GATE EC 2014 Set 1 — Question 52

Step-by-step Derivations:

1. Analyze the original circuit:The circuit consists of two D flip-flops with outputs $Q_1$ and $Q_2$. The state of the system can be represented by $(Q_1, Q_2)$.The inputs to the flip-flops are:

• $D_1 = A \oplus S$
• $D_2 = Q_1$

The next-state equations are:

• $Q_1(t+1) = D_1 = A \oplus S$
• $Q_2(t+1) = D_2 = Q_1$

The problem states that the given state diagram is satisfied when input A is connected to $Q_1$. So, we set $A = Q_1$.The next-state equations for the original system are:

• $Q_1(t+1) = Q_1 \oplus S$ (Equation 1)
• $Q_2(t+1) = Q_1$ (Equation 2)

2. Analyze the modified circuit:The XOR gate is replaced by an XNOR gate. The circuit structure for $D_2$ remains the same.The new next-state equations are:

• $Q_1(t+1) = A \odot S$ (Equation 3)
• $Q_2(t+1) = Q_1$ (Equation 4)

We need to find a new connection for input A such that the behavior of the modified circuit is identical to the original circuit. This means the next-state equations must be the same. Equation 4 is already identical to Equation 2. We need to make Equation 3 identical to Equation 1.So, we need to find an input A such that:
$A \odot S = Q_1 \oplus S$3. Test the options:Let's test the given options to see which one satisfies the condition $A \odot S = Q_1 \oplus S$.

• Option A: Input A is connected to $\overline{Q2}$

$A = \overline{Q2}$. The equation becomes $\overline{Q2} \odot S$. This is not generally equal to $Q_1 \oplus S$.

• Option B: Input A is connected to $Q2$

$A = Q2$. The equation becomes $Q2 \odot S$. This is not generally equal to $Q_1 \oplus S$.

• Option C: Input A is connected to $\overline{Q1}$ and S is complemented

$A = \overline{Q1}$ and the input to the gate is $\overline{S}$. The equation becomes $\overline{Q1} \odot \overline{S}$.
Using the identity $X \odot Y = \overline{X \oplus Y}$, we get $\overline{\overline{Q1} \oplus \overline{S}}$.
Using the identity $\overline{X} \oplus \overline{Y} = X \oplus Y$, we get $\overline{Q1 \oplus S}$.
This is the complement of the required expression $Q_1 \oplus S$, so this option is incorrect.

• Option D: Input A is connected to $\overline{Q1}$

$A = \overline{Q1}$. The equation becomes $\overline{Q1} \odot S$.
Using the identity $X \odot Y = \overline{X \oplus Y}$, we get $\overline{\overline{Q1} \oplus S}$.
Now, let's use the identity for XOR with a complemented input: $\overline{X} \oplus Y = \overline{X \oplus Y}$.
Applying this, we get $\overline{Q1} \oplus S = \overline{Q1 \oplus S}$.
Substituting this back into our expression:
$\overline{\overline{Q1} \oplus S} = \overline{(\overline{Q1 \oplus S})} = Q_1 \oplus S$.
This is exactly the next-state equation for $Q_1$ from the original circuit (Equation 1).Since both next-state equations are preserved, the state diagram remains the same.Therefore, connecting input A to $\overline{Q1}$ preserves the state diagram.