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UnkleRhaukus

  • 2 years ago

Laplace transforms \[F(s)=\mathcal L\left\{ f(t)\right\}=\int\limits_0^\infty f(t)e^{-st}\text dt\]

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  1. UnkleRhaukus
    • 2 years ago
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    [1]\[\mathcal L\left\{ 1\right\}=\int\limits_0^\infty e^{-st}\text dt=\left.\frac{e^{-st}}{-s}\right|_0^\infty=\frac{0-1}{-s}=\frac 1s\]

  2. UnkleRhaukus
    • 2 years ago
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    [2]\[\mathcal L\left\{ t^n\right\}=\int\limits_0^\infty t^ne^{-st}\text dt\] \[\qquad\qquad\text{let } t=\frac us\qquad\qquad\text dt=\frac{\text du}s\]\[\qquad\qquad t=0\rightarrow u=0\qquad t=\infty\rightarrow u=\infty\] \[=\int\limits_0^\infty \left(\frac up\right)^ne^{-s\frac us}\frac{\text du}s\]\[=\frac{1}{s^{n+1}}\int\limits_0^\infty u^ne^{-u} {\text du}\]\[=\frac{\Gamma(n+1)}{s^{n+1}}\][3]\[=\frac{n!}{s^{n+1}},\qquad n\in \mathbb N\]

  3. UnkleRhaukus
    • 2 years ago
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    [4]\[\mathcal L\left\{ e^{-bt}\right\}=\int\limits_0^\infty e^{-bt}e^{-st}\text dt=\int\limits_0^\infty e^{-(s+b)t}\text dt=\left. \frac{e^{-(s+b)t}}{-(s+b)}\right|_0^\infty=\frac 1{s+b}\]

  4. UnkleRhaukus
    • 2 years ago
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    [5]\[\mathcal L\left\{ \sin(nt)\right\}=\int\limits_0^\infty \sin(nt)e^{-st}\text dt\]\[=\int\limits_0^\infty {\frak {I}}\left(e^{int}\right)e^{-st}\text dt\]\[={\frak {I}}\left(\int\limits_0^\infty e^{int}e^{-st}\text dt\right)=\frak I\left(\int\limits_0^\infty e^{-(s-in)t}\text dt\right)= {\frak {I}}\left(\left.\frac{e^{-(s-in)t}}{-(s-in)}\right|_0^\infty\right)\]\[= {\frak I}\left(\frac{1}{s-in}\right)= {\frak I}\left(\frac{1}{s-in}\times\frac{s+in}{s+in}\right)= {\frak I}\left(\frac{s+in}{s^2+n^2}\right)= \frac{n}{s^2+n^2}\]

  5. UnkleRhaukus
    • 2 years ago
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    [6]\[\mathcal L\left\{\cos (nt)\right\}=\int\limits_0^\infty \cos(nt)e^{-st}\text dt\]\[=\int\limits_0^\infty {\frak R}\left(e^{int}\right)e^{-st}\text dt\]\[={\frak R}\left(\int\limits_0^\infty e^{int}e^{-st}\text dt\right)={\frak R}\left(\int\limits_0^\infty e^{-(s-in)t}\text dt\right)= {\frak R}\left(\left.\frac{e^{-(s-in)t}}{-(s-in)}\right|_0^\infty\right)\]\[= {\frak R}\left(\frac{1}{s-in}\right)= {\frak R}\left(\frac{1}{s-in}\times\frac{s+in}{s+in}\right)= {\frak R}\left(\frac{s+in}{s^2+n^2}\right)= \frac{s}{s^2+n^2}\]

  6. UnkleRhaukus
    • 2 years ago
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    [7]\[\mathcal L\left\{\sinh (nt)\right\}=\int\limits_0^\infty \sinh(nt)e^{-st}\text dt\]\[=\int\limits_0^\infty \frac{e^{nt}-e^{-nt}}2e^{-st}\text dt\]\[=\frac 12\int\limits_0^\infty{e^{-(s-n)t}-e^{-(s+n)t}}\text dt\]\[=\frac 12\left(\left.\frac {e^{-(s-n)t}}{-(s-n)}-\frac{e^{-(s+n)t}}{-(s+n)}\right)\right|_0^\infty\]\[=\frac12\left(\frac {1}{s-n}-\frac{1}{s+n}\right)\]\[=\frac12\left(\frac {(s+n)-(s-n)}{s^2-n^2}\right)\]\[=\frac{n}{s^2-n^2}\] [8]\[\mathcal L\left\{\cosh (nt)\right\}=\int\limits_0^\infty \cosh(nt)e^{-st}\text dt\]\[=\int\limits_0^\infty \frac{e^{nt}+e^{-nt}}2e^{-st}\text dt\]\[=\frac 12\int\limits_0^\infty{e^{-(s-n)t}+e^{-(s+n)t}}\text dt\]\[=\frac 12\left(\left.\frac {e^{-(s-n)t}}{-(s-n)}+\frac{e^{-(s+n)t}}{-(s+n)}\right)\right|_0^\infty\]\[=\frac12\left(\frac {1}{s-n}+\frac{1}{s+n}\right)\]\[=\frac12\left(\frac {(s+n)+(s-n)}{s^2-n^2}\right)\]\[=\frac{s}{s^2-n^2}\]

  7. UnkleRhaukus
    • 2 years ago
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    **in [2] p=s

  8. experimentX
    • 2 years ago
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    all right ... find the Laplace transform of |sin(x)|

