How to convert a transfer function into state space representation?

I was trying to convert a transfer function into state space representation, but the matrices in the output are not quiet correct. The numbers are flipped like how in B 0 should be up and 1 should be down, or in C where 2 should be right and 3 should be left. How can I fix this issue?

Answer 1

[Paul]

The state space realization of a transfer function is not unique. In fact, there are infinitely many state space realizations to choose from.

num = [0 3 2];
den = [1 4 4];

The Control System Toolbox uses one methodology

G = ss(tf(num, den))

G =
 
  A = 
       x1  x2
   x1  -4  -2
   x2   2   0
 
  B = 
       u1
   x1   2
   x2   0
 
  C = 
        x1   x2
   y1  1.5  0.5
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

and the Signal Processing Toolbox uses another

[A,B,C,D] = tf2ss(num,den)

A = 2×2

-4 -4 1 0

B = 2×1

1 0

C = 1×2

3 2

D = 0

Each realization has the same transfer function

tf(G)

ans =
 
     3 s + 2
  -------------
  s^2 + 4 s + 4
 
Continuous-time transfer function.

[b,a] = ss2tf(A,B,C,D)

b = 1×3

0 3.0000 2.0000

a = 1×3

1 4 4

Any other realization can be obtained via similarity transformation. The CST provides a function ss2ss to do that

ss2ss(ss(A,B,C,D),[0 1;1 0])

ans =
 
  A = 
       x1  x2
   x1   0   1
   x2  -4  -4
 
  B = 
       u1
   x1   0
   x2   1
 
  C = 
       x1  x2
   y1   2   3
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

tf(ans)

ans =
 
     3 s + 2
  -------------
  s^2 + 4 s + 4
 
Continuous-time transfer function.

[Paul]

It appears that the Control System Toolbox finds a realization with a balance(d) A-matrix, at least in this case. I speculate that such a realization has better numerical properties for operations that are typically performed on such models.

Define a simple transfer function

num = [1,2,3];
den = [1,11,12,13];

The Signal Processing Toolbox returns a realization in a canonical form (offhand, I forget what this form is called). It's easy to convert a transfer function into a canonical form and canonical forms are useful from a theoretical perspective, but, IIRC, they are not preferred numerically.

[a,b,c,d] = tf2ss(num,den);a,b,c

a = 3×3

-11 -12 -13 1 0 0 0 1 0

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b = 3×1

1 0 0

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c = 1×3

1 2 3

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Now balance the a-matrix and get the corresponding transformation matrix.

[T,a] = balance(a);
a

a = 3×3

-11.0000 -3.0000 -1.6250 4.0000 0 0 0 2.0000 0

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Apply the transformation to the b- and c-matrix

b = T\b

b = 3×1

1 0 0

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c = c*T

c = 1×3

1.0000 0.5000 0.3750

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Show that the transfer function is unaffected (as must be the case with a similarity transformation)

[bb,aa] = ss2tf(a,b,c,0)

bb = 1×4

0 1.0000 2.0000 3.0000

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aa = 1×4

1.0000 11.0000 12.0000 13.0000

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The CST returns the same result.

sys = ss(tf(num,den))

sys =
 
  A = 
           x1      x2      x3
   x1     -11      -3  -1.625
   x2       4       0       0
   x3       0       2       0
 
  B = 
       u1
   x1   1
   x2   0
   x3   0
 
  C = 
          x1     x2     x3
   y1      1    0.5  0.375
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

Answer 2

[Walter Roberson]

num = [0 3 2];
den = [1 4 4];
G = tf(num, den);
S = ss(G)

S =
 
  A = 
       x1  x2
   x1  -4  -2
   x2   2   0
 
  B = 
       u1
   x1   2
   x2   0
 
  C = 
        x1   x2
   y1  1.5  0.5
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

S.A

ans = 2×2

-4 -2 2 0

S.B

ans = 2×1

2 0

S.C

ans = 1×2

1.5000 0.5000

S.D

ans = 0

[A, B, C, D] = tf2ss(num, den)

A = 2×2

-4 -4 1 0

B = 2×1

1 0

C = 1×2

3 2

D = 0

At the moment I do not kow why the values do not match.

[Jan]

I guess ss(tf()) prescales the system and tf2ss(tf()) does not