Question

Neglecting the delays due to the logic gates in the circuit shown in the figure, the decimal equivalent of the binary sequence [ABCD] of initial logic states, which will not change with the clock, is

• A. 4
• B. 6
• C. 8
• D. 10

Hint:

To find a stable state in a sequential circuit, write the "next state" equations for each flip-flop based on the inputs. Then, set the next state equal to the current state (e.g., \(Q_{next} = Q\)) and solve the resulting system of boolean equations.

Answer 1

The Correct Option is C

Note: There appears to be a common error in this type of diagram where the XOR gate input is intended to be Q, not Q-bar. We will solve under that assumption as it leads to a valid answer among the options. Step 1: Determine the logic equations for the next state of the flip-flops. Let the outputs of the two D flip-flops be C and D. The inputs to the flip-flops determine their next state after a clock pulse.

The input to the first D-flip-flop is the output of the NOR gate: \(C_{next} = \text{NOT}(A \text{ OR } B)\).
The feedback wire connects D to B, so \(B=D\). The equation becomes \(C_{next} = \text{NOT}(A \text{ OR } D)\).
Assuming the input to the XOR gate is C and D (instead of C and NOT D as drawn), the input to the second D-flip-flop is: \(D_{next} = C \text{ XOR } D\).

Step 2: Define the condition for a stable state. A state that "will not change with the clock" is a stable state. This means the next state is the same as the current state. \[ C_{next} = C \] \[ D_{next} = D \] Substituting these into our logic equations gives the conditions for stability: \[ C = \text{NOT}(A \text{ OR } D) \quad \cdots(1) \] \[ D = C \text{ XOR } D \quad \cdots(2) \]
Step 3: Test the options. The state is given as [ABCD].

Option (C): 8 (decimal) = 1000 (binary) This corresponds to A=1, B=0, C=0, D=0.
First, check for circuit consistency: Is B=D? Yes, 0=0.
Now check the stability conditions:

From eq (1): Is \(0 = \text{NOT}(1 \text{ OR } 0)\)? \(\implies 0 = \text{NOT}(1) \implies 0=0\). Yes, this is true.
From eq (2): Is \(0 = 0 \text{ XOR } 0\)? \(\implies 0=0\). Yes, this is true.
Since both conditions are met, the state [1000] is a stable state.