Give step-by-step solution with explanation and final answer:4.4.2 Highly mismatched lines in digital products can cause what appears to be
“ringing” on the signal output from the line. This is often

Question

Give step-by-step solution with explanation and final answer:4.4.2 Highly mismatched lines in digital products can cause what appears to be
“ringing” on the signal output from the line. This is often referred to as
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FIGURE P4.4.2.

“overshoot” or “undershoot” and can cause digital logic errors. To simulate
this we will investigate the problem shown in Fig. P4.4.2 Two CMOS gates
are connected by a transmission line as shown. A 5-V step function voltage
of the first gate is applied. Sketch the output voltage of the line (the input
voltage to the load CMOS gate) for 0

VIPSolutions · Accepted Answer

Let's break down the transmission line problem step by step:

### **Given:**
- **Source Voltage $ V_s(t) = 5u(t) $ V** (step input)
- **Source Resistance $ R_s = \frac{1}{5}Z_0 $**
- **Load Resistance $ R_L = 5Z_0 $**
- **Transmission line with characteristic impedance $ Z_0 $ and delay $ T_D = T $**
- **We are to find $ V_L(t) $ at the load for $ 0 < t < 9T $.**

---

## **Step 1: Initial Voltage at t = 0+ (before any reflection returns)**

### **Initial Step:**
The first voltage step at the input to the line is set by the voltage divider:

$$
V_{\text{step}} = V_s \cdot \frac{Z_0}{R_s + Z_0}
$$
$$
= 5 \cdot \frac{Z_0}{\frac{1}{5}Z_0 + Z_0}
= 5 \cdot \frac{Z_0}{\frac{6}{5}Z_0}
= 5 \cdot \frac{5}{6}
= \frac{25}{6} \approx 4.167 \text{ V}
$$

But **this is the initial step launched onto the line.**

---

## **Step 2: Reflection Coefficients**

- **At the load (right end):**
  $$
  \Gamma_L = \frac{R_L - Z_0}{R_L + Z_0} = \frac{5Z_0 - Z_0}{5Z_0 + Z_0} = \frac{4}{6} = \frac{2}{3}
  $$

- **At the source (left end):**
  $$
  \Gamma_S = \frac{R_s - Z_0}{R_s + Z_0} = \frac{\frac{1}{5}Z_0 - Z_0}{\frac{1}{5}Z_0 + Z_0} = \frac{-\frac{4}{5}Z_0}{\frac{6}{5}Z_0} = -\frac{4}{6} = -\frac{2}{3}
  $$

---

## **Step 3: Step-by-Step Voltages at the Load**

### **Time Intervals:**
- $ t = 0 $ to $ T $: First step travels, no voltage at load yet.
- $ t = T $: First step arrives at the load.

### **Let's tabulate the voltages:**

#### ### **First Step Arrives at Load ($ t = T $)**

- The initial step at the load: **0 V** (since nothing has arrived yet).

---

#### ### **Second Step ($ t = T $ to $ 3T $)**

- At $ t = T $, the first voltage step arrives at the load:
  - Value: **$ \frac{25}{6} = 4.167 $ V**

- The load sees a reflection:
  - Additional voltage: $ \frac{2}{3} $ of the incident wave is reflected and adds to the voltage at the load.

- The reflected wave travels back to the source, gets reflected with $ -\frac{2}{3} $, travels back to the load, and so on.

Let's keep track of each "bounce":

#### **Step 1:**
- **Incident at load:** $ V_1 = 4.167 $ V at $ t = T $

#### **Step 2:**
- **Reflection from load:** $ V_2 = V_1 \cdot \Gamma_L = 4.167 \times \left(\frac{2}{3}\right) = 2.778 $ V
  - This travels back to the source, arrives at $ t = 2T $

#### **Step 3:**
- **Reflection from source:** $ V_3 = V_2 \cdot \Gamma_S = 2.778 \times \left(-\frac{2}{3}\right) = -1.852 $ V
  - This travels back to the load, arrives at $ t = 3T $

#### **Step 4:**
- **Reflection from load:** $ V_4 = V_3 \cdot \Gamma_L = -1.852 \times \frac{2}{3} = -1.234 $ V
  - Returns at $ t = 4T $

#### **Step 5:**
- **Reflection from source:** $ V_5 = V_4 \cdot \Gamma_S = -1.234 \times (-\frac{2}{3}) = +0.823 $ V
  - Returns at $ t = 5T $

#### **Step 6:** 
- **Reflection from load:** $ V_6 = 0.823 \times \frac{2}{3} = 0.549 $ V
  - Returns at $ t = 6T $

#### **Step 7:** 
- **Reflection from source:** $ V_7 = 0.549 \times (-\frac{2}{3}) = -0.366 $ V
  - Returns at $ t = 7T $

#### **Step 8:** 
- **Reflection from load:** $ V_8 = -0.366 \times \frac{2}{3} = -0.244 $ V
  - Returns at $ t = 8T $

#### **Step 9:** 
- **Reflection from source:** $ V_9 = -0.244 \times (-\frac{2}{3}) = 0.163 $ V
  - Returns at $ t = 9T $

---

## **Step 4: Cumulative Sum at Each Step**

| Interval         | Time          | Voltage at Load ($ V_L $) |
|------------------|--------------|-----------------------------|
| $ 0 < t < T $  | $ 0 < t < T $  | 0 V                        |
| $ T < t < 3T $ | $ T < t < 3T $ | $ 4.167 $ V              |
| $ 3T < t < 5T $| $ 3T < t < 5T $| $ 4.167 + (-1.852) = 2.315 $ V |
| $ 5T < t < 7T $| $ 5T < t < 7T $| $ 2.315 + 0.823 = 3.138 $ V    |
| $ 7T < t < 9T $| $ 7T < t < 9T $| $ 3.138 + (-0.366) = 2.772 $ V |
| Steady state     | $ t \to \infty $ | 5 V (as given)             |

But the problem gives specific voltages for the intervals:
- $ 0 < t < T $: 0 V
- $ T < t < 3T $: 6.944 V
- $ 3T < t < 5T $: 3.858 V
- $ 5T < t < 7T $: 5.23 V
- $ 7T < t < 9T $: 4.62 V
- Steady state: 5 V

Let's see why these differ—**they're using the cumulative total voltage at the load (including all previous bounces up to that point).**

## **Let's Recalculate the Cumulative Sums:**

1. **At $ t = T $:**
   $$
   V_{L,1} = V_1 = 4.167 \text{ V}
   $$

2. **At $ t = 3T $:**
   $$
   V_{L,2} = V_1 + V_3 = 4.167 + (-1.852) = 2.315 \text{ V}
   $$

3. **At $ t = 5T $:**
   $$
   V_{L,3} = V_1 + V_3 + V_5 = 2.315 + 0.823 = 3.138 \text{ V}
   $$

4. **At $ t = 7T $:**
   $$
   V_{L,4} = V_1 + V_3 + V_5 + V_7 = 3.138 + (-0.366) = 2.772 \text{ V}
   $$

5. **At $ t = 9T $:**
   $$
   V_{L,5} = V_1 + V_3 + V_5 + V_7 + V_9 = 2.772 + 0.163 = 2.935 \text{ V}
   $$

But the **given answers are different**, which means they are summing up differently. Let's try the alternate approach based on the problem's hint:

## **Direct Calculation Using Infinite S

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