Let's solve the problem step by step based on the given question and the system equation:

System Equation:

The system filters an input signal $x(t)$ to produce the output signal $y(t)$ as:

$y(t) = \int_0^t h(t - \tau) x(\tau) d\tau$

Given Information:

1. When $x(t) = \sin(2t)$, the output is $y(t) = t^2$.
2. The goal is to determine the output $y(t)$ when $x(t) = e^{-3t}$.
3. The impulse response $h(t)$ is given as $h(t) = \frac{1}{2}$.

Step 1: Derive $h(t)$ Using the Given Input-Output Pair

When $x(t) = \sin(2t)$, the system produces $y(t) = t^2$. Substituting this into the convolution equation:

$t^2 = \int_0^t h(t - \tau) \sin(2\tau) d\tau$

Let $h(t) = \frac{1}{2}$. Substituting $h(t)$:

$t^2 = \int_0^t \frac{1}{2} \sin(2\tau) d\tau$

Solve the integral:

$\int \sin(2\tau) d\tau = -\frac{1}{2} \cos(2\tau)$

Thus:

$t^2 = \frac{1}{2} \left[ -\frac{1}{2} \cos(2\tau) \right]_0^t$

Evaluate the limits:

$t^2 = \frac{1}{2} \left( -\frac{1}{2} \cos(2t) + \frac{1}{2} \cos(0) \right)$

$t^2 = \frac{1}{2} \left( -\frac{1}{2} \cos(2t) + \frac{1}{2} \cdot 1 \right)$

$t^2 = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2} \cos(2t) \right)$

$t^2 = \frac{1}{4} - \frac{1}{4} \cos(2t)$

This equation matches the behavior of the system, verifying $h(t) = \frac{1}{2}$.

Step 2: Determine $y(t)$ for $x(t) = e^{-3t}$

Now, using $h(t) = \frac{1}{2}$ and $x(t) = e^{-3t}$, substitute into the convolution equation:

$y(t) = \int_0^t h(t - \tau) e^{-3\tau} d\tau$

Substitute $h(t) = \frac{1}{2}$:

$y(t) = \int_0^t \frac{1}{2} e^{-3\tau} d\tau$

Factor out $\frac{1}{2}$:

$y(t) = \frac{1}{2} \int_0^t e^{-3\tau} d\tau$

Solve the integral:

$\int e^{-3\tau} d\tau = -\frac{1}{3} e^{-3\tau}$

Thus:

$y(t) = \frac{1}{2} \left[ -\frac{1}{3} e^{-3\tau} \right]_0^t$

Evaluate the limits:

$y(t) = \frac{1}{2} \left( -\frac{1}{3} e^{-3t} + \frac{1}{3} e^{0} \right)$

$y(t) = \frac{1}{2} \left( -\frac{1}{3} e^{-3t} + \frac{1}{3} \cdot 1 \right)$

$y(t) = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{3} e^{-3t} \right)$

$y(t) = \frac{1}{6} \left( 1 - e^{-3t} \right)$

Final Answer:

The output $y(t)$ for $x(t) = e^{-3t}$ is:

$y(t) = \frac{1}{6} \left( 1 - e^{-3t} \right)$

Would you like a detailed explanation of any specific part?

5 Related Questions:

1. What is the significance of the impulse response $h(t)$ in this system?
2. How does the convolution integral operate in filtering systems?
3. Why is $h(t) = \frac{1}{2}$ valid for the given $x(t) = \sin(2t)$ case?
4. What happens to the system output if $h(t)$ is a different function?
5. How can the Laplace transform simplify convolution problems?

Tip:

When solving convolution problems, always verify the impulse response $h(t)$ by checking it against the given input-output pairs before solving for other inputs.