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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 $ ?

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

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