*4.3. Transients of the Fractional Integrator*

Consider the Laplace transform of (28):

$$\begin{cases} sZ(\omega, \mathbf{s}) - z(\omega, 0) = -\omega Z(\omega, \mathbf{s}) + V(\mathbf{s})\,\omega \in [0, \infty) \\ \qquad X(\mathbf{s}) = \int \mu\_n(\omega) Z(\omega, \mathbf{s}) d\omega \end{cases} \tag{30}$$

where *z*(*ω*, 0) is the initial value of *z*(*ω*, *t*) at *t* = 0. This means that *Z*(*ω*,*s*) = *<sup>z</sup>*(*ω*,0) *<sup>s</sup>*+*<sup>ω</sup>* <sup>+</sup> <sup>V</sup>(*s*) *s*+*ω*

$$\text{and } X(s) = \int\_0^\infty \mu\_n(\omega) \frac{z(\omega, 0)}{s + \omega} d\omega + \int\_0^\infty \mu\_n(\omega) \frac{\mathcal{V}(s)}{s + \omega} d\omega \tag{31}$$

Since <sup>1</sup> *<sup>s</sup><sup>n</sup>* <sup>=</sup> # ∞ 0 *μn*(ω) <sup>1</sup> *<sup>s</sup>*+<sup>ω</sup> *d*ω 0 < *n* < 1

we can write in the time domain:

$$x(t) = \int\_0^\infty \mu\_n(\omega) z(\omega, 0) e^{-\omega t} d\omega \, + \,\_0I\_t^n(v(t))\tag{32}$$

where:


Previously, using the definition of the Caputo derivative "initial condition", we wrote (20) *x*(*t*) = *I<sup>n</sup> <sup>t</sup>* (*v*(*t*)) + *x*(0).

The conclusion is that this expression of the free response is wrong, since # ∞ 0 μ*n*(ω)*z*(ω, 0)*e*−*ω<sup>t</sup> d*ω is the initialization function of the integrator. In fact, ∞

*x*(0) = # 0 μ*n*(ω)*z*(ω, 0)*d*ω, and Equation (15) is correct only at *t* = 0 and is wrong for *t* > 0.

The conclusion is that the fractional-integrator transients require to refer to its distributed model.

Basically, the initial condition of the differential system *D<sup>n</sup> <sup>t</sup>* (*x*(*t*)) = *f*(*x*(*t*), *u*(*t*)) is related to the initial condition of the fractional integrator <sup>1</sup> *<sup>s</sup><sup>n</sup>* used for the integration of the FDE/FDS, not to the pseudo-initial condition of any fractional derivative. Notice that if this FDE/FDS is related to a real system, its dynamics must not depend on the fractional derivative definition choice of the user.
