VIPSolutions2) (30 pts) Design a synchronous counter that operates according to the state diagram given below. Your design should involve at least 1 D and 1 JK flip flop and minimum number of components. Show all the steps clearly. ``` 110 | v 001 | v 010 | v 101 | v 111 | v 000 ```
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2) (30 pts) Design a synchronous counter that operates according to the state diagram given below. Your design should involve at least 1 D and 1 JK flip flop and minimum number of components. Show all the steps clearly. ``` 110 | v 001 | v 010 | v 101 | v 111 | v 000 ```
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Asked by: Ruhail Khan
Created at: 2025-07-02 07:33:31
Asked by: Ruhail Khan
Created at: 2025-07-02 07:33:31
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Let's break down the design step by step as required:
---
### **1. List the States and Assign Binary Values**
From the state diagram:
| Present State | Next State |
|:-------------:|:---------:|
| 110 | 001 |
| 001 | 010 |
| 010 | 101 |
| 101 | 111 |
| 111 | 000 |
| 000 | 110 |
Let’s denote the flip-flop outputs as \( Q_2, Q_1, Q_0 \) (from MSB to LSB).
---
### **2. Table of State Transitions**
| Present State | \( Q_2 Q_1 Q_0 \) | Next State | \( Q_2^+ Q_1^+ Q_0^+ \) |
|:-------------:|:----------------:|:----------:|:-----------------------:|
| 110 | 1 1 0 | 001 | 0 0 1 |
| 001 | 0 0 1 | 010 | 0 1 0 |
| 010 | 0 1 0 | 101 | 1 0 1 |
| 101 | 1 0 1 | 111 | 1 1 1 |
| 111 | 1 1 1 | 000 | 0 0 0 |
| 000 | 0 0 0 | 110 | 1 1 0 |
There are 6 states. The other 2 (011, 100) can be considered as don't-cares.
---
### **3. Choose Flip-Flops**
As per the question, use at least **one D and one JK flip-flop**.
Let’s assign:
- Q2: **JK**
- Q1: **D**
- Q0: **JK**
---
### **4. Excitation Tables**
#### **For JK flip-flop (Q2 and Q0):**
| Current Q | Next Q+ | J | K |
|-----------|---------|---|---|
| 0 | 0 | 0 | X |
| 0 | 1 | 1 | X |
| 1 | 0 | X | 1 |
| 1 | 1 | X | 0 |
#### **For D flip-flop (Q1):**
| Current Q | Next Q+ | D |
|-----------|---------|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
---
### **5. Build the Flip-Flop Input Tables**
| Present State | Next State | Q2 | Q1 | Q0 | Q2+ | Q1+ | Q0+ |
|:-------------:|:----------:|:--:|:--:|:--:|:---:|:---:|:---:|
| 110 | 001 | 1 | 1 | 0 | 0 | 0 | 1 |
| 001 | 010 | 0 | 0 | 1 | 0 | 1 | 0 |
| 010 | 101 | 0 | 1 | 0 | 1 | 0 | 1 |
| 101 | 111 | 1 | 0 | 1 | 1 | 1 | 1 |
| 111 | 000 | 1 | 1 | 1 | 0 | 0 | 0 |
| 000 | 110 | 0 | 0 | 0 | 1 | 1 | 0 |
Let’s fill in the required inputs for each flip-flop.
---
#### **JK for Q2**
| Q2 | Q2+ | J2 | K2 |
|----|-----|----|----|
| 1 | 0 | X | 1 |
| 0 | 0 | 0 | X |
| 0 | 1 | 1 | X |
| 1 | 1 | X | 0 |
| 1 | 0 | X | 1 |
| 0 | 1 | 1 | X |
---
#### **D for Q1**
D1 = Q1+ (simply the next value for Q1):
| Q1 | Q1+ | D1 |
|----|-----|----|
| 1 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 1 |
---
#### **JK for Q0**
| Q0 | Q0+ | J0 | K0 |
|----|-----|----|----|
| 0 | 1 | 1 | X |
| 1 | 0 | X | 1 |
| 0 | 1 | 1 | X |
| 1 | 1 | X | 0 |
| 1 | 0 | X | 1 |
| 0 | 0 | 0 | X |
---
### **6. Karnaugh Maps for Flip-Flop Inputs**
Let’s assign the present state as inputs: \( Q_2, Q_1, Q_0 \).
