ECE 602 Fall 2010

State-Space Realization Examples

by

Stanislaw H. Żak

Example 1 We find a state-space realization of the transfer function matrix,

G(s) = \frac{1}{s^{3} + 3s^{2} + 2s} [ \begin{matrix}s + 1 & 2s^{2} + s − 1 & s^{2} − 1 \\ −s^{2} − s & −s^{2} + s & s \end{matrix} ] .  (1)

The above transfer function is strictly proper; therefore

D = G(∞) = [ \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ] .

We proceed to represent G(s) as

G(s) = \frac{1}{s^{3} + 3s^{2} + 2s} ([ \begin{matrix}0 & 2 & 1 \\ −1 & −1 & 0 \end{matrix} ] s^{2} + [ \begin{matrix}1 & 1 & 0 \\ −1 & 1 & 1 \end{matrix} ] s + [ \begin{matrix}1 & −1 & −1 \\ 0 & 0 & 0 \end{matrix} ])
= \frac{1}{s^{3} + 3s^{2} + 2s} (N_{2}s^{2} + N_{1}s + N_{0}) .

Using realization formulas found in the proof of Theorem 1, we obtain

A = [ \begin{matrix}O & I_{3} & O \\ O & O & I_{3} \\ O & −2I_{3} & −3I_{3} \end{matrix} ]

= [ \begin{matrix}0 & 0 & 0 & ⋮ & 1 & 0 & 0 & ⋮ & 0 & 0 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & 1 & 0 & ⋮ & 0 & 0 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & 0 & 1 & ⋮ & 0 & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 & ⋮ & 0 & 0 & 0 & ⋮ & 1 & 0 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & 0 & 0 & ⋮ & 0 & 1 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & 0 & 0 & ⋮ & 0 & 0 & 1 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 & ⋮ & −2 & 0 & 0 & ⋮ & −3 & 0 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & −2 & 0 & ⋮ & 0 & −3 & 0 \\ 0 & 0 & 0 & ⋮ & 0 & 0 & −2 & ⋮ & 0 & 0 & −3 \end{matrix} ] .  (2)

The input matrix B is

B = [ \begin{matrix}O \\ O \\ I & 3 \end{matrix} ] = [ \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ ⋯ & ⋯ & ⋯ \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ] ,  (3)

and the matrix C has the form

C = [ N_{0} N_{1} N_{2} ] = [ \begin{matrix}1 & −1 & −1 & ⋮ & 1 & 1 & 0 & ⋮ & 0 & 2 & 1 \\ 0 & 0 & 0 & ⋮ & −1 & 1 & 1 & ⋮ & −1 & −1 & 0 \end{matrix} ] .  (4)

The dimension of this realization is 9.

We can obtain another realization of G(s) using Theorem 1. We begin by constructing a realization for G(s)^{⊤} using realization formulas from the proof of Theorem 1 to obtain

G(s)^{⊤} = C̃(sI − Ã)^{−1}B̃ + D̃.  (5)

We next transpose both sides of (5) to obtain

(G(s)^{⊤})^{⊤} = G(s)
= (C̃(sI − Ã)^{−1}B̃ + D̃)^{⊤}
= B̃^{⊤} (sI − Ã^{⊤})^{−1} C̃^{⊤} + D̃^{⊤}.

The resulting realization has the form

\begin{matrix}˙ ; x̃ & = & [ \begin{matrix}O & O & O & ⋯ & O & −α_{0}I_{p} \\ I_{p} & O & I_{m} & ⋯ & O & −α_{1}I_{p} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ O & O & O & ⋯ & O & −α_{r−2}I_{p} \\ O & O & O & ⋯ & I_{p} & −α_{r−1}I_{m} \end{matrix} ] x̃ + [ N_{0} ; N_{1} ; ⋮ ; N_{r−2} ; N_{r−1} ] u \\ y & = & [ O O O ⋯ O I_{p} ] x̃ + G(∞)u. \end{matrix} \}  (6)

Note that although the realization (6) is based on the realization found in the proof of Theorem 1, their dimensions will be different if p ≠ m. Example 2 We use the above method to construct a state-space realization of the transfer function matrix given by (1). We first apply realization formulas of Theorem 1 to G(s)^{⊤}, where

G(s)^{⊤} = \frac{1}{s^{3} + 3s^{2} + 2s} [ \begin{matrix}s + 1 & −s^{2} − s \\ 2s^{2} + s − 1 & −s^{2} + s \\ s^{2} − 1 & s \end{matrix} ]

= \frac{1}{s^{3} + 3s^{2} + 2s} ([ \begin{matrix}0 & −1 \\ 2 & −1 \\ 1 & 0 \end{matrix} ] s^{2} + [ \begin{matrix}1 & −1 \\ 1 & 1 \\ 0 & 1 \end{matrix} ] s + [ \begin{matrix}1 & 0 \\ −1 & 0 \\ −1 & 0 \end{matrix} ])

= \frac{1}{s^{3} + 3s^{2} + 2s} (N_{2}^{⊤}s^{2} + N_{1}^{⊤}s + N_{0}^{⊤}) .

