**Abbreviations**

The following abbreviations are used in this manuscript:


### **Appendix A. Example 1**

To compute the time-response of the multivalued transfer function in (5), we consider the following definition of the inverse Laplace transform:

$$y(t) = \mathcal{X}^{\varepsilon - 1} \left[ \frac{1}{\sqrt{s + k}} \right] = \frac{1}{2i\pi} \int\_{\varepsilon - \infty}^{\varepsilon + \infty} G(s) \varepsilon^{st} ds. \tag{A1}$$

To solve (A1), let the contour *C* be defined as the sum of *Ci*, *i* ∈ [1, 6], as shown in Figure A1. Note that the election of the integration path should be one which avoids branch points, namely, point −*k* in this scenario.

**Figure A1.** Integration path of Example 1.

According to the residue theorem

$$\int\_{\mathbb{C}} G(\mathbf{s}) \boldsymbol{\epsilon}^{\rm st} d\mathbf{s} = 0.\tag{A2}$$

Then, we can express (A1) as

$$\int\_{Br} \mathbf{G}(\mathbf{s}) e^{\mathbf{s}t} d\mathbf{s} = -\sum\_{i=1}^{5} \int\_{C\_{i}} \mathbf{G}(\mathbf{s}) e^{\mathbf{s}t} d\mathbf{s}.\tag{A3}$$

Observe that, for *R* → ∞

$$
\int\_{C\_1} = \int\_{C\_2} = 0,\tag{A4}
$$

and, on the other hand, for *ρ* → 0 we have

$$\int\_{C\_3} = 0.\tag{A5}$$

Thus, it suffices to compute the integration along *C*2 and *C*4 to obtain *y*(*t*). In order to perform such an operation, consider the parameterisation *s* + *k* = <sup>−</sup>*r*, which in polar form can be rewritten as *re*±*iπ*, where the positive sign corresponds to *C*2 and the negative sign corresponds to *C*4. This permits us describe the path *r* ∈ (*k* + *ρ*, ∞), with *δ*, *ρ* → 0, i.e.,

$$\int\_{C\_2 + C\_4} = \int\_{\infty}^{\rho} \frac{e^{-(r+k)t} e^{i\pi}}{\sqrt{r} e^{i\pi/2}} dr + \int\_{\rho}^{\infty} \frac{e^{-(r+k)t} e^{-i\pi}}{\sqrt{r} e^{-i\pi/2}} dr,\tag{A6}$$

since *ρ* → 0

$$\int\_{\mathbb{C}\_2 + \mathbb{C}\_4} = \left(e^{i\pi} + e^{-i\pi}\right) \int\_0^\infty \frac{e^{-(r+k)t}}{\sqrt{r}} dr$$

$$= -2i\sin\left(\pi/2\right) e^{-kt} \int\_0^\infty \frac{e^{-rt}}{\sqrt{r}} dr$$

$$= -2i\sin\left(\pi/2\right) e^{-kt} t^{-1/2} \Gamma(1/2).$$

Now, as we know that

$$
\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)},
$$

with *x* = 1/2, from (A1) we have,

$$\mathcal{L}^{-1}\left[\frac{1}{\sqrt{s+k}}\right] = \frac{e^{-kt}}{\sqrt{t\pi}}.\tag{A7}$$

This finishes the proof of statement (6).

### **Appendix B. Example 2**

Following the ideas used to find the inverse Laplace Transform of √ 1*s*+*k* shown in Appendix A, we now perform the time-response of (7) (given in Equation (8)) by computing

$$y(t) = \mathcal{L}^{\rho^{-1}} \left[ \frac{1}{\sqrt{s^2 + k}} \right] = \frac{1}{2i\pi} \int\_{c-\infty}^{c+\infty} G(s) e^{st} ds. \tag{A8}$$

Observe that for *k* > 0, we have a complex conjugate branch point, while for *k* < 0 we have two real points. Under these observations, we consider two different integration paths, shown in Figure A2, where Figure A2a and Figure A2b correspond to the case of *k* negative and positive, respectively.

**Figure A2.** Integration path of Example 2. (**a**) *k* < 0. (**b**) *k* > 0.

We first consider the case *k* < 0; thus, we shall consider the path shown in Figure A2a and create a path such that

$$\int\_{C} G(\mathbf{s})e^{\mathbf{s}\mathbf{f}}d\mathbf{s} = 0.$$

Under the assumption that *ρ* → 0,

$$\int\_{\mathcal{C}\_1 + \mathcal{C}\_2} = 0.\tag{A9}$$

Now, *C*3 and *C*4 *s* vary from √*k* − *ρ* to − √*k* + *ρ*, and then we have

$$\int\_{C\_2 + C\_3} \frac{e^{st}}{\sqrt{s^2 - k}} = -\mathbf{i} \int\_{-\sqrt{k} + \rho}^{\sqrt{k} - \rho} \frac{e^{st}}{\sqrt{k - s^2}} + \mathbf{i} \int\_{\sqrt{k} - \rho}^{-\sqrt{k} + \rho} \frac{e^{st}}{\sqrt{k - s^2}} = -2\mathbf{i} \int\_{-\sqrt{k} + \rho}^{\sqrt{k} - \rho} \frac{e^{st}}{\sqrt{k - s^2}}.\tag{A10}$$
  $\text{Since } \rho \to 0,$ 

*ρ*

$$y(t) = \frac{1}{2\pi\mathbf{i}} \int\_{Br} G(\mathbf{s}) \mathbf{e}^{\mathbf{s}t} d\mathbf{s} = \frac{1}{\pi} \int\_{-\sqrt{k}}^{\sqrt{k}} \frac{\mathbf{e}^{\mathbf{s}t}}{\sqrt{k-s^2}}.\tag{A11}$$

Then, making *s* = *a* cos(*u*), the integral becomes

$$\frac{1}{\pi} \int\_0^\pi e^{kt\cos(u)} du = I\_0(\sqrt{k}t). \tag{A12}$$

We can express the modified first Bessel function in terms of the first Bessel function if − *π* < arg( √*kt*) ≤ *π* 2 : =*e α*

*π*

2 *Iα*( √*kt*)

with *α* = 0:

$$I\_0(\mathfrak{i}\sqrt{k}t) = I\_0(\sqrt{k}t).$$

*Jα*(*i*

√*kt*)

Now, we consider the case where *k* > 0. The inverse Laplace transform of *G*(*s*) can be written in two different forms (see for instance [27]):

$$y(t) = \frac{2}{\pi} \int\_{\sqrt{k}}^{\infty} \sin(st) \frac{1}{\sqrt{s^2 - k}} ds,\tag{A13}$$

$$y(t) = \frac{2}{\pi} \int\_0^{\sqrt{k}} \cos(st) \frac{1}{\sqrt{k - s^2}} ds. \tag{A14}$$

The first Bessel function is given as

$$J\_0(\mathbf{x}) = \frac{1}{\pi} \int\_0^\pi \cos(-\mathbf{x}\sin(\tau))d\tau.$$

Thus, by taking *s* = √*k* sin(*θ*) in (A14), with *θ* ∈ (0, *π* 2) we have

$$y(t) = \frac{2}{\pi} \int\_0^{\frac{\pi}{2}} \cos \left(\sqrt{k}t \sin(\theta)\right) d\theta = \frac{2}{\pi} \frac{\pi}{2} J\_0(\sqrt{k}t). \tag{A15}$$
