I going to estimate the polyomial $R^*$ and $S^*$ from $$ y(t)= \frac{R^*}{A_o^*(z^{-1}) A_m^*(z^{-1})}u(t) + \frac{S^*}{A_o^*(z^{-1}) A_m^*(z^{-1})}y(t)$$

$A_o^*(z^{-1})$, $A_m^*(z^{-1})$ polynomals and $u(t)$ and $y(t)$ vectors are know.

$$\frac{1}{A_o^*(z^{-1}) A_m^*(z^{-1})}$$ and $$\frac{1}{A_o^*(z^{-1}) A_m^*(z^{-1})}$$ can be interprented as a filter.

I going to find the control law:

$$ R^*u(t) = T^*u_c(t) - S^*y(t)$$

Where $$T^* = A_o^*A_m^*(1)$$

If you wonder what the * comes from. This "normal" transfer function is written as a forward operator transfer function.

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

This transfer function is written as a backward operated transfer function:

$$H^*(z^{-1}) = \frac{b_0 z^{-1} + b_1z{^-2}}{1 + a_0 z^{-1} + a_1z^{-2}}$$

The reason is that we need to have the denominator as a monic polynomial. Or else, we cannot use Recursive Least Square algorithm.


If I have a polynomial equation $K$ and I multiply it with another vector $v$.

How should I write the difference equation then $Kv$?

How would I write the difference equation of:

$$ y(t)= \frac{R^*}{A_o^*(z^{-1}) A_m^*(z^{-1})}u(t) + \frac{S^*}{A_o^*(z^{-1}) A_m^*(z^{-1})}y(t)$$


  • $\begingroup$ So you want to do parameter estimation of a discrete transfer function? Do you want to do this online, so recursively, or once all the data has been collected? $\endgroup$ – Kwin van der Veen Jun 20 '18 at 10:24
  • $\begingroup$ Recursively. I using RLS algoritm. $\endgroup$ – Daniel Mårtensson Jun 20 '18 at 20:37
  • $\begingroup$ @KwinvanderVeen I changed the equestion. Edit: Now you're trapped. Haha. Look at the book "Adaptive Control" by Karl Johan Åström second edition. Page 133. The book can be found at internet. $\endgroup$ – Daniel Mårtensson Jun 20 '18 at 20:57
  • $\begingroup$ So the discrete transfer function you are considering is $$H(z^{-1})=\frac{R^*}{T^* - S^*}$$ $\endgroup$ – Kwin van der Veen Jun 21 '18 at 3:29
  • $\begingroup$ No. Sorry. Do you have the book ? $\endgroup$ – Daniel Mårtensson Jun 21 '18 at 6:38

Recursive least squares tries to find parameters $\theta$ recursively which best fit the relation $z=\theta^\top \phi$, where both $z$ and $\phi$ are known. A discrete transfer function defined as follows

$$ H(z) = \frac{b_0 + b_1\,z + \cdots + b_{n-1}\,z^{n-1} + b_n\,z^n}{a_0 + a_1\,z + \cdots + a_{m-1}\,z^{m-1} + z^m} = \frac{b_0\,z^{-m} + b_1\,z^{1-m} + \cdots + b_{n-1}\,z^{n-1-m} + b_n\,z^{n-m}}{a_0\,z^{-m} + a_1\,z^{1-m} + \cdots + a_{m-1}\,z^{-1} + 1} \tag{1} $$

with $n \leq m$. The corresponding difference equation can then be written as

$$ y_k = b_0\,u_{k-m} + b_1\,u_{k+1-m} + \cdots + b_{n-1}\,u_{k+n-1-m} + b_n\,u_{k+n-m} \\ - a_0\,y_{k-m} - a_1\,y_{k+1-m} - \cdots - a_{m-1}\,y_{k-1}. \tag{2} $$

Equation $(2)$ can be written into the form of $z=\theta^\top \phi$ by for example defining $z$, $\theta$ and $\phi$ as

$$ z = y_k, $$

$$ \theta^\top = \begin{bmatrix} b_0 & b_1 & \cdots & b_{n-1} & b_n & -a_0 & -a_1 & \cdots & -a_{m-1} \end{bmatrix}, $$

$$ \phi = \begin{bmatrix} u_{k-m} \\ u_{k+1-m} \\ \vdots \\ u_{k+n-1-m} \\ u_{k+n-m} \\ y_{k-m} \\ y_{k+1-m} \\ \vdots \\ y_{k-1} \end{bmatrix}. $$

If some parameters of equation $(2)$ are assumed to be known, then those terms could be taken to the left hand side. Thus $z$ would then gain some terms, while $\theta$ and $\phi$ would lose some. And from here you can just apply recursive least squares.

  • $\begingroup$ Yes. I Know how RLS works. $\endgroup$ – Daniel Mårtensson Jun 21 '18 at 6:37

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