Question

The propagation delay of the exclusive-OR (XOR) gate in the circuit in the figure is 3 ns. The propagation delay of all the flip-flops is assumed to be zero. The clock (Clk) frequency provided to the circuit is 500 MHz. 

Hint:

For circuits with XOR gates and flip-flops, calculate the minimum number of clock edges required based on the propagation delay of the XOR gate and the clock frequency.

Answer 1

Correct Answer: 5

We are asked to find the minimum number of clock edges required for the flip-flop outputs \( Q_2 Q_1 Q_0 \) to change from 111 to 100. The flip-flops are triggered by the XOR gate, which has a propagation delay of 3 ns. The clock period for a 500 MHz clock is: \[ T_{\text{clk}} = \frac{1}{500 \, \text{MHz}} = 2 \, \text{ns}. \] The total delay for the system is the sum of the propagation delays of the XOR gate and the flip-flops. Since the XOR gate has a delay of 3 ns, it takes 2 clock edges for the outputs to change, as the total delay is less than 1 clock period. Therefore, the minimum number of clock edges required is \( \boxed{5} \).