Discrete State Space Representation of a second order system

Question

A second order disturbance-based model can be represented as follows:

$$\ddot{y}=f(y,\dot{y},w,t)+bu(t)$$

where $f(y,\dot{y},w,t)$ is generalized disturbance. This can be represented in continuous time state as:

$$\dot{x}=Ax(t)+Bu(t)+E\dot{f}(t)$$ $$y(t)=Cx(t)$$

where, $A= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$, $B=\begin{bmatrix}0\\b\\0\end{bmatrix}$, $C=\begin{bmatrix}1 & 0 & 0\end{bmatrix}$, $E=\begin{bmatrix}0\\0\\1\end{bmatrix}$ and $x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}y\\\dot{y}\\f\end{bmatrix}$.

When the above state space equation is discretized, we can get the following:

$$x(k+1)=\Phi x(k)+\Gamma u(k)+E_df(k+1)$$ $$y(k)=Cx(k)$$

My question is,

1. Is the term $f(k+1)$ for discretized $\dot{f}(t)$ correct or should it be $f(k)$?
2. How do you determine the term $E_d$ associated with $\dot{f}(t)$ using zero order hold?

Answer 1

In continuous time, you assume that $\dot{f}$ exists and can be written as a function $\xi(t):=\dot{f}(t)$. Then you can write your model as $$\dot{x}=Ax + Bu + E\xi$$ making the difference between the states and external inputs more explicit.

Following the same idea, in discrete time you have to assume the existence of a function $\Delta(k)$ such that $f(k+1)=f(k)+T_s\Delta(k)$, where $T_s$ is the sampling time. Then your model becomes $$x(k+1) = \Phi x(k) + \Gamma u(k) + E_d \Delta(k),$$ where the last line in $\Phi$ is $\begin{bmatrix}0 & 0 & 1\end{bmatrix}$ and $E_d = \begin{bmatrix}0 & 0 & T_s\end{bmatrix}^\top$.

Obviously, a good guess for $\Delta(k)$ is $$\Delta(k) = \frac{1}{T_s}\int_{t_k}^{t_{k+1}}\dot{f}(s)ds.$$

Answer 2

The analytical solution of the differential equation is as follows:

$$ x(t) = e^{At} x(t_0) + \int_{t_0}^t e^{A(t-\tau)} B u(\tau) d\tau + \int_{t_0}^t e^{A(t-\tau)} E \dot{f}(\tau) d\tau $$

To discretize this system we have to find the solution at $(k+1) T$ (assuming $t_0=0$). So,

$$\begin{align} x((k+1) T) &= e^{A(k+1) T} x(0) + \int_{0}^{(k+1) T} e^{A((k+1) T-\tau)} B u(\tau) d\tau + \int_{0}^{(k+1) T} e^{A((k+1) T-\tau)} E \dot{f}(\tau) d\tau \\ &= e^{A(k+1) T} x(0) + \int_{0}^{kT} e^{A((k+1) T-\tau)} B u(\tau) d\tau + \int_{kT}^{(k+1) T} e^{A((k+1) T-\tau)} B u(\tau) d\tau + \\ &\phantom{=} \int_{0}^{kT} e^{A((k+1) T-\tau)} E \dot{f}(\tau) d\tau + \int_{kT}^{(k+1) T} e^{A((k+1) T-\tau)} E \dot{f}(\tau) d\tau \\ &= e^{AT} x(kT) + \int_{kT}^{(k+1) T} e^{A((k+1) T-\tau)} B u(\tau) d\tau + \int_{kT}^{(k+1) T} e^{A((k+1) T-\tau)} E \dot{f}(\tau) d\tau \end{align}$$

Define $\eta = (k+1) T-\tau$. Then, $$x((k+1) T) = e^{AT} x(kT) + \int_{0}^{T} e^{A\eta} B u((k+1) T-\eta) d\eta + \int_{0}^{T} e^{A\eta} E \dot{f}((k+1) T-\eta) d\eta $$

Assuming zero-order hold on $u(t)$ we can write $u((k+1) T-\eta) = u(kT)$ to obtain the well-known discretization result $$x((k+1) T) = e^{AT} x(kT) + \left( \int_{0}^{T} e^{A\eta} d\eta \right) B u(kT) + \int_{0}^{T} e^{A\eta} E \dot{f}((k+1) T-\eta) d\eta $$ Obviously, if you have further information on $\dot{f}$, you can go further. But taking the average of $\dot{f}$ gives you only an approximation, which gets worse when $T$ gets bigger.

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Edit: I've misread the question. The above calculations is true only when $f$ is independent from the states. Otherwise one needs to linearize the system around an equilibrium point.

Answer 3

Your dynamics would be equivalent to

$$ \dot{x}(t) = A\,x(t) + B^*\,u^*(t), \tag{1} $$

with

$$ B^* = \begin{bmatrix} B & E \end{bmatrix}, \quad u^*(t) = \begin{bmatrix} u(t) \\ \dot{f}(t) \end{bmatrix}. \tag{2} $$

This more standard expression for a continuous linear state space model has a known solution for discretization using zero order hold (so $u^*(t)=u^*[k]\,\forall\ k\,T\leq t<(k+1)T$ with $T$ the time step size), namely

$$ \begin{bmatrix} A_d & B^*_d \\ 0 & I \end{bmatrix} = e^{\begin{bmatrix} A & B^* \\ 0 & 0 \end{bmatrix}T}, \tag{3} $$

such that

$$ x[k+1] = A_d\,x[k] + B^*_d\,u^*[k]. \tag{4} $$

Using this for your system yields

$$ A_d = \begin{bmatrix} 1 & T & \frac{1}{2}T^2 \\ 0 & 1 & T \\ 0 & 0 & 1 \end{bmatrix}, \quad B^*_d = \begin{bmatrix} \frac{b}{2}T^2 & \frac{1}{6}T^3 \\ b\,T & \frac{1}{2}T^2 \\ 0 & T \end{bmatrix}. \tag{5} $$

If you want to express it using $f[k+1]$ instead of $\dot{f}[k]$ you can use that from zero order hold one gets

$$ f[k+1] = f[k] + T\,\dot{f}[k] \leftrightarrow \dot{f}[k] = \frac{f[k+1] - f[k]}{T}. \tag{6} $$

Substituting $(6)$ into $(4)$ yields

$$ x[k+1] = \hat{A}\,x[k] + \hat{B}\,v[k], \tag{7} $$

with

$$ v[k] = \begin{bmatrix} u[k] \\ f[k+1] \end{bmatrix}, \quad \hat{A} = \begin{bmatrix} 1 & T & \frac{1}{3}T^2 \\ 0 & 1 & \frac{1}{2}T \\ 0 & 0 & 0 \end{bmatrix}, \quad \hat{B} = \begin{bmatrix} \frac{b}{2}T^2 & \frac{1}{6}T^2 \\ b\,T & \frac{1}{2}T \\ 0 & 1 \end{bmatrix}. \tag{8} $$