(1)The solution of state space linear fractional system of commensurate order with complex eigenvalues using regular exponential and trigonometric functions D

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(1)The solution of state space linear fractional system of commensurate order with complex eigenvalues using regular exponential and trigonometric functions D. Boucherma, A. Charef & H. Nezzari. International Journal of Dynamics and Control ISSN 2195-268X Volume 5 Number 1 Int. J. Dynam. Control (2017) 5:79-94 DOI 10.1007/s40435-015-0185-y. 1 23.

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(3) Author's personal copy Int. J. Dynam. Control (2017) 5:79–94 DOI 10.1007/s40435-015-0185-y. The solution of state space linear fractional system of commensurate order with complex eigenvalues using regular exponential and trigonometric functions D. Boucherma1 · A. Charef2 · H. Nezzari1. Received: 14 November 2014 / Revised: 24 May 2015 / Accepted: 26 May 2015 / Published online: 12 June 2015 © Springer-Verlag Berlin Heidelberg 2015. Abstract In a previous work, we have derived the general solution of the state space linear fractional system of commensurate order for real simple and multiple eigenvalues of the state space matrix. The obtained solutions of the homogeneous and non-homogeneous cases have been expressed as a linear combination of introduced fundamental functions. In this paper, the above work has been extended to solve the state space linear fractional system of commensurate order for complex eigenvalues of the state space matrix. First, suitable fundamental functions corresponding to the different types of complex eigenvalues of the state space matrix are introduced. Then, the derived formulations of the resolution approach are presented for the homogeneous and the nonhomogeneous cases. The solutions are expressed in terms of a linear combination of the proposed fundamental functions which are in the form of exponentials, sine, cosine, damped sine and damped cosine functions depending on the commensurate fractional order. The results are validated by solving an illustrative example to demonstrate the effectiveness of the proposed analytical tool for the solution of the state space linear fractional system of commensurate order.. B. A. Charef [email protected] D. Boucherma [email protected] H. Nezzari [email protected]. 1. Welding and NDT Research Centre (CSC), BP 64, Cheraga, Algiers, Algeria. 2. Laboratoire de Traitement du signal, Département d’Electronique, Université Constantine 1, Route Ain El-Bey, 25000 Constantine, Algeria. Keywords Complex eigenvalues · Fractional differential equation · Fundamental functions · Modal decomposition · Rational function · State space representation. 1 Introduction In the recent decades the concept of fractional calculus has been arisen in the areas of applied science and engineering where various processes have been found to be properly described by fractional order derivatives and integrals which provide an excellent description of their properties leading to the formulation of fractional differential equations [1– 4]. Consequently, considerable attention has been given to such differential equations to establish reliable and efficient techniques for their solutions and analysis so they may be accessible to the general engineering community. In spite of their resolution was much involved; exact solutions cannot be found. So approximation and numerical techniques must be used extensively. So far, there have been several fundamental works on fractional differential equations that provide a good understanding of their analysis such as the existence, the uniqueness and some numerical and analytical methods for their solutions [5–8]. More recently, there have been more works devoted to fractional differential equations including the development of efficient methods and techniques to solve them. These developed methods can be broadly classified into two classes, analytical and numerical. The purpose of the analytical methods is to obtain an explicit expression for the general solution of the fractional order differential equations [9–12]. Yet, the goal of the numerical techniques is the development of efficient computational solutions of the fractional order differential equations [13– 19].. 123.

(4) Author's personal copy 80. D. Boucherma et al.. This paper deals with the analytical solutions of the statespace linear fractional systems of commensurate order given as: Dm x (t) = Ax (t) + Be(t). (1). where Dm x (t) is the Caputo fractional derivative of order m such that 0 < m < 1, x(t) is the n-state vector, A is the (n × n) state matrix, e(t) is the input and B is the (n × 1) input matrix. In this context, the solution of the state-space equation of (1) is considered only for complex eigenvalues of the state matrix A as an extension of a previous work for real simple and multiple eigenvalues of the state matrix A [12]. The main objective of this work is to treat the linear fractional order systems as it is done with regular linear systems in order to establish the explicit expressions of the homogeneous and non-homogeneous solutions of the statespace linear fractional systems of commensurate order of (1); through the use of suitable fundamental functions corresponding to the different types of complex eigenvalues of the state matrix A and the range of the parameter m. The solutions are derived using the modal decomposition technique [20] where the derived expressions of the homogeneous and the non-homogeneous cases will be given as a linear combination of the proposed suitable fundamental functions which are linear combination of regular exponential and trigonometric functions. An illustrative example is included to demonstrate the validity, applicability and effectiveness of the proposed analytical tool for the solution of the state space linear fractional system of commensurate order.. 2 Fundamental functions In this context, the solution of the state-space linear fractional order systems of (1) is considered only for complex eigenvalues of the state matrix A. So, a pair of complex eigenvalues λ and λ∗ of the matrix A is given as: . λ = −ζ ωn + jωn 1 − ζ 2 and λ∗ = −ζ ωn − jωn 1 − ζ 2. In this case, ζ = 0, the eigenvalues of the state matrix A are in the form of pairs of complex eigenvalues λ = jωn and λ∗ = − jωn with λ2 = ωn2 . The two fundamental functions corresponding to these eigenvalues are given as: λ2 sm and Ff 2 (s) = Ff 1 (s) = 2 2m 2m s + λ s + λ2 (3) Let’s define the fractional sine function fsin(t, λ, m) and the fractional cosine function fcos(t, λ, m) as the inverse Laplace transform of the functions Ff1 (s) and Ff2 (s), respectively. Hence, we can write that: λ2 −1 fsin(t, λ, m) = L and s 2m + λ2 m s (4) fcos(t, λ, m) = L−1 s 2m + λ2 These two fundamental functions are called, respectively, the fractional sine and fractional cosine functions because that if m = 1, the inverse Laplace transform of the functions Ff1 (s) and Ff2 (s) of (3) are, respectively, sin (ωn t) and cos (ωn t) functions. We will also introduce two other functions hfsin(t, λ, m) and hfcos(t, λ, m) which will be used in the solution of the state-space linear fractional order systems of Eq. (1). These two functions are defined, respectively, as the convolution of the fractional sine function fsin(t, λ, m) and the fractional cosine function fcos(t, λ, m) with the unity step function u(t). Hence, the functions hfsin(t, λ, m) and hfcos(t, λ, m) are given as: hfsin(t, λ, m) = [fsin(t, λ, m)] ∗ u(t) and hfcos(t, λ, m) = [fcos(t, λ, m)] ∗ u(t). (5). From (4), we can then write that: (2). where ωn > 0 and 0 ≤ ζ < 1. Hence, the suitable fundamental functions which will be used in the solution are the inverse Laplace transform of the functions defined according to the type of the complex eigenvalues of the state matrix A and the range of the fractional derivative m.. 123. 2.1 Fundamental functions for a pair of complex eigenvalues with null real parts. −1. hfsin(t, λ, m) = L. −1. hfcos(t, λ, m) = L. λ2 s2m + λ2 sm s2m + λ2. . 1 s 1 s. and (6).

