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Oct26-08, 10:54 PM
P: 2,251
Quote Quote by Mårten View Post
Before we go on - thank you very much for this personal class I get here in control theory! It helps me a lot!

Okey, sorry. That was to signal that they could be matrices or vectors. Bold face may be better...

Okey, I think I got it there. y(0) would just have been a transformation (made by the C-matrix) of the vector x(0).
yeah, and it would be nice if i practice what i preach. i should have said x(t) for the state vector instead of X(t).

Then comes the ultimate question: Why don't always use the state-space model in preference to the Laplace transform model? What benefits does the Laplace transform model have, which the state-space model doesn't have?
simplicity. if a system is known to be linear and time-invariant, and if you don't care about what's going on inside the black box, but only on how the system interacts (via its inputs and outputs) with the rest of the world that it is connected to, the input-output transfer function description is all that you need. if you have an Nth-order, single-input, single-output system, then 2N+1 numbers fully describe your system, from an input-output POV. with the state variable description, an Nth-order system (single input and output) has N2 numbers, just for the A matrix. and 2N+1 numbers for the B, C, and D matrices. so there are many different state-variable systems that have the same transfer function. all of these different systems behave identically with the outside world until some internal state saturates or goes non-linear and that's where they are modeled differently in practice. you can even have internal state(s) blow up and not even know it (if the system is "not completely observable") until the state(s) that are unstable hit the "rails", the maximum values they can take before clipping or some other non-linearity. when something like this happens, there is pole-zero cancellation, as far as the input-output description is concerned. so maybe some zero killed the pole and they both disappeared in the transfer function, G(s), but inside that bad pole still exists.