IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VER

Question

IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentencesFigure 5 shows the circuit of a register.
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Figure 5

(a) Initially, M=0, Do=0, Dy; =0,D>=1, D3 =0 and the switch is at position 1. Then
M changes from logic 0 to logic 1. The switch SW is kept at position 1 for the first
two clock cycles after M changes to 1. On the falling edge of the second clock cycle,
SW is switched to position 2 (connected) while M remains at logic 1. Determine the
sequence of output bits generated at So for the first eight clock cycles when M = 1.
(8 Marks)
(b) Each two-input AND gate has a minimum propagation delay of 2 ns and a
maximum propagation delay of 7 ns, and each two-input OR gate has a minimum
propagation delay of 3 ns and a maximum propagation delay of 8 ns. Each D flip-
flop has a setup time of 9 ns and a hold time of 6 ns. The minimum an

VIPSolutions · Accepted Answer

Let's solve each part step by step as requested:

---

### **Step 1: Introduction to the Concept**

This question is about a **register circuit** using D flip-flops, AND/OR gates, and a multiplexer structure. The circuit stores and shifts data, controlled by a switch (SW) and a mode signal (M). The timing and output depend on the state of the inputs and the clock (CLK). Understanding propagation delays and timing margins is key for reliable circuit operation.

---

### **Step 2: Solving Part (a): Output Sequence at So**

#### **Given Initial Conditions:**
- $ M = 0 $
- $ D_0 = 0, D_1 = 0, D_2 = 1, D_3 = 0 $
- SW at position 1
- $ M $ switches to 1, SW stays at 1 for 2 clock cycles, then moves to 2.

#### **Step-by-step explanation:**

1. **When $ M = 0 $:** The circuit loads the data $ D_0, D_1, D_2, D_3 $ into the flip-flops.
2. **At $ M = 1 $, SW at position 1:** The register operates as a shift register, shifting the bits right each clock cycle.
3. **After two cycles, SW moves to position 2:** Now, the output is taken from $ Q_3 $.

#### **Cycle-by-cycle state evolution:**

| Cycle | SW pos | Q0 | Q1 | Q2 | Q3 | So |
|-------|--------|----|----|----|----|----|
| 0     | 1      | 0  | 0  | 1  | 0  | 0  |
| 1     | 1      | 0  | 0  | 0  | 1  | 1  |
| 2     | 1      | 0  | 0  | 0  | 0  | 0  |
| 3     | 2      | -  | -  | -  | 0  | 0  |
| 4     | 2      | -  | -  | -  | 0  | 0  |
| 5     | 2      | -  | -  | -  | 0  | 0  |
| 6     | 2      | -  | -  | -  | 0  | 0  |
| 7     | 2      | -  | -  | -  | 0  | 0  |

**Final output sequence at So:** **0, 1, 0, 0, 0, 0, 0, 0**

---

### **Step 3: Solving Part (b): Setup/Hold Time Margins and Max Clock Frequency**

#### **Given:**
- **AND gate delay:** min 2 ns, max 7 ns
- **OR gate delay:** min 3 ns, max 8 ns
- **D Flip-Flop:** setup = 9 ns, hold = 6 ns, prop delay min = 4 ns, max = 12 ns
- $ f_{CLK} = 25 $ MHz ($ T_{CLK} = 40 $ ns)

#### **Setup Time Margin Calculation:**

- **Data path max delay (worst-case):**  
  - AND (7 ns) + OR (8 ns) + DFF (12 ns) = **27 ns**
- **Setup time required:** 9 ns

- **Clock period:** 40 ns

- **Setup time margin:**  
  $ = T_{CLK} - \text{(Data path max delay + setup time)} $  
  $ = 40~\text{ns} - (27~\text{ns} + 9~\text{ns}) = 4~\text{ns} $

#### **Hold Time Margin Calculation:**

- **Data path min delay (best-case):**  
  - AND (2 ns) + OR (3 ns) + DFF (4 ns) = **9 ns**
- **Hold time required:** 6 ns

- **Hold time margin:**  
  $ = \text{(Data path min delay)} - \text{hold time} $  
  $ = 9~\text{ns} - 6~\text{ns} = 3~\text{ns} $

#### **Maximum Clock Frequency Calculation:**

- **Total time required per cycle:**  
  $ = \text{max data path delay} + \text{setup time} $  
  $ = 27~\text{ns} + 9~\text{ns} = 36~\text{ns} $

- **Max frequency:**  
  $ f_{max} = \frac{1}{36~\text{ns}} \approx 27.78~\text{MHz} $

---

### **Step 4: Solving Part (c): Maximum Clock Skew before Hold Time Violation**

- **Hold time margin:** 3 ns (from above)
- **Allowed clock skew = Hold time margin**

**Maximum tolerated clock skew:** **3 ns**

---

### **Step 5: Final Summary Answer (Few Sentences)**

- The output sequence at So for the first eight clock cycles is **0, 1, 0, 0, 0, 0, 0, 0**.
- The setup time margin is **4 ns**, hold time margin is **3 ns**, and the maximum clock frequency is approximately **27.78 MHz**.
- The circuit can tolerate up to **3 ns** of clock skew before hold time violation occurs.

---

### **Step 6: What We Did in Each Step**

1. Introduced the register circuit and its functional concept.
2. Analyzed the register's behavior and derived the output sequence for given initial conditions.
3. Calculated setup/hold time margins and determined the maximum safe clock frequency.
4. Evaluated the maximum clock skew the circuit can tolerate.
5. Summarized all answers concisely.
6. Explained the logical flow and calculation steps for full clarity.

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