(5) Author's personal copy The solution of state space linear fractional system of commensurate order with complex…. Ff1 (s) and Ff2 (s) of (3) are irrational functions; hence, to be used in the development of the dynamical behavior of the solutions of the state-space linear fractional order systems of Eq. (1), they have to be approximated by rational ones. The approximation will be considered according to the range of the fractional derivative m. We also note that the poles of the derived rational function approximation are the best way to explain and analyze the behavior of the responses of the fractional order system. 2.1.1 Case 1: 0 < m < 0.5 In this case, 2m < 1, so the approximation of the fundamental function Ff1 (s) of (3) by a rational one, in the frequency band of interest [0, ωH ], is given as [21]: λ2. ∼ Ff 1 (s) = = s 2m + λ2. N1 i=1. k1i s + p1i. (7). Hence, the fractional sine function fsin(t, λ, m) is given by: 2 λ fsin(t, λ, m) = L−1 s 2m + λ2 N N1 1 k 1i −1 ∼ k1i exp −p1i t = L = s + p 1i i=1 i=1 (8) The rational function approximation of the function Ff2 (s) of (3), in the frequency band of interest [ωL , ωH ], is also given as [22,23]: Ff 2 (s) = . s 2m. sm ∼ = + λ2. N2 i=1. k2i s + p2i. (9). So, the fractional cosine function fcos(t, λ, m) is given by: m s fcos(t, λ, m) = L−1 s 2m + λ2 N N2 2 k 2i −1 ∼ k2i exp −p2i t = L = s + p 2i i=1 i=1. 81. hfcos(t, λ, m) . sm s2m + λ2. = L−1 = L−1. ⎧ ⎨ N2 ⎩. k2i s + p2i. i=1. 1 s ⎫ N2 1 ⎬ = s ⎭. . i=1. . k2i p2i. 1-exp −p2i t. . (12) 2.1.2 Case 2: m = 0.5 In this case the fundamental function Ff1 (s) of (3) is given by: λ2 Ff 1 (s) = s + λ2. (13). and the fractional sine function fsin(t, λ, 0.5) will be: fsin(t, λ, 0.5) = L. −1. . = λ2 exp − λ2 t 2 s + λ λ2. (14) The rational function approximation of the function Ff2 (s) of (3), in the frequency band of interest [ωL , ωH ], is given as follows [22]: k20 s 0.5 ∼ + Ff 2 (s) = = 2 s + λ s + λ2. N2 i=1. k2i s + p2i. (15). and the fractional cosine function fcos(t, λ, 0.5) will be: s 0.5 −1 fcos(t, λ, 0.5) = L s + λ2 N2 k k 20 2i ∼ + = L−1 s + p2i s + λ2 i=1 = k20 exp − λ2 t +. N2. k2i exp −p2i t i=1. (16). (10) In this case, from (8) and (10), the functions hfsin(t, λ, m) and hfcos(t, λ, m) are then given as: hfsin(t, λ, m) . λ2 s2m + λ2. = L−1 = L−1. ⎧ ⎨ N1 ⎩. i=1. k1i s + p1i. . . 1 s ⎫ N1 1 ⎬ = s ⎭. i=1. So, the functions hfsin(t, λ, 0.5) and hfcos(t, λ, 0.5) are given, respectively, from (13) and (15), as: hfsin(t, λ, 0.5) = L−1. λ2. . s + λ2. 1 s. . = 1-exp − λ2 t (17). k1i p1i. 1-exp −p1i t. (11). hfcos(t, λ, 0.5) =L. −1. s 0.5 s + λ2. . 1 s. . 123.

(6) Author's personal copy 82. D. Boucherma et al.. −1. =L. k20 λ2. =. N2. N. 2 k20 1 + s s + λ2 i=1 1-exp − λ2 t. k2i p2i. + i=1. k2i s + p2i. 1 − exp (− p2i t). 1 s. . Then, the function fcos(t, λ, m) is given by the following Eq. [24,25]: fcos(t, λ, m) . (18). 2.1.3 Case 3: 0.5 < m < 1 In this case, 2m > 1, hence the approximation of the fundamental function Ff1 (s) of (3) by a rational one, in the frequency band of interest [0, ωH ], is given as [21]: λ2 Ff 1 (s) = s 2m + λ2 ⎡ ⎤ A s + B 1 1 ∼ ⎦ =⎣ 1 2 2 m m s + 2α λ s + λ N 1 k1i + s + p1i. −1. (19). . N1. +. k1i exp −p1i t. (20). i=1. The rational function approximation of the function Ff2 (s) of (3), in the frequency band of interest [ωL , ωH ], is also given as [21–23]: Ff 2 (s) sm = s 2m + λ2 ⎧⎡ ⎤ ⎫ N2 ⎨ ⎬ A s + B k 2 2 2i ∼ ⎦ + = L−1 ⎣ 1 2 ⎩ s 2 + 2α λ m s + λ m s + p2i ⎭ i=1. (21). 123. . (22). λ2. . 1 s. . s 2m + λ2 ⎧⎡⎛ ⎞ ⎤ ⎨ s + B 1 A 1 1 ⎦ ⎠ = L−1 ⎣⎝ 1 2 ⎩ s s 2 + 2α λ m s + λ m N 1 k1i 1 =1 + s + p1i s i=1 1 1 + C11 exp −α λ m t sin 1 − α 2 λ m t + 11 =L. s 2m + λ2 ⎧⎡ ⎤ ⎨ s + B A 1 1 ⎦ = L−1 ⎣ ⎩ s 2 + 2α λ m1 s + λ m2 N 1 k1i + s + p1i i=1 1 1 1 − α 2 λ m t + 1 = C1 exp −α λ m t sin =L. . λ2. k2i exp −p2i t i=1. fsin(t, λ, m) . So, the function fsin(t, λ, m) is given as [21]:. −1. N2. +. In this case, the functions hfsin(t, λ, m) and hfcos(t, λ, m) are given from (20) and (22) as follows:. i=1. fsin(t, λ, m) . sm s 2m + λ2 ⎧⎡ ⎤ ⎫ N2 ⎨ ⎬ A s + B k 2 2 2i ⎦ + = L−1 ⎣ 1 2 ⎩ s 2 + 2α λ m s + λ m s + p2i ⎭ i=1 1 1 1 − α 2 λ m t + 2 = C2 exp −α λ m t cos. = L−1. N1. − i=1. k1i p1i. fcos(t, λ, m) −1. =L. . exp −p1i t . . sm + λ2. k2i p2i. exp −p2i t. . (23). 1 s. . s 2m ⎧⎡⎛ ⎞ ⎤ ⎨ 1 A2 s + B2 −1 ⎣⎝ ⎦ ⎠ =L 1 2 ⎩ s s 2 + 2α λ m s + λ m N 2 k2i KD 1 = + s + p2i s λ2 i=1 1 1 + C21 exp −α λ m t sin 1 − α 2 λ m t + 21 N2. − i=1. (24). 2.2 Fundamental functions for a pair of complex eigenvalues with negative real parts In this case, 0 < ζ < 1, the eigenvalues of the state matrix A are in the form of pairs of complex eigenvalues λ = −ζ ωn + jωn 1 − ζ 2 and λ∗ = −ζ ωn − jωn 1 − ζ 2 . The two fundamental functions corresponding to these eigen-.

