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0. Question

tl;dr: just see section 2. c)

I'd like to know how a non-linear system of first order ODEs (non-linear dynamical system) which was linearized at a point which is not the equilibrium of the dynamical system, can be discretized exactly.

I know that for a linear system linearized at an equilibrium state but struggle with the exact discretization of the $0$-order term of the Taylor approximation.

The resources I could find so far, all cover only the case of discretizing a (linearized) linear system without the $0$-order term of the Taylor linearization.

Below I have composed where I am and where I struggle at the linearization.

1. Findings: General Linearization

Be $f$ a non-linear time-invariant dynamical system, $x\in \mathbb{R}^n$ the state vector, $u\in \mathbb{R}^m$ an arbitrary input vector: $$ \dot{x}(t)=f(x(t),u(t)) $$ Let the tuple $(\bar{x},\bar{u})$ be the linearization point and $x_{\delta}(t) = x(t) - \bar{x}$ and $u_{\delta}(t) = u(t) - \bar{u}$ be the delta-state resp. the delta-input, the linearization of $f$ (ignoring Taylor terms of order 2 and higher) is given as $$ \dot{x}(t) = A x_{\delta}(t) + B u_{\delta}(t) + \bar{f} \\ \text{with} \quad A = \left. \frac{\partial}{\partial x} f(x,u) \right|_{\substack{x = \bar{x} \\ u = \bar{u}}}, \quad B = \left. \frac{\partial}{\partial u} f(x,u) \right|_{\substack{x = \bar{x} \\ u = \bar{u}}} \quad \text{and} \quad \bar{f} = f(\bar{x},\bar{u}) \text{.} \\ $$ Note: $\bar{f} \neq 0$ as $(\bar{x},\bar{u})$ is not constrained to be an equilibrium point.

2. a) Findings: Exact Discretization of linear system

A linear system of the form $$ \dot{x}(t) =A x(t) + B u(t) $$ can be linearized exactly as $$ x[k+1]=A_d x[k] + B_d u[k]\\ \text{with} \quad A_d = e^{A_{l}T} \quad \text{and} \quad B_d = \int\limits_{0}^{T} e^{A_{l}\tau} B_{l} d \tau \text{.} $$

2. b) Struggling: Exact Discretization of linearized system

Suppose I have this linearized system, $$ \dot{x}(t) =A x(t) + B u(t) + \bar{f} $$ I now struggle to with finding a term $f_d$ analogous to $A_d$ and $B_d$ for a discrete linear system of the form $$ x[k+1]=A_d x[k] + B_d u[k] + f_d $$

2. c) Guess for approach: Solve differential equation of linear system??

I also know the solution of differential equation of the linear state space system $$ \dot{x} = Ax(t) + Bu(t), \quad x(t_0) = x_0 $$ which is $$ x(t) = e^{A(t-t_0)}x_0 + \int_{t_0}^{t} e^{A(t-\tau)}Bu(\tau) d\tau \text{.} $$ and very likely must have been used in the exact linearization of the system in section 2a.

I'm not sure, but maybe solving the differential equation of this system $$ \dot{x}(t) = A x(t) + B u(t) + \bar{f} $$ might yield to a term for $f_d$? Would this be the right approach (and if yes, how would you approach to solve this ODE)?


Another course for mistakes might be messing up the notation of the delta-states and -inputs because when discretizing a linearized system, this linearized system is in its delta-notation as written in 1.

I appreciate any comments and thank you for your help!

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closed as unclear what you're asking by Did, Error 404, José Carlos Santos, Shailesh, Cyclohexanol. Oct 22 '17 at 3:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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The solution of the initial value problem $$ \dot x= Ax+f(t),\; x(t_0)=x_0 $$ is equal to $$ x(t)=e^{A(t-t_0)}x_0+\int\limits_{t_0}^te^{A(t-\tau)} f(\tau)\,d\tau. $$ Using this fact, we can obtain $$ x[k+1]=e^{AT}x[k]+\int\limits_{t_0}^te^{A(t-\tau)} (Bu[k]+\bar f)\,d\tau $$ $$ =e^{AT}x[k]+\int\limits_{0}^Te^{A(T-\tau)} B\,d\tau\,\cdot u[k]+ \int\limits_{0}^Te^{A(T-\tau)}\bar f \,d\tau $$ $$ =e^{AT}x[k]-\int\limits_{0}^Te^{A(T-\tau)} B\,d(T-\tau)\,\cdot u[k]- \int\limits_{0}^Te^{A(T-\tau)} \bar f \,d(T-\tau)\, $$ $$ =e^{AT}x[k]+\int\limits_{0}^Te^{A\tau} B\,d\tau\,\cdot u[k]+ \int\limits_{0}^Te^{A\tau} \bar f\,d\tau. $$ Hence, $$ f_d= \int\limits_{0}^Te^{A\tau} \bar f\,d\tau. $$

(Let me also note that this result is likely unuseful to control the original nonlinear system. Since $\bar f\ne 0$, $x[k+1]$ may differ a lot from $x[k]$ and $\bar x$, thus $A_d$, $B_d$, $f_d$ that we have found for the point $\bar x$ can become inadequate)

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  • $\begingroup$ Thanks, AVK, for your answer! While going through it, I have to admit, I unfortunately have not encountered the change of $\int_{\tau=0}^{\tau=T} \dots d\tau$ to $-\int_{T-\tau=0}^{T-\tau=T} \dots d(T-\tau)$. (Are this integral limit labels correct?) So far I would guess, that it is possible to replace $d\tau$ with $-d(T-\tau)=d(\tau-T)$ as $T$ is a constant and does not influence the infinitesimal small nature of $d\tau$? May I ask you to give me a hint or term to look up on that? Thanks again! $\endgroup$ – hrd Oct 21 '17 at 15:25
  • $\begingroup$ @hrd It is an integration by substitution. Denote $T-\tau=z$, $\int_0^Te^{A(T-\tau)}\bar f\, d\tau=\int_{T-z=0}^{T-z=T} e^{Az}\bar f\, d(T-z)=\int_{z=T}^{z=0} e^{Az}\bar f\, \frac{d(T-z)}{dz}\,dz= -\int_{z=T}^{z=0} e^{Az}\bar f\, dz=\int_{z=0}^{z=T} e^{Az}\bar f\, dz$ $\endgroup$ – AVK Oct 21 '17 at 18:04

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