  9. experimentX
    • 2 years ago
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    * \( |\sin (t) |\)

  10. UnkleRhaukus
    • 2 years ago
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    \[\mathcal L\left\{ \left|\sin(t)\right|\right\}=\int\limits_0^\infty \left|\sin(t)\right|e^{-st}\text dt\]\[\qquad\qquad\qquad=\]

  11. experimentX
    • 2 years ago
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    yep ... I'll give you more general problem. Question directly from my Question paper. if f(x+l) = f(x), how that it's Laplace Transform is \[ \huge { \int_0^L e^{-st}f(t) dt \over 1 - e^{sL} }\]

  12. experimentX
    • 2 years ago
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    \( f(t+L) = f(t) \)

  13. UnkleRhaukus
    • 2 years ago
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    what is \(L\) ?

  14. UnkleRhaukus
    • 2 years ago
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    of course

  15. experimentX
    • 2 years ago
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    periodic functions are represented as f(x+T) = f(x) .. T would be period.

  16. UnkleRhaukus
    • 2 years ago
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    \[\mathcal L\left\{t\sin(nt)\right\}=\int\limits_0^\infty t\sin(nt)e^{-pt}\text dt\] \[=\int_0^\infty t\left(\frac{e^{int}-e^{-int}}{2i}\right)e^{-pt}\text dt\] \[=\frac 1{2i}\int_0^\infty t({e^{int}-e^{-int}})e^{-pt}\text dt\] \[=\frac 1{2i}\left[\int_0^\infty t\cdot e^{-(p-in)t}\text dt-\int_0^\infty t\cdot e^{-(p+in)t}\text dt\right]\] \[=\frac 1{2i}\left[\left(\left.\frac{t\cdot e^{-(p-in)}}{-(p-in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p-in)t}}{-(p-in)}\text dt\right)\right.\]\[\qquad\qquad\left.-\left(\left.\frac{t\cdot e^{-(p+in)}}{-(p+in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p+in)t}}{-(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\left(0+\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt\right)-\left(0+\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt-\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right]\] \[=\frac 1{2i}\left[\left.\frac{e^{-(p-in)t}}{-(p-in)^2}-\frac{e^{-(p+in)t}}{-(p+in)^2}\right|_0^\infty\right]\] \[=\frac 1{2i}\left[\frac{1}{(p-in)^2}-\frac{1}{(p+in)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p+in)^2-(p+in)^2}{(p-in)^2(p-in)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p^2-2ipn-n^2)-(p^2+2ipn-n^2)}{\left((p-in)(p+in)\right)^2}\right]\] \[=\frac 1{2i}\left[\frac{-4ipn}{\left(p^2+n^2\right)^2}\right]\] \[=\frac{2pn}{\left(p^2+n^2\right)^2}\]

  17. UnkleRhaukus
    • 2 years ago
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    \[\mathcal L\left\{t\cos(nt)\right\}=\int\limits_0^\infty t\cos(nt)e^{-pt}\text dt\] \[=\int\limits_0^\infty t\left(\frac{e^{-int}+e^{int}}2\right)e^{-pt}\text dt\] \[=\frac12\left[\int\limits_0^\infty t{e^{-(p+in)t}\text dt+\int\limits_0^\infty te^{-(p-in)t}}\text dt\right]\] \[=\frac12\left[\left(\left.\frac{t\cdot e^{-(p+in)t}}{-(p+in)}\right|_0^\infty-\int\limits_0^\infty\frac{e^{-(p+in)t}}{-(p+in)}{\text dt}\right)\right.\]\[\qquad\qquad \left.+\left(\left.\frac{t\cdot e^{-(p-in)t}}{-(p-in)}\right|_0^\infty-\int\limits_0^\infty\frac{e^{-(p-in)t}}{-(p-in)}{\text dt}\right)\right]\] \[=\frac12\left[\left(0+\int\limits_0^\infty\frac{e^{-(p+in)t}}{(p+in)}{\text dt}\right)+\left(0+\int\limits_0^\infty\frac{e^{-(p-in)t}}{(p-in)}{\text dt}\right)\right]\] \[=\frac12\left[\int\limits_0^\infty\frac{e^{-(p+in)t}}{(p+in)}{\text dt}+\int\limits_0^\infty\frac{e^{-(p-in)t}}{(p-in)}{\text dt}\right]\] \[=\frac12\left[\left.\frac{e^{-(p+in)t}}{(p+in)^2}+\frac{e^{-(p-in)t}}{(p-in)^2}\right|_0^\infty\right]\] \[=\frac12\left[\frac{1}{(p+in)^2}+\frac{1}{(p-in)^2}\right]\] \[=\frac12\left[\frac{(p-in)^2+(p+in)^2}{(p+in)^2(p-in)^2}\right]\] \[=\frac12\left[\frac{(p^2-2inp-n^2)+(p^2+2inp-n^2)}{\left((p+in)(p-in)\right)^2}\right]\] \[=\frac12\left[\frac{2p^2-2n^2)}{\left(p^2+n^2\right)^2}\right]\] \[=\frac{p^2-n^2}{\left(p^2+n^2\right)^2}\]

  18. UnkleRhaukus
    • 2 years ago
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    ta da

  19. UnkleRhaukus
    • 2 years ago
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    p=s, n= \(\omega\)

  20. experimentX
    • 2 years ago
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    there you go http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

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