h(t) = u(t +1)−u(t−1) ; x(t) = (1/3)t(u(t)−u(t−3)), calculate convolution.

To solve the convolution of \( h(t) \) and \( x(t) \), we follow the steps for calculating convolution, which is defined by:

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

Let's break down the two functions first:

### 1. **For \( h(t) \):**
\[
h(t) = u(t + 1) - u(t - 1)
\]
Where \( u(t) \) is the unit step function, which is 0 for \( t < 0 \) and 1 for \( t \geq 0 \).

- \( u(t + 1) \) shifts the unit step function left by 1, so it turns on at \( t = -1 \).
- \( u(t - 1) \) shifts the unit step function right by 1, so it turns on at \( t = 1 \).

Thus, \( h(t) \) is a pulse that is 1 for \( -1 \leq t < 1 \) and 0 elsewhere. It can be written as:

\[
h(t) = \text{rect}\left(\frac{t}{2}\right)
\]
where \( \text{rect}(t/2) \) is the rectangular pulse of width 2 centered at \( t = 0 \).

### 2. **For \( x(t) \):**
\[
x(t) = \frac{1}{3}t \left( u(t) - u(t - 3) \right)
\]
This is a ramp function, but limited to the interval \( 0 \leq t < 3 \). It's zero for \( t < 0 \) and \( t \geq 3 \), and within the interval \( 0 \leq t < 3 \), it behaves as \( \frac{1}{3} t \).

Now we calculate the convolution \( y(t) = (h * x)(t) \).

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

Since \( h(t) \) is only non-zero in the interval \( -1 \leq t < 1 \), the integral reduces to:

\[
y(t) = \int_{-1}^{1} h(\tau) x(t - \tau) \, d\tau
\]

### Step-by-Step Convolution:

- For \( t < 0 \), both \( h(t) \) and \( x(t) \) are zero, so \( y(t) = 0 \).
- For \( 0 \leq t < 4 \), we need to evaluate the convolution by substituting \( h(\tau) \) and \( x(t - \tau) \) in the integral. This is the main region where the convolution is non-trivial.

I will now perform the integration over this interval.The convolution of \( h(t) \) and \( x(t) \) is given by the following expression:

\[
y(t) = \frac{1}{3} t \min(1, \max(-1, t)) - \frac{1}{3} t \min(1, \max(-1, t - 3)) - \frac{1}{6} \min(1, \max(-1, t))^2 + \frac{1}{6} \min(1, \max(-1, t - 3))^2
\]

This formula combines the behavior of both functions \( h(t) \) and \( x(t) \) over the intervals of interest.

Would you like further details on how this result applies over specific time intervals or how it behaves graphically?

Here are 5 follow-up questions:
1. How does the convolution behave for negative \( t \)?
2. Can you explain the role of unit step functions in the convolution process?
3. What changes in the convolution if \( h(t) \) had a different width?
4. How does the convolution reflect the piecewise nature of \( x(t) \)?
5. Could this convolution be solved using Laplace transforms for faster results?

**Tip:** Always analyze the regions where the functions are non-zero when computing convolutions.

h(t) = u(t +1)−u(t−1); x(t) = (1/3)t(u(t)−u(t−3)), calculate convolution.

Learn how to calculate the convolution of h(t) = u(t +1)−u(t−1) and x(t) = (1/3)t(u(t)−u(t−3)) in this detailed solution. This problem covers key concepts in convolution, unit step functions, and piecewise functions. Suitable for students at the undergraduate level studying signal processing or systems theory.

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