00:02
Hello students, here the circuit given we have to write the output of this flip -flop circuit given that q is defined as t plus q and q of t plus 1 is defined as t plus q of t.
00:25
So, here in the circuit we can consider that t naught, t1 and t2 are equal to 1.
00:32
In the first circuit the output that is for q naught of t plus 1 from this equation we can write in the place of t, t naught plus q of t plus q naught of t.
00:49
Here we have taken t naught is equal to 1 then 1 plus q naught 1 plus q naught will become q naught bar.
01:00
So, this is the output of first circuit then it is given to next that is q1 of t plus 1 again we can write in the place of t, t1 and in the place of q, q1 of t.
01:19
Then again t1 is equal to 1, q1 of t, q1 output is q1 bar.
01:29
Then at last we need to write just q2 which is equal to t plus q hence in the place of t, t2 in the place of q, q2 of t.
01:40
From this we get 1 plus q2 which is equal to q2 bar.
01:45
So, these are the output of flip -flop.
01:50
Further the timing diagram is given like this.
01:54
Here the lower state shows 0, upper state shows 1...