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?
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!