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70 votes
70 votes

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 $180°$?

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53 votes
53 votes

Ans- C.

 

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17 votes
B and D are inverting f and hence cannot be the answer.

In A, the output is activated by CLK on the final D flip flop. So, the output will have the same phase as f.

In C, the output is activated by CLK', and since CLK is having 50% duty cycle, this should mean the output will now have a phase difference of 180 degrees.
8 votes
8 votes

Explanation: We assume the D flip-flop to be negative edge triggered.

 
In option (A), during the negative edge of the clock, first flip-flop inverts complement of ‘f’. But, the output of first flip-flop has the same phase as ‘f’. Now, we give this output as input to the second flip-flop, which is enabled by ‘clk’.

Thus, we get a double inverted output having same phase as the input. So, A is not the correct option.

In option (B) and (D), the output is inverted ‘f’. But, we want ‘f’ as the output.
So, (B) and (D) can’t be the answer.

In option (C), the first flip-flop is activated by ‘clk’. So, the output of first flip-flop has the same phase as ‘f’. But, the second flip-flop is enabled by complement of ‘clk’. Since the clock ‘clk’ has a duty cycle of 50% , we get the output having phase delay of 180 degrees.

 
Therefore, (C) is the correct answer.

4 votes
4 votes

The inverter at f, and Q' will flip the waveform.
The inverter at CLK will delay the waveform.

Option B flips waveform via an inverter at f.

Option D flips waveform via making it pass through Q'.

So, these both can't be the answer.

 

In Option A, we delay however much we want in the beginning, but the output of the first FF will supply a continuous stream of waveform to the second FF. And, the second FF is "in sync" with CLK. Since only inverted CLK can delay the waveform here, and we don't have it at the second FF, we don't get any delay.

In Option C, we don't delay at all through the first FF. But at second FF, we invert the CLK, which delays the output waveform. By how much would it be delayed? Since the duty cycle is 50%, well get a delay of 50% of 360° => 180°.

Option C.

Answer:

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