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What is $mu$ - synthesis analysis? Uncertainty modelling

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0

$begingroup$

I wonder what $mu$ - synthesis analysis is? I have heard that is an uncertainty modelling.

I think it's an extra help for the $H_infty$ controller because the $mu$ - synthesis analysis make sure that the $H_infty$ controller can stand against nonlinearities.

So $mu$ + $H_infty$ = Robust nonlinear control.

Am I right?

The reason why I'm asking this simple question is that the books which teach robust control, cannot explain why we are going to use $mu$ - synthesis analysis. They only teach math, not the purpose.

optimization control-theory nonlinear-system optimal-control linear-control

share|cite|improve this question

edited Aug 24 '17 at 6:08

Daniel Mårtensson

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

$endgroup$

add a comment | 

0

$begingroup$

I wonder what $mu$ - synthesis analysis is? I have heard that is an uncertainty modelling.

I think it's an extra help for the $H_infty$ controller because the $mu$ - synthesis analysis make sure that the $H_infty$ controller can stand against nonlinearities.

So $mu$ + $H_infty$ = Robust nonlinear control.

Am I right?

The reason why I'm asking this simple question is that the books which teach robust control, cannot explain why we are going to use $mu$ - synthesis analysis. They only teach math, not the purpose.

optimization control-theory nonlinear-system optimal-control linear-control

share|cite|improve this question

edited Aug 24 '17 at 6:08

Daniel Mårtensson

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

$endgroup$

add a comment | 

$begingroup$

I wonder what $mu$ - synthesis analysis is? I have heard that is an uncertainty modelling.

I think it's an extra help for the $H_infty$ controller because the $mu$ - synthesis analysis make sure that the $H_infty$ controller can stand against nonlinearities.

So $mu$ + $H_infty$ = Robust nonlinear control.

Am I right?

The reason why I'm asking this simple question is that the books which teach robust control, cannot explain why we are going to use $mu$ - synthesis analysis. They only teach math, not the purpose.

optimization control-theory nonlinear-system optimal-control linear-control

share|cite|improve this question

edited Aug 24 '17 at 6:08

Daniel Mårtensson

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

$endgroup$

share|cite|improve this question

edited Aug 24 '17 at 6:08

Daniel Mårtensson

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

share|cite|improve this question

edited Aug 24 '17 at 6:08

Daniel Mårtensson

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

asked Aug 23 '17 at 21:19

Daniel MårtenssonDaniel Mårtensson

984419

2 Answers
2

active

oldest

votes

1

$begingroup$

$H_infty$ deals with minimizing the influence of uncertainty in your plant, but the uncertainty is unstructured, i.e. each uncertainty in your plant couples with every other. Generally this is not the case in most problems. For a harmonic oscillator, the uncertainty in the mass should not couple with the uncertainty in the spring constant. H-inf assumes this is the case.

$mu$-synthesis attempts to deal with structured uncertainty.

Therefore, $H_infty$ produces more conservative controllers that might not be able to meet the design specs. Using u-synthesis you can increase the performance of the system while still meeting the requirement that the induced disturbances from the uncertainty remain below a certain level.

share|cite|improve this answer

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it practical or theoretical?
  $endgroup$
  – Daniel Mårtensson
  Mar 14 at 0:39
  

  
  
  
  

add a comment | 

2

$begingroup$

$H_infty$ deals with the problem of finding a controller $F(s)$ for a known system $G(s)$ such that the gain (in $H_infty$ sense) from an external signal to an output is minimized.

$mu$-synthesis extends this to the case when $G(s)$ is uncertain, and tries to minimize the worst-case gain given the uncertainty description.

share|cite|improve this answer

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So μ-synthesis is very good if $G(s)$ contains nonlinearities, so the worst-case gain in $F(s)$ will be minimized ? Summary: μ-synthesis is just an extension to the control law $F(s)$ to make it better?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry! If $G(s)$ does not contains nonlinearities.
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any difference between μ-synthesis and μ-analysis? Or are them both the same?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As always, analysis simply analyses a given setup, while synthesis creates something. In this case, a controller is created.
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily nonlinear uncertainties. Any uncertainty which can be framed in the whole mathematical setup (parametric uncertainties, dynamic uncertainties, static nonlinearities etc)
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:31
  

  
  
  
  

add a comment | 

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1

$begingroup$

$H_infty$ deals with minimizing the influence of uncertainty in your plant, but the uncertainty is unstructured, i.e. each uncertainty in your plant couples with every other. Generally this is not the case in most problems. For a harmonic oscillator, the uncertainty in the mass should not couple with the uncertainty in the spring constant. H-inf assumes this is the case.

$mu$-synthesis attempts to deal with structured uncertainty.

