*6.4. Computation of the Distributed Exponential Function*

The interest of the distributed exponential function, either in scalar or matrix form, is essentially theoretical. Practically, in order to compute it, the distributed exponential function requires frequency discretization, i.e., a discretization of the continuous distributed integrator model:

$$\begin{cases} \frac{\partial z(\omega, t)}{\partial t} = -\omega z(\omega, t) + v(t) \,\omega \in [0, \infty) \\ \qquad x(t) = \int \mu(\omega) z(\omega, t) d\omega \end{cases} \tag{81}$$

Direct discretization is possible, but indirect discretization is more efficient [34,35]. It is based on the approximation of the frequency response of <sup>1</sup> *<sup>s</sup><sup>n</sup>* 0 < *n* < 1, on a frequency interval [ωmin, ωmax] with J first-order cells, associated with an integer-order integrator at =0. So, the global model is composed of *J* + 1 cells (or modes) such as:

$$\begin{cases} \frac{dz\_j(t)}{dt} = -\omega\_j z\_j(t) + v(t) & j = 0 \text{ to } J\\ \qquad \mathbf{x}(t) = \sum\_{j=0}^{l} c\_j z\_j(t) \end{cases} \tag{82}$$

and the elementary system *Dn*(*x*(*t*)) = *ax*(*t*) becomes:

$$\begin{cases} \frac{dz\_j(t)}{dt} = -\omega\_j z\_j(t) + a \sum\_{j=0}^{l} c\_j z\_j(t) \; j = 0 \; to \; l \\\ x(t) = \sum\_{j=0}^{l} c\_j z\_j(t) \end{cases} \tag{83}$$

where ω*<sup>j</sup>* varies from ω<sup>0</sup> = 0 to ω*J*.

Let us define:

$$\begin{aligned} \underline{z}\_{I}^{T}(t) &= \begin{bmatrix} z\_{0}(t) \dots z\_{j}(t) \dots z\_{I}(t) \end{bmatrix} \quad \text{dim}(\underline{z}\_{I}) = J + 1\\ A\_{I} &= \begin{bmatrix} -\omega\_{0} \\ & -\omega\_{j} \\ & & -\omega\_{I} \end{bmatrix} \quad \underline{B}\_{I}^{T} = \begin{bmatrix} 1 \dots 1 \dots 1 \end{bmatrix} \quad \underline{\mathbb{C}}\_{I} = \begin{bmatrix} \varepsilon\_{0} \dots \varepsilon\_{j} \dots \varepsilon\_{I} \end{bmatrix} \end{aligned} \tag{84}$$

Then, the previous system (83) is transformed into the following one

$$\begin{cases} \frac{d\underline{\boldsymbol{\tau}}\_{l}(t)}{dt} = A\_{I}\underline{\boldsymbol{\tau}}\_{l}(t) + a\underline{\boldsymbol{B}}\_{I}\underline{\boldsymbol{\zeta}}\_{I}\underline{\boldsymbol{\tau}}\_{l}(t) = A\_{\text{syst}}\underline{\boldsymbol{\tau}}\_{l}(t) \\\ \boldsymbol{x}(t) = \underline{\boldsymbol{\zeta}}\_{I}\underline{\boldsymbol{\tau}}\_{l}(t) \end{cases} \\ \text{with } A\_{\text{syst}} = A\_{I} + a\underline{\boldsymbol{B}}\_{I}\underline{\boldsymbol{\zeta}}\_{I} \tag{85}$$

Thus, the solution of system (85) with the initial condition *z<sup>T</sup> <sup>I</sup>* (0) =  *z*0(0)... *zj*(0)... *zJ*(0) is expressed as

$$
\underline{Z}(t) = e^{A\_{\text{sy}t}t} \underline{Z}(0) = \phi\_d(t) \underline{Z}(0), \tag{86}
$$

where φ*d*(*t*) = *eAsystt* is the matrix exponential corresponding to the frequency discretization of the distributed exponential function <sup>φ</sup>(*t*) = exp*t* # ∞ ψ(ω, ξ)*d*ξ . .

0 Numerical examples of the distributed exponential function are available in chapter 9 volume 1 of [34].
