*5.3. Solution Derived from the Caputo Derivative Definition*

This free response can also be expressed using Equation (15) at *t*<sup>0</sup> = *T* with *x*(*t*0) = *x*(*T*), i.e., *x*(*t*) = *I<sup>n</sup> <sup>t</sup>* (*u*(*t*)) + *x*(*T*) *f or t* ≥ *T*.

Thus (see Figure 2),

$$\mathbf{x}(t) = \mathbf{x}(T) \text{ for } t \ge T \tag{36}$$

This result is obviously in complete contradiction with Equation (35), i.e., *x*(*t*) = *<sup>U</sup>* Γ(*n*+1) *t <sup>n</sup>* <sup>−</sup> (*<sup>t</sup>* <sup>−</sup> *<sup>T</sup>*) *<sup>n</sup> f or t* <sup>≥</sup> *<sup>T</sup>*.

Notice that, for *n* = 1, we obtain:

*x*(*T*) = *UT f or t* = *T* and *x*(*t*) = *U*[*t* − (*t* − *T*)] = *UT f or t* ≥ *T*.

Thus, we verify that *x*(*t*) = *x*(*T*) *f or t* ≥ *T*, i.e., Equation (36) is only correct in the integer-order case.

With this very simple example, we have demonstrated that Equation (15) is wrong in the fractional-order case. So, what is the reason of this basic error?

*5.4. Solution Derived from the Distributed Frequency Model of the Fractional Integrator* Consider again the elementary example (33):

$$D^n(\mathbf{x}(t)) = \boldsymbol{\mu}(t) \quad 0 < \boldsymbol{\mu} < 1$$

This system is supposed at rest at *t* = 0, i.e., *z*(ω, 0) ∀ω ∈ [0, ∞). So, using (30) we obtain

$$Z^{+}\left(\omega,s\right) = \frac{\mathcal{U}}{s\left(s+\omega\right)}\text{ for }\mu(t) = \mathcal{U}H(t).$$

Thus, according to (29), *X*(*s*) = # ∞ 0 μ*n*(ω) <sup>1</sup> (*s*+ω) *U <sup>s</sup> <sup>d</sup>*<sup>ω</sup> <sup>=</sup> *<sup>U</sup> sn*+<sup>1</sup> and *<sup>x</sup>*+(*t*) = *<sup>t</sup> nU* <sup>Γ</sup>(*n*+1) *H*(*t*)

*<sup>x</sup>*−(*t*) = <sup>−</sup> *<sup>U</sup>*(*t*−*T*) *n* <sup>Γ</sup>(*n*+1) *H*(*t* − *T*) .

Obviously, we recover the same result as (35) using the distributed model. Moreover, this model allows us to express *z*(ω, *t*), i.e.,

$$\begin{aligned} \mathbf{z}^+(\omega, t) &= \frac{\mathbf{U}}{\omega} (1 - e^{-\omega \mathbf{t}}) H(t) \\\\ \text{So} \begin{cases} \mathbf{x}^+(t) = \operatorname{lIH}(t) \int\_0^\infty \frac{\mu\_n(\omega)}{\omega} (1 - e^{-\omega \mathbf{t}}) d\omega \\ \mathbf{x}^-(t) = -\operatorname{lIH}(t - T) \int\_0^\infty \frac{\mu\_n(\omega)}{\omega} \left(1 - e^{-\omega (t - T)}\right) d\omega \end{cases} \end{aligned} \tag{37}$$

Consequently, the free response is expressed as:

$$\ln x(t) = \mathcal{U} \int\_0^\infty \frac{\mu(\omega)}{\omega} \left( e^{-\omega \left( t - T \right)} - e^{-\omega \omega t} \right) d\omega \text{ for } t \ge T\_\prime \tag{38}$$

which is the distributed equivalent of Equation (35).

Moreover, we can verify that it is now possible to calculate the response of the integrator for *t* ≥ *T* using the expression (32).

So, consider the response initialized at *t* = *T* with *u*(*t*) = 0 for *t* ≥ *T*.

Since *z*(ω, *T*) is the initial condition at *t* = *T*, with

$$z(\omega, T) = z^+(\omega, T) = \frac{\mathcal{U}}{\omega} \left( 1 - e^{-\omega \cdot T} \right),\tag{39}$$

and since *I<sup>n</sup> <sup>t</sup>* (0) = 0, we obtain

$$\alpha(t) = \mathcal{U} \int\_0^\infty \mu\_n(\omega) z(\omega, T) e^{-\omega(t-T)} d\omega \quad \text{for } t \ge T$$

So

$$\alpha(t) = \mathcal{U} \int\_0^\infty \frac{\mu\_n(\omega)}{\omega} \left(1 - e^{-\omega \cdot T}\right) e^{-\omega \cdot (t - T)} d\omega \quad \text{for } t \ge T$$

And we obtain the same result as previously (38), i.e.,

$$\varphi(t) = U \int\_0^\infty \frac{\mu\_n(\omega)}{\omega} \left( e^{-\omega(t-T)} - e^{-\omega t} \right) d\omega \quad \text{for } t \ge T$$

We can conclude that the distributed state-space model provides the exact expression of the free response using the usual tools of linear system theory. Consequently, this distributed model is the necessary tool to express transients of the fractional integrator.

Notice that numerical simulations corresponding to this counter example are available in [33].
