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)

To solve this problem, we need to compute the output \( y(t) \), which is the convolution of the input \( x(t) \) with the impulse response \( h(t) \). The output can be expressed as:

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

### Step 1: Analyze the input and impulse response functions
We start by analyzing both the input \( x(t) \) and the impulse response \( h(t) \):

- The **impulse response** is given as:
  \[
  h(t) = u(2 - t) + u(t - 2)
  \]
  This is a combination of step functions \( u(t) \). Specifically:
  - \( u(2 - t) \) is a step function that equals 1 for \( t \leq 2 \) and 0 for \( t > 2 \).
  - \( u(t - 2) \) is a step function that equals 0 for \( t < 2 \) and 1 for \( t \geq 2 \).
  
  So, \( h(t) \) is 1 everywhere, as the sum of these two step functions results in:
  \[
  h(t) = 1 \quad \text{for all} \, t.
  \]

- The **input** \( x(t) \) is given as:
  \[
  x(t) = t[u(t - 2) - u(t - 4)]
  \]
  This is a piecewise function, nonzero only between \( t = 2 \) and \( t = 4 \). It can be expressed as:
  \[
  x(t) =
  \begin{cases}
  t, & 2 \leq t \leq 4, \\
  0, & \text{otherwise}.
  \end{cases}
  \]

### Step 2: Set up the convolution integral
We now compute the convolution \( y(t) = x(t) \ast h(t) \), which simplifies due to \( h(t) = 1 \) for all \( t \). The convolution becomes:
\[
y(t) = \int_{-\infty}^{\infty} x(\tau) d\tau
\]

Since \( x(\tau) \) is nonzero only in the interval \( [2, 4] \), the integral reduces to:
\[
y(t) = \int_2^4 \tau \, d\tau
\]

### Step 3: Perform the integration
Let's now compute this integral:
\[
y(t) = \int_2^4 \tau \, d\tau = \left[\frac{\tau^2}{2}\right]_2^4
\]
Substituting the limits:
\[
y(t) = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6
\]

### Step 4: Final Output
The output \( y(t) \) will be a constant value \( y(t) = 6 \) for the time interval where \( x(t) \) and \( h(t) \) overlap. Since \( x(t) \) is nonzero only between \( t = 2 \) and \( t = 4 \), the convolution will be nonzero in this region.

### Plotting \( y(t) \)
The output \( y(t) \) is a piecewise function:
\[
y(t) =
\begin{cases}
6, & 4 \leq t \leq 6, \\
0, & \text{otherwise}.
\end{cases}
\]

I can generate a plot for you as well if you'd like! Would you like me to provide more details or help you with a plot?

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Here are some related questions you might explore:

1. How does the convolution change if the impulse response is not constant?
2. What would happen if the input function \( x(t) \) was shifted by a different amount?
3. How does the unit step function impact the behavior of linear time-invariant (LTI) systems?
4. What are the properties of the convolution integral?
5. How would the output change if the limits of the input \( x(t) \) were altered?

**Tip:** In convolution problems, identifying where the functions are nonzero simplifies the integration significantly.

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).

Determine and plot the output of an analog LTI system with impulse response h(t) = u(2 − t) + u(t − 2) and input x(t) = t[u(t − 2) − u(t − 4)]. Learn the convolution process, integration steps, and the final output, including a plot for visualization. Suitable for undergraduate students studying LTI systems and signal processing.

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