hinfnorm command

Hello I am using the command 'hinfnorm' to calculate the infinity norm of a descriptor linear system described by a ltisys command... Sometimes, hinfnorm command says that the descriptor linear system has closed-right-half plane poles while when i check the eigen values of that same system by eig(sys_descriptor)... the real part of the all the eigens values are negative. What is going wrong with this. is hinfnorm not reliable? or eig doesn't give eigen values for descriptor system?

Thanks in advance.

[Oleg Komarov]

Negative by how much?

[M.Mohsin Siraj]

by a decent value, i mean may be around -0.369...

secondly, for a SISO system i checked the bode diagram and the peak value is -260dB so the h infinity norm must be very small but the hinfnorm shows it somewhere around 60. which is extremely weird.

[Jan]

This became the ugliest thread in the forum, because necessary information about the problem are distributed to different answers and related comments are posted as further answers. Please follow the layout:

1. All information, which define the problem, are found in the question.
2. Clarifications of the problem are added by editing the question. The [EDITED] tag helps to recognize changes.
3. Answers are posted as answers.
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5. Questions for clarifications are posted as comments to the question.

Thanks. The readability of the threads are important for all users.

[Walter Roberson]

(This message should not be closed, just cleaned up.)

Answer 1

[Suat Gumussoy]

Could you send me or post your plant that you used for norm computation? I suggest you to use norm(sys,inf,tol) with desired tolerance tol. norm function implements the state-of-the-art numerical methods and it is likely to give reliable answers.

[M.Mohsin Siraj]

for example, the system below:

A = [-1 0 0 0; 0 -2 0 0;0 0 -0.4172 -0.07477;0 0 -0.1495 -0.6489];

B = [4;2;3 8.718;7.071];

C = [19 25 -4.166 -2.946]

D = 0 ;

E = [1 0 0 0;0 1 0 0; 0 0 0.4172 0.07477;0 0 0.07477 0.3244];

sys = dss(A,B,C,D,E);

sys_lti = ltisys(A,B,C,D,E);

figure, bode(sys)

hinfnorm(sys_lti)

The bode diagram(equivalent to Sigma in SISO as here) shows that the peak is at extremely low value while the hinfnorm command shows the hinfinity norm to be 69.555.

Isn't is strange?

[M.Mohsin Siraj]

it*

Answer 2

[Suat Gumussoy]

Some other things... assuming that plant matrices E,A,B,C,D, you can create and ss object by

sys = dss(A,B,C,D,E); and eigenvalues can be computed by pole(sys)

In order to compute eigenvalues of descriptor system, you need to do eig(A,E) since the system is descriptor and don't forget to exclude eigenvalues at infinity since these corresponds to the null space of E.

Answer 3

[Suat Gumussoy]

B data has incompatible dimensions.

Answer 4

[Suat Gumussoy]

As I pointed out earlier, svd plot (in this case bode) and norm(sys,inf) are consistent.

>> [f,w]=norm(sys,inf)

f =

    0.0086

w =

     0

Note that in 2012b, all norm computations are done via getPeakGain function which is the function norm relies on. I'll get back for hinfnorm.

Answer 5

[Suat Gumussoy]

Which matlab version do you have?

Answer 6

[Suat Gumussoy]

I double checked the computation in your version of MATLAB R2011a, norm(sys,inf) gives 0.0086 at frequency 0 and this is consistent with bode plot. Again getting back to hinfnorm.

Answer 7

[Suat Gumussoy]

About hinfnorm... this function is not implemented to be used with descriptor systems. see line 45 in hinfnorm function

[a,b,c,d] = unpck(sys);

it passes a,b,c,d data not e. In fact, you may not pass e, since it gives an error as

K>> [a,b,c,d,e] = unpck(sys); Error using unpck Too many output arguments.

On the other hand, if you want to compute norm without e,

sys = ss(A,B,C,D); sys_lti = ltisys(A,B,C,D);

they give similar norm norm(sys,inf)

ans =

   34.6214

>> hinfnorm(sys_lti,1e-6) norm between 34.6214 and 34.6215 achieved near 0.79022

If you want to use hinfnorm, since your E is not singular, you may write as ss object and then compute norm

sys1=ss(E\A,E\B,C,D); sys_lti1=pck(sys1.A,sys1.B,sys1.C,sys1.D) hinfnorm(sys_lti1,1e-6)

hinfnorm(sys_lti1,1e-6) norm between 0.0086415 and 0.0086415 achieved near 0

which gives similar result as norm(.,inf).

