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A study of non-commensurate fractional linear system is done in a parallel way to the commensurate case. A partial fraction decomposition is accomplished using a recursive procedure. Each partial fraction is inverted in two different ways. The decomposition integer/fractional is done also. Some examples are presented.
Journal of Advanced Research 25 (2020) 11–17 Contents lists available at ScienceDirect Journal of Advanced Research journal homepage: www.elsevier.com/locate/jare Non-commensurate fractional linear systems: New results Manuel D Ortigueira a,⇑, Gabriel Bengochea b a b CTS-UNINOVA and NOVA School of Science and Technology of NOVA University of Lisbon, Campus da FCT da UNL, Quinta da Torre, 2829-516 Caparica, Portugal Academia de Matemáticas, Universidad Autónoma de la Ciudad de México, Ciudad de México, Mexico h i g h l i g h t s g r a p h i c a l a b s t r a c t It presents a partial fraction decomposition of non commensurate systems Suitable inversion of each fraction is done in two ways: series and integer/ fractional decomposition a r t i c l e i n f o Article history: Received 19 December 2019 Revised 24 January 2020 Accepted 27 January 2020 Available online February 2020 Keywords: Non-commensurate order Fractional Calculus Partial fraction decomposition Laplace transform a b s t r a c t A study of non-commensurate fractional linear system is done in a parallel way to the commensurate case A partial fraction decomposition is accomplished using a recursive procedure Each partial fraction is inverted in two different ways The decomposition integer/fractional is done also Some examples are presented Ó 2020 The Authors Published by Elsevier B.V on behalf of Cairo University This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) Introduction The last 30 years of Fractional Calculus [5,14,15] brought a remarkable progress and became popular in many scientific and technical areas [4,6–9,16] due to its ability to better describe many natural phenomena The fact that fractional models represent systems which require lower number of parameter than those of integer order is a point in favor of fractional systems (see [2]) This is due to their capacity of supplying us with more reliable time and frequency representations Peer review under responsibility of Cairo University ⇑ Corresponding author E-mail addresses: mdo@fct.unl.pt (M.D Ortigueira), gabriel.bengochea@uacm edu.mx (G Bengochea) We cannot say that there many works on non-commensurate systems The first meaningful study was presented in [13], based on a manipulation of the transfer function and the use of the properties of Laplace transform Another one described in [12] was based on a series expansion of the transfer function Much of the research in fractional