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 solve the given problem step by step.

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

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

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

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

   \( h(t) = u(2 - t) + u(t - 2) \) can be interpreted as:

   - \( u(2 - t) \) is a unit step function that is 1 when \( t \leq 2 \) and 0 when \( t > 2 \).
   - \( u(t - 2) \) is a unit step function that is 0 when \( t < 2 \) and 1 when \( t \geq 2 \).

   So, \( h(t) \) represents a pulse from \( t = 0 \) to \( t = 2 \).

   Therefore:
   \[
   h(t) = 
   \begin{cases}
   1 & \text{for } 0 \leq t \leq 2 \\
   0 & \text{otherwise}
   \end{cases}
   \]

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

   \( x(t) = t[u(t - 2) - u(t - 4)] \) 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, \( x(t) \) represents a ramp function that starts at \( t = 2 \) and ends at \( t = 4 \).

   Therefore:
   \[
   x(t) = 
   \begin{cases}
   t & \text{for } 2 \leq t \leq 4 \\
   0 & \text{otherwise}
   \end{cases}
   \]

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

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

Since both \( x(t) \) and \( h(t) \) are zero outside their respective intervals, we can limit the integration within the non-zero intervals.

- \( h(t - \tau) \) is non-zero for \( 0 \leq t - \tau \leq 2 \), or equivalently \( t - 2 \leq \tau \leq t \).
- \( x(\tau) \) is non-zero for \( 2 \leq \tau \leq 4 \).

So, the limits of integration are determined by the overlap of these intervals.

\[
y(t) = \int_{\max(2, t-2)}^{\min(4, t)} \tau d\tau
\]

This can be broken into three cases depending on the value of \( t \):

1. **Case 1: \( 0 \leq t < 2 \)**

   Here, \( h(t - \tau) \) does not overlap with \( x(\tau) \), so:
   \[
   y(t) = 0
   \]

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

   The integration limits are 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 integration limits are from \( t-2 \) to \( 4 \):
   \[
   y(t) = \int_{t-2}^{4} \tau d\tau = \left[\frac{\tau^2}{2}\right]_{t-2}^{4} = \frac{16}{2} - \frac{(t-2)^2}{2} = 8 - \frac{(t-2)^2}{2}
   \]

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

   There is no overlap between \( h(t - \tau) \) and \( x(\tau) \), so:
   \[
   y(t) = 0
   \]

Therefore, the complete expression for \( y(t) \) 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 \( y(t) \)

To visualize \( y(t) \), the function will exhibit a parabolic rise from \( t = 2 \) to \( t = 4 \), followed by a parabolic decline from \( t = 4 \) to \( t = 6 \), and zero outside this interval.

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

A system is BIBO (Bounded Input Bounded Output) stable if, for every bounded input \( x(t) \), the output \( y(t) \) is also bounded.

The impulse response \( h(t) \) is given by:
\[
h(t) = 
\begin{cases}
1 & \text{for } 0 \leq t \leq 2 \\
0 & \text{otherwise}
\end{cases}
\]

Since \( h(t) \) is bounded (its maximum value is 1), and the output is the convolution of a bounded input with this bounded impulse response, the output \( y(t) \) will also be bounded for any bounded input.

**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 \( t \leq t_0 \)).

The given impulse response \( h(t) \) is non-zero only for \( 0 \leq t \leq 2 \). Since the system output \( y(t) \) depends on the convolution of the input \( x(t) \) with \( h(t) \), and \( h(t) \) does not anticipate future inputs, the system is causal.

**Conclusion**: The system is causal.

---

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

Here are 5 related questions:
1. How does changing the duration of the impulse response \( h(t) \) affect the stability and causality of the system?
2. What would happen to the system output \( y(t) \) if the input \( x(t) \) were a sinusoidal signal?
3. How can we determine the frequency response of the system given \( h(t) \)?
4. What are the implications of BIBO stability in practical systems?
5. How would you design an LTI system with a desired impulse response?

**Tip**: Convolution in time domain corresponds to multiplication in the frequency domain, which can simplify analyzing systems with complex inputs.

Explore the analysis of an analog LTI system with a specified impulse response function and input signal. Includes determination and plotting of the system output, assessment of BIBO stability, and causality. Ideal for advanced undergraduate students and researchers.

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