(7) Author's personal copy The solution of state space linear fractional system of commensurate order with complex…. 83. fdcos(t, λ, m) . values are given as:. s m + λ ζ =L s 2m + 2ζ λ s m + λ2 N 4 a s + ζ 4i = L−1 k4i 2 s + 2βωi s + ωi2 i=1. λ2 and Ff 3 (s) = s 2m + 2ζ λ s m + λ2 s m + λ ζ Ff 4 (s) = s 2m + 2ζ λ s m + λ2. (25). N4. We define the fractional damped sine function fdsin(t, λ, m) and the fractional damped cosine function fdcos(t, λ, m) as the inverse Laplace transform of the functions Ff3 (s) and Ff4 (s), respectively, because that if m = 1, the inverse Laplace transform of the functions Ff3 (s) and Ff4 (s) of (25) are, respectively, the damped sine function exp(−ζ ωn t) sin(ωn t) and the damped cosine function exp(−ζ ωn t) cos(ωn t) functions. Hence, we have: fd sin(t, λ, m) = L. −1. . fd cos(t, λ, m) = L. −1. s 2m. . s 2m. . −1. λ2 + 2ζ λ s m + λ2. . s m + λ ζ + 2ζ λ s m + λ2. and. ωi k4i exp (−βωi t) cos ωi. =. $ 1 − β2 t. i=1. +. m−1 φ m. (30). The functions hfdsin(t, λ, m) and hfdcos(t, λ, m) which will be used in the solution of the state-space linear fractional order systems of Eq. (1) are defined, respectively, as the convolution of the fractional damped sine function fdsin(t, λ, m) and the fractional damped cosine function fdcos(t, λ, m) with the unity step function u(t). Hence, they are given as: hfdsin(t, λ, m) = [fdsin(t, λ, m)] ∗ u(t) and. . (26). hfdcos(t, λ, m) = [fdcos(t, λ, m)] ∗ u(t). (31). From (28) and (30), we can then write that: For 0 < m < 1, the rational functions approximating the fundamental functions Ff3 (s) and Ff4 (s) of (25), in the frequency band of interest [0, ωH ], are given as [24,25]: Ff 3 (s). ⎡. ⎤ 2 1 − ζ ⎦ = k3i ∼ =⎣ s 2m + 2ζ λ s m + λ2 s 2 + 2βωi s + ωi2 i=1 λ2. N3. a3i s +. . (27) Ff 4 (s). ⎡. ⎤. N. 4 s m + λ ζ a4i s + ζ ∼ ⎣ ⎦ = k4i = 2 2 2m m 2 s + 2ζ λ s + λ s + 2βω s + ω i i i=1. (28) So, the functions fdsin(t, λ, m) and fdcos(t, λ, m) will be: fdsin(t, λ, m) = L−1. λ2. N3. ωi k3i exp (−βωi t) sin ωi m−1 φ m. i=1. (32) hfdcos(t, λ, m) & % s m + λ ζ 1 = L−1 s s 2m + 2ζ λ s m + λ2 N 4 1 a4i s + ζ −1 =L k4i 2 s s + 2βωi s + ωi2 i=1 N $ 4 ζ = − C4 k4i exp (−βωi t) cos ωi 1 − β 2 t + φ4 λ. (33). . 3 Solution of the state space linear fractional order system 3.1 Modal decomposition. $ 1 − β2 t. Let’s consider the following regular state space linear system of order n:. i=1. +. λ2 1 s s 2m + 2ζ λ s m + λ2 N 3 1 a3i s + 1 − ζ 2 −1 =L k3i 2 s s + 2βωi s + ωi2 i=1 N $ 3 2 = 1 − C3 k3i exp (−βωi t) sin ωi 1 − β t + φ3 = L−1. i=1. s 2m + 2ζ λ s m + λ2 N 3 2 a s + 1 − ζ 3i = L−1 k3i 2 s + 2βωi s + ωi2 i=1 =. hfdsin(t, λ, m) . (29) Dx (t) = Ax (t) + Be(t). (34). 123.

(8) Author's personal copy 84. D. Boucherma et al.. It is well known that the modes of the {A, B} realization are a useful description of its behavior. These modes are determined by the eigenvalues λi (i = 1, 2, . . ., n) of the (nxn) A matrix, i.e., the roots of the characteristic equation (s) = det (sI - A) = 0. For simplicity of the modal decomposition discussion, we shall assume that all the eigenvalues are distinct. Now consider the homogenous state space linear system [20]:. where D m x (t) is the Caputo fractional derivative of order m (0 < m < 1), x(t) is the state vector, A is the (n × n) state matrix, e(t) is the input and B is the (n × 1) input matrix. In this work, all the eigenvalues of the matrix A are only complexes numbers with null real part and with negative real part; hence n is an even number. Taking the Laplace transform of Eq. (42), we obtain:. Dx (t) = Ax (t) , x (0) = x0. X (s) = s (m−1) s m I − A. (35). Taking the Laplace transform of (35), we can obtain the following equation: X (s) = (s I − A)−1 x0. (36). Making the partial fraction expansion of (36), we can write that: X (s) = (s I − A)−1 x0 =. n i=1. Ri x0 s − λi. (37). where the residue matrices Ri (i = 1, 2, . . ., n) are given by: Ri = pi qi. (38). with the vectors pi and qi (i = 1, 2, . . ., n) are, respectively, the right (or column) and the left (or row) eigenvectors of the matrix A associated with the eigenvalues λi (i = 1, 2, . . ., n). They are obtained as follows: A pi = λi pi , qi A = λi qi and qi pi = 1. (39). Therefore we have the modal decomposition: n. X (s) = i=1. qi x0 pi = s − λi. n i=1. αi pi s − λi. (40). where αi = qi x0 . By taking the inverse Laplace transform of (40), the time response of the homogenous state space linear system of (35) is given by: n. x (t) =. αi eλi t pi. Let’s now consider the state-space linear fractional order system of Eq. (1):. −1. x0 + s m I − A. −1. x (t) = L −1 {X (s)} = L −1 s (m−1) s m I − A −1 ∗ Be(t) +L −1 s m I − A. BE(s) (43). −1. x0 (44). Let λ0k = jb0k (b0k > 0) and λ0(k+1) = λ∗0k , for k = 1, 3, k odd, . . . , (n1 − 1) and n1 an even number, be the complex eigenvalues of the matrix A with null real part and let λk = ak + jbk (ak < 0 and bk > 0) and λ(k+1) = λ∗k , for k = (n1 + 1) , (n1 + 3), k odd, . . . , (n − 1), be the complex eigenvalues of real part. Also, we the matrix A with negative let X 1 (s) = s (m−1) (s m I − A)−1 x0 . Then, from Eq. (37) we can write that: X 1 (s) = s (m−1) s m I − A % n 1 −1 (m−1) = s k=1,k odd n−1. +. s. −1. x0. & Rk∗ Rk x0 + m s m − λ0k s − λ∗0k. (m−1). %. & Rk∗ Rk x0 + m s m − λk s − λ∗k (45). (41). 3.2 Solution. (42). where x0 = x(0) is the initial state called the Caputo initial conditions. From the literature, we note that the initial conditions from Riemann–Liouville and Caputo definitions for fractional differentiation do not predict correctly the homogenous solution in the general case. However in [26], the authors have proposed coherent initial conditions to a fractional system to take into account initial conditions in a convenient way from a physical point of view. So, the expression of the state vector x(t) is determined by taking the inverse Laplace transform of Eq. (43); we will then have:. k=n 1 +1,k odd. i=1. 123. Dm x (t) = Ax (t) + Be(t). %. Rk = pk qk where from Eq. (38) ∗ and from Eq. (40) Rk∗ = pk qk % αk = qk x0 . So, we will have: αk∗ = qk∗ x0.

