Exam Question
A JK flip-flop has two inputs J1, J2 and K1, K2 combined through AND gates as follows:
J = J1 · J2, K = K1 · K2. The flip-flop is initially at Q = 0.
The input sequences applied (rightmost bit first) are:
| Step | J1 | J2 | K1 | K2 |
|---|---|---|---|---|
| 1 | 0 | 1 | 0 | 1 |
| 2 | 1 | 0 | 1 | 1 |
| 3 | 1 | 0 | 1 | 0 |
| 4 | 0 | 1 | 0 | 1 |
| 5 | 1 | 1 | 1 | 1 |
| 6 | 1 | 0 | 0 | 0 |
| 7 | 0 | 1 | 0 | 1 |
| 8 | 1 | 1 | 1 | 1 |
Step 1: Compute J and K using AND gates
| Step | J = J1·J2 | K = K1·K2 |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 0 | 1 |
| 3 | 0 | 0 |
| 4 | 0 | 0 |
| 5 | 1 | 1 |
| 6 | 0 | 0 |
| 7 | 0 | 0 |
| 8 | 1 | 1 |
Step 2: Apply JK flip-flop rules
JK Flip-Flop rules:
| J | K | Q(next) |
|---|---|---|
| 0 | 0 | No change |
| 0 | 1 | Reset → 0 |
| 1 | 0 | Set → 1 |
| 1 | 1 | Toggle → Q̅ |
Step 3: Compute Q & Q̅ step by step
| Step | J | K | Q(next) | Q̅ |
|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 0 | 0 | 1 |
| 4 | 0 | 0 | 0 | 1 |
| 5 | 1 | 1 | 1 | 0 |
| 6 | 0 | 0 | 1 | 0 |
| 7 | 0 | 0 | 1 | 0 |
| 8 | 1 | 1 | 0 | 1 |
Step 4: Serial output (Q during clock)
Take Q before each clock pulse, rightmost first:
| Step | Q(output) |
|---|---|
| 1 | 0 |
| 2 | 0 |
| 3 | 0 |
| 4 | 0 |
| 5 | 0 |
| 6 | 1 |
| 7 | 1 |
| 8 | 1 |
Step 5: Rightmost first applied → final output
Final Output: 01110000
Waveform Diagram
Clock: ┌─┐ ┌─┐ ┌─┐ ┌─┐ ┌─┐ ┌─┐ ┌─┐ ┌─┐
└─┘ └─┘ └─┘ └─┘ └─┘ └─┘ └─┘ └─┘
1 2 3 4 5 6 7 8
Q : ────────╭─┐───────╭─┐───────
0 0 0 0 0 1 1 1 0
Q̅ : ─╮───────┐╮───────┐╮───────
1 1 1 1 1 0 0 0 1