anonymous
  • anonymous
The Inverse laplace transform of \[ \fract{2 e^{- \pi s} }{s^2 + 2s +2} \] Is the following but I cant figure out how to get there, can someone help me? Thanks \[ -2 e^{\pi-t} sin(t) \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[ \frac{2e^{−πs}}{s2+2s+2} \] Sorry
experimentX
  • experimentX
\[ \frac{2 e^{-\pi s}}{s^2 + 2s+ 2} \]
anonymous
  • anonymous
First complete the square in the denominator. Doesn't look like it factors nicely.

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experimentX
  • experimentX
use convolution ... not sure if I used correct formula. \[ \frac{2 e^{-\pi s}}{s^2 + 2s+ 2} = \frac{2 e^{-\pi s}}{(s+1)^2 + 1} = \int_0^t u(\tau - \pi ) \sin (\tau) e^{-\tau} d\tau \]
experimentX
  • experimentX
woops!! looks like o
anonymous
  • anonymous
I'm not so sure convolution is necessary here. Given some Laplace transform \(e^{-as}F(s)\), its inverse transform would have the form \(f(t-a)\). Given some transform \(F(s+c)\), its inverse would have the form \(e^{-ct}f(t)\). Use these two properties for this problem.
experimentX
  • experimentX
|dw:1370903567555:dw| |dw:1370903666617:dw|
anonymous
  • anonymous
@SithsAndGiggles That works and lets me solve it pretty easy. Thanks! @experimentX I am not sure how I would integrate the unit step function. I learned what it's laplace transform is but not how to integrate or derive the function itself. And that integral seems pretty ugly so I think I'd rather just use what Stihs said.
experimentX
  • experimentX
of course you can do it by looking at table ... convolution is just alternative. still something is not working. I am fixing it.
experimentX
  • experimentX
The answer should be \( -2 e^{\pi - t} \sin(t) u(t-\pi )\) just plug this integral into Mathematica `Integrate[UnitStep[Tau - Pi] (Cos[t - Tau] - Sin[t - Tau]) E^(-(t - Tau)), {Tau, 0, t}, Assumptions :> t > 0]`
experimentX
  • experimentX
\[-2 e^{\pi-t} \sin(t)\] is incorrect
anonymous
  • anonymous
@experimentX I see that now. The full answer in the book is 'x(t)= 0 if ) <= t < pi, x(t) = -2e^(-(t-pi))sin(t) if t>= pi' I still am not sure how you would integrate the unit step function but thinking I need to go back and look at some of the early chapters in this book.
experimentX
  • experimentX
\[ \int_0^t u(\tau - \pi) \sin(t-\tau ) e^{-(t-\tau) }d\tau = \int_{\pi}^t \sin(t-\tau ) e^{-(t-\tau) }d\tau = u(t-\pi) \int_{\pi}^t \sin(t-\tau ) e^{-(t-\tau) }d\tau\] you can evaluate it. good luck .. I am also not very fond of step functions.
experimentX
  • experimentX
\[ \int_0^b u(t-a) f(t) dt = \int_a^b f(t)dt\] step function works like a switch.
anonymous
  • anonymous
Ahh, alright thanks. :)
experimentX
  • experimentX
yw

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