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The circuit given in the figure below is

  1. An oscillating circuit and its output is square wave
  2. The one whose output remains stable in $\text{‘1'}$ state
  3. The one having output remains stable in $\text{‘0'}$ state
  4. has a single pulse of three times propagation delay
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2 Answers

Best answer
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23 votes
A is Ans.

If we give the i/p 0 then we get O/P 1

And then it (1) feedbacks to circuit again as I/P which will generate 0.

So it keep on oscillating between 0-1-0-1....

Which is equivalent to square wave.

 

Vice versa if I/p is 1 initially. Means we get 1-0-1-0-...etc

So Ans is Option A.
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Given circuit is logically equivalent to this one. There's a feedback loop.

If x is the input, x' comes out

Then x' is the input which gives x again

Repeat forever.

Hence, Option A

Option D is incorrect because it clearly won't have a single pulse. It'll keep on having pulses (ups-and-downs)


Option D would've been correct without the feedback loop.

Answer:

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