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
1, 0, 1, 0, 1......
0, 0, 0, 0, 0......
1, 1, 1, 1, 1......
0, 1, 0, 1, 0......
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Srishti said:   2 weeks ago
The figure shows a JK flip-flop where the K input is directly connected to Q (the current output), and the J input is connected to a constant logic high (1).

Initially, Q = 1.
On each clock pulse, the state transitions depend on the values of J and K.

State Analysis:
J = 1 at all times.
K = Q (feedback connection).

For Q = 1:
J = 1, K = 1 (since Q = 1)
A JK flip-flop with J = 1, K = 1 will toggle its output.

So, after the first clock pulse, Q becomes 0.
For Q = 0:
J = 1, K = 0 (since K = Q = 0)
JK flip-flop with J = 1, K = 0 sets Q to 1 on the clock edge.
Q goes to 1 again.

This pattern repeats with each clock pulse.

Sequence of Q:
The output Q will alternate between 1 and 0 on every clock pulse:
1, 0, 1, 0, 1, 0, ...

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.
(1)

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.
(2)

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.

Vishnu said:   4 months ago
Final Result:

The output Q toggles between 1 and 0 on every clock pulse.
Hence, the subsequent states of Q will be:
1 → 0 → 1 → 0 → 1 → 0 → ...
This forms a T-flip flop (toggle behaviour) using a JK flip-flop where:
J = Q̅
K = Q.

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.

Amar.B said:   10 years ago
If we assume j=0 and for the given condition answer should be 'B'?

Guru said:   10 years ago
Qn+1 = JQ'+K'Q.

Now put k = Q.

Qn+1 = JQ'+1 = 1;

LOKNATH REDDY D said:   1 decade ago
J=? not given and for J=0 or 1 C is wrong answer.

D Venkat said:   3 years ago
I think it should be option D.


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