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### Example of State-Space Diagonalization

For the example of Eq.(G.7), we obtain the following results:

>> % Initial state space filter from example above:
>> A = [-1/2, -1/3; 1, 0]; % state transition matrix
>> B = [1; 0];
>> C = [2-1/2, 3-1/3];
>> D = 1;
>>
>> eig(A) % find eigenvalues of state transition matrix A

ans =
-0.2500 + 0.5204i
-0.2500 - 0.5204i

>> roots(den) % find poles of transfer function H(z)

ans =
-0.2500 + 0.5204i
-0.2500 - 0.5204i

>> abs(roots(den)) % check stability while we're here

ans =
0.5774
0.5774

% The system is stable since each pole has magnitude < 1.


Our second-order example is already in real form, because it is only second order. However, to illustrate the computations, let's obtain the eigenvectors and compute the complex modal representation:

>> [E,L] = eig(A)  % [Evects,Evals] = eig(A)

E =

-0.4507 - 0.2165i  -0.4507 + 0.2165i
0 + 0.8660i        0 - 0.8660i

L =

-0.2500 + 0.5204i        0
0            -0.2500 - 0.5204i

>> A * E - E * L   % should be zero (A * evect = eval * evect)

ans =
1.0e-016 *
0 + 0.2776i        0 - 0.2776i
0                  0

% Now form the complete diagonalized state-space model (complex):

>> Ei = inv(E); % matrix inverse
>> Ab = Ei*A*E % new state transition matrix (diagonal)

Ab =
-0.2500 + 0.5204i   0.0000 + 0.0000i
-0.0000            -0.2500 - 0.5204i

>> Bb = Ei*B   % vector routing input signal to internal modes

Bb =
-1.1094
-1.1094

>> Cb = C*E    % vector taking mode linear combination to output

Cb =
-0.6760 + 1.9846i  -0.6760 - 1.9846i

>> Db = D      % feed-through term unchanged

Db =
1

% Verify that we still have the same transfer function:

>> [numb,denb] = ss2tf(Ab,Bb,Cb,Db)

numb =
1.0000             2.0000 + 0.0000i   3.0000 + 0.0000i

denb =
1.0000             0.5000 - 0.0000i   0.3333

>> num = [1, 2, 3]; % original numerator
>> norm(num-numb)

ans =
1.5543e-015

>> den = [1, 1/2, 1/3]; % original denominator
>> norm(den-denb)

ans =
1.3597e-016


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