systems is developed for commensurate orders in a way that is a direct generalization of traditional formalism However, most of the methods used to solve commensurate fractional linear systems cannot be easily extended to noncommensurate case In such situation, we find the partial fraction decomposition very useful in inverting Laplace and Z transforms currently used in the study of linear systems, when performing the computation of the impulse response from the transfer function (TF) The implementation of such inversion using the decomposition of the TF in partial fractions, not only simplifies the https://doi.org/10.1016/j.jare.2020.01.015 2090-1232/Ó 2020 The Authors Published by Elsevier B.V on behalf of Cairo University This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) 12 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 11–17 procedure, but gives more insight into the characteristics of the system, namely, stability and existing vibration modes The procedures in [1,12,13] are not suitable to display such characteristics, mainly to perform the modal decomposition In this paper, we look for obtaining for non-commensurate order systems such kind of decomposition, provided we know the pseudo-pole/zero factorization We start from the simplest case where we have only two orders and two pseudo-poles and decompose it into a sum of two fractions From it, we turn to the case of three pseudo-poles Finally, we deduce the general case and show how add pseudo-zeros For each term we obtain the inverse LT by using the operational method presented in [1] The paper is organized as follows Firstly, we present our results related to simple fraction decomposition with non-commensurate order Then, we resolve several examples of lineal fractional systems with non-commensurate order We continue with the decomposition of transfer function in two parts, a part of integer order and the other one of fractional order Finally, the conclusions are presented orders does not have a factorization as referred Relation (3) can serve as guide for obtaining the factorization of polynomials with a few factors Two pseudo-poles case The simple fraction decomposition is a widely used tool in several areas of science In the case of one variable, a well known simple result is that 1 ẳ bỵa bỵa ; z aịz bị ðz À aÞ ðz À bÞ a – b: ð4Þ Our goal is the decomposition of a fraction of the type: ; ðsa1 À c1 Þðsa2 À c2 Þ but it is a simple task to show that it is not possible to obtain a result equal to (4) However, we can obtain a similar decomposition a using a trick: if we define in (4) the parameters z ¼ 1; a ¼ sc and a b ¼ sc , with c1 ; c2 be different non-zero complex numbers, then Partial fraction decomposition we obtain the result stated in next Theorem Non-commensurate transfer function Theorem Let