Therefore, $H_infty$ produces more conservative controllers that might not be able to meet the design specs. Using u-synthesis you can increase the performance of the system while still meeting the requirement that the induced disturbances from the uncertainty remain below a certain level.

share|cite|improve this answer

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it practical or theoretical?
  $endgroup$
  – Daniel Mårtensson
  Mar 14 at 0:39
  

  
  
  
  

add a comment | 

1

$begingroup$

$H_infty$ deals with minimizing the influence of uncertainty in your plant, but the uncertainty is unstructured, i.e. each uncertainty in your plant couples with every other. Generally this is not the case in most problems. For a harmonic oscillator, the uncertainty in the mass should not couple with the uncertainty in the spring constant. H-inf assumes this is the case.

$mu$-synthesis attempts to deal with structured uncertainty.

Therefore, $H_infty$ produces more conservative controllers that might not be able to meet the design specs. Using u-synthesis you can increase the performance of the system while still meeting the requirement that the induced disturbances from the uncertainty remain below a certain level.

share|cite|improve this answer

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it practical or theoretical?
  $endgroup$
  – Daniel Mårtensson
  Mar 14 at 0:39
  

  
  
  
  

add a comment | 

$begingroup$

$H_infty$ deals with minimizing the influence of uncertainty in your plant, but the uncertainty is unstructured, i.e. each uncertainty in your plant couples with every other. Generally this is not the case in most problems. For a harmonic oscillator, the uncertainty in the mass should not couple with the uncertainty in the spring constant. H-inf assumes this is the case.

$mu$-synthesis attempts to deal with structured uncertainty.

Therefore, $H_infty$ produces more conservative controllers that might not be able to meet the design specs. Using u-synthesis you can increase the performance of the system while still meeting the requirement that the induced disturbances from the uncertainty remain below a certain level.

share|cite|improve this answer

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 13 at 17:16

Mefitico

1,091218

edited Mar 13 at 17:16

Mefitico

1,091218

answered Mar 13 at 15:43

JeremyJeremy

261

New contributor

Jeremy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 13 at 15:43

JeremyJeremy

261

2

$begingroup$

$H_infty$ deals with the problem of finding a controller $F(s)$ for a known system $G(s)$ such that the gain (in $H_infty$ sense) from an external signal to an output is minimized.

$mu$-synthesis extends this to the case when $G(s)$ is uncertain, and tries to minimize the worst-case gain given the uncertainty description.

share|cite|improve this answer

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So μ-synthesis is very good if $G(s)$ contains nonlinearities, so the worst-case gain in $F(s)$ will be minimized ? Summary: μ-synthesis is just an extension to the control law $F(s)$ to make it better?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry! If $G(s)$ does not contains nonlinearities.
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any difference between μ-synthesis and μ-analysis? Or are them both the same?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As always, analysis simply analyses a given setup, while synthesis creates something. In this case, a controller is created.
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily nonlinear uncertainties. Any uncertainty which can be framed in the whole mathematical setup (parametric uncertainties, dynamic uncertainties, static nonlinearities etc)
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:31
  

  
  
  
  

add a comment | 

2

$begingroup$

$H_infty$ deals with the problem of finding a controller $F(s)$ for a known system $G(s)$ such that the gain (in $H_infty$ sense) from an external signal to an output is minimized.

$mu$-synthesis extends this to the case when $G(s)$ is uncertain, and tries to minimize the worst-case gain given the uncertainty description.

share|cite|improve this answer

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So μ-synthesis is very good if $G(s)$ contains nonlinearities, so the worst-case gain in $F(s)$ will be minimized ? Summary: μ-synthesis is just an extension to the control law $F(s)$ to make it better?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry! If $G(s)$ does not contains nonlinearities.
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there any difference between μ-synthesis and μ-analysis? Or are them both the same?
  $endgroup$
  – Daniel Mårtensson
  Aug 24 '17 at 16:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As always, analysis simply analyses a given setup, while synthesis creates something. In this case, a controller is created.
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily nonlinear uncertainties. Any uncertainty which can be framed in the whole mathematical setup (parametric uncertainties, dynamic uncertainties, static nonlinearities etc)
  $endgroup$
  – Johan Löfberg
  Aug 24 '17 at 18:31
  

  
  
  
  

add a comment | 

$begingroup$

$H_infty$ deals with the problem of finding a controller $F(s)$ for a known system $G(s)$ such that the gain (in $H_infty$ sense) from an external signal to an output is minimized.

$mu$-synthesis extends this to the case when $G(s)$ is uncertain, and tries to minimize the worst-case gain given the uncertainty description.

share|cite|improve this answer

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

$endgroup$

share|cite|improve this answer

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811

answered Aug 24 '17 at 6:41

Johan LöfbergJohan Löfberg

5,4151811