# 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?

Feb 19, 2010
3,518
515
3. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
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
565
33
Working backwards:

$

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

$

Transforming to S-Domain:

$

\Delta F(s) = \frac{-K|P_{d}|}{s} + \frac{K|P_{d}|}{s + \frac{1}{T}}

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

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

$

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.