c1 ; c2 be different non-null complex numbers and a1 ; a2 be positive real numbers Then Consider a linear system with TF given by Hsị ẳ sb1 f1 Þðsb2 À f2 Þ Á Á Á ðsbm À fm Þ ; ðsa1 À c1 Þðsa2 À c2 Þ Á Á Á ðsan À cn Þ ð1Þ where the ci ; i ¼ 1; 2; ; n; fj ; j ¼ 1; ; m, are non-null pseudopoles and pseudo-zeroes that are, not necessarily different, complex numbers The derivative orders, bm ; an are real numbers in the interP P val ð0; 1Þ, and for stability reasons, m bm n an In applications, we have a problem not easily solvable: the obtention of the factorization To understand the difficulties we consider the relation between the factorization and the pseudopolynomial Consider a pseudo-polynomials with format Pn sị ẳ sa1 c1 ịsa2 c2 Þ Á Á Á ðsan À cn Þ; where the ci ’s are different a ¼ ða1 ; a2 ; ; an ị Rỵ n If we define k¼ complex numbers Let a k j1 ¼ ak ; ak ; j1 – j2 ; j1 ; j2 ½1; n; Remark If c1 ¼ c2 ¼ 0, the theorem does not apply, because we have no pseudo-pole, but only a branchcut point For c1 – and Àa c2 ¼ 0, we observe that sa1 1c ịsa2 ẳ sas1 2c ị Therefore, we invert 1 ðsa1 Àc1 Þ Àa2 to s and afterwards perform the anti-derivation corresponding Remark It is important to note that the term ðc1 sa2 À c2 sa1 Þ has no zeroes in the first Riemann sheet, in the non-commensurate case we are dealing Therefore, each term in the right hand side in (1) only has a pseudo-pole If the orders commensurate, we can continue the decomposition as we in the classic procedure In the next theorem, we tackle the case with three simple pseudo-poles kj1 ;j2 ; ;jnÀ1 ¼ Theorem Let c1 ; c2 ; c3 be different non-null complex numbers, and a1 ; a2 ; a2 be positive real numbers Then ajn ; kj1 ;j2 ; ;jn ¼ 0; sa1 c1 ịsa2 c2 ịsa3 c3 ị Pn sị ẳ sk À n X j1 ¼0 n X j1 ;j2 ;j3 ẳ0 cj1 skj1 ỵ n X j1 ;j2 ẳ0 j1 –j2 c21 a a a a a s Àc1 s Þðc3 s Àc1 s Þðs Àc1 Þ ¼ ðc c22 a a a a a s Àc2 s Þðc3 s Àc2 s Þðs Àc2 Þ then (2) can be written as À : General decomposition n X kẳ1 kj1 ;j2 .ẳ c2 sa2 c2 ịc1 sa2 c2 sa1 ị j1 ẵ1; n; kẳ1 kj1 kj1 ;j2 ẳ 2ị n X k; kẳ1 n X c1 ẳ sa1 c1 ịsa2 c2 Þ ðsa1 À c1 Þðc1 sa2 À c2 sa1 Þ þ ðc c23 a a a a a s Àc3 s Þðc1 s Àc3 s ịs c3 ị ỵ c cj2 cj1 skj1 ;j2 cj3 cj2 cj1 skj1 ;j2 ;j3 ỵ þ ðÀ1Þnþ1 cjn Á Á Á cj2 cj1 ; Proof From Theorem 1, we obtain that ð3Þ j1 –j2 –j3 which shows that there are many non-factorizable pseudopolynomials For example, sa ỵ asb ỵ b, with non-commensurate sa1 c1 Þðsa2 Àc2 Þðsa3 Àc3 Þ ¼ À ðsa1 Àc Þðs c1 a3 Àc Þðc sa1 Àc sa2 Þ À ðsa2 Àc Þðsa3 Àcc2Þðc sa2 Àc sa1 Þ : Applying again the Theorem 1, it follows that : 13 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 11–17 c21 a a a a a s Àc1 s Þðc3 s Àc1 s Þðs Àc1 Þ sa4 h1 sa1 c1 ịsa2 c2 ịsa3 c3 ị ẳ c sa1 c1 ịsa2 c2 ịsa3 c3 ị ỵ c22 a a a a a s Àc2 s Þðc3 s Àc2 s Þðs c2 ị ỵ c c23 sa4 a a a a a s Àc3 s Þðc1 s Àc3 s Þðs Àc3 Þ c1 