0
$\begingroup$

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.

Question:

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:

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

??

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.