Inverse Laplace Transform

Discussion in 'Homework Help' started by jegues, Feb 19, 2014.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Evening gents,

    Can someone explain to me how,

    \Delta f(s) = \frac{-K}{1+sT} \Delta P_{d}(s)

    becomes

    \Delta f(t) = -K (1-e^{\frac{-t}{T}})|P_{d}(t)|

    when removed from the Laplace domain.

    I would have expected the following result,

    \Delta f(t) = \frac{-K}{T} e^{\frac{-t}{T}}P_{d}(t)

    I'm guessing maybe there was an initial condition imposed such that,

    \Delta f(t=0) = 0

    Any ideas?
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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  3. WBahn

    Moderator

    Mar 31, 2012
    17,716
    4,788
    What's the relationship between the function ΔP_d and the function P_d?

    In the s-domain you have two functions multiplied together. When you take the inverse Laplace transform that ends up being convolution in the time domain.
     
  4. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    Working backwards:

    <br />
<br />
\Delta f(t) = -K|P_{d}| + K|P_{d}|e^{-\frac{t}{T}}<br />
<br />

    Transforming to S-Domain:

    <br />
<br />
\Delta F(s) = \frac{-K|P_{d}|}{s} + \frac{K|P_{d}|}{s + \frac{1}{T}}<br />
<br />
\Delta F(s) = \frac{-K|P_{d}|}{s} + \frac{TK|P_{d}|}{1 + sT}<br />
<br />
\Delta F(s) = \frac{-K|P_{d}| - sTK|P_{d}| + sTK|P_{d}|}{s(1 + sT)} = \frac{-K|P_{d}|}{s(1 + sT)}<br />
<br />

    By this logic and assuming |P_d| is a constant I can only assume that Δf is some sort of step response to a given step function P_d(t), though the Δ notation is somewhat confusing.
     
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