The outputs Q and Q̅ of the master-slave SR flip-flop are connected to its R and S inputs respectively. The output Q when clock pulses are applied will be:
The outputs Q and Q̅ of the master-slave SR flip-flop are connected to its R and S inputs respectively. The output Q when clock pulses are applied will be: Correct Answer complementing with every clock pulse
CHECK
Explanation:
The general structure of master slave SR flipflop is shown below:
[ alt="F6 Shubham 26-10-2020 Swati D7" src="//storage.googleapis.com/tb-img/production/20/10/F6_Shubham_26-10-2020_Swati_D7.png" style="width: 558px; height: 193px;">
Qm, Q̅m : Output of master flipflop
Qs, Q̅s : Output of slave flip flop
Analysis:
When CLK = 1 the only master is active and the slave is inactive.
When CLK = 0, the slave is active and the master is inactive.
Given that output of the master flipflop Qm Q̅m are connected to R and S respectively, i.e.
[ alt="F6 Shubham 26-10-2020 Swati D8" src="//storage.googleapis.com/tb-img/production/20/10/F6_Shubham_26-10-2020_Swati_D8.png" style="width: 339px; height: 174px;">
Truth Table of SR flip slop will be:
|
S R |
Q(n + 1) |
|
0 0 |
Q(n) |
|
0 1 |
0 |
|
1 0 |
1 |
|
1 1 |
Xx (Indeterminent) |
Characteristic table is:
|
S |
R |
Q(n) |
Q(n + 1) |
|
|
0 |
0 |
0 1 |
0 10 |
Q(n) |
|
0 |
1 |
0 1 |
0 0 |
Reset |
|
1 |
0 |
0 1 |
1 1 |
Set |
|
1 |
1 |
0 1 |
X x |
Indeterminate |
Let the initial value of Q be ‘0’ then Q̅ = 1
|
|
Qm |
Q = Q̅m |
R = Qm |
Q(m + 1) |
|
Initial |
0 |
- |
- |
- |
|
CLK 1 |
|
s = 1 |
R = 0 |
1 |
|
CLK 2 |
1 |
s = 0 |
R = 1 |
0 |
|
CLK 3 |
0 |
s = 1 |
R = 0 |
1 |
|
CLK 4 |
1 |
s = 0 |
R = 1 |
0 |
|
CLK 5 |
0 |
s = 1 |
R = 0 |
1 |
|
CLK 6 |
1 |
s = 0 |
R = 1 |
0 |
This will be repeated for further operation.
Conclusion:
The output (Qm) when the clock applied will be like 1 → 0 → 1 → 0 → 1 → 0 - - -
Complementing with every clock pulse.