Laplace

Problem 02 | Laplace Transform by Integration

Problem 02
Find the Laplace transform of   $f(t) = e^{at}$.
 

Solution 02
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

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

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

$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-(s - a)t} \, dt$

$\displaystyle \mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \int_0^\infty e^{-(s - a)t} \, [ \, -(s - a) \, dt \, ]$

Problem 01 | Laplace Transform by Integration

Problem 01
Find the Laplace transform of   $f(t) = 1$   when   $t > 0$.
 

Solution 01
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} (1) \, dt$

$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} \, dt$

$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \int_0^\infty e^{-st} \, (-s \, dt)$

$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \left[ e^{-st} \right]_0^\infty$

Laplace Transform by Direct Integration

To get the Laplace transform of the given function   $f(t)$,   multiply   $f(t)$   by   $e^{-st}$   and integrate with respect to   $t$   from zero to infinity. In symbol,
 

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

 

Laplace Transform

Definition of Laplace Transform

Let   $f(t)$   be a given function which is defined for   $t \ge 0$. If there exists a function   $F(s)$  so that
 

$\displaystyle F(s) = \int_0^\infty e^{-st} \, f(t) \, dt$,

 

then   $F(s)$   is called the Laplace Transform of   $f(t)$, and will be denoted by   $\mathcal{L} \left\{f(t)\right\}$.   Notice the integrator   $e^{-st} \, dt$   where   $s$   is a parameter which may be real or complex.
 

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