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10.4: La Matriz Exponencial a través de la Transformación de Laplace

  • Page ID
    113053
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    Puede recordar del módulo Transformar de Laplace que puede lograr\(e^{at}\) a través de

    \[e^{at} = \mathscr{L}^{-1}(\frac{1}{s-a}) \nonumber\]

    Por lo tanto, la definición de matriz natural es

    \[e^{At} = \mathscr{L}^{-1} ((sI-A)^{-1}) \nonumber\]

    donde\(I\) es la matriz de identidad n-por-n.

    Ejemplo\(\PageIndex{1}\)

    El caso más fácil es el caso diagonal, e.g.

    \[A = \begin{pmatrix} {1}&{0}\\ {0}&{2} \end{pmatrix} \nonumber\]

    para entonces

    \[(sI-A)^{-1} = \begin{pmatrix} {\frac{1}{s-1}}&{0}\\ {0}&{\frac{1}{s-2}} \end{pmatrix} \nonumber\]

    y así

    \[e^{At} = \begin{pmatrix} {\mathscr{L}^{-1} (\frac{1}{s-1})}&{0}\\ {0}&{\mathscr{L}^{-1} (\frac{1}{s-2})} \end{pmatrix} \nonumber\]

    Ejemplo\(\PageIndex{2}\)

    Como segundo ejemplo supongamos

    \[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]

    y computar, en matlab,

    >> inv(s*eye(2)-A)  
    	
    	   ans = [ s/(s^2+1),  1/(s^2+1)]
    	         [-1/(s^2+1),  s/(s^2+1)]
    
    	>> ilaplace(ans)
    
    	   ans = [ cos(t),  sin(t)]
    	         [-sin(t),  cos(t)]
    	       
    Ejemplo\(\PageIndex{3}\)

    Si

    \[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]

    entonces

    >> inv(s*eye(2)-A)  
    	
    	   ans = [ 1/s,  1/s^2]
    	         [   0,    1/s]
    
    	>> ilaplace(ans)
    
    	   ans = [ 1,  t]
    	         [ 0,  1]	       
    

    This page titled 10.4: La Matriz Exponencial a través de la Transformación de Laplace is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by Steve Cox via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.