Hope this helps.

Answer 8

[Suat Gumussoy]

You're welcome. If you have singular E, I suggest to use dss and create ss object. bode, sigma and norm are defined for descriptor systems. These functions are continuously updated and improved with each release whereas in general hinfnorm is not.

Answer 9

[Suat Gumussoy]

Bode, sigma plot and norm commands work for descriptor systems. Assuming system matrices are a,b,c,d,e. Define your descriptor system as sys=dss(a,b,c,d,e) and you can use them as bode(sys), sigma(sys), norm(sys,.).

Answer 10

[M.Mohsin Siraj]

for example, the system below:

    A = [-1 0 0 0; 0 -2 0 0;0 0 -0.4172 -0.07477;0 0 -0.1495 -0.6489]; 
    B = [4;2;3 8.718;7.071]; 
    C = [19 25 -4.166 -2.946] 
    D = 0 ; 
    E = [1 0 0 0;0 1 0 0; 0 0 0.4172 0.07477;0 0 0.07477 0.3244]; 
    sys = dss(A,B,C,D,E); 
    sys_lti = ltisys(A,B,C,D,E); 
    figure, bode(sys) ;
    hinfnorm(sys_lti)

The bode diagram(equivalent to Sigma in SISO as here) shows that the peak is at extremely low value while the hinfnorm command shows the hinfinity norm to be 69.555. Isn'tsit strange?

Answer 11

[M.Mohsin Siraj]

Sorry for the typo...

B = [4; 2; 8.718; 7.071]

That's the correct B.

Answer 12

[M.Mohsin Siraj]

the norm command as follows:

norm(sys,inf)

gives the h infinity norm of 3.1974e-014.

Answer 13

[M.Mohsin Siraj]

I have MATLAB 7.12.0 R2011a.

[Jan]

Please post comments to an answer as a comment, not as an additional answer. Details about your question should be added by editing the original question, such that readers get all required information in one message, not distributed over the question, some answers without a fixed order and some comments.

Answer 14

[M.Mohsin Siraj]

Thank you so much for your detailed answers.... One thing, are 'bode' or 'sigma' and 'norm' defined for descriptor systems as well? Actually in this particular example, my E is not singular otherwise generally it is singular.

Thanks a lot again.

[Suat Gumussoy]

Yes, they do support descriptor systems.

Answer 15

[M.Mohsin Siraj]

Thanks, how can i confirm that the bode and sigma plots and norm does consider the descriptor systems? Also I read in one of the User Guide about Control system Toolbox ver 4.2, that dss doesn't support a non singular E but a badly conditioned E (nearer to singularity) can be considered. I can send you the user guide, if required.

[Craig]

See the following release notes: http://www.mathworks.com/help/toolbox/control/rn/f0-82252.html

Release Notes: Version 7.0 (R2006a) Control System Toolbox Software

LTI Objects

Descriptor and Improper State-Space Models Fully Supported

There is now full support for descriptor state-space models with a singular E matrix. This now lets you build state-space representations, such as PID, and manipulate improper models with the superior accuracy of state-space computations. In previous versions, only descriptor models with a nonsingular E matrix were supported.

[M.Mohsin Siraj]

Thanks for your answer... Can you kindly further comment on my first question about the use of bode, sigma plot and norm command for descriptor systems? Do they support descriptor systems or they are only designed to work with ordinary state-space realization.

Answer 16

[M.Mohsin Siraj]

Thanks for your answer. But now consider the following simple first order descriptor system

      a = 
                x1
         x1  0.04178

b = 
           u1
   x1  -4.697

c = 
          x1
   y1  4.166

d = 
       u1
   y1   0

e = 
            x1
   x1  0.03067
sys = dss(0.04178,-4.697,4.166,0,0.03067);
[hinf f] = norm(sys);
ans =

    468.3365

it is an unstable system, but the norm command still calculates its h-infinity norm. Isn't it strange?

[Suat Gumussoy]

You're right. It should be called as L-infinity norm since it does not check whether the system is stable. However, when system is stable, L-infinity and H-infinity norms are equal. So user needs to check system stability himself. By the way, what you compute is h2 norm for h-infinity norm, you need

[hinf,f]=norm(sys,inf);

[Suat Gumussoy]

Stability can be easily checked by isstable(sys).