h1 sa1 þ ðc c1 c3 þ ðc a a a a a s Àc1 s Þðc1 s Àc3 s Þðs Àc3 Þ c2 c3 a a a a a s Àc2 s Þðc2 s Àc3 s Þðs Àc3 Þ þ ðc c21 sa4 a a a a a s Àc1 s Þðc3 s Àc1 s Þðs Àc1 Þ c22 sa4 ðc1 sa2 Àc2 sa1 ịc3 sa2 c2 sa3 ịsa2 c2 ị ẳ c À ðc a a a a a s Àc1 s Þðc3 s Àc1 s Þðs Àc1 Þ c2 h1 sa2 : À ðc a a a a a s Àc2 s Þðc3 s Àc2 s Þðs Àc2 Þ c3 h1 sa3 Finally, simplifying we get the result h From Theorems and 2, we deduce the general result À ðc a a a a a s Àc3 s Þðc1 s Àc3 s Þðs Àc3 Þ Theorem Let c1 ; c2 ; ; cn be different non-null complex numbers and a1 ; a2 ; ; an positive real numbers Then ðsa1 À c1 Þðsa2 À c2 Þ Á Á Á ðsan À cn Þ n X cn1 i ẳ 1ịnỵ1 : Y iẳ1 sai ci Þ nðcj sai À ci saj Þ Now, we can deduce the next Theorem Theorem Let c1 ; c2 ; ; cn ; h1 , be different non-null complex numbers and a1 ; a2 ; ; an ; anỵ1 , be real numbers Then sanỵ1 h1 sa1 c1 ịsa2 c2 ịsan cn ị ẳ 1ịnỵ1 Example Suppose that we want to apply the simple fraction decomposition to transfer function À c1 Þðsa2 À c2 Þ Y nðcj sai Àci saj Þ : j¼1 j–i For adding another pseudo-zero h2 , with order anỵ2 , we apply the previous Theorem and get sanỵ1 h1 ịsanỵ2 h2 ị sa1 c1 ịsa2 c2 ịsan cn ị ẳ 1ịnỵ1 n X : cn1 sanỵ1 ỵanỵ2 i Y a iẳ1 s i ci ị ncj sai ci saj ị jẳ1 ji n X cn2 h1 sai ỵanỵ2 i a iẳ1 s i ci ị c1 ẳ a c1 ịsa2 c2 ị ðs À c1 Þðc2 sa1 À c1 sa2 Þ Y ncj sai ci saj ị jẳ1 ji c2 a : ðs À c2 Þðc1 sa2 À c2 sa1 ị Then cn2 h1 sai i a iẳ1 s i Àci Þ By the Theorem 1, we have that ðsa1 nðcj sai Àci saj Þ j–i n X À ji sa1 cn1 sanỵ1 i Y jẳ1 For the case when we have multiple pseudo-poles we only need to apply several times the Theorems To illustrate the procedure, we present the next example Hsị ẳ n X a iẳ1 s i ci ị jẳ1 : n X 1ịnỵ1 6ị h2 cn1 sanỵ1 i a iẳ1 s i ci ị Y ncj sai ci saj ị jẳ1 sa1 c1 ịsa2 c2 ị ẳ sa2 1c 2ị ðsa1 Àc c1 a a Þðc2 s Àc1 s ị c1 ẳ sa1 c a a a Þðs Àc2 Þðc2 s Àc1 s Þ À ðsa2 Àc À c2 j–i n X ỵ a a ịc1 s c2 s Þ c2 ðsa2 Àc2 Þ ðc1 sa2 Àc2 sa1 ị : cn2 h1 h2 sai i a iẳ1 s i ci ị Y jẳ1 ji Again by applying Theorem 1, we get that ðsa1 Àc1 Þðsa2 c2 ị ẳ The same procedure can be applied to case of more pseudo-zeros In the next section we present some examples of our decomposition with zeros c21 ðsa1 Àc1 Þðc2 sa1 Àc1 sa2 Þ c1 c2 a a a a Þðc1 s Àc2 s Þðc2 s c1 s ị ỵ sa2 c c2 ðsa2 Àc2 Þ ðc2 sa2 Àc2 sa1 Þ : Commensurate case Remark There is an eventually simpler approach to this example that consists in taking the decomposition of Theorem and compute the order derivative relatively to c2 in both sides of the relation In this subsection, we present some particular cases with which we verify some known results Consider a1 ¼ a2 in Theorem Then c2 c1 ị c1 c2 ị ẳ ðsc1a=ð À ðsc2a=ð : Àc Þsa1 Àc Þsa1 ðsa1 Àc1 Þðsa2 Àc2 Þ Simple pseudo-poles/zeroes cases c1 sa3 a a a s Àc1 s Þðs Àc1 Þ À ðc h1 sa1 a a a s Àc2 s Þðs Àc1 Þ À ðc ¼ À ðc À ðc The previous