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UnkleRhaukus

  • 2 years ago

laplace transform of a periodic functions

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

  3. UnkleRhaukus
    • 2 years ago
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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

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

  5. Jonask
    • 2 years ago
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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

  6. UnkleRhaukus
    • 2 years ago
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    i have been using a program called TeXShop

  7. tkhunny
    • 2 years ago
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    How sure are you that it converges?

  8. UnkleRhaukus
    • 2 years ago
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    i dont know

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

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

  11. malevolence19
    • 2 years ago
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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.

  12. UnkleRhaukus
    • 2 years ago
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    geometric series?

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