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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 '1' state
3. The one having output remains stable in '0' state
4. has a single pulse of three times propagation delay
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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.
by Boss (23.8k points)
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why option D is wrong?
+3
D is not correct because input and output are not same ..... delay should have been ok if output would have been same with input. 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.

ago by Active (2.1k points)