relation can be rewritten as Now, we add pseudo-zeros to transfer function (1) We suppose that the number of pseudo-poles is bigger than the number of pseudo-zeroes We procedure as in Theorem 1, but we add a pseudo-zero h1 of order a3 A simple computation yields sa3 Àh1 ðsa1 Àc1 Þðsa2 Àc2 Þ : nðcj sai Àci saj Þ c2 sa3 a a a s Àc2 s Þðs Àc2 Þ h1 sa2 a a a s Àc1 s Þðs Àc2 Þ : ð5Þ For the case of three pseudo-poles and one pseudo-zero we obtain that ðsa1 Àc1 Þðsa2 c2 ị ẳ cc1c ẳ c1 sa1 Àc1 Þ 1 c2 Àc1 ðsa1 Àc1 Þ À À sa11 c À c cÀ2c c2 ðsa1 Àc2 Þ À sa21 c c1 Àc2 ðsa1 Àc2 Þ Now, let < a1 < 1; < a2 < 1, and set a1 ¼ ma; a2 ¼ na; < a < 1, where m; n N We want to see if c1 sa2 À c2 sa1 has zeroes Let s ¼ qeih We can show easily that with c qmnịa ẳ ; c1 14 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 11–17 and h¼ Using the method presented in Appendix A, the solution associated to basic element argðc2 Þ À argðc1 Þ ; ðm À nÞa H1 ðsÞ ¼ À we have a zero, if jhj < p For example, with c2 and c1 real numbers with the same sign, there is a zero and consequently the term c1 sa2 À c2 sa1 will contribute with another pseudo-pole to (1), but having different signs there will be no pseudo-pole Consider a1 ¼ a2 ¼ a3 in Theorem Then ðsa1 Àc1 Þðsa2 Àc2 Þðsa3 Àc3 Þ c21 ðc3 Àc1 Þðc2 Àc1 Þ s2a1 ðsa1 c1 ị ẳ : c2 ỵ ðc Àc Þð3c Àc Þ s2a1 ðs1a1 Àc Þ 3 ðsa1 Àc1 Þðsa2 Àc2 Þðsa3 c3 ị ẳ c3 c1 ịc2 c1 ị sa1 c1 ỵ c c ị1 c c ị ỵ c c3 ịc2 Àc3 Þ ðsa1 Àc1 Þðsa1 Àc2 Þðsa1 Àc3 Þ ¼ ðc Àc1 Þðc2 Àc1 Þ c1 sa11 s2ca21 sa1 c3 sa11 ỵ ðc Àc Þð1 c Àc Þ þ ðc Àc Þð1 c Àc Þ 3 k¼0 l¼0 & LÀ1 À À s2ca31 : & LÀ1 À 1 sa1 Àc2 sa1 Àc3 ðsa1 Àc1 Þðsa2 Àc2 Þ ¼ ðsa1 Àc ðsa2 ' 2sa3 ¼ y2 ðtÞ: À 2Þðsa2 À 2sa1 Þ 2Þ 1 sa1 c1 p ỵ 2p c c ỵ c ị 2 kẳ0 lẳ0 X X ỵ 1p sa1 c2 1p sa1 ỵ c2 kẳ0 lẳ0 : 2lkỵ1ịa ỵklỵ1ịa a t 22lk C2lkỵ1ị a2 ỵklỵ1ịa1 a3 ỵ1ị etị: Example Consider the system associated to transfer function Computing the impulse response of some fractional linear systems In this section, in order to illustrate how to use our decomposition, we solve several fractional linear systems using the simple fraction decomposition introduced in the previous section We show how compute the inverse Laplace transform of our basic elements To it, we use the results presented in Appendix A to invert each term of (1) to obtain ' c1 ðsa1 Àc1 Þðc2 sa1 Àc1 sa2 Þ X X t 2lkỵ1ịa1 ỵklỵ1ịa2 a3 2kl C2lkỵ1ị a1 ỵklỵ1ịa2 a3 ỵ1ị etị ytị ẳ 8ị Now, if we have that xtị ẳ etị in (7), then we only need calculate the integral (omitting the sum of constant) to (8) Therefore the solution of (7) with xtị ẳ etị is given by pffiffiffiffi a pffiffiffiffi : c2 Þðs þ c2 Þ ffi pffiffiffiffi þ À2pffiffiffi c2 ðc1 c2 ị kẳ0 lẳ0 sa1 ' sa3 ẳ y1 tị; 1ị2sa1 sa2 ị ytị ẳ y1 tị ỵ y2 tị: a ịs ẳ c21ỵc Hsị ẳ : sa1 c1 ịsa2 c2 ị 2lkỵ1ịa ỵklỵ1ịa 1 t c2lk ckl C2lkỵ1ịa1 ỵklỵ1ịa2 ị etị; Hsị ẳ c1 sa1 À c1 