• Can every possible nonlinear time-invariant Volterra series be represented by a Volterra series?

  • If not, which ones can?

  • Also, are there any time-variant systems that can be represented by Volterra series? I don't see how from the equation, but want to be sure I'm not missing something.


1 Answer 1


The system must be time invariant and smooth in the functional derivative sense. That doesn't guarantee that the Volterra series converges (like with Taylor series, there are pathological counter examples), but almost all systems that have these properties have a convergent Volterra series.

The problem in practice is that the required expansion order for good approximations of your system is often far too great to provide for a useful implementation of the system. A condition for quick convergence is, that the functional derivatives are well bounded and decay quickly with order.

A little more detail:

Consider a non-linear system $f:V\to W$, where $W$ is an inner product space with induced norm and $V$ is a normed vector space. With this, look at a map

$$\begin{align} W\times V&\to K_W \\ (w,x) &\mapsto \langle w,f(v+\lambda x)\rangle \end{align}$$

This map is anti-linear in $w$ and non-linear in $x$. We treat $\lambda \in K_V$ and $v \in V$ as parameters. Further below, $v$ will be the center of the Volterra expansion and $\lambda$ will reveal the different expansion orders.

If we hold $w$ constant, we get a functional

$$ \begin{align} F_w:V&\to K_W\\ x&\mapsto F_w[x]=\langle w,f(v+\lambda x)\rangle \end{align} $$

For bounded $x$, this functional can be expanded into a series of powers of $\lambda$ around $v$: $$F[x]=\sum_{n=0}^\infty \frac{\lambda^n}{n!}F^{(n)}(x)$$

where $F^{(n)}(x)$ is the $n$-th (functional) derivative of $F$ around $v$ in direction $x$.

The assumption that all $F^{(n)}$ exist and the series converges for all finite norm choices of $w$ is what we mean, when we say: "the system $f$ is smooth".

Going from here to the full Volterra series is not too difficult. The key is understanding, how the functional derivatives relate to $w$ and $x$. Since $F$ is anti-linear in $w$, every single term in the series expansion must be, and therefore $F^{(n)}$ as the only factor with dependence on the choice of $w$. It is also clear by the option to absorb $\lambda$ into $x$, that the derivative $F^{(n)}$ must be of $n$-th order in $x$.

Hence we can write $F^{(n)}(x)=\langle w, G_n(x) \rangle$ with multilinear $\hat{G}_n:V^n \to W$ as well as an embedding $E_n:V \to V^n$ with $E_n(v)=(v,v,\dots,v)$ composed to $G_n=\hat{G}_n \circ E_n$.

In infinite dimensional vector spaces with uncountable bases, these multilinear maps are expressible through nested integrals over kernel functions. If you use the time basis for this expansion and make use time-invariance, you get nested convolution integrals and arrive at the Volterra series. I'll leave this mostly notational part to you. If you need help with it, I can also elaborate more.

As the last step, you use the properties of the inner product to draw $w$ out of the series expression and simply drop it, as the resulting equation needs to hold for all proper $w$.

  • $\begingroup$ When you say "smooth" in this context, what do you mean? How do you extrapolate from the definition of a smooth function to a "smooth system?" $\endgroup$ Commented Aug 6, 2016 at 17:15
  • $\begingroup$ "Smooth" usually means that you have infinitely many continuous derivatives. Depending on the context it can also be a little stronger or weaker, but that's the idea. It's the general property that allows power series to meaningfully converge. Here, the derivatives are taken in the functional sense. A functional derivative is the generalisation of a multi-dimensional derivative. You can consider the nonlinear system as a functional that takes one signal to another signal. The derivative is then in the direction of a certain disturbing signal. $\endgroup$
    – Jazzmaniac
    Commented Aug 6, 2016 at 20:02
  • $\begingroup$ In other words, if you have a functional $y(t)=F[x(t)]$ that describes your signal, then if you can write $F[x(t)+\lambda g(t)]=y(t)+\lambda h(t) + O(\lambda^2)$ for all bounded probing signals $g(t)$, your system has a derivative in the functional sense. Extend that to a higher order for $\lambda$ and you have asserted the existence of higher derivatives. Add in time invariance to relate shifted test signals with shifted response signals by convolution and you already get the very definition of the Volterra series. $\endgroup$
    – Jazzmaniac
    Commented Aug 6, 2016 at 20:08
  • $\begingroup$ the definition of a functional that I know is that it's a function $f: V \to K$ from a vector space $V$ to its field of scalars $K$. If $V$ is a space of complex-valued functions of a real variable, as in this case, then $K$ would be $\mathbb C$. But the sense in which you're using the term "functional" is a function $f: V \to V$. So that's why I'm confused. $\endgroup$ Commented Aug 6, 2016 at 22:36
  • $\begingroup$ @MikeBattaglia, yes, that is true. The usual functional you will encounter is of that form and will produce a scalar. But you can easily generalise it to map to any other space. The idea of the functional derivative stays the same, which is the important bit here. So if you don't want to call it functional, fine. A stricter way to get a true functional would be to introduce a second test function that is applied to the system's output by means of an inner product. But then you'd have to handle two independent test functions and things tend to get more complicated. If you want, I can show you. $\endgroup$
    – Jazzmaniac
    Commented Aug 7, 2016 at 9:35

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