7.3: La Transformación Inversa de Laplace- Integración Compleja
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)La transformación inversa de Laplace
Si\(q\) es una función racional con polos\(\{\lambda_{j} | j = \{1, \cdots, h\}\}\) entonces la transformada inversa de Laplace de\(q\) es
\[\mathscr{L}^{-1}(q)(t) \equiv \frac{1}{2 \pi i} \int q(z) e^{zt} dz \nonumber\]
donde\(C\) hay una curva que encierra cada uno de los polos de\(q\)
\[\mathscr{L}^{-1}(q)(t) = \sum_{j = 1}^{h} res(\lambda_{j}) \nonumber\]
Pongamos a prueba esta preciosa fórmula. Tomamos nuestros ejemplos de la discusión de la Transformada de Laplace y la Transformada inversa de Laplace. Primero calculemos la transformada inversa de Laplace de
\[q(z) = \frac{1}{(z+1)^2} \nonumber\]
Según la Ecuación es simplemente el residuo de\(q(z)e^{zt}\) at\(z = -1\) i.e.,
\[res(-1) = \lim_{z \rightarrow -1} de^{zt} dz = te^{-t} \nonumber\]
Esto cierra el círculo sobre el ejemplo iniciado en la discusión de la Transformación de Laplace y continuó en el ejercicio uno para el capítulo 6.
Para nuestro siguiente ejemplo recordamos
\[\mathscr{L} (x_{1}(s)) = \frac{0.19(s^2+1.5s+0.27)}{(s+1/6)^{4}(s^3+1.655s^2+0.4978s+0.0039)} \nonumber\]
de la Transformada Inversa de Laplace. Usando numde
, sym2poly
y residuo
, vea fib4.m
para más detalles, devoluciones
\[r_{1} = \begin{pmatrix} {0.0029}\\ {262.8394}\\ {-474.1929}\\ {-1.0857}\\ {-9.0930}\\ {-0.3326}\\ {211.3507} \end{pmatrix} \nonumber\]
y
\[p_{1} = \begin{pmatrix} {-1.3565}\\ {-0.2885}\\ {-0.1667}\\ {-0.1667}\\ {-0.1667}\\ {-0.1667}\\ {-0.0100} \end{pmatrix} \nonumber\]
Se te pedirá en los ejercicios que demuestres que esto efectivamente conforma con el
\[x_{1}(t) = 211.35e^{\frac{-t}{100}}-(0.0554t^3+4.5464t^2+1.085t+474.19)e^{\frac{-t}{6}}+e^{\frac{-329t}{400}}(262.842 \cosh (\frac{\sqrt{73}t}{16})+262.836 \sinh (\frac{\sqrt{73}t}{16})) \nonumber\]
logrado en la Transformación de Laplace vía ilaplace
.