(9) Author's personal copy The solution of state space linear fractional system of commensurate order with complex…. X 1 (s) = s (m−1) s m I − A % n 1 −1 = s (m−1) k=1,k odd n−1. +. s. −1. We have λ0k = b0k , λk = − λakk , Eq. (51) will then be:. x0. α∗ αk pk + m k ∗ pk∗ m s − λ0k s − λ0k. (m−1). %. k=n 1 +1,k odd. &. α∗ αk pk + m k ∗ pk∗ m s − λk s − λk. &. (46) Hence, for 1 ≤ i ≤ n, the ith element X1i (s) of the vector X1 (s) is given by: n 1 −1. s (m−1). X 1i (s) = k=1,k odd. n−1. %. ∗ αk∗ pki αk pki + s m − λ0k s m − λ∗0k. s (m−1). +. %. k=n 1 +1,k odd. &. α∗ p∗ αk pki + m k ki ∗ m s − λk s − λk. (47) &. Equation (47) can be rewritten as: n 1 −1. s (m−1). X 1i (s) = k=1,k odd. n−1. (Aki s m + B1ki ) 2 s 2m + b0k. s (m−1). + k=n 1 +1,k odd. (Aki s m + B2ki ) (s m − ak )2 + bk2. (48). Aki = 2Re[αk pik ] B1ki = −2Re[λ∗0k αk pik ]. X 1i (s) =. s. −1. k=1,k odd n−1. (49). k=n 1 +1,k odd. Aki s 2m + B2ki s m − ak ). 2. + bk2. (50). B1ki s m E ki + 2m X 1i (s) = Aki + 2m 2 2 s + b0k s + b0k k=1,k odd n−1 1 Cki (s m − ak ) + Aki + s (s m − ak )2 + bk2 k=n 1 +1,k odd Dki + (51) (s m − ak )2 + bk2 n 1 −1. 1 s. . 1 sm 1 Aki +B1ki 2m s s s + λ0k 2 k=1,k odd λ0k 2 E ki 1 + λ0k 2 s s 2m + λ0k 2 n−1 s m + λk ζk 1 1 Aki + Cki + 2m s s s + 2ζk λk s m + λk 2 k=n 1 +1,k odd λk 2 Dki 1 + (53) λk 2 s s 2m + 2ζk λk s m + λk 2. From Sect. 2, the inverse Laplace transform of 2 1 sm 1 0k the four functions s s 2m +λ 2 , s s 2mλ 2 , +λ 0k 0k λk 2 s m +λk ζk 1 1 and s 2m are s s 2m +2ζk λk s m +λk 2 s +2ζk λk s m +λk 2 given, respectively, by the four expressions hfcos(t, λ0k , m), hfsin(t, λ0k , m), hfdcos(t, λk , m) and hfdsin(t, λk , m). Hence, x1i (t), for 1 ≤ i ≤ n, the inverse Laplace transform of X 1i (s) is given as: n 1 −1. %. =. Aki u (t) + B1ki hfcos(t, λ0k , m) k=1,k odd. & E ki hfsin(t, , m) λ 0k λ0k 2 % n−1 Aki u (t) + Cki hfdcos(t, λk , m) + Dki + hfdsin(t, λk , m) λk 2. 2 s 2m + b0k. (s m. . n 1 −1. =. k=n 1 +1,k odd. Aki s 2m + B1ki s m. s −1. +. & (54). Now, let X 2 (s) = (s m I − A)−1 BE(s). Also, from Eq. (37) we can write that: −1. X 2 (s) = s m I − A BE (s) % & n 1 −1 Rk∗ Rk BE (s) + m = s m − λ0k s − λ∗0k k=1,k odd. n−1. + k=n 1 +1,k odd. %. & Rk∗ Rk BE (s) + s m − λk s m − λ∗k. (55). %. where the coefficients Eki , Cki , and Dki are given by: 2 Aki , Cki = (2ak Aki + B2ki ) , E ki = −b0k Dki = Aki ak2 − bk2 + ak B2ki. $ ak2 + bk2 and for ζk =. +. Rearranging Eq. (48) we will get: n 1 −1. X 1i (s). x1i (t). where the coefficients Aki , B1ki and B2ki are given as follows: B2ki = −2Re[λ∗k αk pik ]. 85. (52). Rk = pk qk where from Eq. (38) ∗ and from Eq. (40) Rk∗ = pk qk % ᾱk = q k B . So, we will have: ᾱk∗ = qk∗ B. 123.

