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How can I transpose the output of state space model in Mathematica? trying the following code would transpose the top right symbol s instead of the matrices.

Transpose @ StateSpaceModel @ TransferFunctionModel[{{a^2/(s^2 + b s + c)}}, s]

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  • $\begingroup$ Normal the model, operate on parts as desired, StateSpaceModel the result... $\endgroup$
    – ciao
    Jul 17, 2015 at 5:31
  • $\begingroup$ @ciao, I don't understand what you mean. $\endgroup$
    – ar2015
    Jul 17, 2015 at 5:34
  • $\begingroup$ You'll find all the details of those functions here $\endgroup$
    – ciao
    Jul 17, 2015 at 5:44
  • $\begingroup$ @ciao, i uesd normal. It tries to give me norm!! $\endgroup$
    – ar2015
    Jul 17, 2015 at 5:50
  • $\begingroup$ I'd venture you used Norm, instead of Normal? $\endgroup$
    – ciao
    Jul 17, 2015 at 6:09

2 Answers 2

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I am not sure what a valid transpose of a StateSpaceModel is but here is an attempt:

ssm = StateSpaceModel @ TransferFunctionModel[{{a^2/(s^2 + b s + c)}}, s]

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Transpose /@ #[[{1, 3, 2, 4}]] & /@ ssm

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  • $\begingroup$ Looks like perhaps they snuck some changes in either structure or made transpose aware of structure in 10.x: Latter code blows up on 9.x (I used StateSpaceModel@(Transpose /@ #[[{1, 3, 2, 4}]] &@(Normal@ssm)) for it to work). While that's a valid model, it's patently unclear what OP is after. Congratulations on 150, btw, and +1 here. $\endgroup$
    – ciao
    Jul 17, 2015 at 10:13
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To not be at the mercy of the internal representation of StateSpaceModel use DualSystemsModel.

ssm = StateSpaceModel @ TransferFunctionModel[{{a^2/(s^2 + b s + c)}}, s];
DualSystemsModel[ssm]

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Simplify[%, Element[a, Reals] && Element[b, Reals] && Element[c, Reals]]

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It will also work for descriptor systems.

ssm1 = StateSpaceModel@TransferFunctionModel[{{a^2 s^3/(s^2 + b s + c)}}, s];
DualSystemsModel[ssm1]

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