Question

Find the Laplace transform of e-at u(t) and its ROC.​

Answer 1

Answer:

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Answer 2

Laplace transform of  $e^{at}u(t)$  is  $\frac{1}{(s-a)}$ with ROC of $s > a$

Step by step explanation:

To Find:

Laplace transform of  $e^{at}u(t)$ and its ROC

Formula used:

$X(s) = \int\limits^{\infty}_{-\infty} {x(t)}*e^{-st} \, dt$

Solution:

Here $x(t) = e^{at}u(t)$ so we apply the formula of Laplace transform. where u(t) is a unit step signal from 0 to $\infty$ with an amplitude of 1

$X(s) = \int\limits^{\infty}_{-\infty} {e^{at}u(t)}*e^{-st} \, dt$

now we remove u(t) by changing the limit

$X(s) = \int\limits^{\infty}_{0} {e^{at}}*e^{-st} \, dt$

$X(s) = \int\limits^{\infty}_{0} {e^{at-st} \, dt$

Taking the power common

$X(s) = \int\limits^{\infty}_{0} {e^{-(s-a)t} \, dt$

Applying integration

$X(s) =\frac{e^{-(s-a)t}}{-(s-a)} ]^{\infty}_{0}$

On substitution

$X(s) = (\frac{e^{-(s-a)\infty}}{-(s-a)} ) - (\frac{e^{-(s-a)0}}{-(s-a)} )$

$X(s) =\frac{{0-1}}{-(s-a)}$

$X(s) =\frac{{1}}{(s-a)}$

Since the integral is from 0 to $\infty$  , that is t>0.

which means $s - a > 0$

                      $s > a$

                     

so, the ROC is $s > a$

Therefore, Laplace transform of  $e^{at}u(t)$  is  $\frac{1}{(s-a)}$ with ROC of $s > a$