(10) Author's personal copy 86. D. Boucherma et al.. X 2 (s) = s m I − A n 1 −1 % = k=1,k odd. −1. BE(s). n−1. ᾱk∗ ᾱk p + p ∗ E(s) k s m − λ0k s m − λ∗0k k %. n−1. + k=n 1 +1,k odd. & ᾱk∗ ᾱk ∗ E(s) p + p k s m − λk s m − λ∗k k (56). then, for 1 ≤ i ≤ n, the ith element X2i (s) of the vector X2 (s) is given by: %. n 1 −1. ∗ & ᾱk∗ pki ᾱk pki + E(s) s m − λ0k s m − λ∗0k. X 2i (s) = k=1,k odd. %. n−1. + k=n 1 +1,k odd. ∗ & ᾱk∗ pki ᾱk pki + E(s) (57) s m − λk s m − λ∗k. Equation (57) can be rewritten as: n 1 −1. Āki s m + B̄1ki. X 2i (s) = k=1,k odd. s 2m. n−1. + k=n 1 +1,k odd. + λ0k 2. (s m + λk ζk ) E(s) ⎩ s 2m +2ζ k λk s m + λk 2 k=n 1 +1,k odd ⎫ ⎬ λ0k 2 D̄ki + E(s) (60) ⎭ λ0k 2 s 2m +2ζ k λk s m + λk 2. +. &. ⎧ ⎨. Āki. where the coefficient D̄ki is given by: Dki = B̄2ki − Āki λk ζk. (61). From Sect. inverse transform of the four func Laplace 2, the λ0k 2 s m +λk ζk sm , , tions 2 2m s 2m+λ0k 2 s 2m +2ζk λk s m +λk 2 s +λ0k 2 λk and 2m are given, respectively, by the s +2ζk λk s m +λk 2 four expressions fcos(t, λ0k , m), fsin(t, λ0k , m), fdcos(t, λk , m) and fdsin(t, λk , m). Hence, x2i (t), for 1 ≤ i ≤ n, the inverse Laplace transform of X 2i (s) is given as: n 1 −1 % Āki fcos(t, λ0k , m) ∗ e(t) x2i (t) = k=1,k odd. E(s). Āki s m + B̄2ki s 2m +2ζ k λk s m + λk 2. E(s). & B̄1ki + fsin(t, λ0k , m) ∗ e(t) λ0k 2 % n−1 + Āki fdcos(t, λk , m) ∗ e(t) k=n 1 +1,k odd. (58) where the coefficients Āki , B̄1ki and B̄2ki are given as follows: Āki = 2Re [ᾱk pik ] B1ki = −2Re λ∗0k ᾱk pik B2ki = −2Re λ∗k ᾱk pik. (59). +. D̄ki fdsin(t, λk , m) ∗ e(t) λk 2. & (62). We have the state vector x(t) = x1 (t) + x2 (t); therefore, its ith state variable xi (t), for 1 ≤ i ≤ n, is given by xi (t) = x1i (t) + x2i (t) as follows:. ⎧ ⎫ ⎪ ⎨ Aki u (t) +B1ki hfcos(t, λ0k , m) + λE ki2 hfsin(t, λ0k , m) + ⎪ ⎬ 0k xi (t) = ⎪ ⎪ B̄1ki ⎭ fsin(t, λ0k , m) ∗ e(t) k=1,k odd ⎩ Āki fcos(t, λ0k , m) ∗ e(t) + λ0k 2 ⎧ ⎫ ⎪ n−1 ⎨ Aki u (t) +C ki hfdcos(t, λk , m) + λDki2 hfdsin(t, λk , m) + ⎪ ⎬ k + ⎪ ⎪ D̄ki ⎭ fdsin(t, λk , m) ∗ e(t) k=n 1 +1,k odd ⎩ Āki fdcos(t, λk , m) ∗ e(t) + λ 2 n 1 −1. (63). k. Rearranging Eq. (58) we will get: X 2i (s) n 1 −1. . sm Āki = E(s) s 2m + λ0k 2 k=1,k odd λ0k 2 B̄1ki + E(s) λ0k 2 s 2m + λ0k 2. 123. 4 Illustrative example In this section, we will present an example with three different domains of the commensurate order simulated on a PC using MATLAB to show the effectiveness and the usefulness of the proposed approach for the solution of the state space linear fractional system of commensurate order with complex eigenvalues of the state space matrix..

(11) Author's personal copy The solution of state space linear fractional system of commensurate order with complex…. Let us consider the following state-space linear fractional system of commensurate order: ⎡. −2 d m x (t) ⎢ −6 =⎢ ⎣ m −8 dt −8. 1 0 0 0. 0 1 0 0. ⎡ ⎤ ⎤ 1 0 ⎢0⎥ 0⎥ ⎥ x(t) + ⎢ ⎥ e(t) ⎣1⎦ 1⎦ 0 0. (64). 87. For e(t) = u(t) the unit step and from Eqs. (5) and (31) we will have the following: ⎧ fcos(t, λ01 , m) ∗ u(t) = hfcos(t, λ01 , m) ⎪ ⎪ ⎨ fsin(t, λ01 , m) ∗ u(t) = hfsin(t, λ01 , m) fdcos(t, ⎪ λ3 , m) ∗ u(t) = hfdcos(t, λ3 , m) ⎪ ⎩ fdsin(t, λ3 , m) ∗ u(t) = hfdsin(t, λ3 , m) then, the state variables xi (t), for 1 ≤ i ≤ 4, of Eq. (65) will be:. ⎫ ⎧ ⎨ A1i u (t) +B11i hfcos(t, λ01 , m) + E 1i 2 hfsin(t, λ01 , m) + ⎬ λ01 xi (t) = ⎩ Ā hfcos(t, λ , m) + B̄11i hfsin(t, λ , m) ⎭ 1i 01 01 2 λ01 ⎧ ⎫ ⎨ A3i u (t) +C 3i hfdcos(t, λ3 , m) + D3i 2 hfdsin(t, λ3 , m) + ⎬ λ3 + ⎩ Ā hfdcos(t, λ , m) + D̄3i hfdsin(t, λ , m) ⎭ 3i 3 3 2 λ3 ⎧ ⎫ ⎨ (A1i +A3i ) u(t)+ B11i + Ā1i hfcos(t, λ01 , m) + E 1i 2 + B̄11i 2 hfsin(t, λ01 , m) ⎬ λ01 λ01 xi (t) = ⎩ C3i + Ā3i hfdcos(t, λ01 , m) + D3i 2 + D̄3i 2 hfsin(t, λ01 , m) ⎭ λ λ 3. (66). (67). 3. ⎡. ⎤ x1 (t) ⎢ x2 (t) ⎥ ⎥ where x(t) = ⎢ ⎣ x3 (t) ⎦ is the state vector, e(t) is the input. x4 (t) The A matrix eigenvalues are λ01 = (2j), λ02 = (−2j), λ3 =. Once all the numerical values of the coefficients have been calculated, the state vector x(t) solution of the linear fractional system of commensurate order of Eq. (64) is given as:. ⎛. ⎞ −1 + 2.2 hfcos(t, λ01 , m) − 0.2 hfsin(t, λ01 , m) + 1.8 hfdcos(t, λ3 , m) + 0.3 hfdsin(t, λ3 , m) ⎜ 1 + 3.6 hfcos(t, λ01 , m) − 2.6 hfsin(t, λ01 , m) + 2.4 hfdcos(t, λ3 , m) − 0.6 hfdsin(t, λ3 , m) ⎟ ⎟ x (t) = ⎜ ⎝ 2.8 hfcos(t, λ01 , m) − 4.8 hfsin(t, λ01 , m) + 7.2 hfdcos(t, λ3 , m) + 1.2 hfdsin(t, λ3 , m) ⎠ 1 − 1.6 hfcos(t, λ01 , m) − 4.4 hfsin(t, λ01 , m) + 9.6 hfdcos(t, λ3 , m) − 2.4 hfdsin(t, λ3 , m) (−1 and λ⎡4 = (−1 ⎡ + j) ⎤ ⎤ − j). The initial state vector is x(0) = x1 (0) −1 ⎢ x2 (0) ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ x3 (0) ⎦ = ⎣ 0 ⎦. x4 (0) 1 The state variables xi (t), for 1 ≤ i ≤ 4, of the state vector x(t) solution of the above linear fractional system of commensurate order are obtained from Eq. (63) as:. (68). 4.1 Case 1: 0 < m < 0.5 For m= 0.3, the state vector x(t) solution of the linear fractional system of commensurate order of Eq. (64) is given from Eq. (68) as follows:. ⎫ ⎧ ⎨ A1i u (t) +B11i hfcos(t, λ01 , m) + E 1i 2 hfsin(t, λ01 , m) + ⎬ λ01 xi (t) = ⎩ Ā fcos(t, λ , m) ∗ e(t) + B̄11i fsin(t, λ , m) ∗ e(t) ⎭ 1i 01 01 λ01 2 ⎫ ⎧ ⎨ A3i u (t) + C3i hfdcos(t, λ3 , m) + D3i 2 hfdsin(t, λ3 , m) + ⎬ λ3 + ⎩ A fdcos(t, λ , m) ∗ e(t) + D̄3i fdsin(t, λ , m) ∗ e(t) ⎭ 3i 3 3 2 λ . (65). 3. 123.

