Hello MIMO Control Friends,

PURPOSE
The purpose of this note is to illustrate how the singular value
decomposition may be used to study the steady state behavior of MIMO LTI
dynamical systems to sinusoidal inputs.

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NOTATION
Let H denote an m x m transfer function matrix for a MIMO LTI system.
Assuming zero initial conditions, the output vector y is related to the
input vector 
u by an equation of the form:
                              y(s) = H(s) u(s).

Suppose that
                              H(jw) = U SIGMA V^H
where  
       u_kj exp(j theta_kj) is the kjth entry of U,
 
       SIGMA = diag(sigma_1, ... , sigma_m ),

       sigma_i is the ith singular value of H(jw), and

       v_{ij} exp(j phi_ij ) is the ijth entry of V.

Note that the ith column of V is the right singular vector of H(jw) which
corresponds to the ith singular value sigma_i of H(jw). Similarly, note
that the ith column of U is the left singular vector of H(jw) which
corresponds to the ith singular value sigma_i of H(jw). 

Suppose further that the system H is stable and that H is being excited by
a vector input u consisting of sinusoids of frequency w. The stability
assumption on H guarantees that all transients will decay and that the
steady state output vector y will also consist of sinusoidal signals having
frequency w.  

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MAIN RESULT
Specifically, suppose that the ith input is chosen to be

       u_i = v_ij cos(wt + phi_ij).

Note that it has been constructed using information from the jth column of
V; i.e. using information from the jth right singular vector of H(jw).

Given the above, the steady state output in the kth output channel can be
shown to be given by

       y_k = sigma_j u_kj cos(wt + theta_kj).

Note that this output has been constructed using information from the jth
column of U; i.e. using information from the jth left singular vector of
H(jw).

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An example illustrating how the above result may be used is now provided.

EXAMPLE: SVD AND STEADY STATE SINUSOIDAL ANALYSIS
Suppose that 
             H(j10) = U SIGMA V^H
                    = [ 0.9501 + 0.8913i   0.6068 - 0.4565i
                       -0.2311 - 0.7621i   0.4860 + 0.0185i ]
where
             U = [ 0.6654 + 0.6461j   0.1999 + 0.3160j
                  -0.1333 - 0.3494j   0.1187 + 0.9198j ]

               = [ 0.9275 exp( j  44.1583 deg)   0.3739 exp(j 57.6869 deg)
                   0.3739 exp(-j 110.8821 deg)   0.9275 exp(j 82.6465 deg)

             SIGMA = [   1.5956      0.0000
                         0.0000      0.7736 ]

             V = [ 0.9433            -0.3320          
                   0.0235 + 0.3312j   0.0669 + 0.9409j ]

               = [ 0.9433 exp(j   0.0000 deg)   0.3320 exp(j 180.0000 deg)
                   0.3320 exp(j  85.9345 deg)   0.9433 exp(j  85.9345 deg)

Given the above singular value decomposition for H(j10), it follows that if
the
input u is selected as follows

             u(t) = [ 0.9433 cos(10t + 0.0000 deg)
                      0.3320 cos(10t + 85.9345 deg)]

i.e. "in the direction of the right singular vector corresponding to the
maximum singular value," then the steady state output will be "in the
direction of the left singular vector associated with the maximum singular
value" and given by

            y(t) = 1.5956 [ 0.9275 cos(10t +  44.1583 deg)
                            0.3739 cos(10t - 110.8821 deg) ].

Similarly, if the input u is selected as follows

             u(t) = [ 0.3320 cos(10t + 180.0000 deg)
                      0.9433 cos(10t -  85.9345 deg)]

i.e. "in the direction of the right singular vector corresponding to the
minimum singular value," then the steady state output will be "in the
direction of the left singular vector associated with the minimum singular
value" and given by

            y(t) = 0.7736 [ 0.3739 cos(10t + 57.6869)
                            0.9275 cos(10t + 82.6465 deg) ].

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Hope the above is helpful.
AAR