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UnkleRhaukus

inverse laplace transform

  • one year ago
  • one year ago

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  1. math>philosophy
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    ok

    • one year ago
  2. UnkleRhaukus
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    \[\begin{align*} F(p)&=\frac{7p-13}{p^2(p^2+6p+13)}\\ &=\frac{7p-13}{p^2(p^2+6p+9+4)}\\ &=\frac{7p-13}{p^2\big((p+3)^2+4\big)}\\ &=\frac{7}{p\big((p+3)^2+4\big)}-\frac{13}{p^2\big((p+3)^2+4\big)}\\ &=\frac{7}{p}\times\frac1{\big((p+3)^2+4\big)}-\frac{13}{p^2}\times\frac1{\big((p+3)^2+4\big)}\\ f(t)&=\frac72\mathcal L^{-1}\left\{\frac1{p}\times\frac2{\big((p+3)^2+2^2\big)}\right\}-\frac{13}2\mathcal L^{-1}\left\{\frac{1}{p^2}\times\frac2{\big((p+3)^2+2^2\big)}\right\}\\ &=\frac72\int\limits_0^t1\Big|_{t\rightarrow t-u}\times e^{-3u}\sin(2u)\text du-\frac{13}2\int\limits_0^tt\Big|_{t\rightarrow t-u}\times e^{-3u}\sin(2u)\text du\\ &=\frac72\int\limits_0^t e^{-3u}\sin(2u)\text du-\frac{13}2\int\limits_0^t(t-u)e^{-3u}\sin(2u)\text du\\&=\dots \end{align*}\]

    • one year ago
  3. UnkleRhaukus
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    am i doing this right/

    • one year ago
  4. UnkleRhaukus
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    how am i to evaluate the integral on the right ?

    • one year ago
  5. UnkleRhaukus
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    is there an easier method (using the convolution theorem)

    • one year ago
  6. AccessDenied
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    Integrals in the form \(\displaystyle \int e^{\alpha x} \sin \beta x \; \text{d}x \) have a general form that can be found through two integration by parts and some algebra with the original integral For \(\displaystyle \int x e^{\alpha x} \sin \beta x \; \text{d}x \), you can find it with the above's formula + cosine's and integration by parts: http://mathbin.net/114299 Only problem is that they seem fairly tedious to apply here...

    • one year ago
  7. UnkleRhaukus
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    agh,

    • one year ago
  8. mahmit2012
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    =1/p2+1/p-(p+3)/[(p+3)2+4]-2/[(p+3)2+4]->t+1-e^-3t(cost+sint)

    • one year ago
  9. UnkleRhaukus
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    ,

