Otóż mam sobie zadanko, w którym trzeba policzyc odwrotna transformate Laplace'a czegoś takiego:
\(\displaystyle{ \overline{f}}\)\(\displaystyle{ (s)=\frac{2s^{3}+s^{2}+2s+2}{s^{4}+1}}\)\(\displaystyle{ -\frac{1}{s-1}}\)
I o ile jeszcze policzenie drugiej czescie to banal (wystarczy do tablic zajrzec), o tyle co do pierwszej to nie mam pojecia jak sie za to zabrac. Z góry dzieki za wszelka pomoc.
odwrotna transformata Laplace'a
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kuch2r
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odwrotna transformata Laplace'a
\(\displaystyle{ s^4+1=(s^2-\sqrt{2}s+1)(s^2+\sqrt{2}s+1)}\)
\(\displaystyle{ \frac{2s^3+s^2+2s+2}{s^4+1}=\frac{As+B}{s^2-\sqrt{2}s+1}+\frac{Cs+D}{s^2+\sqrt{2}s+1}\\
\begin{cases}A=1-\frac{1}{2\sqrt{2}}\\B=1\\C=1+\frac{1}{2\sqrt{2}}\\D=1\end{cases}}\)
Stad:
\(\displaystyle{ \frac{s(1-\frac{1}{2\sqrt{2}})+1}{s^2-\sqrt{2}+1}=\frac{s-\frac{s\sqrt{2}}{4}+1}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{1+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}s}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}s-1}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}(s-\frac{\sqrt{2}}{2})+\frac{3\sqrt{2}}{8}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\\=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}(s-\frac{\sqrt{2}}{2})}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{3}{4}\cdot\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=F(s)}\)
\(\displaystyle{ \mathcal{L}^{-1}[F(s)]=(1-\frac{\sqrt{2}}{4})e^{\frac{\sqrt{2}}{2}t}\cdot \cos{\frac{\sqrt{2}}{2}t}+(1-\frac{3}{4})e^{\frac{\sqrt{2}}{2}t}\cdot \sin{\frac{\sqrt{2}}{2}t}}\)
Druga czesc analogicznie...
\(\displaystyle{ \frac{2s^3+s^2+2s+2}{s^4+1}=\frac{As+B}{s^2-\sqrt{2}s+1}+\frac{Cs+D}{s^2+\sqrt{2}s+1}\\
\begin{cases}A=1-\frac{1}{2\sqrt{2}}\\B=1\\C=1+\frac{1}{2\sqrt{2}}\\D=1\end{cases}}\)
Stad:
\(\displaystyle{ \frac{s(1-\frac{1}{2\sqrt{2}})+1}{s^2-\sqrt{2}+1}=\frac{s-\frac{s\sqrt{2}}{4}+1}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{1+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}s}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}s-1}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}(s-\frac{\sqrt{2}}{2})+\frac{3\sqrt{2}}{8}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=\\=\frac{s-\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}+\frac{\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{\sqrt{2}}{4}(s-\frac{\sqrt{2}}{2})}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}-\frac{\frac{3}{4}\cdot\frac{\sqrt{2}}{2}}{(s-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=F(s)}\)
\(\displaystyle{ \mathcal{L}^{-1}[F(s)]=(1-\frac{\sqrt{2}}{4})e^{\frac{\sqrt{2}}{2}t}\cdot \cos{\frac{\sqrt{2}}{2}t}+(1-\frac{3}{4})e^{\frac{\sqrt{2}}{2}t}\cdot \sin{\frac{\sqrt{2}}{2}t}}\)
Druga czesc analogicznie...