Applying to the above (6), we obtain another state-space realization of the transfer function (1), where

Â = [ \begin{matrix}O & O & O \\ I_{2} & O & −2I_{2} \\ O & I_{2} & −3I_{3} \end{matrix} ]

= [ \begin{matrix}0 & 0 & ⋮ & 0 & 0 & ⋮ & 0 & 0 \\ 0 & 0 & ⋮ & 0 & 0 & ⋮ & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 1 & 0 & ⋮ & 0 & 0 & ⋮ & −2 & 0 \\ 0 & 1 & ⋮ & 0 & 0 & ⋮ & 0 & −2 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & ⋮ & 1 & 0 & ⋮ & −3 & 0 \\ 0 & 0 & ⋮ & 0 & 1 & ⋮ & 0 & −3 \end{matrix} ] .  (7)

The input matrix is

B̂ = [ \begin{matrix}N & 0 \\ N & 1 \\ N & 2 \end{matrix} ] = [ \begin{matrix}1 & −1 & −1 \\ 0 & 0 & 0 \\ ⋯ & ⋯ & ⋯ \\ 1 & 1 & 0 \\ −1 & 1 & 1 \\ ⋯ & ⋯ & ⋯ \\ 0 & 2 & 1 \\ −1 & −1 & 0 \end{matrix} ] ,  (8)

and the output matrix has the form

Ĉ = [ O O I_{2} ] = [ \begin{matrix}0 & 0 & ⋮ & 0 & 0 & ⋮ & 1 & 0 \\ 0 & 0 & ⋮ & 0 & 0 & ⋮ & 0 & 1 \end{matrix} ] .  (9)

The obtained realization is of lower dimension than the previous one. This does not mean that the alternative method will always yield lower dimension realizations than the first method.

We now present yet another method of constructing a state-space realization of a given transfer function matrix. This method is being used by the MATLAB’s function tf2ss. We begin by denoting the i-th column of a proper transfer function matrix G(s) as g_{i}(s). We observe that the input-output relation of an MIMO system can be expressed as

Y (s) = G(s)U(s)
= g_{1}(s)U_{1}(s) + g_{2}(s)U_{2}(s) + ⋯ + g_{m}(s)U_{m}(s),

where U_{i}(s) is the i-th component of the input vector U(s). We construct a state-space realization of each column g_{i}(s) and then combine these realizations to obtain the statespace realization of G(s) as follows. If the quadruple (A_{i}, b_{i}, C_{i}, d_{i}) is the realization of g_{i}(s), then a realization (A, B, C, D) of G(s) has the form,

\begin{matrix}A & = & [ \begin{matrix}A_{1} & O & ⋯ & O \\ O & A_{2} & ⋯ & O \\ ⋮ & ⋮ & ... & ⋮ \\ O & O & ⋯ & A_{m} \end{matrix} ] , B = [ \begin{matrix}b_{1} & 0 & ⋯ & 0 \\ 0 & b_{2} & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & b_{m} \end{matrix} ] , \\ C & = & [ C_{1} C_{2} ⋯ C_{m} ] , and D = [ d_{1} d_{2} ⋯ d_{m} ] . \end{matrix} \}  (10)

Example 3 We use the column-by-column method to construct a realization of the transfer function matrix,

G(s) = [ \begin{matrix}\frac{1}{s+1} & \frac{2}{s+1} \\ \frac{−1}{s^{2}+3s+2} & \frac{1}{s+2} \end{matrix} ] .  (11)

This transfer function matrix is strictly proper, therefore

D = [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ] .

There are two columns, that is,

G(s) = g_{1}(s) + g_{2}(s) = [ \frac{1}{s+1} ; \frac{−1}{s^{2}+3s+2} ] + [ \frac{2}{s+1} ; \frac{1}{s+2} ] .

We apply the realization formulas from the proof of Theorem 1 to

g_{1}(s) = [ \frac{1}{s+1} ; \frac{−1}{s^{2}+3s+2} ] = \frac{1}{s^{2} + 3s + 2} [ \begin{matrix}s + 2 \\ −1 \end{matrix} ] = [ \begin{matrix}1 \\ 0 \end{matrix} ] s + [ \begin{matrix}2 \\ −1 \end{matrix} ] = N_{1}s + N_{0}.

We obtain

A_{1} = [ \begin{matrix}0 & 1 \\ −2 & −3 \end{matrix} ] , b_{1} = [ \begin{matrix}0 \\ 1 \end{matrix} ] , C_{1} = [ \begin{matrix}2 & 1 \\ −1 & 0 \end{matrix} ] , d_{1} = [ \begin{matrix}0 \\ 0 \end{matrix} ] .

