Question

The following Flip Flops JK fix implements a binary counter; assuming that at time t1, all outputs Q are ZERO, it indicates the value of Q2, Q1 and Q0 at time t4.

Q2 = LOW . . Q1 = HIGH . QO = LOW 1 J Q2 J Q1 J QO CLK CLK CLK к Q2 K Q K Q. *All PRE and CLR are HIGH t1 Input clock pulses

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Answer #1

Here Q0 flip-flop is connected directly to clock. Also it's negative edge triggered flip-flop. Thus it will change on falling edge of clock.

Q1 changes when Q0 changes from 1 to 0 as it's clock is connected to Q0.

Q2 will change when Q1 changes from 1 to 0 as it's clock is connected to Q1.

Also all flip-flop will only toggle as it's input J=K=1.

Let's see with diagram

ta ta tu clock . | & a 8. Qz o @ t= 4, & a low Q, High Q. High

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