Define the convolution for an LTI system 2M
Consider the signal x(t) = 𝛼
𝑡
𝑢(𝑡)
b) Calculate and sketch the signal g(𝑡) = 𝑥(𝑡) − 𝛼𝑥(𝑡 − 2). 4M
c) Use the result of part (a) in conjunction with properties of convolution in order to
determine a sequence h(t) such that 4M
x(t) ∗ h(t) = (
1
4
)
𝑡
[𝑢(𝑡 + 4) − 𝑢(𝑡 − 4)]

### Part (a) - Define the Convolution for an LTI System

The convolution of two signals \( x(t) \) and \( h(t) \) in the context of a Linear Time-Invariant (LTI) system is defined as:

\[
(y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau )
\]

Here:

- \( x(t) \): The input signal.
- \( h(t) \): The impulse response of the system.
- \( y(t) \): The output signal (response of the system).

Convolution is a fundamental operation for LTI systems because it describes how the output of the system can be determined by the input signal and the system's impulse response.

### Part (b) - Calculate and Sketch \( g(t) = x(t) - \alpha x(t - 2) \)

Given \( x(t) = \alpha t u(t) \), where \( u(t) \) is the unit step function, we need to compute \( g(t) \):

\[
g(t) = x(t) - \alpha x(t - 2)
\]

First, express \( x(t) \) and \( x(t - 2) \):

1. \( x(t) = \alpha t u(t) \)
2. \( x(t - 2) = \alpha (t - 2) u(t - 2) \)

Now, calculate \( g(t) \):

\[
g(t) = \alpha t u(t) - \alpha [t - 2] u(t - 2)
\]

This expression can be broken down into two regions:

1. **For \( t < 2 \):**
   - \( u(t) = 1 \), \( u(t - 2) = 0 \)
   - \( g(t) = \alpha t \)

2. **For \( t \geq 2 \):**
   - Both \( u(t) = 1 \) and \( u(t - 2) = 1 \)
   - \( g(t) = \alpha t - \alpha (t - 2) = 2\alpha \)

**Sketching**:
- The graph is a linear function \( \alpha t \) starting from \( t = 0 \) to \( t = 2 \).
- At \( t = 2 \), the function levels off at \( g(t) = 2\alpha \).

### Part (c) - Determine \( h(t) \) Such That \( x(t) * h(t) = \frac{1}{4^t} [u(t + 4) - u(t - 4)] \)

To find \( h(t) \), use the properties of convolution:

Given \( x(t) * h(t) = \frac{1}{4^t} [u(t + 4) - u(t - 4)] \), decompose the right-hand side:

1. The right-hand side represents a signal that starts at \( t = -4 \) and ends at \( t = 4 \).
2. A natural choice for \( h(t) \) would be a signal that, when convolved with \( x(t) \), produces this behavior.

Assume:

\[
h(t) = \frac{1}{4^t} [u(t + 4) - u(t - 4)]
\]

The convolution will give the desired signal. The range over which the function is active depends on the properties of \( x(t) \) and the scaling factor ensures the exponential decay matches.

---

### Do you need further details or have any questions?

Here are some related questions for practice:

1. How would you compute the convolution of two discrete signals?
2. What are the properties of convolution in LTI systems?
3. How does the unit step function affect the convolution of signals?
4. Can you explain the graphical interpretation of convolution?
5. What is the significance of the impulse response in an LTI system?

**Tip:** In convolution, the integral or summation accounts for the interaction of the entire signal history. Keep track of the limits!

Explore the definition of convolution in Linear Time-Invariant (LTI) systems through a detailed example. Calculate and sketch signals, and determine sequences using convolution properties. Suitable for advanced undergraduate students and engineers.

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