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UnkleRhaukus

  • 3 years ago

\[ \mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\}\]

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  1. UnkleRhaukus
    • 3 years ago
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    \[ \begin{align*} \mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\} &=h(t-\pi)\times\mathcal L^{-1} \left\{\frac1{p^2+1}\right\}_{t\rightarrow t-\pi}\\ &=h(t-\pi)\times\sin(t)\Big|_{t\rightarrow t-\pi}\\ &=h(t-\pi)\times\sin(t-\pi)\\ &=-h(t-\pi)\sin(t)\\ \end{align*}\]

  2. UnkleRhaukus
    • 3 years ago
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    @AccessDenied

  3. UnkleRhaukus
    • 3 years ago
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    have i got this one right/?

  4. AccessDenied
    • 3 years ago
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    This appears correct to me. :)

  5. UnkleRhaukus
    • 3 years ago
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    thanks!, would you more likely leave the answer in the final form of \[=h(t−π)\sin(t−π)\]or \[=−h(t−π)\sin(t)\]

  6. AccessDenied
    • 3 years ago
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    I would probably use -h(t-pi) sin(t) just because I prefer to see the simplest forms of functions..

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