# Develop linear parametric model of a third order plant

Consider we have the third order plant described by the equation:

$$y = G(s)u$$

where $$\ G(s) = \frac{b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_2s^2+a_1s+a_0}$$

If we assume that all parameters $$\ a_0,a_1,a_2,b_0,b_1,b_2$$ are unknown then we can develos a linear parametric model of the form:

$$z = θ^{*Τ}φ$$

where

$$\ θ^*=[a_2\ a_1\ a_0\ b_2\ b_1\ b_0]^T$$, $$\ φ = [-\frac{s^2}{Λ(s)}y \ \ \ -\frac{s}{Λ(s)}y \ \ \ -\frac{1}{Λ(s)}y \ \ \ -\frac{s^2}{Λ(s)}u \ \ \ -\frac{s}{Λ(s)}u \ \ \ -\frac{1}{Λ(s)}u]^T$$ ,

$$\ z = \frac{s^3}{Λ(s)}y$$

Now, let's suppose that the parameters $$\ a_2,a_1,a_0$$ are known and their values are:

$$\ a_2 = 3$$

$$\ a_1 = 1$$

$$\ a_0 = 2$$

How can we develop a linear parametric model of the plant considering these parameters known and as a result the vector $$\ θ$$ should be $$\ θ^* = [b_2 \ \ \ b_1 \ \ \ b_0]^T$$ ?

• Can't you just move the now considered known terms to the other side of the equals sign and thus change your definition of $z$? Jan 30, 2020 at 0:20
• I thought what you said but wasn’t sure if it is the correct way of doing the parameterization. Jan 30, 2020 at 1:03

As fibonatic commented the proper linear parameterization in terms of the unknown $$\ b_2,b_1,b_0$$ parameters is as follows:

$$\ y = G(s)u$$ is in terms of a differential equation:

$$\ \dddot{y} + a_2\ddot{y}+a_1\dot{y}+a_0y = b_2\ddot{u}+b_1\dot{u}+b_0u$$

Due to the fact that we only have measurements of the input $$\ u$$ and output $$\ y$$ of the system, we can't use their derivatives. As a result, we filter each term with a third order stable filter (poles of filter need to be negative) $$\ Λ(s) = s^3 + λ_1s^2 + λ_2s + λ_3$$ where the coefficients $$\ λ_1,λ_2,λ_3$$ are chosen though the poles of the filter. Filtering the above differential equation results in no use of differentiators and produces the following equation:

$$\ \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ = b_2\frac{s^2}{Λ(s)}u \ +b_1\frac{s}{Λ(s)}u \ + b_0\frac{1}{Λ(s)}u$$

Let's define $$\ z = \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ \$$ since $$\ Λ(s),a_2,a_1,a_0,y$$ are known. Our equation now becomes:

$$\ z = [b_2 \ \ b_1 \ \ b_0] [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T$$ which is in the form: $$\ z = θ^{*Τ}φ$$ with $$\ θ^{*} = [b_2 \ \ b_1 \ \ b_0]^T$$ and $$\ φ = [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T$$. So, now we have the linear parametric model in terms of the unknown parameters $$\ b_2, b_1,b_0$$.

Correspondingly we can come up with the linear parametric model in terms of the parameters $$\ a_2,a_1,a_0$$ being unknown and the parameters $$\ b_2,b_1,b_0$$ being known. Following the same procedure, the linear parametric model is:

$$\ z = θ^{*Τ}φ$$ where $$\ z = \frac{s^3}{Λ(s)}y \ - b_2\frac{s^2}{Λ(s)}u \ - b_1\frac{s}{Λ(s)}u - b_0\frac{1}{Λ(s)}u \ \$$ since $$\ Λ(s),b_2,b_1,b_0,y$$ are known, $$\ θ^{*} = [a_2 \ \ a_1 \ \ a_0]^T$$ and $$\ φ = [-\frac{s^2}{Λ(s)}y \ \ -\frac{s}{Λ(s)}y \ \ -\frac{1}{Λ(s)}y]^T$$. So, now we have the linear parametric model in terms of the unknown parameters $$\ a_2, a_1,a_0$$.