Laplace transform

Problem 02 | Second Shifting Property of Laplace Transform

Problem 01
Find the Laplace transform of   $g(t) = \begin{cases} f(t - 2)^3 & t \gt 2 \\ 0 & t \lt 2 \end{cases}$
 

Problem 01 | Second Shifting Property of Laplace Transform

Problem 01
Find the Laplace transform of   $g(t) = \begin{cases} f(t - 1)^2 & t \gt 1 \\ 0 & t \lt 1 \end{cases}$
 

Second Shifting Property | Laplace Transform

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
 

then,

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$

 

Problem 03 | First Shifting Property of Laplace Transform

Problem 03
Find the Laplace transform of   $f(t) = e^{-3t} \cos t$.
 

Solution 03

Problem 02 | First Shifting Property of Laplace Transform

Problem 02
Find the Laplace transform of   $f(t) = e^{-5t} \sin 3t$.
 

Solution 02

First Shifting Property | Laplace Transform

First Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   when   $s > a$   then,
 

$\mathcal{L} \left\{ e^{at} \, f(t) \right\} = F(s - a)$

 

In words, the substitution   $s - a$   for   $s$   in the transform corresponds to the multiplication of the original function by   $e^{at}$.
 

Problem 02 | Linearity Property of Laplace Transform

Problem 02
By using the linearity property, show that

$\mathcal{L}(\cosh at) = \dfrac{s}{s^2 - a^2}$

 

Solution 02
$f(t) = \cosh at$

$\displaystyle \mathcal{L}\left\{ f(t) \right\} = \int_0^\infty e^{st} f(t) \, dt$

$\displaystyle \mathcal{L}(\cosh at) = \int_0^\infty e^{st} \cosh at \, dt$
 

But
$\cosh at = \dfrac{e^{at} + e^{-at}}{2}$
 

Thus,
$\displaystyle \mathcal{L}(\cosh at) = \int_0^\infty e^{st} \left( \dfrac{e^{at} + e^{-at}}{2} \right) \, dt$

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