An analog LTI system with impulse response function h(t) = u(2 − t) + u(t − 2) is 
presented with an input x(t) = t[u(t − 2) − u(t − 4)].
a) Determine and plot the system output y(t) = x(t) ∗ h(t). 3M
b) Is the system BIBO-stable? Justify your answer. 4M
c) Is the system causal? Justify. 

Let's go through each part of the problem to determine the solution.

### Given:
- Impulse response: \( h(t) = u(2 - t) + u(t - 2) \)
- Input signal: \( x(t) = t[u(t - 2) - u(t - 4)] \)

### Part (a) Determine and plot the system output \( y(t) = x(t) * h(t) \)

#### Step 1: Interpret the functions \( h(t) \) and \( x(t) \)

1. **Impulse Response \( h(t) \):**

   The function \( h(t) = u(2 - t) + u(t - 2) \) can be simplified as:

   - \( u(2 - t) \) is 1 when \( t \leq 2 \) and 0 when \( t > 2 \).
   - \( u(t - 2) \) is 0 when \( t < 2 \) and 1 when \( t \geq 2 \).

   Therefore:
   \[
   h(t) = 1 \text{ for all } t
   \]

   This implies \( h(t) \) is a unit step function with a value of 1 across all \( t \).

2. **Input Signal \( x(t) \):**

   The function \( x(t) = t[u(t - 2) - u(t - 4)] \) can be interpreted as:

   - \( u(t - 2) \) is 0 when \( t < 2 \) and 1 when \( t \geq 2 \).
   - \( u(t - 4) \) is 0 when \( t < 4 \) and 1 when \( t \geq 4 \).

   So, \( x(t) = t \) between \( t = 2 \) and \( t = 4 \), and 0 elsewhere:
   \[
   x(t) =
   \begin{cases}
   t & \text{for } 2 \leq t \leq 4 \\
   0 & \text{otherwise}
   \end{cases}
   \]

#### Step 2: Convolution \( y(t) = x(t) * h(t) \)

The output \( y(t) \) is the convolution of \( x(t) \) and \( h(t) \):
\[
y(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau
\]

Given that \( h(t) = 1 \), the convolution simplifies to integrating \( x(\tau) \) over the region where both functions are non-zero.

Since \( h(t) \) is constant and equal to 1:
\[
y(t) = \int_{-\infty}^{\infty} x(\tau) d\tau
\]

We can break this into different cases based on \( t \).

1. **Case 1: \( t < 2 \)**

   - \( x(\tau) = 0 \) for \( \tau < 2 \).
   - Therefore, \( y(t) = 0 \) for \( t < 2 \).

2. **Case 2: \( 2 \leq t \leq 4 \)**

   - The convolution range will be from \( 2 \) to \( t \):
   \[
   y(t) = \int_{2}^{t} \tau d\tau = \left[\frac{\tau^2}{2}\right]_{2}^{t} = \frac{t^2}{2} - 2
   \]

3. **Case 3: \( 4 < t \leq 6 \)**

   - The convolution range will be from \( t - 2 \) to \( 4 \):
   \[
   y(t) = \int_{t-2}^{4} \tau d\tau = \left[\frac{\tau^2}{2}\right]_{t-2}^{4} = 8 - \frac{(t-2)^2}{2}
   \]

4. **Case 4: \( t > 6 \)**

   - The output is 0 as there is no overlap between \( x(t) \) and \( h(t) \).

Therefore, the output \( y(t) \) is:
\[
y(t) =
\begin{cases}
0 & \text{for } t < 2 \\
\frac{t^2}{2} - 2 & \text{for } 2 \leq t \leq 4 \\
8 - \frac{(t-2)^2}{2} & \text{for } 4 < t \leq 6 \\
0 & \text{for } t > 6
\end{cases}
\]

#### Plot of \( y(t) \)

The function \( y(t) \) will start at 0, increase parabolically from \( t = 2 \) to \( t = 4 \), reach a peak, and then decrease parabolically until \( t = 6 \), after which it will return to 0.

### Part (b) Is the system BIBO-stable? Justify your answer.

A system is BIBO-stable (Bounded Input Bounded Output stable) if every bounded input produces a bounded output.

The impulse response \( h(t) = u(2 - t) + u(t - 2) \) is bounded (it is always equal to 1), and the output \( y(t) \) is the convolution of a bounded input \( x(t) \) with a bounded impulse response \( h(t) \).

Since convolution of two bounded functions results in a bounded function, the output \( y(t) \) will also be bounded for any bounded input \( x(t) \).

**Conclusion**: The system is BIBO-stable.

### Part (c) Is the system causal? Justify.

A system is causal if the output at any time \( t_0 \) depends only on the input at the present time or past times (i.e., \( x(t) \) for \( t \leq t_0 \)).

Given the impulse response \( h(t) = 1 \), the output \( y(t) \) depends on the values of the input \( x(t) \) only up to the current time \( t_0 \), because convolution inherently respects causality (i.e., it integrates over past values of the input).

**Conclusion**: The system is causal.

---

Would you like more details on any part, or have any questions?

Here are 5 related questions:
1. What would happen if \( h(t) \) was a decaying exponential instead of a unit step function?
2. How would you analyze the system if \( x(t) \) were a periodic function?
3. Can you derive the system's frequency response using Fourier transforms?
4. How does the causality of a system relate to its impulse response?
5. What are the implications of BIBO stability in the design of filters?

**Tip**: When analyzing the stability of a system, always check the impulse response, as it provides key insights into the system's behavior.

Explore the analysis of an analog LTI system with a given impulse response function and input signal. Understand convolution, BIBO stability, and causality in the context of system analysis. Suitable for college-level students and professionals in electrical engineering and signal processing.

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