VALUE=_(alpha )BY SIGN CRITERION FIND FROM THE LAPLACE FUNCTION BY THE CONDITION Select one: Phi (z_(alpha ))=alpha Phi (z_(alpha ))=(1-2alpha )/(2) Phi (z_(alpha ))=(alpha )/(2) Phi (z_(alpha ))=(1-alpha )/(2) Phi (z_(alpha ))=1-(alpha )/(2)

To find the Laplace transform of the function $f(t) = e^{-at}u(t-a)$, we can use the Laplace transform of the exponential function.<br /><br />The Laplace transform of the exponential function $e^{-at}$ is given by:<br /><br />$\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}$<br /><br />where $s$ is the complex frequency variable.<br /><br />Now, let's consider the given function $f(t) = e^{-at}u(t-a)$, where $u(t-a)$ is the unit step function.<br /><br />The unit step function $u(t-a)$ is defined as:<br /><br />$u(t-a) = \begin{cases} 0 & \text{if } t < a \\ 1 & \text{if } t \geq a \end{cases}$<br /><br />To find the Laplace transform of $f(t)$, we can use the convolution theorem:<br /><br />$\mathcal{L}\{f(t)\} = \mathcal{L}\{e^{-at}u(t-a)\} = \mathcal{L}\{e^{-at}\} \cdot \mathcal{L}\{u(t-a)\}$<br /><br />The Laplace transform of the unit step function $u(t-a)$ is given by:<br /><br />$\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}$<br /><br />Now, we can substitute the Laplace transforms into the equation:<br /><br />$\mathcal{L}\{f(t)\} = \frac{1}{s+a} \cdot \frac{e^{-as}}{s} = \frac{e^{-as}}{s(s+a)}$<br /><br />Therefore, the correct answer is:<br /><br />$\Phi(z_{\alpha}) = \frac{1-2\alpha}{2}$