Admiral, unfortunately I never took a few EE courses when I still remembered math. Maybe somebody would have to be real smart to figger it out self-taught, without having it pounded into one's dense skull by a grumpy perfessor.

However, here are some "very basic" pointers to the s-domain stuff, not that I understand it well enough to use.

http://en.wikipedia.org/wiki/S-domain
The parameter s is a complex number:

s = σ + iω , with real numbers σ and ω.

A complex number has real and imaginary components. The first squiggle in the above equation is sigma (the real component).

The second squiggle that looks like a 'w' is omega. The i in this case is the imaginary multiplier, the root of -1.

'i' * omega is the imaginary component.

In many AC electrical engineering calculations, they assume sigma equals zero, so it degrades down to this in many filter transfer function examples--

s = iωNow here is another piece of the puzzle-- 'i' is considered by mathematicians to be the square root of -1. HOWEVER, 'j' is considered by electrical engineers to be the square root of -1. THEREFORE, in electrical engineer speak, you would see it written thisaway--

s = jωNow, OMEGA is the Angular Frequency--

http://en.wikipedia.org/wiki/Angular_frequency ω = (2 * pi) / T = 2 * pi * f

where:

ω is the angular frequency or angular speed (measured in radians per second),

T is the period (measured in seconds),

f is the ordinary frequency (measured in hertz) (sometimes symbolised with ν).

So in all of transfer function equations with the s variables scattered in them, you could substitute for s, for instance

(-1^.5) * 2 * pi * f
In other words, you could plug in any particular frequency and solve the damn thing and get the phase and amplitude response AT THAT FREQUENCY.

Here is JOS being typically opaque explaining the thang--

https://ccrma.stanford.edu/~jos/filters ... lysis.html
So anyway, maybe that is a "starter" if you want to try to puzzle it out. I'm gonna try to puzzle it out further, one of these days.