Repeated uncertanity Blocks in robust control

for calculating the SSV using the mussv command, I am required to define the BlocKStructure. How to define the Blockstructure as defined in the above image?

[Paul]

Similar, if not exactly the same, question here: How can I compute structured singular value with repeated full complex blocks as allowable structured uncertainty - MATLAB Answers - MATLAB Central

Answer 1

[Paul]

Hi Amit,

As far as I can see, mussv does not support repeated, full, complex blocks. Out of curiosity, in what sort of problem does one encounter such an uncertainty structure?

However, the block structure can be rearranged to solve a different problem that might (or might not) be of interest.

I'll show an example with Ny = 2 and only two repeated blocks to keep things simpler (and less typing).

Define the full block

delta = sym('delta',[6,2]);

And the full uncertainty block

Delta = blkdiag(delta,delta);

The input to Delta, i.e., the output from M, are two, 2 x 1 vectors.

z1 = sym('z1',[2,1]);
z2 = sym('z2',[2,1]);
z = [z1;z2];

The output of Delta is then

w = Delta*z

w = 

Rearrange the elements of Delta so that instead having two, repeated 6 x 2 blocks we have 6 blocks, where each block is a repeated scalar of one of the elements of delta.

bDelta = [delta(:),delta(:)].';
bDelta = diag(bDelta(:))

bDelta = 

Now we define a right-side pre-multiplication and left-side post-multiplication on bdelta such that:

L*bdelta*R*z = Delta*z.

R and L are (there's probably a smart way to automate this)

R = zeros(24,4);
R(1:2:12,1) = 1;  % z11
R(2:2:12,3) = 1;  % z21
R(13:2:end,2) = 1; % z12
R(14:2:end,4) = 1; % z22

L = zeros(12,24);
L(1,[1,13]) = 1;
L(2,[3,15]) = 1;
L(3,[5,17]) = 1;
L(4,[7,19]) = 1;
L(5,[9,21]) = 1;
L(6,[11,23]) = 1;
L(7,[2,14]) = 1;
L(8,[4,16]) = 1;
L(9,[6,18]) = 1;
L(10,[8,20]) = 1;
L(11,[10,22]) = 1;
L(12,[12,24]) = 1;

Show the equivalence

sympref('AbbreviateOutput',false)

ans = logical
   1

[L*bDelta*R*z , Delta*z]

ans = 

all(isAlways(L*bDelta*R*z == Delta*z))

ans = logical
   1

Now we can analyze how large bdelta can be to maintain loop stability of R*M*L*bdelta instead of how large Delta can be to maintain loop stability of M*Delta.

For example

rng(100);
M = rand(4,12) + 1j*rand(4,12);

Run mussv on R*M*L

[bounds,muinfo] = mussv(R*M*L,repmat([2,0],12,1));
bounds

bounds = 1×2

16.4127 16.3555

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Extract the peturbation with the structure of bdelta that corresponds to the lower bound

bdelta = mussvextract(muinfo);

Verify that bdelta makes the loop unstable (eigenvalue at 1)

eig(R*M*L*bdelta)

ans = 
   1.0000 + 0.0000i
   0.0088 + 0.1163i
   0.0006 + 0.0011i
   0.0004 + 0.0004i
   0.0000 - 0.0000i
   0.0000 + 0.0000i
   0.0000 + 0.0000i
   0.0000 - 0.0000i
  -0.0000 + 0.0000i
  -0.0000 + 0.0000i
   0.0000 + 0.0000i
   0.0000 - 0.0000i
  -0.0000 + 0.0000i
   0.0000 - 0.0000i
   0.0000 - 0.0000i
  -0.0000 + 0.0000i
  -0.0000 - 0.0000i
  -0.0000 + 0.0000i
   0.0000 - 0.0000i
  -0.0000 + 0.0000i
  -0.0000 + 0.0000i
   0.0000 + 0.0000i
  -0.0000 - 0.0000i
   0.0000 - 0.0000i

Compute Delta from bdelta and verify that Delta makes the loop unstable

Delta = L*bdelta*R;
eig(M*Delta)

ans = 
   1.0000 + 0.0000i
   0.0088 + 0.1163i
   0.0006 + 0.0011i
   0.0004 + 0.0004i

However, the norm of bdelta is quite a bit smaller than the norm of Delta and, I believe, it's possible that there might exist a different Delta with smaller norm that also makes the M*Delta loop unstable.

[max(svd(bdelta)), 1/bounds(2)]

ans = 1×2

0.0611 0.0611

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max(svd(Delta))

ans = 0.2118

Like I said, we're solving a different problem than posed, but maybe it can still provide useful information for the problem of interest. Maybe.

[AMIT]

This uncertainity structure is related to fluid flow problem