Please find laplace and inverse Laplace

Name: Please find laplace and inverse Laplace
Uploaded: 2022-02-23T17:22:57+00:00
Description: Please find laplace and inverse Laplace

Question

RizerMix · Accepted Answer

To find the Laplace transform of x(t), we use the linearity property of the Laplace transform and the fact that the Laplace transform of $e^{- a t} u (t)$ is $\frac{1}{s + a}$ , where u(t) is the unit step function.

Thus, we have:

$L {x (t)} = L {\frac{13}{5} e^{- 3 (t + 3)} u (t + 3)} - L {9 (t + 2) u (t + 2)} + L {2 \cos (t - 3) u (t - 3)}$

$= (\frac{13}{5}) L {e^{- 3 (t + 3)} u (t + 3)} - 9 L {(t + 2) u (t + 2)} + 2 L {\cos (t - 3) u (t - 3)}$

Using the time-shifting property of the Laplace transform, we have:

$L {e^{- 3 (t + 3)} u (t + 3)} = \frac{e^{- 9 s}}{s + 3}$

$L {(t + 2) u (t + 2)} = (\frac{1}{s^{2}}) + (\frac{2}{s^{3}})$

To find $L {\cos (t - 3) u (t - 3)}$ , we first use the fact that $\cos (t - 3) = (\frac{1}{2}) (e^{i (t - 3)} + e^{- i (t - 3)})$ , where i is the imaginary unit. Then, using the Laplace transform of the unit step function and the Laplace transform of $e^{i a t}$ , we have:

$L {\cos (t - 3) u (t - 3)} = (\frac{1}{2}) [L {e^{i (t - 3)} u (t - 3)} + L {e^{- i (t - 3)} u (t - 3)}]$

$= (\frac{1}{2}) [\frac{1}{s - i} + \frac{1}{s + i}]$

$= (\frac{1}{2}) [\frac{s + i}{s^{2} + 1} + \frac{s - i}{s^{2} + 1}]$

$= \frac{s}{s^{2} + 1}$

Therefore, we have:

$L {x (t)} = (\frac{13}{5}) \frac{e^{- 9 s}}{s + 3} - 9 [(\frac{1}{s^{2}}) + (\frac{2}{s^{3}})] + 2 \frac{s}{s^{2} + 1}$

To find the inverse Laplace transform of $L {x (t)}$ , we use partial fraction decomposition and inverse Laplace transforms of basic functions. After some algebraic manipulation, we have:

$x (t) = (\frac{13}{5}) e^{- 3 (t + 3)} u (t + 3) - 9 t u (t + 2) - 18 u (t + 2) + \cos (t - 3) u (t - 3) - \sin (t - 3) u (t - 3)$

Therefore, the Laplace transform of x(t) is $(\frac{13}{5}) \frac{e^{- 9 s}}{s + 3} - 9 [(\frac{1}{s^{2}}) + (\frac{2}{s^{3}})] + \frac{2 s}{s^{2} + 1}$ , and the inverse Laplace transform of X(s) is $x (t) = (\frac{13}{5}) e^{- 3 (t + 3)} u (t + 3) - 9 t u (t + 2) - 18 u (t + 2) + \cos (t - 3) u (t - 3) - \sin (t - 3) u (t - 3)$ .

Please find laplace and inverse Laplace

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