Question

In the circuit, the present value of \( Z \) is 1. Neglecting the delay in the combinational circuit, the values of \( S \) and \( Z \), respectively, after the application of the clock will be:
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• A. \( S = 0, Z = 0 \)
• B. \( S = 0, Z = 1 \)
• C. \( S = 1, Z = 0 \)
• D. \( S = 1, Z = 1 \)

Hint:

For circuits involving D flip-flops: - The output \( Z \) of the flip-flop is updated to the value of the input \( S \) on the clock edge. - Use XOR truth tables for combinational circuit calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the circuit.
The circuit consists of:
1. A combinational logic circuit with inputs \( X \), \( Y \), and \( Z \) generating \( S \) using the logic: \[ S = X \oplus Y \oplus Z, \] where \( \oplus \) denotes the XOR operation. 2. A D flip-flop where the input is \( S \), and the output \( Z \) is updated on the rising edge of the clock. Step 2: Current values of inputs.
From the problem, the initial conditions are: \[ X = 1, \, Y = 0, \, Z = 1. \] Step 3: Calculate \( S \).
Using the formula for \( S \): \[ S = X \oplus Y \oplus Z = 1 \oplus 0 \oplus 1. \] Using XOR logic: 1. \( 1 \oplus 0 = 1 \), 2. \( 1 \oplus 1 = 0 \). Thus, \( S = 0 \). Step 4: Determine the updated \( Z \).
The D flip-flop updates \( Z \) to the value of \( S \) after the clock edge. Therefore: \[ Z = S = 0. \] Step 5: Re-evaluating \( S \) after clock update.
After the clock updates \( Z \), its new value \( Z = 0 \) is used in the XOR calculation for \( S \). Substituting the updated \( Z \): \[ S = X \oplus Y \oplus Z = 1 \oplus 0 \oplus 0. \] Using XOR logic: 1. \( 1 \oplus 0 = 1 \), 2. \( 1 \oplus 0 = 1 \). Thus, \( S = 1 \). Step 6: Final values after clock application. After the clock edge: \[ S = 1, \, Z = 0. \] Conclusion: The final values are \( \mathbf{S = 1, Z = 0} \), which corresponds to option (C).