#### **a) D1 (Q1's D input)**
From the table:
| Q2 | Q1 | Q0 | Q1+ |
|----|----|----|-----|
| 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 |
So, D1 = Q1+.
Let’s write out the minterms where D1 = 1:
- (0,0,1): Q2' Q1' Q0
- (1,0,1): Q2 Q1' Q0
- (0,0,0): Q2' Q1' Q0'
- (Others are 0)
Thus,
D1 = Q1' (Q2' Q0 + Q2' Q0' + Q2 Q0)
= Q1' (Q2' (Q0 + Q0') + Q2 Q0)
= Q1' (Q2' *1 + Q2 Q0)
= Q1' (Q2' + Q2 Q0)
**So:**
\[
\boxed{D1 = Q1' (Q2' + Q2 Q0)}
\]
---
#### **b) J2 and K2 (JK for Q2)**
From above:
| Q2 | Q2+ | J2 | K2 |
|----|-----|----|----|
| 1 | 0 | X | 1 |
| 0 | 0 | 0 | X |
| 0 | 1 | 1 | X |
| 1 | 1 | X | 0 |
| 1 | 0 | X | 1 |
| 0 | 1 | 1 | X |
So, collect where J2 = 1:
- (0,1,0): Q2=0, Q2+=1: J2=1 (row 3)
- (0,0,0): Q2=0, Q2+=1: J2=1 (row 6)
So, for present state (Q2 Q1 Q0):
Rows where J2=1:
- (0,1,0): Q2' Q1 Q0'
- (0,0,0): Q2' Q1' Q0'
So,
J2 = Q2' Q0'
For K2:
Rows where K2=1:
- (1,1,0): Q2=1, Q2+=0: K2=1 (row 1)
- (1,1,1): Q2=1, Q2+=0: K2=1 (row 5)
So, (1,1,0): Q2 Q1 Q0', (1,1,1): Q2 Q1 Q0
K2 = Q2 Q1
---
#### **c) J0 and K0 (JK for Q0)**
From above:
Rows where J0=1:
- (1,1,0): Q0=0, Q0+=1: J0=1 (row 1)
- (0,1,0): Q0=0, Q0+=1: J0=1 (row 3)
So, (1,1,0): Q2 Q1 Q0', (0,1,0): Q2' Q1 Q0'
So,
J0 = Q1 Q0'
Rows where K0=1:
- (0,0,1): Q0=1, Q0+=0: K0=1 (row 2)
- (1,1,1): Q0=1, Q0+=0: K0=1 (row 5)
So, (0,0,1): Q2' Q1' Q0, (1,1,1): Q2 Q1 Q0
K0 = Q0 (Q2' Q1' + Q2 Q1)
---
### **7. Final Circuit Equations**
- **Q2 (JK):**
- \( J_2 = Q2' Q0' \)
- \( K_2 = Q2 Q1 \)
- **Q1 (D):**
- \( D_1 = Q1' (Q2' + Q2 Q0) \)
- **Q0 (JK):**
- \( J_0 = Q1 Q0' \)
- \( K_0 = Q0 (Q2' Q1' + Q2 Q1) \)
---
### **8. Circuit Diagram**
- **Q2:** JK flip-flop, inputs as above.
- **Q1:** D flip-flop, input as above.
- **Q0:** JK flip-flop, inputs as above.
- All flip-flops are clocked **synchronously**.
---
### **9. Handling Unused States**
States 011 and 100 are not used. You can design the counter to reset to 000 if these states are entered (optional for robustness).
---
## **Summary Table**
| Flip-Flop | Type | Input(s) | Equation |
|-----------|------|-------------------|--------------------------------------|
| Q2 | JK | J2, K2 | \( J_2 = Q2' Q0' \), \( K_2 = Q2 Q1 \) |
| Q1 | D | D1 | \( D_1 = Q1' (Q2' + Q2 Q0) \) |
| Q0 | JK | J0, K0 | \( J_0 = Q1 Q0' \), \( K_0 = Q0 (Q2' Q1' + Q2 Q1) \) |
---
## **10. State Diagram (for reference)**
```
110 --> 001 --> 010 --> 101 --> 111 --> 000 --> 110
```
---
**Show all these steps in your solution, draw the circuit using 1 D and 2 JK flip-flops as above, and connect the logic as per the equations. This is the minimum-component solution as required.**
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