 01->00, 10->01, 11->10 (saturating down count).

Step 2: Write the minterms where \(D_1=1\). \(D_1=1\) at \((P,Q_1,Q_0)\) = 001, 010, 011, 111.

Step 3: Simplify \(D_1\). Group 010 and 011: \(\bar P Q_1\bar{Q_0} + \bar P Q_1 Q_0 = \bar P Q_1\). Group 011 and 111: \(\bar P Q_1 Q_0 + P Q_1 Q_0 = Q_1 Q_0\). Group 001 and 011: \(\bar P \bar{Q_1} Q_0 + \bar P Q_1 Q_0 = \bar P Q_0\). Together these three groups cover all four minterms, giving \(D_1 = \bar P Q_1 + \bar P Q_0 + Q_1 Q_0\).

Step 4: Write the minterms where \(D_0=1\). \(D_0=1\) at \((P,Q_1,Q_0)\) = 000, 010, 011, 110.

Step 5: Simplify \(D_0\). Group 000 and 010: \(\bar P \bar{Q_1}\bar{Q_0} + \bar P Q_1 \bar{Q_0} = \bar P \bar{Q_0}\). Group 010 and 011: \(\bar P Q_1\bar{Q_0}+\bar P Q_1 Q_0 = \bar P Q_1\). Group 010 and 110: \(\bar P Q_1 \bar{Q_0} + P Q_1\bar{Q_0} = Q_1\bar{Q_0}\). Together this gives \(D_0 = \bar P \bar{Q_0} + \bar P Q_1 + Q_1 \bar{Q_0}\).

Step 6: Verify. Substituting all 8 combinations of \((P,Q_1,Q_0)\) into both expressions exactly reproduces the given next-state table.

Final Answer: \(\boxed{D_1 = \bar P Q_1 + \bar P Q_0 + Q_1 Q_0,\ D_0 = \bar P \bar{Q_0} + \bar P Q_1 + Q_1 \bar{Q_0} \text{ (Option B)}}\)