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Given system of ODE:

$\begin{cases} \dot{x}=G+u_1 \\ \dot{z}=-z+\frac{df}{dt} \\ \dot{G}=-G+z \cdot u_2 \end{cases}$

where $f=-x^2$,

$u_1=\frac{d}{dt}(\alpha \sin(\omega \cdot t))$ and $u_2=\alpha \sin(\omega \cdot t)$

f = -x[t]^2;
asys = AffineStateSpaceModel[{x'[t] == G[t] + u1[t], z'[t] + z[t] == D[f, t], G'[t] + G[t] == z[t]  u2[t]}, {{x[t], 1}, {G[t], 0}, {z[t], 0}}, {u1[t], u2[t]}, {x[t], z[t], G[t]}, t];

And here is the result of the numerical calculation:

pars = {\[Alpha] = 0.5, \[Omega] = 2 Pi 5, xs = 1, xe = 0};
s1 = \[Alpha] Sin[\[Omega] t];
or = OutputResponse[asys, {D[s1, t], s1}, {t, 0, 500}]
Plot[{or[[1]] - s1, xe}, {t, 0, 25}, PlotRange -> All, PlotPoints -> 100]

enter image description here

Next, I try to get the transfer function of the system:

tf = TransferFunctionModel[asys]

And results:

or = OutputResponse[tf, {D[s1, t], s1}, {t, 0, 500}]
Plot[{or[[1]]}, {t, 0, 5}, PlotRange -> All, PlotPoints -> 100]

enter image description here


Should the results be the same, and if not, why not?

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1 Answer 1

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Of course they are not going to be the same, because the linearization removed all the nonlinear terms.

The responses are being computed for these two different systems.

NonlinearStateSpaceModel /@ {asys, StateSpaceModel[asys]}

enter image description here

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  • $\begingroup$ Suba Thomasб hello! Thank you for your answer! I came across this question while trying to linearize a non-autonomous differential equation. I figured out the current question, but not yet with linearization. mathematica.stackexchange.com/questions/253133/… $\endgroup$
    – dtn
    Aug 10, 2021 at 13:15
  • $\begingroup$ In this form, the system does not lend itself to linearization, because it has no equilibrium points. Maybe the original $\dot{x}=-x^2+\frac{1}{t+1}+1$ can be linearized? A comment was left there, it is not an option, because non-autonomy can be more complex and some more or less universal approach is needed. $\endgroup$
    – dtn
    Aug 10, 2021 at 13:16

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