Is this a correct way to convert an convolution equation into differential/difference equation?

Question

1. For functions $f,g,h$ that are defined over $\mathbb{R}$, suppose we have a convolution equation: $$ f = g * h. $$

   I would like to convert it into a differential equation. Is it correct that $$ \frac{df}{dt} = \frac{dg}{dt} * h $$ under some conditions (unclear to me yet, differential under integral sign?)?

   Why when the Laplace transform G of g is a rational function, the convolution is generally converted to a higher order differential equation, instead of first order one like above?

   Why when the Laplace transform G of g is not a rational function, the convolution is generally converted to a infinite order differential equation?

2. Similarly, for functions $f,g,h$ that are defined over $\mathbb{Z}$, suppose we have a convolution equation: $$ f = g * h. $$

   I would like to convert it into a difference equation. Is it correct that $$ df = dg * h $$ where $$ df(n) := f(n+1) - f(n), $$ $$ dg(n) := g(n+1) - g(n) $$ under some conditions (unclear to me yet)?

   Why when the Z-transform G of g is a rational function, the convolution is generally converted to a higher order difference equation, instead of first order one like above?

   Why when the Z-transform G of g is not a rational function, the convolution is generally converted to a infinite order difference equation?

The above questions arose from my difficulty understanding a reply by leonbloy. Thanks and regards!

Answer 1

You wrote

Why when the Laplace transform $G$ of $g$ is a rational function, the convolution is generally converted to a higher order differential equation, instead of first order one like above?

The equation you wrote, namely,

$$f' = g' * h$$

is the convolution of the derivative of $g$ with $h$. Yes, there is a derivative involved, but there is the convolution integral, too. Not exactly a typical ODE, is it? It would be instructive to write the actual convolution integral, which is the following

$$\displaystyle f' (x) = \int_{\mathbb{R}} h (y) \, g' (x - y) \, dy$$

You want a 1st order ODE for all convolutions, which would be too good to be true. Note that:

• If $g$ has "low descriptive complexity", i.e., if $g$ is the superposition of a "few" exponential functions of the form $e^{s x}$, where $s \in \mathbb{C}$, then the Laplace transform of $g$, which we denote by $G$, will be a rational function in which the degrees of the numerator and denominator will be "small". In other words, there will be "few" poles and zeros in $G$.

• If $g$ has "high descriptive complexity", i.e., it the superposition of "many" exponential functions of the form $e^{s x}$, where $s \in \mathbb{C}$, then $G$ will have "many" poles and zeros.

The number of poles and zeros in $G$ is thus a measure of how hard it is to describe the signal $g$. It is a measure of the Kolmogorov complexity of the signal, so to speak. The poles and zeros in $G$ give you an ODE, which is a "blueprint" for you to build a system that can generate the signal $g$. For example, a sinusoid is of very low complexity, and a single sinusoidal signal generator can generate it. By contrast, a square wave is impossible to replicate exactly using a finite number of sinusoidal signal generators, but one can get close to the ideal square wave using "many" sinusoidal signal generators.