Question

Sketch signals given by following equations : 1. x(t) = u(t) - u(t-1) 2. x(t) = δ(t+1) - δ(t-1)

Answer 1

x(t) = u(t) - u(t-1)

Answer 2

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## **Understanding the Functions**
Before sketching, you need to understand two basic functions:

### 1. **Unit Step Function \( u(t) \):**
- **What is it?**
  The unit step function "turns on" at a specific time.
  - \( u(t) = 0 \) for \( t < 0 \) (off before \( t = 0 \)).
  - \( u(t) = 1 \) for \( t \geq 0 \) (on starting from \( t = 0 \)).

- **How to shift it?**
  - \( u(t-1) \): The function "turns on" at \( t = 1 \) instead of \( t = 0 \).
  - \( u(t+2) \): The function "turns on" at \( t = -2 \).

### 2. **Impulse Function \( \delta(t) \):**
- **What is it?**
  The impulse function is like a "spike" that occurs only at a single instant of time:
  - At \( t = 0 \), it is infinitely tall (but with an area = 1).
  - Everywhere else, \( \delta(t) = 0 \).

- **How to shift it?**
  - \( \delta(t+1) \): The spike happens at \( t = -1 \).
  - \( \delta(t-2) \): The spike happens at \( t = 2 \).

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## **Equation (i): \( x(t) = u(t) - u(t-1) \)**

Let’s analyze this step by step:

### **Step 1: Understand \( u(t) \)**
- \( u(t) \) is the unit step function.
- It "turns on" at \( t = 0 \), so:
  - For \( t < 0 \): \( u(t) = 0 \).
  - For \( t \geq 0 \): \( u(t) = 1 \).

### **Step 2: Understand \( u(t-1) \)**
- \( u(t-1) \) is the unit step function shifted to the right by 1.
- It "turns on" at \( t = 1 \), so:
  - For \( t < 1 \): \( u(t-1) = 0 \).
  - For \( t \geq 1 \): \( u(t-1) = 1 \).

### **Step 3: Subtract \( u(t-1) \) from \( u(t) \)**
Now subtract the two step functions:
- For \( t < 0 \): Both \( u(t) = 0 \) and \( u(t-1) = 0 \), so \( x(t) = 0 - 0 = 0 \).
- For \( 0 \leq t < 1 \): \( u(t) = 1 \) and \( u(t-1) = 0 \), so \( x(t) = 1 - 0 = 1 \).
- For \( t \geq 1 \): Both \( u(t) = 1 \) and \( u(t-1) = 1 \), so \( x(t) = 1 - 1 = 0 \).

### **Sketch for (i):**
1. From \( t < 0 \), the signal is 0.
2. From \( t = 0 \) to \( t = 1 \), the signal is 1 (a flat line at height 1).
3. For \( t > 1 \), the signal goes back to 0.

This creates a rectangular pulse from \( t = 0 \) to \( t = 1 \).

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## **Equation (ii): \( x(t) = \delta(t+1) - \delta(t-1) \)**

Let’s analyze this step by step:

### **Step 1: Understand \( \delta(t+1) \)**
- \( \delta(t+1) \) is the impulse function shifted to \( t = -1 \).
- It creates a "spike" at \( t = -1 \).

### **Step 2: Understand \( \delta(t-1) \)**
- \( \delta(t-1) \) is the impulse function shifted to \( t = 1 \).
- It creates a "spike" at \( t = 1 \).

### **Step 3: Subtract \( \delta(t-1) \) from \( \delta(t+1) \)**
Now subtract the two impulse functions:
- At \( t = -1 \): There is a positive spike because of \( \delta(t+1) \).
- At \( t = 1 \): There is a negative spike because of \( -\delta(t-1) \).
- Everywhere else, the signal is 0 because impulse functions are 0 everywhere except at their specific locations.

### **Sketch for (ii):**
1. Draw a positive spike at \( t = -1 \).
2. Draw a negative spike at \( t = 1 \).

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## **Key Points for Sketching**
- For **step functions**, focus on when they "turn on" and how their subtraction affects the signal.
- For **impulse functions**, focus on where the spikes occur and whether they are positive or negative.

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Answer 3

x(t) = δ(t+1) - δ(t-1)