Why can't an improper transfer function be realized?

Question

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.

Answer 1

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?

Answer 2

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.