diff options
Diffstat (limited to 'content/know/concept/laplace-transform')
| -rw-r--r-- | content/know/concept/laplace-transform/index.pdc | 12 |
1 files changed, 9 insertions, 3 deletions
diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc index bd7673b..5e91a04 100644 --- a/content/know/concept/laplace-transform/index.pdc +++ b/content/know/concept/laplace-transform/index.pdc @@ -33,6 +33,12 @@ This is solved by restricting the domain of $\tilde{f}(s)$ to $s$ where $\mathrm{Re}\{s\} > s_0$, for an $s_0$ large enough to compensate for the growth of $f(t)$. +The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration, +and is therefore a lot more difficult to calculate. +Fortunately, it is usually avoidable by rewriting a given $s$-space expression +using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/), +and then looking up the individual terms. + ## Derivatives @@ -42,7 +48,7 @@ This is especially useful for transforming ODEs with variable coefficients: $$\begin{aligned} \boxed{ - \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\} + \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\} } \end{aligned}$$ @@ -107,9 +113,9 @@ $$\begin{aligned} \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\} &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t} \\ - &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t} + &= \Big[ f^{(n - 1)}(t) \exp\!(- s t) \Big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t} \\ - &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t} + &= - f^{(n - 1)}(0) + s \Big[ f^{(n - 2)}(t) \exp\!(- s t) \Big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t} \end{aligned}$$ And so on. |