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To solve this convolution problem, we need to find the convolution of two signals given as:

1. \( x(t) = e^{-3t} u(t) \)
2. \( h(t) = e^{-2t} u(t - 2) \)

where \( u(t) \) is the unit step function.

### Step 1: Set Up the Convolution Integral
The convolution of two signals \( x(t) \) and \( h(t) \) is defined as:
\[
y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau.
\]

Given \( x(t) = e^{-3t} u(t) \), this signal is zero for \( t < 0 \). So:
\[
x(\tau) = e^{-3\tau} u(\tau).
\]

The second signal \( h(t) = e^{-2t} u(t - 2) \) also involves a delay by 2, which shifts the function to start at \( t = 2 \). Thus:
\[
h(t - \tau) = e^{-2(t - \tau)} u(t - \tau - 2).
\]

### Step 2: Define the Limits of Integration
Since both functions are causal (involving unit steps), the integral limits will depend on the time \( t \) and the delay introduced by the unit step function in \( h(t) \). This will lead to piecewise definitions based on the ranges of \( t \) where the product \( x(\tau) h(t - \tau) \) is non-zero.

### Step 3: Solve the Convolution Integral
To proceed further, let's apply the convolution in two cases based on time intervals.

1. **For \( t < 2 \):** Here, \( h(t - \tau) = 0 \) for all \( \tau \) since \( t - \tau - 2 < 0 \). Hence, \( y(t) = 0 \) in this interval.
   
2. **For \( t \ge 2 \):** Both \( x(\tau) \) and \( h(t - \tau) \) contribute to the integral:
   \[
   y(t) = \int_0^t e^{-3\tau} e^{-2(t - \tau)} u(\tau) u(t - \tau - 2) \, d\tau.
   \]
   The step functions imply that \( \tau \) ranges from \( 2 \) to \( t \):
   \[
   y(t) = \int_2^t e^{-3\tau} e^{-2(t - \tau)} \, d\tau.
   \]
   Simplifying inside the integral:
   \[
   y(t) = \int_2^t e^{-3\tau} e^{-2t + 2\tau} \, d\tau = e^{-2t} \int_2^t e^{-\tau} \, d\tau.
   \]
   Integrate \( e^{-\tau} \) from \( 2 \) to \( t \):
   \[
   y(t) = e^{-2t} \left[ -e^{-\tau} \right]_2^t = e^{-2t} \left( -e^{-t} + e^{-2} \right).
   \]
   Simplifying further:
   \[
   y(t) = e^{-2t} \cdot e^{-2} - e^{-3t} = e^{-2(t+1)} - e^{-3t}.
   \]

So the final solution is:
\[
y(t) = \begin{cases} 
      0, & t < 2 \\
      e^{-2(t+1)} - e^{-3t}, & t \ge 2.
   \end{cases}
\]

Would you like further details on each step or additional explanations?

### Related Questions:
1. What is the Laplace transform of the convolution of two signals?
2. How does the unit step function affect the limits of integration in convolution problems?
3. What are common methods for finding convolutions without integration?
4. How does time-shifting a function affect its convolution?
5. Can we use Fourier transform properties to solve convolution problems?

### Tip:
Remember, convolving with delayed signals often requires adjusting limits carefully due to the unit step function constraints.

Convolve the following signals x(t) = e^(-3t)u(t) and x(t) = e^(-2t)u(t-2).

Step-by-step solution for convolving two signals: x(t) = e^(-3t)u(t) and h(t) = e^(-2t)u(t-2). This problem explores convolution integral and unit step function effects, suitable for undergraduate signal processing students.

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