0 votes 0 votes the given answer will be satisfied only if we considere 5 NOT gates here.. but why not to consider 6 NOT gate?? how is D answer? S Ram asked Jan 11, 2017 S Ram 370 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Mandeep Singh commented Jan 11, 2017 reply Follow Share Here feedback has been taken after 5 not gate. So we get the inverted output here ( 1 half cycle) which is again fed to 1 not gate. So after traversing till 5 not gate again, 1 complete cycle will be there. So total time = 10 * 2 = 20 ns. Note here 6 gate is just for inverting the output. So will not be considered. 2 votes 2 votes S Ram commented Jan 11, 2017 reply Follow Share what will be the answer if we use another NOT gate at the end... not inside cycle or feedback loop? 0 votes 0 votes Mandeep Singh commented Jan 11, 2017 reply Follow Share Still it will be same. Remember cycle time is only dependent on feedback. These additional gates will increase propagation delay but won't have any impact on cycle time. 1 votes 1 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes every odd no of Not gate always complement the input(1->0, 0->1) u can also think this problem as fo 6th not gate i/p is complemented after 2*5=10 ns so for 6th not gate it is in 1 for 10 ns nd in 0 for 10 ns hence time period =10+10=20 ns saurabh rai answered Jan 11, 2017 • selected Jan 11, 2017 by S Ram saurabh rai comment Share Follow See all 0 reply Please log in or register to add a comment.