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alosnomolina1122

15.11.2022

Mathematics

$\large\begin{array}{l} \textsf{Find the inverse Laplace transform of}\\\\ \mathsf{F(s)=\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}} \end{array}$

$\large\begin{array}{l} \textsf{First, split F(s) into partial fractions:}\\\\ \begin{array}{lll} \mathsf{F(s)}&\!\!=\!\!&\mathsf{\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}=\dfrac{A}{s+\sqrt{2}}+\dfrac{B}{s-\sqrt{3}}}\\\\ &&\mathsf{\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}=\dfrac{A(s-\sqrt{3})}{(s+\sqrt{2})(s-\sqrt{3})}+\dfrac{B(s+\sqrt{2})}{(s+\sqrt{2})(s-\sqrt{3})}}\\\\ &&\mathsf{\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}=\dfrac{A(s-\sqrt{3})+B(s+\sqrt{2})}{(s+\sqrt{2})(s-\sqrt{3})}}\\\\ &&\mathsf{\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}=\dfrac{As-\sqrt{3}\,A+Bs+\sqrt{2}\,B}{(s+\sqrt{2})(s-\sqrt{3})}}\\\\&&\mathsf{\dfrac{1}{(s+\sqrt{2})(s-\sqrt{3})}=\dfrac{(A+B)s-\sqrt{3}\,A+\sqrt{2}\,B}{(s+\sqrt{2})(s-\sqrt{3})}} \end{array} \end{array}$

$\large\begin{array}{l} \textsf{Numerators must be equal to each other:}\\\\ \mathsf{1=(A+B)s+\big(\!-\!\sqrt{3}\,A+\sqrt{2}\,B\big)}\\\\ \mathsf{0s+1=(A+B)s+\big(\!-\!\sqrt{3}\,A+\sqrt{2}\,B\big)}\\\\\\ \textsf{So, you get a linear system:}\\\\ \left\{\! \begin{array}{rrrrr} \mathsf{A}&\!\!+\!\!&\mathsf{B}&\!\!=\!\!&\mathsf{0}\\ \mathsf{-\sqrt{3}\,A}&\!\!+\!\!&\mathsf{\sqrt{2}\,B}&\!\!=\!\!&\mathsf{1}\\ \end{array} \right. \end{array}$

$\large\begin{array}{l}\textsf{Solving it through subtitution:}\\\\ \mathsf{B=-A}\\\\\\ \mathsf{-\sqrt{3}\,A+\sqrt{2}\,(-A)=1}\\\\ \mathsf{-\sqrt{3}\,A-\sqrt{2}\,A=1}\\\\ \mathsf{-(\sqrt{3}+\sqrt{2})A=1}\\\\ \mathsf{A=-\,\dfrac{1}{\sqrt{3}+\sqrt{2}}} \end{array}$

$\large\begin{array}{l} \mathsf{A=-\,\dfrac{1}{\sqrt{3}+\sqrt{2}}\cdot \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\\\ \mathsf{A=-\,\dfrac{\sqrt{3}-\sqrt{2}}{\big(\!\sqrt{3}+\sqrt{2}\big)\cdot \big(\!\sqrt{3}-\sqrt{2})}}\\\\ \mathsf{A=\dfrac{-\big(\!\sqrt{3}-\sqrt{2})}{\big(\!\sqrt{3}\big)^{\!2}-\big(\!\sqrt{2}\big)^{\!2}}}\\\\ \mathsf{A=\dfrac{-\sqrt{3}+\sqrt{2}}{3-2}}\\\\ \mathsf{A=\dfrac{\sqrt{2}-\sqrt{3}}{1}}\\\\ \boxed{\begin{array}{l}\mathsf{A=\sqrt{2}-\sqrt{3}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Substitute into one of those equations to find B:}\\\\ \mathsf{B=-(\sqrt{2}-\sqrt{3})}\\\\ \mathsf{B=-\sqrt{2}+\sqrt{3}}\\\\ \boxed{\begin{array}{c}\mathsf{B=\sqrt{3}-\sqrt{2}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Now that we have already }\mathsf{F(s)}\textsf{ split into partial fractions,}\\\\ \mathsf{F(s)=\dfrac{\sqrt{2}-\sqrt{3}}{s+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{s-\sqrt{3}}}\\\\ \mathsf{F(s)=\big(\!\sqrt{2}-\sqrt{3}\big)\cdot \dfrac{1}{s+\sqrt{2}}+\big(\!\sqrt{3}-\sqrt{2}\big)\cdot \dfrac{1}{s-\sqrt{3}}}\\\\\\ \textsf{we can easily get the inverse Laplace transform}\\ \textsf{(from a table of Laplace transforms)}\\\\ \mathsf{\mathcal{L}_s^{-1}\left\{\dfrac{1}{s-p}\right\}=e^{pt},\qquad t\ge 0.} \end{array}$

$\large\begin{array}{l} \textsf{So,}\\\\ \mathsf{f(t)=\mathcal{L}_s^{-1}\left\{F(s)\right\}}\\\\ \mathsf{f(t)=\mathcal{L}_s^{-1}\left\{\big(\!\sqrt{2}-\sqrt{3}\big)\cdot \dfrac{1}{s+\sqrt{2}}+\big(\!\sqrt{3}-\sqrt{2}\big)\cdot \dfrac{1}{s-\sqrt{3}}\right\}}\\\\ \mathsf{f(t)=\big(\!\sqrt{2}-\sqrt{3}\big)\cdot \mathcal{L}_s^{-1}\left\{\dfrac{1}{s+\sqrt{2}}\right\}+\big(\!\sqrt{3}-\sqrt{2}\big)\cdot \mathcal{L}_s^{-1}\left\{\dfrac{1}{s-\sqrt{3}}\right\}} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{l}\mathsf{f(t)=\big(\!\sqrt{2}-\sqrt{3}\big)\cdot e^{-\sqrt{2}\,t}+\big(\!\sqrt{3}-\sqrt{2}\big)\cdot e^{\sqrt{3}\,t}} \end{array}}\\\\\\ \textsf{for }\mathsf{t\ge 0.} \end{array}$

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$\large\textsf{I hope this helps. :-)}$

Tags: inverse laplace transform partial fraction frequency function

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