# Exact discretization of non-linear dynamical system which was linearized at a non-equilibrium point. [closed]

## 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.

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)
• 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! – hrd Oct 21 '17 at 15:25
• @hrd It is an integration by substitution. Denote $T-\tau=z$, $\int_0^Te^{A(T-\tau)}\bar f\, d\tau=\int_{T-z=0}^{T-z=T} e^{Az}\bar f\, d(T-z)=\int_{z=T}^{z=0} e^{Az}\bar f\, \frac{d(T-z)}{dz}\,dz= -\int_{z=T}^{z=0} e^{Az}\bar f\, dz=\int_{z=0}^{z=T} e^{Az}\bar f\, dz$ – AVK Oct 21 '17 at 18:04