Control theory: Laplace versus state space representationby Mårten Tags: control theory, laplace transform, state space 

#19
Nov1408, 10:57 PM

P: 2,265

Marten,
i just now saw this thread was left hanging. i just received a few infractions from Evo (from a totally different thread) and i need some time to think about what you were asking in post #17. there is a straightforward way to go from statespace representation to transfer function. it's in the text books but it's pretty easy to derive. of course, there is loss of information (since many different statespace representations can turn into a single transferfunction representation). it's maybe not as straightforward, but you can combine the states of one box and the states of another box (in series or parallel) into a single box where the number of states is the sum of the two and where the states are defined exactly as they were before. 



#20
Nov1508, 10:58 AM

P: 127

The reason I'm so eager to stick to state space representations, is that I would like to be able to handle, and to fully understand, systems with multiple inputs/outputs and with arbitrary initial values, and still have pretty simple calculations (for instance, arbitrary initial values is not good for transfer functions cause then simplicity disappears; multiple inputs/outputs only possible to handle with state space representations). u > [itex]G_1[/itex] > v > [itex]G_2[/itex] > y where [itex]G_1: a_1v'' + b_1v' + c_1v = u[/itex] and [itex]G_2: a_2y'' + b_2y' + c_2y = v[/itex], i.e. G_1 has u and v as input/output and G_2 has v and y as input/output. For simplicity, [itex]a_1 = a_2 = 1[/itex]. The state space representation for this then becomes (as far as I can see) [tex] \left( \begin{array}{c} x'_1 = v' \\ x'_2 = v'' \\ x'_3 = y' \\ x'_4 = y'' \\ \end{array} \right) = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ c_1 & b_1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & c_2 & b_2 \\ \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ u \\ 0 \\ 0 \\ \end{array} \right). [/tex] Okey, it seems to work! The output from G_1 becomes the input to G_2 through the number 1 in the Amatrix' lower left corner (element [itex]a_{41}[/itex], sort of the "chain element" which connects the two boxes). Now comes the question of how to solve this, ehhh... Using eigenvalue techniques or [itex]e^{At}[/itex]matrices, and so forth, maybe leading to unsolvable nth degree equations (n>4). Or simply using a computer... 



#21
Nov1508, 03:05 PM

P: 1,070

yeah, it's been a good 12 or 13 years since i modelled state space systems, which is why i haven't tried to help. but yeah, computing the discrete statespace matrix from the continuous time transfer function is done with a computer. something like Matlab or Scilab, maybe Octave, should have the functions available. Matlab i know for sure does, because that's what i used to use.




#22
Nov1808, 07:25 PM

P: 127

Of course, MATLAB would do the trick! I should know that.
Okey, I guess I won't get out any more in this particular area. Thanks for all help! Don't hesitate to comment further, what I've said above. I've now started a new thread on the topic of frequency response, see here. 



#23
Nov1908, 03:29 PM

P: 1,070

well, it requires computing a matrix exponential, doesn't it? iirc, there's really no good way to do that by hand.



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