    • one year ago
  10. UnkleRhaukus
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    \[\begin{align*} F(p)&=\frac{7p-13}{p^2(p^2+6p+13)}\\ &=\frac{7p-13}{p^2(p^2+6p+9+4)}\\ &=\frac{7p-13}{p^2\big((p+3)^2+4\big)}\\ &=\frac{7}{p\big((p+3)^2+4\big)}-\frac{13}{p^2\big((p+3)^2+4\big)}\\ &=\frac{7}{p}\times\frac1{\big((p+3)^2+4\big)}-\frac{13}{p^2}\times\frac1{\big((p+3)^2+4\big)}\\ \end{align*}\]\[ \begin{align*} f(t)&=\frac72\mathcal L^{-1}\left\{\frac1{p}\times\frac2{\big((p+3)^2+2^2\big)}\right\}-\frac{13}2\mathcal L^{-1}\left\{\frac{1}{p^2}\times\frac2{\big((p+3)^2+2^2\big)}\right\}\\ &=\frac72\int\limits_0^t1\Big|_{t\rightarrow t-u}\times e^{-3u}\sin(2u)\text du-\frac{13}2\int\limits_0^tt\Big|_{t\rightarrow t-u}\times e^{-3u}\sin(2u)\text du\\ &=\frac72\int\limits_0^t e^{-3u}\sin(2u)\text du-\frac{13}2\int\limits_0^t(t-u)e^{-3u}\sin(2u)\text du\\ &=\left[\frac72\times\frac{e^{-3t}}{2^2+(-3)^2}\Big(-3\sin(2u)-2\cos(2u)\Big)\Big|_0^t\right]-\frac{13t}2\int\limits_0^te^{-3u}\sin(2u)\text du+\frac{13}2\int\limits_0^tue^{-3u}\sin(2u)\text du\\ &=\left[\frac7{26}\times{e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]-\left[\frac{13t}2\times\frac{e^{-3t}}{2^2+(-3)^2}\Big(-3\sin(2u)-2\cos(2u)\Big)\Big|_0^t\right]\\ &\qquad\qquad\qquad\qquad\qquad-\frac{13}2\left[u\int\limits_0^te^{-3u}\sin(2u)\text du\Big|_0^t-\int\int_0^te^{-3u}\sin(2u)\text du\text du\right]\\ &=\left[\frac7{26}{e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]-\left[\frac{t}2{e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]\\ &\qquad\qquad\qquad-\frac{13}2\left[u{e^{-3t}}\Big(-3\sin(2u)-2\cos(2u)\Big)\Big|_0^t-\int{e^{-3t}}\Big(-3\sin(2u)-2\cos(2u)\Big)\text du\right]\\ &=\left[\left(\frac7{26}-\frac t2\right){e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]\\ &\qquad\qquad\qquad-\frac{13}2\left[t{e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)-2\Big)-\int{e^{-3t}}\Big(-3\sin(2u)-2\cos(2u)\Big)\text du\right]\\ &=\left[\left(\frac7{26}-\frac t2-\frac{13t}{2}\right){e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]\\ &\qquad\qquad\qquad-3\int\limits_0^t{e^{-3t}}\sin(2u)\text du-2\int\limits_0^te^{-3t}\cos(2u)\Big)\text du\\ &=\left[\left(\frac7{26}-\frac t2-\frac{13t}{2}\right){e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)\right]\\ &\qquad-3\times\frac{e^{-3t}}{2^2+(-3)^2}\Big(-3\sin(2u)-2\cos(2u)\Big)\Big|_0^t-2\times\frac{e^{-3u}}{2^2+(-3)^2}\left(-3\cos(2u)+2\sin(2u)\right)\Big|_0^t\\ &=\left(\frac7{26}-\frac t2-\frac{13t}{2}+\frac{3}{13}\right){e^{-3t}}\Big(-3\sin(2t)-2\cos(2t)+2\Big)-\frac{2}{13}e^{-3t}\left(-3\cos(2t)+2\sin(2t)-3\right)\\ &=\left[\left(\frac12-7t\right)\Big(-3\sin(2t)-2\cos(2t)+2\Big)-\frac{2}{13}\Big(-3\cos(2t)+2\sin(2t)-3\Big)\right]e^{-3t}\\ &=\left[\Big(\frac12-7t-\frac4{13}\Big)\sin(2t)+\Big(-1+14t\Big)\cos(2t)+\Big(1-14t+\frac6{13}\Big)\right]e^{-3t}\\ &=\left[\Big(\frac5{26}-7t\Big)\sin(2t)+\Big(-1+14t\Big)(1-2\sin^2t)+\Big(\frac{19}{13}-14t\Big)\right]e^{-3t}\\ &=\left[\Big(\frac5{26}-7t\Big)\sin(2t)+14t\sin^2t+\Big(\frac{6}{13}\Big)\right]e^{-3t}\\ \\ \end{align*}\]

    • one year ago
  11. mahmit2012
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    =1/p2+1/p-(p+3)/[(p+3)2+4]-2/[(p+3)2+4]->t+1-e^-3t(cos2t+sin2t) I guess you wrong.

    • one year ago
  12. mahmit2012
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    |dw:1353823095307:dw|

    • one year ago
  13. mahmit2012
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    |dw:1353823272401:dw|

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  14. mahmit2012
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    |dw:1353823506027:dw|

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  15. mahmit2012
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    |dw:1353823547140:dw|

    • one year ago
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