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UnkleRhaukus

  • one year ago

Can someone please check this; \[\begin{equation*} f(t)=\cos(t)\big(h(t-\pi)-h(t)\big) \end{equation*}\] \[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t \end{align*} \]

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  1. UnkleRhaukus
    • one year ago
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    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\ &=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\ &=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\ &=\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_\pi^\infty-\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_0^\infty\\ &=\frac{-e^{-p\pi}\big(-p\cos \pi+\sin \pi\big)}{p^2+1}+\frac{\big(-p\cos 0+\sin 0\big)}{p^2+1}\\ &=\frac{-pe^{-p\pi}}{p^2+1}-\frac{p}{p^2+1}\\ \\ &=\frac{-p\big(1+e^{-p\pi }\big)}{p^2+1}\\ \end{align*}\]

  2. whpalmer4
    • one year ago
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    I haven't had my coffee yet, but I don't quite see how you managed to make h(t) vanish

  3. phi
    • one year ago
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    I think h is the step function

  4. whpalmer4
    • one year ago
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    ah, okay, sure, I didn't realize that we could assume anything about h(t)

  5. experimentX
    • one year ago
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    isn't that Laplace transform? .. let p = s

  6. UnkleRhaukus
    • one year ago
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    The unit heaviside step function \[h(t-a) =\begin{cases}0&t<a\\ 1&t≥a\end{cases} \]

  7. experimentX
    • one year ago
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    |dw:1358356261290:dw|

  8. UnkleRhaukus
    • one year ago
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    h=H=u=θ

  9. experimentX
    • one year ago
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    just integrate negative of cos(x) e^(-px) from 0 to pi

  10. experimentX
    • one year ago
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    yes!! multiple ways of representing unit step function.

  11. UnkleRhaukus
    • one year ago
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  12. experimentX
    • one year ago
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    think (h(t−π)−h(t)) as a switch (toward negative) that get's on at t=0 and off at t=pi .

  13. experimentX
    • one year ago
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    yes that's correct!!

  14. experimentX
    • one year ago
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    http://www.wolframalpha.com/input/?i=Integrate+e^%28-pt%29+cos%28t%29+from+0+to+pi

  15. experimentX
    • one year ago
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    there's a short formula.for int e^(ax) cos(bx) dx ...also note that you can shift integral for step function.

  16. experimentX
    • one year ago
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    http://www.wolframalpha.com/input/?i=Integrate+e^%28at%29+cos%28bt%29

  17. experimentX
    • one year ago
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    |dw:1358358498070:dw|

  18. experimentX
    • one year ago
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    |dw:1358358737716:dw| this should be simpler.

  19. UnkleRhaukus
    • one year ago
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    Ah yeah I can see that now, one integral is simpler than two

  20. UnkleRhaukus
    • one year ago
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    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\ &=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\ &=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\ &=\intl\pi0{\cos(t)e^{-pt}}t\\ &=\left.\frac{e^{-pt}\big(\sin (t)+p\cos (t)\big)}{(-p)^2+1^2}\right|_\pi^0\\ &=\frac{\sin 0+p\cos 0-e^{-p\pi}\big(\sin \pi+p\cos \pi\big)}{p^2+1^2}\\ &=\frac{p+pe^{-p\pi}}{p^2+1^2}\\ &=\frac{p\left(1+e^{-p\pi}\right)}{p^2+1^2}\\&\color{red}\checkmark \end{align*} \]

  21. UnkleRhaukus
    • one year ago
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    thanks experimentX

  22. experimentX
    • one year ago
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    yw!! you work hard man!!

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