Final value theorem for non-rational transfer functions

Question

One version of the Final Value Theorem often seen in controls textbooks:

Suppose $f(t)$ has (one-sided) Laplace Transform $F(s)$ and further suppose that every pole of $s F(s)$ is in the open left-half plane. Then the limit $\lim_{s\to 0} s F(s)$ exists, and $\lim_{t\to\infty} f(t) = \lim_{s\to 0} s F(s)$.

Critically, there is no a priori explicit assumption that the limit $\lim_{t\to\infty} f(t)$ exists; it follows as a consequence of the assumptions made on $s F(s)$.

The standard proof uses partial fraction decomposition, which requires assuming $F(s)$ is a rational function. Is there a more general version of this result that can handle the cases where:

• $F(s)$ may have infinitely many poles
• $F(s)$ may not be rational

I want to preserve the property that no assumptions are made about $f(t)$ or its limiting behavior.

I suspect the result should still hold if we replace "every pole of $sF(s)$ is in the open left-half plane" with something like "there exists $\varepsilon > 0$ such that $sF(s)$ is analytic in the set of $s$ satisfying $\mathrm{Re}(s)>-\varepsilon$". Has anybody seen such a result or have ideas on how to prove such a thing?

Answer 1

It sounds like you are looking for the generalized final value theorem. You may want to check the paper (and their references):

Chen, Jie, et al. "The final value theorem revisited-Infinite limits and irrational functions." IEEE Control Systems Magazine 27.3 (2007): 97-99.

They discuss a generalized final value theorem that also works for irrational functions:

$$ \lim_{t\rightarrow \infty}\frac{f(t)}{t^{\lambda}}=\frac{1}{\Gamma(\lambda+1)}\lim_{s\downarrow 0}s^{\lambda+1}F(s) $$

with $f(t)$ Laplace transformable, $\lambda>-1$, both limits exist, $\Gamma$ is the gamma function and $s\downarrow 0$ means "that $s$ approaches $0$ through the positive numbers".

For $\lambda=0$ this is just the normal final value theorem.

Answer 2

This is not a full answer, yet. It is getting late here and will add more things when I will have some time.

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Let us start with this remark:

Since we have that

$$sF(s)=\int_0^\infty f'(t)e^{-st}dt,$$ then we get $$\lim_{s\to0}sF(s)=\left.\int_0^\infty f'(t)e^{-st}dt\right|_{s=0}=\lim_{t\to\infty}f(t),$$ when the limit exists and where we have assumed that $f(0)=0$.

So, the only difficulty is that the $f'$ should go to zero sufficiently fast so that $\lim_{t\to\infty}f(t)$ exists. This formulation shows that it does not really matter what the form of $F(s)$ is; i.e. rational or not.

So, in the end, if we can show that $f'$ is exponentially decaying to, then we have our convergence. Of course, this excludes a lot of signals which are decaying at least as fast as $1/t^2$, but anyway...

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A first idea is to assume that the signal $y$ is the output of an infinite-dimensional dynamical $(A,B,C,D)$ where $A$ is the generator of a $C_0$-semigroup $T(t)$. We also assume that the input/output operators $B,C,$ and $D$ are such that the system is a Pritchard-Salamon system (admissible operators). See e.g. Logemann, "Stabilization and Regulation of Infinite-Dimensional Systems Using Coprime Factorizations".

The impulse response of that system is given by $CT(t)B+D\delta(t)$ and the associated transfer function (or characteristic function) is simply given by $C(sI-A)^{-1}B+D$ where $s\in\rho(A)$ where $\rho(A)$ is the resolvent set of $A$ and where $(sI-A)^{-1}$ is the resolvent of $A$.

In that case, the exponential stability of the semigroup is equivalent to the fact that the transfer function be analytic in the closed right-half plane under the assumption that the system is admissible stabilizable and detectable. Of course, in the current context, we just need the input-output exponential convergence, so we can drop the stabilizability and detectability conditions. What is important here is that stability follows from the analyticity of the function. Note that this may not be the case, as discussed below.

Under the necessary and sufficient condition of the Hille-Yosida-Miyadera-Feller-Philips theorem, $A$ generates a $C_0$ semigroup $T$ and we have that $$||T(t)||\le M e^{\omega t},\ t\ge0,$$ for some real $M,\omega$. Then, $A$ has the following property $$C_\omega:=\{\lambda\in\mathbb{C}:\Re[\lambda]>\omega\}\subset\rho(A).$$

So, this means that if the semigroup is exponentially stable, then the spectrum of $A$ lies in the open left half-plane. The converse does not hold in general but it does in certain cases such as when $A$ is a bounded operator, $T$ is a differentiable semigroup, $T$ is nilpotent, $T$ is analytic, etc.

This means that we can look at the location of the "poles" or the analyticity of the resolvent in the closed right half plane for establishing the stability of the process and the convergence to zero of the impulse response.

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The case of time-delay system is a corollary of the above discussion as the semigroup is eventually differentiable. For instance, consider the following time-delay system

$$\dot{x}(t)=-x(t)-2x(t-h)+u,\ y(t)=x(t),$$ where $h>0$ is sufficiently small so that the system is stable. The transfer function associated with this system is given by

$$H(s)=\dfrac{1}{s+1+2e^{-sh}}$$

and since the system is stable the spectral abscissa is negative (so the system is BIBO stable). If the input is a step function, then we have that $Y(s)=H(s)/s$ and we have that

$$\lim_{t\to\infty}y(t)=\lim_{s\to0}sY(s)=\lim_{s\to0}H(s)=1/3.$$

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Some reading:

1. Curtain and Zwart, "An Introduction to Infinite-Dimensional Linear Systems Theory"
2. Benoussan, Da Prato, Delfour, and Mitter, "Representation and Control of Infinite Dimensional Systems"