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Synthetic parametric model: Difference between revisions

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The numerical values for the different variables and <math> i = 1,\ldots,k </math> are
The numerical values for the different variables and <math> i = 1,\ldots,k </math> are


* the residues <math> r_i </math> are real and equally spaced in <math>[10^{-3}, 1]</math>, with <math> r_1 = 1 </math> and <math> r_k = 10^{-3} </math>.
* <math> r_i </math> equally spaced in <math>[10^{-3}, 1]</math>, with <math> r_1 = 1 </math> and <math> r_k = 10^{-3} </math>.


* <math>a_i </math> linearly spaced between <math> [10^{-1}, 10^3]</math>,
* <math>a_i </math> equally spaced in <math> [10^{-1}, 10^3]</math>,


* <math>b_i </math> linearly spaced between <math>[10, 10^3]</math>,
* <math>b_i </math> equally spaced in <math>[10, 10^3]</math>,


* <math>\varepsilon \in [1,20]</math>.
* <math>\varepsilon \in [1,20]</math>.

Revision as of 13:12, 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, pi=εai+jbi. For a system in pole-residue form


H(s)=i=1nrispi=i=1nris(εai+jbi),


we can then write down the state-space realisation


A^=ε[a1an]+[jb1jbn]=εA^ε+A^0,

B^=[1,,1]T,C^=[r1,,rn],D=0.


Notice that the system matrices have complex entries.

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

p1=εa1+jb1,p2=εa1jb1,,pn1=εak+jbk,pn=εakjbk,

which also implies that the residues form complex conjugate pairs r1,r¯1,,rk,r¯k.

Then a realization with matrices having real entries is given by


A=TA^T*,B=TB^,C=C^T*,D=0,


with T=[T0T0], for T0=12[1j1j].

Numerical values

The numerical values for the different variables and i=1,,k are

  • ri equally spaced in [103,1], with r1=1 and rk=103.
  • ai equally spaced in [101,103],
  • bi equally spaced in [10,103],
  • ε[1,20].



In MATLAB this is easily done as follows test