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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.

Rodrigo de Azevedo
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ITA
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  • Do you see the intuitive problem in the special case $G(s)=1$ (in which case $g(x)=\delta(x)$)? – Ian Aug 26 '16 at 16:26

3 Answers3

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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.

Pait
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    Not entirely buying this. An integrator is proper, but can also not provide infinite gain at low frequencies. – Henk van der Laak Jan 04 '23 at 00:20
  • 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.

     – Pait Feb 06 '23 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?

Rodrigo de Azevedo
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  • 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. – Pait Aug 27 '16 at 10:22
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    @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. – Rodrigo de Azevedo Aug 27 '16 at 12:15
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    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. – Pait Aug 27 '16 at 17:56
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I did not have the space to write my comment, so I am adding this as answer to contribute to the debate.

It is important to distinguish the concepts of realizability in terms of "admitting a state-space realization" and "admitting a physical realization." The OP asked about former, and the answer by Rodrigo de Azevedo is correct (it is only proving one side of the implication; the general proof can be found in, e.g., Theorem 3.3, Chapter 5, of Linear Systems by Antsaklis and Michel). Sure, you can hack your state-space formulation by adding input derivatives, but then it is not a state-space realization, it is another type of system description.

The answer by Pait is more on "admitting a physical realization". It is not correct, because there are several devices compute derivatives. E.g., Ogata's Modern Control Engineering, Chap. 3, shows several circuits using resistors, capacitors, and operational amplifiers to build PD or PID controllers, which have improper transfer functions. A pure differentiator can be build with a capacitor, a resistor, and an OpAmp, see here.

Now, notice that every improper transfer function will be the sum of a few derivatives and a proper transfer function (just do polynomial division). The general issue with why, in practice, time-derivatives are avoided, is because they amplify noise. Noise is often modeled as a stochastic (aka random) process, the purest of it being the so-called white noise. One of the main "features" of the white noise is that its power spectral density is constant, meaning that all frequencies are present. Differentiating white noise gives infinite gain at infinite frequency, meaning that, in practice, you immediately obtain immense noise. Integrating white noise generates the Wiener process (or the Brown noise, depending on the formalism you pick), which generally drifts towards infinity as time goes to infinity... so there is a long-term problem with integration as well. Hence, Pait is right in the sense that differentiating creates an instantaneous problem, while integrating does not.

Gabriel Gleizer
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