Plant Identification

Second Order LTI

Full State Access

We recall that the state space representation for a discrete time second order LTI plant is :

$$\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}_{k+1} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix} \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}_{k} + \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix} \begin{pmatrix} u \end{pmatrix}_{k}$$

or

$$\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}_{k+1} = \begin{pmatrix} a_{11} & a_{12} & b_1 \\ a_{21} & a_{22} & b_2 \end{pmatrix} \begin{pmatrix}x_1 \\ x_2 \\ u \end{pmatrix}_{k}$$

We notice that this is also the equation for a single layer ANN with 3 inputs, 2 outputs, no bias and a linear activation function.

keras code

Fig1. - SO LTI test trajectory, Full State Plant Identification.

Input/Output Access

$$\theta_{k+1} = f(\theta_k, \theta_{k-1}, u_k, u_{k-1})$$

As our plant is linear, $f$ can be written

$$\theta_{k+1} = -b_1.\theta_k -b_0.\theta_{k-1} + a_1.u_k + a_0.u_{k-1}$$

which leads to the following transfer function

$$\frac{\theta(z)}{u(z)} = \frac{a_1.z+a_0}{z^2+b_1.z+b_0}$$

keras code

Fig1. - SO LTI test trajectory, Input/Output Plant Identification with sklearn.