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  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!

1 Answers 1

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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.

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    Added to (1), usually when using ODE(s) to describe a LTI system, is the ODE(s) for input or output signals?2012-09-30