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+13 votes

Consider the circuit in the diagram. The $\oplus$ operator represents Ex-OR. The D flip-flops are initialized to zeroes (cleared).

The following data: 100110000 is supplied to the “data” terminal in nine clock cycles. After that the values of $q_{2}q_{1}q_{0}$ are:

- 000
- 001
- 010
- 101

+20 votes

Best answer

Please tell the approach used to calculate this... Q0 can be calculated easily since it depends on D0 only which is given initially. How to calculate Q2 and Q1.. they depend on D2 and D1 which in turn depend on old values od Q1 and Q2 which are not available initially

in the XOR operation why is Q0 taking the old value?

new value of Q0 is already available by the time XOR is calculated?

new value of Q0 is already available by the time XOR is calculated?

@Prerna+Chauhan That's what we do in synchronous circuits, when a clock is applied output by the application of this clock cycle should be present at each $FF$, which is then used as input in next clock.

yes i totally agree, but when the first clock cycle is applied the output of Q0 is 1. so, when we have this new output why is older one used?

Initially, $Q_{2}Q_{1}Q_{0} = 000$ (reset), and when first clock is applied $D_{0} = data = 1$, $Q_{0n} = 1$, $Q_{1n} = Q_{0}$xor$Q_{2} = 0$xor$0 = 0$ and $Q_{2n} = Q{1} = 0$

0 votes

The question is simple, but just because of delays of XOR gate q1 & q2 are working old datas of q0.. At a particular instance, value present at q0 is from data but q1 is getting from XOR gate, so the value calculated from previous bit. Thats it.

Just follow the data given to q0,

q0 value is the data bit.

q1,q2 will be evaluated from previous state.

C is the ans.

Just follow the data given to q0,

q0 value is the data bit.

q1,q2 will be evaluated from previous state.

C is the ans.

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