inverse transform

Problem 05 | Inverse Laplace Transform

Problem 05
Find the inverse transform of   $\dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15}$

Problem 04 | Inverse Laplace Transform

Problem 04
Perform the indicated operation:   $\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right]$
 

Problem 03 | Inverse Laplace Transform

Problem 03
Find the inverse transform of   $\dfrac{7}{s^2 + 6}$.
 

Problem 02 | Inverse Laplace Transform

Problem 02
Find the inverse transform of   $\dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9}$.
 

Solution 02
$\mathcal{L}^{-1} \left[ \dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9} \right]= 5\mathcal{L}^{-1}\left( \dfrac{1}{s - 2} \right) - 4\mathcal{L}^{-1}\left( \dfrac{s}{s^2 + 9} \right)$

$\mathcal{L}^{-1} \left[ \dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9} \right]= 5e^{2t}\mathcal{L}^{-1}\left( \dfrac{1}{s} \right) - 4\mathcal{L}^{-1}\left( \dfrac{s}{s^2 + 3^2} \right)$

Problem 01 | Inverse Laplace Transform

Problem 01
Find the inverse transform of   $\dfrac{8 - 3s + s^2}{s^3}$.
 

The Inverse Laplace Transform

Definition
From   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   the value   $f(t)$   is called the inverse Laplace transform of   $F(s)$. In symbol,
 

$\mathcal{L}^{-1}\left\{ F(s) \right\} = f(t)$

where   $\mathcal{L}^{-1}$   is called the inverse Laplace transform operator.
 

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