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Consider the system:

$$ y(s)=b\frac{e^{−sτ}}{s+a}u(s) $$

where a,b are unknown, constant parameters and $\ τ>0 $ is an unknown delay constant. How can I obtain a plant parametric model for this system in the following form ?

$$ z = \theta^{*T}\phi \ + η $$

which satisfies the following condition:

  • $\tau = 0 \Rightarrow η = 0 $
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  • $\begingroup$ So $\theta$ should contain $a$, $b$ and $\tau$? $\endgroup$ – fibonatic Feb 11 '20 at 4:16
  • $\begingroup$ $θ$ should contain $a,b$ and $η$ should contain $τ$. If $τ=0$ then $η$ should be also 0. $\endgroup$ – Teo Protoulis Feb 11 '20 at 8:55
  • $\begingroup$ Do you want to linearize the nonlinear model? If so, why don't use a Taylor series? $\endgroup$ – Ken Grimes Feb 11 '20 at 17:07
  • $\begingroup$ I want to obtain a linear parametric model of the form stated at the question where $θ^{*T} = [a \ b]$. $\endgroup$ – Teo Protoulis Feb 11 '20 at 17:10
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Solution to the linear parameterization of the given system:

$\ y = \frac{b}{s+a}e^{-sτ}u = \frac{b}{s+a}e^{-sτ}u \ + \ \frac{b}{s+a}u \ - \ \frac{b}{s+a}u = \frac{b}{s+a}u \ + \ \frac{b}{s+a}(e^{-s\tau} \ - \ 1)u $

$$ sy = -ay+bu+b(e^{-s\tau}-1)u $$

We filter both sides of the equation with a stable first order filter

$$ Λ(s) = s + l_1, \ l_1 >0 $$

and the final parametric model is now:

$\ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u$

And the last equation can be written in the form:

$$ z = \theta^{*T}\phi + η $$

where $\ z = \frac{s}{Λ(s)}y, \ \theta^* = [a \ b], \ \phi = [-\frac{1}{Λ(s)}y \ \ \frac{1}{Λ(s)}u]^T $ and $\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $.

It is obvious that for $\ \tau = 0 \to η=0. $

The new input vector of the parametric system $\ z $, can be derived directly from the input of the initial system $\ y. $

EDIT: Slightly different approach

$$ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u $$

From this point, we can come up with a linear parametric model which contains the initial input vector $\ y $ following the procedure below:

$$ Λ(s) = s + l_1 \Rightarrow s = Λ(s) - l_1, \ l_1>0$$

$$ \frac{Λ(s) - l_1}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$

$$ \frac{Λ(s)}{Λ(s)}y - \frac{l_1}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$

$$ y = -(a-l_1)\frac{1}{Λ(s)}y +b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u $$

and the last equation can be written in the form:

$$ y = \theta^{*T}\phi + η $$

where $\ \theta^* = [(a-l_1) \ \ b], \ \phi = [-\frac{1}{Λ(s)}y \ \ \frac{1}{Λ(s)}u]^T $ and $\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $.

It is obvious that for $\ \tau = 0 \Rightarrow η=0. $

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  • 1
    $\begingroup$ It is worth nothing that $s\,y$ is not causal (not poper). Therefore, one can also apply a low pass filter to both sides of the equation and since $\theta^*$ is constant this is equivalent to applying the low pass filter to your $z$ and $\phi$. $\endgroup$ – fibonatic Feb 15 '20 at 23:12
  • $\begingroup$ Forgot the filter, I edited the answer. $\endgroup$ – Teo Protoulis Feb 15 '20 at 23:33

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