Synthetic parametric model: Difference between revisions

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Revision as of 12:50, 28 November 2011

Introduction

On this page you will find a purely synthetic parametric model. The goal is to have a simple parametric model which one can use to experiment with different system orders, parameter values etc.

System description

The parameter $ε$ scales the real part of the system poles, that is, $p_{i} = ε a_{i} + j b_{i}$ . For a system in pole-residue form

$H (s) = \sum_{i = 1}^{n} \frac{r_{i}}{s - p_{i}} = \sum_{i = 1}^{n} \frac{r_{i}}{s - (ε a_{i} + j b_{i})},$

we can then write down the state-space realisation

$\hat{A} = ε [\begin{array}{ccc} a_{1} \\ ⋱ \\ a_{n} \end{array}] + [\begin{array}{ccc} j b_{1} \\ ⋱ \\ j b_{n} \end{array}] = ε {\hat{A}}_{ε} + {\hat{A}}_{0},$

$\hat{B} = [1, \dots, 1]^{T}, \hat{C} = [r_{1}, \dots, r_{n}], D = 0 .$

Notice that the system matrices have complex entries.

For simplicity, assume that $n$ is even, $n = 2 k$ , and that all system poles are complex and ordered in complex conjugate pairs, i.e.

$p_{1} = ε a_{1} + j b_{1}, p_{2} = ε a_{1} - j b_{1}, \dots, p_{n - 1} = ε a_{k} + j b_{k}, p_{n} = ε a_{k} - j b_{k},$

which also implies that the residues form complex conjugate pairs $r_{1}, {\bar{r}}_{1}, \dots, r_{k}, {\bar{r}}_{k} .$

Then a realization with matrices having real entries is given by

$A = T \hat{A} T^{*}, B = T \hat{B}, C = \hat{C} T^{*}, D = 0,$

with $T = [\begin{array}{ccc} T_{0} \\ ⋱ \\ T_{0} \end{array}]$ , for $T_{0} = \frac{1}{\sqrt{2}} [\begin{array}{cc} 1 & - j \\ 1 & j \end{array}]$ .

@@ Line 5: / Line 5: @@
 == System description ==
-The parameter <math>\varepsilon</math> scales the real part of the system poles, that is, <math>p_k=\varepsilon a_k+jb_k</math>.
+The parameter <math>\varepsilon</math> scales the real part of the system poles, that is, <math>p_i=\varepsilon a_i+jb_i</math>.
-If the system is in pole-residue form, then
+For a system in pole-residue form
 <math> H(s) =  \sum_{i=1}^{n}\frac{r_i}{s-p_i} =  \sum_{i=1}^{n}\frac{r_i}{s-(\varepsilon a_i+jb_i)} ,</math>
-which has the state-space realisation
+we can then write down the state-space realisation
 <math>\widehat{A} = \varepsilon \left[\begin{array}{ccc} a_1 & & \\ & \ddots & \\ & & a_n\end{array}\right] +\left[\begin{array}{ccc} jb_1 & & \\ & \ddots & \\ & & jb_n\end{array}\right] = \varepsilon \widehat{A}_\varepsilon + \widehat{A}_0,</math>
 <math>\widehat{B} = [1,\ldots,1]^T,\quad \widehat{C} = [r_1,\ldots,r_n],\quad D = 0.</math>
 Notice that the system matrices have complex entries.
@@ Line 20: / Line 25: @@
 For simplicity, assume that <math> n </math> is even, <math> n=2k </math>, and that all system poles are complex and ordered in complex conjugate pairs, i.e.
-<math> p_1 = \varepsilon a_1+jb_1, p_2 = \varepsilon a_1-jb_1, \ldots, p_{n-1} = \varepsilon a_k+jb_k, p_n = \varepsilon a_k-jb_k. </math>
+<math> p_1 = \varepsilon a_1+jb_1, p_2 = \varepsilon a_1-jb_1, \ldots, p_{n-1} = \varepsilon a_k+jb_k, p_n = \varepsilon a_k-jb_k, </math>
-Which also implies that the residues form complex conjugate pairs <math>r_1, \bar{r}_1,\ldots , r_k, \bar{r}_k.</math>
+which also implies that the residues form complex conjugate pairs <math>r_1, \bar{r}_1,\ldots , r_k, \bar{r}_k.</math>
 Then a realization with matrices having real entries is given by
 <math> A = T\widehat{A}T^*, \quad B = T\widehat{B}, \quad C = \widehat{C}T^*, \quad D = 0,</math>
-with
-<math> T = \left[\begin{array}{ccc} T_0 & & \\ & \ddots & \\ & & T_0 \end{array}\right] </math>,
+with <math> T = \left[\begin{array}{ccc} T_0 & & \\ & \ddots & \\ & & T_0 \end{array}\right] </math>,
 for <math>T_0 = \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -j\\ 1 & j \end{array}\right]</math>.
 == Numerical values ==