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UnkleRhaukus

laplace transform of a periodic functions

  • one year ago
  • one year ago

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  1. UnkleRhaukus
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    • one year ago
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  2. Outkast3r09
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    What is this you're doing lol

    • one year ago
  3. UnkleRhaukus
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    i have taken the Laplace transform of some periodic functions i dont know how to check if i have done it right

    • one year ago
  4. UnkleRhaukus
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    i have used \[\mathcal L\{f(x)\}=\int\limits_0^\infty f(x)e^{-px}\text dx\] \[=\int\limits_0^\tau f(x)e^{-px}\text dx+\int\limits_\tau^{2\tau} f(x)e^{-px}\text dx+\int\limits_{2\tau}^{3\tau} f(x)e^{-px}\text dx+\dots\]\[=(1+e^{-p\tau}+e^{-2p\tau}+\dots)\int\limits_0^\tau f(z)e^{-pz}\text dz\]\[=\frac{1}{1-e^{-p\tau}}\int\limits_0^\tau f(z)e^{-pz}\text dz\]

    • one year ago
  5. Jonask
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    how do you type a document like that is it LaTex ,it looks great csry i cant help with the content but the presentation is fine hw do you create such a presentation

    • one year ago
  6. UnkleRhaukus
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    i have been using a program called TeXShop

    • one year ago
  7. tkhunny
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    How sure are you that it converges?

    • one year ago
  8. UnkleRhaukus
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    i dont know

    • one year ago
  9. malevolence19
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    As long as the function grows slower than Me^pt then it converges.

    • one year ago
  10. UnkleRhaukus
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    ok it converges because the slope is only positive or negative 1 , which is less slope than any exponential

    • one year ago
  11. malevolence19
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    Yeah, I'm not sure of the name of the theorem but I just went over that in one of my classes not too long ago.

    • one year ago
  12. UnkleRhaukus
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    geometric series?

    • one year ago
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