Rearrange state space system

I've modelled a system in simulink and used operspec and linearise to extract a state space system. While the activity has worked well, I have a downstream process which requires the state space to be expressed in a different form.

Without modifying the simulink model, is there a matlab command or series of commands that is capable of rearranging a state space system such that some parameters which where previously ouputs become states. For context, my downstream process will also be dropping a number of states.

I've been using simulink, simulink control, system identification and the control system toolbox.

Answer 1

[Paul]

Let's assume that the state space model from Simulink is defined as:

xdot = A*x + B*u

[y1;y2] = [C1; C2] * x + [0;D2] u

where y1 are the outputs that you want to be state variables (you can always order the ouputs this way) and all matrices are of compatible dimensions. Note that D1 == 0.

Now we have at least two options.

The first option is to add additional states (w) with derivatives equal to the derivatives of the outputs

w = y1

wdot = y1dot = C1*xdot = C1*A*x + C1*B*u.

Now we can augment the state vector x witht the new states w. By definition, the new outputs that replace y1 are just w. Hence the new state space model is:

[xdot;wdot] = [A 0;C1*A 0] * [x;w] + [B;C1*B] * u

[ynew;y2] = [0 I;C2 0] * [x;w] + [0;D] * u

where the zero and identity matrices are of approropriate dimension. The result will be a non--minimal state space realization that preseves the original states, retains the input/output transfer function from u to y, and has outputs ynew that are state variables.

Here's an example.

First, define some state space model

>> sys1 = minreal(ss(zpk([tf(1,[1 1]);tf(1,conv([1 1],[1 2]));tf(1,conv([1 1],[1 3]))])))

sys1 =
 
  A = 
            x1       x2       x3
   x1   -2.067  -0.2775  -0.1388
   x2   0.2401   -1.212    0.894
   x3  0.03861   0.5588   -2.721
 
  B = 
           u1
   x1  0.9679
   x2  0.6052
   x3   1.303
 
  C = 
            x1       x2       x3
   y1   0.1629   0.6703   0.3352
   y2  -0.6095   0.4695   0.2348
   y3  0.06517   0.4681  -0.2659
 
  D = 
       u1
   y1   0
   y2   0
   y3   0
 
Continuous-time state-space model.

Let's look at its transfer function matrix in zpk form (note the cancellations in the individual elements):

>> zpk(sys1)

ans =
 
  From input to output...
          (s+2) (s+3)
   1:  -----------------
       (s+2) (s+3) (s+1)
 
             (s+3)
   2:  -----------------
       (s+2) (s+3) (s+1)
 
             (s+2)
   3:  -----------------
       (s+2) (s+3) (s+1)
 
Continuous-time zero/pole/gain model.

Now defne a new state space realization using the augmented state vector and check its transfer function in zpk form:

>> sys2=ss([[sys1.a ; sys1.c(1:2,:)*sys1.a] zeros(5,2)],[sys1.b;sys1.c(1:2,:)*sys1.b],[zeros(2,3) eye(2);sys1.c(3,:) zeros(1,2)],[zeros(2,1);sys1.d(3,:)]);
>> zpk(sys2)

ans =
 
  From input to output...
          s (s+2) (s+3)
   1:  -------------------
       s (s+3) (s+2) (s+1)
 
             s (s+3)
   2:  -------------------
       s (s+3) (s+2) (s+1)
 
             (s+2)
   3:  -----------------
       (s+3) (s+2) (s+1)
 
Continuous-time zero/pole/gain model.

The transfer function matrices are the same (after cancellation), but additional poles at the origin show up because of the non-minimal realization. As must be the case, the first two outputs are state variables.

>> sys2.c

ans =

            0            0            0   1.0000e+00            0
            0            0            0            0   1.0000e+00
   6.5167e-02   4.6814e-01  -2.6593e-01            0            0

Another option is to define a new state vector (z) that is a linear combination of the original state vector (x) such that elements of z are the outputs that we want to be states. In other words

z = Tx = [C1; V] *x

where it is assumed that C1 has full row rank and that a matrix V can be found such that T is invertible. I'm pretty sure (not positive) that such a V can always be found. If we further assume that each row of C1 has a non-zero element in more than one column, then one option is to choose V such that each row of V is zero except in one column (different columns for each row). If this is assumption is not true, then it still should be easy to find a V.

Here we define T such that the first two states are the desired outputs and the third state is preserved.

>> T=[sys1.c(1:2,:);[0 0 1]]

T =

   1.6292e-01   6.7034e-01   3.3517e-01
  -6.0951e-01   4.6951e-01   2.3476e-01
            0            0   1.0000e+00

Verify that T is invertible.

            
>> rank(T)

ans =

     3

This tansformation T is called a similarity transformation and can used to define a new realization in terms of z (instead of x)

>> sys3=ss2ss(sys1,T)

sys3 =
 
  A = 
               x1          x2          x3
   x1          -1  -4.857e-17  -4.441e-16
   x2           1          -2  -4.441e-16
   x3      0.7396      0.1343          -3
 
  B = 
              u1
   x1          1
   x2  2.116e-16
   x3      1.303
 
  C = 
               x1          x2          x3
   y1           1           0  -5.551e-17
   y2           0           1           0
   y3      0.6513     0.06717        -0.5
 
  D = 
       u1
   y1   0
   y2   0
   y3   0
 
Continuous-time state-space model.

>> zpk(sys3)

ans =
 
  From input to output...
          (s+2) (s+3)
   1:  -----------------
       (s+1) (s+2) (s+3)
 
             (s+3)
   2:  -----------------
       (s+1) (s+2) (s+3)
 
             (s+2)
   3:  -----------------
       (s+1) (s+2) (s+3)
 
Continuous-time zero/pole/gain model.

As expected, the first two outputs of sys3 are now state variables and the transfer function matrix of sys3 is the same as that of sys1.

[Adam Evans]

That's brilliant. Thank you very the very detailed response!