Question 9321 – GATE 2006

Duty cycle is the period of time where the signal high, i.e. 1.

50% of duty cycle means, the wave is 1 for half of the time and 0 for the other half of the time. It is a usual digital signal with 1 and 0.

The waveform f changes for every negative edge, that means f value alters from 1 to 0 or 0 to 1 for every negative edge of the clock.
Now the problem is that we need to find the circuit which produces a phase shift of 180, which means the output is 0 when f is 1 and output is 1 when f is 0.

Like the below image.

Now to find the answer we can choose elimination method.

F changes for negative edge, so that output too should change at negative edge. i.e if f becomes 0, then at the same time output should become 1, vice versa.
So, whenever input changes, at the same point of time output too should change. As input changes on negative edge, the output should be changed at negative edge only.
To have the above behaviour, the second D flip-flop which produces the final output should be negative edge triggered. because whatever the 2nd flip-flop produces, that is the output of the complete circuit.
So, we can eliminate option a, d.

Now either b or c can be answer.
How the flip-flop chain works in option b and c is as below.
—> F changes at negative edge.
—> But flip-flop1 responds at next positive edge.
—> After this flip-flop2 responds at next negative edge.

That means flip-flop2 produces the same input which is given to flip-flop now after a positive edge and a negative edge, that means a delay of one clock cycle, which is 180 degrees phase shift for the waveform of f.

Option b) we are giving f’, so that the output is f’ with 180 degrees phase shift.
Option c) we are giving f, so that the output is f with 180 degrees phase shift.

Hence option C is the answer.