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?

 

http://www.youtube.com/watch?v=Y8GXpS31CGI
Why bother guessing?

 

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?

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.

 

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.