# How to obtain plant parametric model of given system?

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:

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

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 $$\ \tau = 0 \Rightarrow η=0.$$

• 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 $\phi$. – fibonatic Feb 15 '20 at 23:12
• Forgot the filter, I edited the answer. – Teo Protoulis Feb 15 '20 at 23:33