Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 24 (Q.No. 2)
2.
In the given figure assume that initially Q = 1 with Clock Pulses being given, the subsequent states of Q will be
Discussion:
11 comments Page 1 of 2.
Arnab Bhattacharya said:
8 years ago
If we assume that J is tied to +Vcc (logic 1) (assumption from the symbol), then K=1 makes the flipflop toggle thus making it Q=0. Next J=1, K=0 (since Q=0). So again Q becomes 1.
This thing repeats and the output should be:
0,1,0,1...
I think 'D' is the right option.
This thing repeats and the output should be:
0,1,0,1...
I think 'D' is the right option.
(2)
Arnab Bhattacharya said:
8 years ago
If we assume, J is tied to Gnd/-ve (i.e. logic 0), then initial Q=1, makes K=1 thus resetting the flipflop making it output Q=0. Thus at next clock pulse Q=0, K=0, J=0 will hold the output making new output (Q=0). This will continue and the output will be:
0,0,0,0,0........
And Q' (inverted) will be:
1,1,1,1,1.......
So, if J=0, The correct option will be 'B' I suppose.
0,0,0,0,0........
And Q' (inverted) will be:
1,1,1,1,1.......
So, if J=0, The correct option will be 'B' I suppose.
(1)
Prana said:
1 decade ago
How please explain?
LOKNATH REDDY D said:
1 decade ago
J=? not given and for J=0 or 1 C is wrong answer.
Guru said:
10 years ago
Qn+1 = JQ'+K'Q.
Now put k = Q.
Qn+1 = JQ'+1 = 1;
Now put k = Q.
Qn+1 = JQ'+1 = 1;
Amar.B said:
10 years ago
If we assume j=0 and for the given condition answer should be 'B'?
K suman said:
10 years ago
None of the answer. We can't estimate the next because of,
Qn+1 = JQ'+K'Q at clock 1.
Q = 1, K = Q then Qn+1 = J(1)'+Q'Q == 0.
At clock 2.
Q = 0, K = 0 then from character equation.
Output Qn+1 = J(0)'+(0)'0 == J so we can't estimate the next state.
Qn+1 = JQ'+K'Q at clock 1.
Q = 1, K = Q then Qn+1 = J(1)'+Q'Q == 0.
At clock 2.
Q = 0, K = 0 then from character equation.
Output Qn+1 = J(0)'+(0)'0 == J so we can't estimate the next state.
K suman said:
10 years ago
Qn+1 = JQ'+K'Q.
So observation Q = 1, K = Q substitute in that Qn+1 = J1'+Q'Q implies that Qn+1 = 0 again same repeats above Qn+1 = J. So answer not match.
So observation Q = 1, K = Q substitute in that Qn+1 = J1'+Q'Q implies that Qn+1 = 0 again same repeats above Qn+1 = J. So answer not match.
Bhanu said:
8 years ago
I think the Answer is D.
D Venkat said:
3 years ago
I think it should be option D.
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