We next concern ourselves with the second column of G(s),

g_{2}(s) = [ \frac{2}{s+1} ; \frac{1}{s+2} ] = \frac{1}{s^{2} + 3s + 2} [ \begin{matrix}2(s + 2) \\ s + 1 \end{matrix} ] = [ \begin{matrix}2 \\ 1 \end{matrix} ] s + [ \begin{matrix}4 \\ 1 \end{matrix} ] = N_{1}s + N_{0}.

We apply realization formulas from the proof of Theorem 1 to obtain

A_{2} = [ \begin{matrix}0 & 1 \\ −2 & −3 \end{matrix} ] , b_{2} = [ \begin{matrix}0 \\ 1 \end{matrix} ] , C_{2} = [ \begin{matrix}4 & 2 \\ 1 & 1 \end{matrix} ] , d_{2} = [ \begin{matrix}0 \\ 0 \end{matrix} ] .

Combining the above realizations gives an overall realization of G(s) of the form,

A = [ \begin{matrix}A_{1} & O \\ O & A_{2} \end{matrix} ] = [ \begin{matrix}0 & 1 & ⋮ & 0 & 0 \\ −2 & −3 & ⋮ & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & ⋮ & 0 & 1 \\ 0 & 0 & ⋮ & −2 & −3 \end{matrix} ] , B = [ \begin{matrix}b_{1} & 0 \\ 0 & b_{2} \end{matrix} ] = [ \begin{matrix}0 & ⋮ & 0 \\ 1 & ⋮ & 0 \\ ⋯ & ⋯ \\ 0 & ⋮ & 0 \\ 0 & ⋮ & 1 \end{matrix} ] ,

C = [ C_{1} C_{2} ] = [ \begin{matrix}2 & 1 & ⋮ & 4 & 2 \\ −1 & 0 & ⋮ & 1 & 1 \end{matrix} ] , and D = [ d_{1} d_{2} ] = [ \begin{matrix}0 & ⋮ & 0 \\ 0 & ⋮ & 0 \end{matrix} ] .

Example 4 We now apply the method resulting in the state-space realization given by (6) to the transfer function (11) from the previous example. We begin by transposing G(s) to obtain

G(s)^{⊤} = [ \begin{matrix}\frac{1}{s+1} & \frac{−1}{s^{2}+3s+2} \\ \frac{2}{s+1} & \frac{1}{s+2} \end{matrix} ] = g̃_{1}(s) + g̃_{2}(s) = [ \frac{1}{s+1} ; \frac{2}{s+1} ] + [ \frac{−1}{s^{2}+3s+2} ; \frac{1}{s+2} ] .

The column g̃_{1}(s) can be represented as

g̃_{1}(s) = \frac{1}{s + 1} [ \begin{matrix}1 \\ 2 \end{matrix} ] .

Hence, we have one-dimensional realization of g̃_{1}(s),

Ã_{1} = [ −1 ] , b̃_{1} = [ 1 ] , C̃_{1} = [ \begin{matrix}1 \\ 2 \end{matrix} ] , nad d̃_{1} = [ \begin{matrix}0 \\ 0 \end{matrix} ] .  (12)

We then find a realization of g̃_{2}(s), which we first represent as

g̃_{2}(s) = [ \frac{−1}{s^{2}+3s+2} ; \frac{1}{s+2} ] = \frac{1}{s^{2} + 3s + 2} [ \begin{matrix}−1 \\ s + 1 \end{matrix} ] = [ \begin{matrix}0 \\ 1 \end{matrix} ] s + [ \begin{matrix}−1 \\ 1 \end{matrix} ] = Ñ_{1}s + Ñ_{0}.

We obtain a second-order realization of g̃_{2}(s), where

Ã_{2} = [ \begin{matrix}0 & 1 \\ −2 & −3 \end{matrix} ] , b̃_{2} = [ \begin{matrix}0 \\ 1 \end{matrix} ] , C̃_{2} = [ \begin{matrix}−1 & 0 \\ 1 & 1 \end{matrix} ] , and d̃_{2} = [ \begin{matrix}0 \\ 0 \end{matrix} ] .  (13)

Combining (12) and (13) and applying the transposition operation we obtain a third-order realization of (11),

Â = [ \begin{matrix}Ã_{1} & 0 \\ 0 & Ã_{2} \end{matrix} ]^{⊤} = [ \begin{matrix}−1 & 0 & 0 \\ 0 & 0 & −2 \\ 0 & 1 & −3 \end{matrix} ] , B̂ = [ C̃_{1} C̃_{2} ]^{⊤} = [ \begin{matrix}1 & 2 \\ −1 & 1 \\ 0 & 1 \end{matrix} ] ,

Ĉ = [ \begin{matrix}b̃_{1} & 0 \\ 0 & b̃_{2} \end{matrix} ]^{⊤} = [ \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} ] , D̂ = [ D̃_{1} D̃_{2} ]^{⊤} = [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ] .

In the above examples, we obtained realizations of different dimensions for the same transfer function. A natural questions arises of how to find a minimal realization for a given transfer function matrix. To answer this question we need notions of controllability and observability. We will prove that a realization is minimal if and only if it is both reachable and observable.