Question #286772

find inverse laplace transorm : se^-2s/(s^2 + pi^2)





1
Expert's answer
2022-01-24T18:51:09-0500

By property of Laplace Transform,if L1(G(s))=g(t),thenL1(easG(s))=u(ta)×g(ta)Now,L(f(t))=se2ss2+π2=F(s)f(t)=L1(F(s))=L1(se2ss2+π2)=L1(e2s×ss2+π2)=L1(e2s×G(s)),where G(s)=ss2+π2But,L1(G(s))=L1(ss2+π2)=cos(πt)=g(t),thusf(t)=L1(e2s×G(s))=u(t2)×g(t2),from the above stated property where a=2=u(t2)×cos(π(t2)),since g(t)=cos(πt)Thus,L1(se2ss2+π2)=u(t2)×cos(π(t2)).\qquad\text{By property of Laplace Transform,}\\ if\ \mathcal{L}^{-1}(G(s))=g(t), then\\ \mathcal{L}^{-1}(e^{-as}G(s))=u(t-a)\times g(t-a)\\ \quad\\ Now,\mathcal{L}(f(t))=\frac{se^{-2s}}{s^2+\pi^2}=F(s)\\ \Rightarrow f(t)=\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}(\frac{se^{-2s}}{s^2+\pi^2})=\mathcal{L}^{-1}(e^{-2s}\times \frac{s}{s^2+\pi^2})\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\mathcal{L}^{-1}(e^{-2s}\times G(s)), where\ G(s)=\frac{s}{s^2+\pi^2}\\ But, \mathcal{L}^{-1}(G(s))=\mathcal{L}^{-1}(\frac{s}{s^2+\pi^2})=\cos(\pi t)=g(t), thus\\ \Rightarrow f(t)=\mathcal{L}^{-1}(e^{-2s}\times G(s))\\ \qquad\quad =u(t-2)\times g(t-2), \text{from the above stated property where a=2}\\ \qquad\quad =u(t-2)\times \cos(\pi(t-2)), since\ g(t)=\cos(\pi t)\\ Thus, \mathcal{L}^{-1}(\frac{se^{-2s}}{s^2+\pi^2})=u(t-2)\times \cos(\pi(t-2)).


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