GATE CSE 2006 | Question: 8

Question

You are given a free running clock with a duty cycle of $50\%$ and a digital waveform $f$ which changes only at the negative edge of the clock. Which one of the following circuits (using clocked D flip-flops) will delay the phase of $f$ by $180°$?

• A.
• B.
• C.
• D.

Comments

@chottu isn't the phase difference for option b, 90 deg ?

---

i think in explanation with waveforms a is wrongly solved......and phase shift in a is 90 not 270

---

YES, @vrajdobariya Even I think there is a problem in option B too.

 

Answer 1

Answer. C

Key lies in noting that there is a 90° phase lag -

1. between f and output of FF1 if it is positive edge triggered, since f changes for every negative edge.

2. between the two FFs as one is positive edge triggered and the other is negative edge triggered.

So,

A. FF1 output = ((f')') = f     and    FF2 output = f$\angle$ 90°

B. FF1 output  = f'$\angle$ 90°     and    FF2 output = f'$\angle$ 180° = f  (since f is a periodic digital signal with 50% duty cycle)

C. FF1 output  = f$\angle$ 90°      and   FF2 output = f$\angle$ 180°

D. FF1 output = f'                 and    FF2 output   = f'$\angle$ 90°

 

This explains the timing diagrams and @Arjun sir's answer.