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A major result in control system theory is that a transfer function, $$G\left( s \right) = \frac{{Y\left( s \right)}}{{U\left( s \right)}}$$ has a state space realization if and only if the degree of $Y(s)$ is less than or equal to the degree of $U(s)$. I cannot find a proof of this fact in most major (undergraduate and introductory graduate) textbooks. If someone knows the proof could they sketch it out for me or point me to references where the proof exists?

There is a related question here but it still does not answer the "why" of state-space realizations being non-existent for improper transfer functions.

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  • $\begingroup$ Do you see the intuitive problem in the special case $G(s)=1$ (in which case $g(x)=\delta(x)$)? $\endgroup$
    – Ian
    Aug 26, 2016 at 16:26

2 Answers 2

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To realize an improper transfer function, derivatives of the input would be needed. The answer above by Rodrigo de Azevedo helps make clear why. The problem is that it is not possible to realize perfect derivatives. A number of arguments are helpful in understanding why.

The modulus of the frequency response of a differentiator increases with frequency. However it is not possible to construct an apparatus whose gain becomes arbitrary large at large frequencies. On the contrary, any device known will have a cutoff frequency after which its response falls.

Or, suppose you feed a discontinuous signal into a perfect differentiator. It will have to compute the derivative of the signal, before noticing that the derivative doesn't exist! So any "differentiator" will be at best an approximation.

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  • $\begingroup$ Not entirely buying this. An integrator is proper, but can also not provide infinite gain at low frequencies. $\endgroup$ Jan 4, 2023 at 0:20
  • $\begingroup$ Over time any physical construction of an integrator will saturate. Yes, a perfect linear integrator also cannot be built. The issue with a perfect differentiator is instantaneous. $\endgroup$
    – Pait
    Feb 6, 2023 at 12:53
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Suppose we have a state-space model

$$\begin{align} \dot{\mathrm x} &= \mathrm A \mathrm x + \mathrm B \mathrm u\\ \mathrm y &= \mathrm C \mathrm x + \mathrm D \mathrm u \end{align}$$

where $\mathrm A \in \mathbb R^{n \times n}$. Laplace-transforming both the state equation and the output equation, we conclude that the transfer function is the following matrix-valued function

$$\mathrm G (s) = \mathrm C (s \mathrm I_n - \mathrm A)^{-1} \mathrm B + \mathrm D$$

Note that

$$(s \mathrm I_n - \mathrm A)^{-1} = \frac{\mbox{adj} (s \mathrm I_n - \mathrm A)}{\det (s \mathrm I_n - \mathrm A)}$$

and that

  • each entry of the adjugate is a polynomial in $s$ of degree at most equal to $n-1$.
  • the determinant of $s \mathrm I_n - \mathrm A$ is a polynomial in $s$ of degree $n$.

Thus, we can conclude that each of the $n^2$ SISO transfer functions in $\mathrm G (s)$ has the property that the degree of the numerator is less than or equal to the degree of the denominator.

How can an improper transfer function have a state-space realization, then?

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  • $\begingroup$ This only shows that an improper transfer function cannot be realized in the particular form you wrote. It could be realized using derivatives of the input. $\endgroup$
    – Pait
    Aug 27, 2016 at 10:22
  • $\begingroup$ @Pait The problem is that pure differentiators do not have state-space realizations. Hence, what you're proposing sounds like a distortion of the problem. You're relabeling the input vector so that the differentiators stay out of the system. $\endgroup$ Aug 27, 2016 at 12:15
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    $\begingroup$ Yes they do! $y = \dot{u}$ is a realization. Just not in the usual form that you mentioned and that we all know and love. The crucial question is why we chose that form without differentiators. It is not written in stone that realizations must follow it; rather, the usual form was chosen because of very practical considerations. $\endgroup$
    – Pait
    Aug 27, 2016 at 17:56

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