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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Properties of the Modal Representation
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Finding the Eigenvalues of A in Practice