Determine the Laplace Transform of the following signals:x(t) = e'al sin wo t u(t)

1. Break Down the Problem

To find the Laplace Transform of the signal $x(t) = e^{\alpha t} \sin(\omega_0 t) u(t)$, we need to recognize that this is a product of an exponential function, a sine function, and the unit step function $u(t)$.

2. Relevant Concepts

The Laplace Transform of a function $x(t)$ is defined as: $X(s) = \int_0^{\infty} x(t) e^{-st} dt$ For the specific case of $e^{\alpha t} \sin(\omega_0 t) u(t)$, the Laplace Transform is given by the formula: $\mathcal{L}\{e^{\alpha t} \sin(\omega_0 t)\} = \frac{\omega_0}{(s - \alpha)^2 + \omega_0^2}$ where $s$ is a complex frequency parameter.

3. Analysis and Detail

1. Identify parameters:

   • $\alpha$ is the coefficient of the exponential decay.
   • $\omega_0$ is the frequency of the sine wave.

2. Apply the formula for the Laplace Transform: $X(s) = \mathcal{L}\{e^{\alpha t} \sin(\omega_0 t)\} = \frac{\omega_0}{(s - \alpha)^2 + \omega_0^2}$

4. Verify and Summarize

• The form of the result corresponds to the standard table of Laplace Transforms. The derived formula is valid under the assumption that $s > \alpha$ to ensure convergence of the integral.

Final Answer

The Laplace Transform of the signal $x(t) = e^{\alpha t} \sin(\omega_0 t) u(t)$ is: $X(s) = \frac{\omega_0}{(s - \alpha)^2 + \omega_0^2}$