(12) Author's personal copy 88. D. Boucherma et al.. ⎛. ⎞ −1 + 2.2 hfcos(t, λ01 , 0.3) − 0.2 hfsin(t, λ01 , 0.3) + 1.8 hfdcos(t, λ3 , 0.3) + 0.3 hfdsin(t, λ3 , 0.3) ⎜ 1 + 3.6 hfcos(t, λ01 , 0.3) − 2.6 hfsin(t, λ01 , 0.3) + 2.4 hfdcos(t, λ3 , 0.3) − 0.6 hfdsin(t, λ3 , 0.3) ⎟ ⎟ x (t) = ⎜ ⎝ 2.8 hfcos(t, λ01 , 0.3) − 4.8 hfsin(t, λ01 , 0.3) + 7.2 hfdcos(t, λ3 , 0.3) + 1.2 hfdsin(t, λ3 , 0.3) ⎠ 1 − 1.6 hfcos(t, λ01 , 0.3) − 4.4 hfsin(t, λ01 , 0.3) + 9.6 hfdcos(t, λ3 , 0.3) − 2.4 hfdsin(t, λ3 , 0.3) From Eqs. (11), (12), (32) and (33), the functions hfsin (t, λ01 , 0.3), hfcos(t, λ01 , 0.3), hfdsin(t, λ3 , 0.3) and hfdcos (t, λ3 , 0.3) can be easily obtained as follows: • hfsin(t, λ01 , 0.3) . . . 1 4 s s 0.6 + 4 N 1 1 k 1i −1 ∼L = s s + p1i i=1. hfsin(t, λ , 0.3) = L−1 01. N1. = i=1. k1i 1-exp −p1i t p1i. p1i = 10.08 ∗ (4)(i−12) , sin[0.4π] 5.04(4)(i−12) k1i = π cosh 0.6 log (4)(12−i) − cos[0.4π]. • hfcos(t, λ01 , 0.3): hfcos(t, λ , 0.3) 01 & % s0.3 1 = L−1 s [s0.6 + λ01 2 ] ⎧ ⎫ N2 ⎨ N2 ⎬ k2i 1 = L−1 = ⎩ s + p2i s ⎭ i=1. with N1 = 23, for 1 ≤ i ≤ 23 the poles p2i and their corresponding residues k2i are given as:. • for 1 ≤ i ≤ 35. • for 36 ≤ i ≤ 58. k2i =. +. . sin[(0.4)π ] 0.16 ∗ 10−2 ∗ (4)(i−47) cosh 0.6 log (4)(47−i) − cos[(0.4)π ] ⎧⎛ ⎞ ⎫ 35 , ⎪ ⎪ −(1.5) ∗ 10−2 ∗ (4)2(i−47) (1−1.39(2.99)( − j) ) ⎪ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎪ j=1 ⎪ ⎪ × ⎪ ⎪ ⎝ ⎠ 35 ⎪ ⎪ , ⎪ ⎪ (i−47) −8 ( −1) − j) ( ⎪ ⎪ −2.04 ∗ 10 (2.99) (1−(2.99) ) ] 35 ⎨ [10 ∗ (4) ⎬ j=1, = j ⎪⎡ ⎪ ⎪ ⎪ ⎣ ⎪ ⎪ ⎩ cosh0.6 log. =1 ⎪ ⎪ ⎪. 123. ⎤ sin[(0.4)π ] ⎦ (4)(47−i) −cos[(0.4)π ]. . 1-exp −p2i t. . with λ01 = 2 and N2 = 58, the poles p2i and their corresponding residues k2i are given as:. p2i = 2.04 ∗ 10−8 ∗ (2.99)(i−1) , ⎧⎛ ⎞ ⎫ 35 , ⎪ ⎪ 1−1.39(2.99)(i− j) (4)( −12) (2.99)(i−1) ⎪ ⎪ ⎪ ⎜ ⎪ ⎟ ⎪ ⎪ j=1 ⎪ × ⎪ ⎝ ⎠ ⎪ , 23 ⎪ 35 ⎨ ⎬ −9 ( −12) 10 −8 (i−1) (i− j) −10 ∗ (4) 2.04 ∗ 10 (2.99) (1−(2.99) ) k2i = j=1,i= j ⎪ ⎪ 8π ⎪ =1 ⎪ ⎪ ⎪ ⎪ ⎪ sin[(0.4)π] ⎪ ⎪ ⎪ ⎪ ⎩ cosh 0.6 log 4(12− ) −cos[(0.4)π ] ⎭. p2i = 10 ∗ (4)(i−47) . i=1. k2i p2i. ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭. .

(13) Author's personal copy The solution of state space linear fractional system of commensurate order with complex…. • hfdsin(t, λ3 , 0.3) hfdsin(t, λ, 0.3) 3 . . λ3 2. = L−1. s[s 0.6 + 2ζ λ3 s 0.3 + λ3 2 ] ⎧ ⎫ ⎨ N3 2 + a s 1⎬ 1 − ζ 3i k3i = L−1 ⎩ s 2 + 2βωi s + ωi2 s ⎭ i=1 ⎡ N3. = ⎣1 − C 3. 89. From Eqs. (17), (18), (32) and (33), the functions hfsin (t, λ01 , 0.5), hfcos(t, λ01 , 0.5), hfdsin(t, λ3 , 0.5) and hfdcos (t, λ3 , 0.5) are given as follows: • hfsin(t, λ01 , 0.5): −1. hfsin(t, λ , 0.5) = L 01. ⎤ $ 2 k3i exp (−βωi t) sin ωi 1 − β t + φ3 ⎦. = 1-exp − λ01 2 t. . λ01 2 s + λ01 2. 1 s. . i=1. • hfcos(t, λ01 , 0.5):. • hfdcos(t, λ3 , 0.3) hfdcos(t, λ, 0.3) 3 = L −1. s 0.3 + λ3 ζ 0.6 s[s + 2ζ λ3 s 0.3 + λ3 2 ]. hfcos(t, λ ,0.5) 01 . . =L. . ⎧ ⎨ N4. ⎫ ζ + a4i s 1⎬ −1 =L k4i ⎩ s 2 + 2βωi s + ωi2 s ⎭ i=1 ⎤ ⎡ N4 $ ζ 2 − C4 k4i exp (−βωi t) cos ωi 1 − β t + φ4 ⎦ =⎣ λ3 . ωi =. 1 (0.31) ∗ (1.10). (363−i). =L =. k20 λ01 2 N2. , a3i = (−0.26) ∗ (1.10)(363−i) ,. + i=1. k3i = k4i. . k2i p2i. (0.60) ∗ (1.10)2(363−i) sin [0.7π] × cosh 0.3 log (1.10)(363−i) − cos[0.7π]. Figure (1) shows the plots of the functions hfcos(t, λ01 , 0.3), hfsin(t, λ01 , 0.3), hfdcos(t, λ3 , 0.3) and hfdsin(t, λ3 , 0.3). Figure (2) shows the plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t). 4.2 Case 2: m = 0.5. 1 s 1 s. . N2. + i=1. k2i s + p2i. 1 s. . . 1-exp − λ01 2 t. 1 − exp (− p2i t). p2i = 1.99 ∗ 10−8 (2.51)(i−1) , ⎡ k2i. ⎢ −10−12 ∗ (6.31)(i−1) ⎢ =⎢ ⎣ (2.51)(i−1) 10−8 ∗ (2.51)(i−1) − 2. 1. =. . with λ01 = 2, N2 = 35, k20 = −2.07 and for 1 < i < 35, the poles p2i and their corresponding residues k2i are by:. (363−i). a4i = 0.17 ∗ (1.10). k20 s + λ01 2. −1. i=1. with λ3 = 1.41, ζ = 0.91, β = 0.14, C3 = 2.41, C4 = 0.71, ϕ3 = ϕ4 = 0.43 and N3 = N4 = 725 and the parameters ωi , a3i , a4i and k3i = k4i are given, for 1 ≤ i ≤ 725, as:. s 0.5 s + λ01 2. −1. ⎤ 1 − 1.59(2.51)(i− j) ⎥ j=1 ⎥ ⎥ 35 ⎦ , j) (i− 1 − (2.51) 35 ,. j=1,i = j. • hdfsin(t, λ01 , 0.5): hfdsin(t, λ, 0.5) 3 . λ3 2 s s + 2ζ λ3 s 0.5 + λ3 2 N 3 1 − ζ 2 + a3i s 1 −1 =L k3i 2 s + 2βωi s + ωi2 s i=1 N. . = L−1. 3. = 1 − C3. k3i exp (−βωi t) sin ωi. $ 2 1 − β t + φ3. i=1. In this case the solution x(t) is: ⎛. ⎞ −1 + 2.2 hfcos(t, λ01 , 0.5) − 0.2 hfsin(t, λ01 , 0.5) + 1.8 hfdcos(t, λ3 , 0.5) + 0.3 hfdsin(t, λ3 , 0.5) ⎜ 1 + 3.6 hfcos(t, λ01 , 0.5) − 2.6 hfsin(t, λ01 , 0.5) + 2.4 hfdcos(t, λ3 , 0.5) − 0.6 hfdsin(t, λ3 , 0.5) ⎟ ⎟ x (t) = ⎜ ⎝ 2.8 hfcos(t, λ01 , 0.5) − 4.8 hfsin(t, λ01 , 0.5) + 7.2 hfdcos(t, λ3 , 0.5) + 1.2 hfdsin(t, λ3 , 0.5) ⎠ 1 − 1.6 hfcos(t, λ01 , 0.5) − 4.4 hfsin(t, λ01 , 0.5) + 9.6 hfdcos(t, λ3 , 0.5) − 2.4 hfdsin(t, λ3 , 0.5). 123.

(14) Author's personal copy 90. D. Boucherma et al.. Fig. 1 Plots of the functionshfcos(t, λ01 , 0.3), hfsin(t, λ01 , 0.3) hfdcos(t, λ3 , 0.3) and hfdsin(t, λ3 , 0.3). 1.2. 1 hfcos 0.8. hfsin. Amplitude. hfdcos 0.6. hfdsin. 0.4. 0.2. 0. -0.2 0. 0.5. 1. 1.5. 2. 2.5. 3. 3.5. 4. 4.5. 5. Time(sec). Fig. 2 Plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t). 6 5 4. Amplitude. 3 2 1. x1 x2. 0. x3 x4. -1 -2. 0. 0.5. 1. • hdfcos(t, λ01 , 0.5): hfdcos(t, λ, 0.5) 3 = L −1. s 0.5 + λ3 ζ. . s s + 2ζ λ3 s 0.5 + λ3 2 ⎧ ⎫ ⎨ N4 ζ + a4i s 1⎬ = L −1 k4i 2 2 ⎩ s + 2βωi s + ωi s ⎭ i=1 ⎤ ⎡ N4 $ ζ − C4 k4i exp (−βωi t) cos ωi 1 − β 2 t + φ4 ⎦ =⎣ λ3 i=1. with λ3 = 1.41, ζ = 0.91, β = 0.66, C3 = 2.41, C4 = 0.71, ϕ3 = ϕ4 = 0.43 and N3 = N4 = 725 and for 1 ≤ i ≤ 725, the parameters ωi , a3i , a4i and k3i = k4i are given as: ωi =. 1 0.5 ∗ (1.10)(363−i). 123. , a3i = (−0.21) ∗ (1.10)(363−i) ,. 1.5. 2. 2.5 Time(sec). 3. 3.5. 4. 4.5. 5. a4i = 0.46 ∗ (1.10)(363−i) 1 k3i = (1.57) ∗ (1.10)2(363−i) sin [0.5π] × cosh 0.5 log (1.10)(363−i) − cos[0.5π]. Figure (3) shows the plots of the functions hfcos(t, λ01 , 0.5), hfsin(t, λ01 , 0.5), hfdcos(t, λ3 , 0.5) and hfdsin(t, λ3 , 0.5). Figure (4) shows the plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t). 4.3 Case 3: 0.5 < m < 1 For m = 0.87, the state vector x(t) solution of the linear fractional system of commensurate order of Eq. (64) is given as:.

(15) Author's personal copy The solution of state space linear fractional system of commensurate order with complex… Fig. 3 Plots of the functions hfcos(t, λ01 , 0.5), hfsin(t, λ01 , 0.5), hfdcos(t, λ3 , 0.5) and hfdsin(t, λ3 , 0.5). 91. 1 0.9. hfcos hfsin. 0.8. hfdcos hfdsin. Amplitude. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0. 0. 0.5. 1. 1.5. 2. 2.5. 3. 3.5. 4. 4.5. 5. Time(sec). Fig. 4 Plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t). 5. 4. x1. 3. x2 Amplitude. x3 2. x4. 1. 0. -1. -2. 0. 0.5. 1. 1.5. 2. 2.5 Time(sec). 3. 3.5. 4. 4.5. 5. ⎛. ⎞ −1 + 2.2 hfcos(t, λ01 , 0.87) − 0.2 hfsin(t, λ01 , 0.87) + 1.8 hfdcos(t, λ3 , 0.87) + 0.3 hfdsin(t, λ3 , 0.87) ⎜ 1 + 3.6 hfcos(t, λ01 , 0.87) − 2.6 hfsin(t, λ01 , 0.87) + 2.4 hfdcos(t, λ3 , 0.87) − 0.6 hfdsin(t, λ3 , 0.87) ⎟ ⎟ x (t) = ⎜ ⎝ 2.8 hfcos(t, λ01 , 0.87) − 4.8 hfsin(t, λ01 , 0.87) + 7.2 hfdcos(t, λ3 , 0.87) + 1.2 hfdsin(t, λ3 , 0.87) ⎠ 1 − 1.6 hfcos(t, λ01 , 0.87) − 4.4 hfsin(t, λ01 , 0.87) + 9.6 hfdcos(t, λ3 , 0.87) − 2.4 hfdsin(t, λ3 , 0.87) From Eqs. (23), (24), (32) and (33), the functions hfsin(t, λ01 , 0.87), hfcos(t, λ01 , 0.87), hfdsin(t, λ3 , 0.87) and hfdcos(t, λ3 , 0.87) can be easily obtained as:. • hfsin(t, λ01 , 0.87): hfsin(t, λ01 , 0.87) = L −1. %. 4 [s 1.74 + 4]. 1 s. &. 123.

(16) Author's personal copy 92. D. Boucherma et al. ⎧⎡⎛ ⎞ ⎤ ⎨ 1 ⎦ A1 s + B1 ⎠ ⎣⎝ 1 2 ⎩ s s 2 + 2α λ01 0.87 s + λ01 0.87 ⎫ N 1 ⎬ k1i 1 + ⎭ s + p1i s. ∼ = L −1. i=1. N1. =1−. 1 k1i exp −p1i t + C11 exp −α λ01 0.87 t. i=1. × sin. 1 1 − α 2 λ01 0.87 t + 11. k1i =. (5.33) ∗ 10 ,. ×. 5.12 ∗ (3.78(i−1) ). 2 3.78(i−1). . − (1.06) ∗ (3.78)(i−1) + 1. 1 − (1.41) (3.78)(i− j). j=1,i= j. λ3 2 =L s[s 1.74 + 2ζ λ3 s 0.87 + λ3 2 ] N 3 1 − ζ 2 + a3i s 1 −1 =L k3i 2 s + 2βωi s + ωi2 s i=1 N $ −1. 1 − (3.78)(i− j). 1 − β2. t + φ3. i=1. • hfdcos(t, λ3 , 0.87): hfdcos(t, λ3 , 0.87) = L −1. s 0.87 + λ3 ζ. . s[s 1.74 + 2ζ λ3 s 0.87 + λ3 2 ] ⎧ ⎫ ⎨ N4 ζ + a4i s 1⎬ −1 =L k4i ⎩ s 2 + 2βωi s + ωi2 s ⎭ i=1 ⎤ ⎡ N4 $ ζ − C4 k4i exp (−βωi t) cos ωi 1 − β 2 t + φ4 ⎦ =⎣ λ3 i=1. • hfcos(t, λ01 , 0.87): hfcos(t, λ01 , 0.87) & % 1 s 0.87 = L−1 [s 1.74 + 4] s ⎧⎡⎛ ⎞ ⎤ ⎨ s + B 1 A 2 2 ⎦ ⎠ = L−1 ⎣⎝ 1 2 ⎩ s s 2 + 2α λ01 0.87 s + λ01 0.87 N 2 k2i 1 + s + p2i s i=1. N. 2 k2i KD exp −p2i t − 2 p2i λ01 i=1 1 1 1−α 2 λ01 0.87 t +21 + C21 exp −α λ01 0.87 t sin. =. with λ01 = 2, K D = 1.72 ∗ 10−5 , C21 = −0.58, α = 0.23, 21 = 0.0011, N2 =11 and, for 1 ≤ i ≤ 11, the poles p2i and their corresponding residues k2i are given as:. with λ3 = 1.41, ζ = 0.71, β = 0.62, C3 = 1.41, C4 = 0.71, φ3 = φ4 = 0.79, N3 = N4 =427 and, for 1 ≤ i ≤ 427, the parameters ωi , a3i , a4i and k3i = k4i are given as: ωi =. 1 (0.67) ∗ (1.10)(214−i). j=1, j=i. , a3i = (−0.08) ∗ (1.10)(214−i) ,. a4i = (0.67) ∗ (1.10)(214−i) 1 k3i = k4i = 2.83 ∗ (1.10)2(214−i) sin [0.13π] × cosh 0.87 log (1.10)(214−i) − cos[0.13π]. Figure (5) shows the plots of the functions hfcos(t, λ01 , 0.87), hfsin(t, λ01 , 0.87), hfdcos(t, λ3 , 0.87) and hfdsin(t, λ3 , 0.87). and hfdsin(t, λ3 , 0.87) Figure (6) shows the plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t).. p2i = 2.75 ∗ 10−5 ∗ (9.58)(i−1) 11.80 ∗ 10−11 ∗ (9.58)(i−1) ∗ 1 − (1.24 ∗ 10−5 ∗ (9.58)(i−1) ) k2i = (1.24 ∗ 10−5 ∗ (9.58)(i−1) )2 − (0.57 ∗10 −5 ∗ (9.58)(i−1) ) + 1 ⎡ 6 ⎤ 5 , , (i−i1) ) (i−i2+6) ) (1 − 7.14 ∗ (9.58) (1 − 1.34 ∗ (9.58) ⎢ ⎥ ⎢ i1=1 ⎥ i2=1 ×⎢ ⎥ 11 ⎣ ⎦ , (1 − (9.58)(i− j) ). 123. k3i exp (−βωi t) sin ωi. = 1 − C3. j=1 10 ,. hfdsin(t, λ, 0.87) 3 . 3. with λ01 = 2, C11 = 1.07, α = 0.23, 11 = 1.57, N1 =10 and, for 1 ≤ i ≤ 10, the poles p1i and their corresponding residues k1i are given as: p1i = 5.12 ∗ (3.78(i−1) ) . • hfdsin(t, λ3 , 0.87):.

(17) Author's personal copy The solution of state space linear fractional system of commensurate order with complex… Fig. 5 Plots of the functions hfcos(t, λ01 , 0.87), hfsin(t, λ01 , 0.87), hfdcos(t, λ3 , 0.87) and hfdsin(t, λ3 , 0.87). 93. 1.2. 1. 0.8. 0.6 Amplitude. hfcos hfsin. 0.4. hfdcos hfdsin. 0.2. 0. -0.2. -0.4 0. 2. 4. 6. 8. 10. 12. 14. Time(sec). Fig. 6 Plots of the state variables x1 (t), x2 (t), x3 (t) and x4 (t). 18 x1 x2. 16. x3 14. x4. 12. Amplitude. 10 8 6 4 2 0 -2 0. 2. 4. 6. 8. 10. 12. 14. Time(sec). 5 Conclusion In this work, we have derived, using the Caputo derivative definition, the solution of the state space linear fractional system of commensurate order for complex eigenvalues of the state space matrix. The obtained solutions of the homogeneous and non-homogeneous cases have been expressed as a linear combination of introduced suitable fundamental functions as it is done with regular state space linear systems using the modal decomposition technique without needing a deep study of the fractional derivatives. The proposed analytic time solutions for fractional systems of commensurate order are in the form of exponentials, sine, cosine, damped sine and damped cosine functions depending on. the commensurate fractional order; but the Mittag-Leffler function is a polynomial function. An important benefit of using exponentials and trigonometric functions leads to the computations of the impulse and step responses in a manner very similar to the classic case; besides, elementary linear RLC circuits can be easily derived and used to synthesize these types of fractional systems. Hence, the method introduces a promising tool for solving fractional differential equations of commensurate order. A numerical example has been solved as illustration to show the effectiveness and the usefulness of the proposed analytical tool for the solution of the state space linear fractional system of commensurate order with complex eigenvalues.. 123.

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