Þðc2 sa1 À c1 sa2 Þ y1 ðtÞ ¼ ð7Þ Suppose that the input xðtÞ ¼ dðtÞ From Theorem we have that Hsị ẳ sa1 sa3 2sa3 À a : a a 2 À 1Þð2s À s Þ ðs À 2Þðsa2 À 2sa1 Þ c2 ðsa2 À c2 Þðc1 sa2 À c2 sa1 Þ : c1 ðsa1 À c1 Þðc2 sa1 À c1 sa2 Þ ¼ sa1 ðsa1 ; À c1 Þ sa2 Àa1 À cc2 is given by Example Consider the system associated to transfer function ðsa1 À Using the method presented in Appendix A, the solution associated to basic element X X c2lÀk c2kÀl k¼0 lẳ0 sa3 : 1ịsa2 2ị 9ị Suppose that the input xtị ẳ dtị From Theorem we have that H1 sị ẳ ẳ where etị is the Heaviside unit step function Hsị ẳ etị: Therefore the solution yðtÞ of system (7) is : Using the case when a1 ¼ a2 ¼ a3 , we get that sa1 c1 ịs2a1 c2 ị t 2lkỵ1ịa2 ỵklỵ1ịa1 a3 C2l k ỵ 1ịa2 ỵ k l þ 1Þa1 À a3 Þ It follows that Consider a2 ¼ 2a1 in Theorem Then X X X X 22lk y2 tị ẳ sa1 Àc2 sa1 Àc 2sa3 sa3 À Á; ¼ À 2Þðsa2 À 2sa1 Þ sa2 ðsa2 À 2Þ sa1 Àa2 À 12 ðsa2 and Simplifying & LÀ1 À k¼0 lẳ0 t2lkỵ1ịa1 ỵklỵ1ịa2 a3 etị; C2l k ỵ 1ịa1 ỵ k l ỵ 1ịa2 a3 ị is À sa1 À s2a1 X X 2kl y1 tị ẳ H2 sị ẳ The previous relation can be rewritten as is given by and for c22 s2a1 ðsa1 Àc2 Þ c2 ịc3 c2 ị ỵ c sa3 sa3 ẳ a a ; a a À 1Þð2s À s Þ s ðs À 1Þðsa2 Àa1 À 2Þ sa1 t2lkỵ1ịa1 ỵklỵ1ịa2 etị; C2l k ỵ 1ịa1 þ ðk À l þ 1Þa2 Þ and for H2 sị ẳ c2 sa2 c2 ịc1 sa2 c2 sa1 ị ẳ sa2 sa2 ; c2 Þ sa1 Àa2 À cc1 is y2 ðtÞ ẳ X X t2lkỵ1ịa2 ỵklỵ1ịa1 kẳ0 lẳ0 C2l k ỵ 1ịa2 ỵ k l ỵ 1Þa1 Þ c2lÀk c1kÀl eðtÞ: 15 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 11–17 It follows that & L1 The term Gsị ẳ ' c1 Qn jẳ1 cj sai ci saj ị in (11) is analytic in the first j–i ¼ y1 ðtÞ; Riemann surface and has no zeroes (of course in the analyticity region that excludes the origin that is the branch cut point) c2 ẳ y2 tị: sa2 c2 Þðc1 sa2 À c2 sa1 Þ yðtÞ ¼ y1 ðtÞ þ y2 ðtÞ: In these conditions we can use the integration path C in Fig 1, [3,10], and we apply the residue theorem Let u Rỵ and consider Feip uÞ and FðeÀip uÞ, the values of FðsÞ immediately above and below the branch cut line Proceeding as in [3] we obtain Example Consider the transfer function f ðtÞ ¼ Ai eci ðsa1 À c1 Þðc2 sa1 À c1 sa2 Þ and & LÀ1 À ' Therefore the solution ytị of system (9) is 1=ai Hsị ẳ sa3 ; sa1 iịsa2 ỵ iị 10ị t etị þ 2pi Z  Àip à Fðe uÞ À Fðeip uÞ eÀut du Á eðtÞ; 1=ai where the constant Ai is the residue of (11) at ci 12ị : b p where i ẳ We want the impulse response for the particular case in which a1 ¼ a2 Applying (5) to transfer function (10), we get that cai i Ai ¼ ci1=ai À1 n Y aj a ỵ1 : cj ci ci i ị jẳ1 ji sa3 sa1 iịsa2 ỵiị ẳ isa3 isa1 ỵisa2 ịsa1 iị ỵ isa3 isa2 ỵisa1 ịsa2 ỵiị a Computing the LT of both sides in (12) we obtain a isa2 ỵis2sa1 1ịsa1 iị ỵ isa1 ỵis2sa2 2ịsa2 ỵiị a a a Fsị ẳ F i sị ỵ F f sị; a ẳ isa1isỵis3 a2s þ ðisa2isþis3 aþ2s : Þðsa1 ÀiÞ Þðsa2 þiÞ where the integer order part is Because a1 ¼ a2 , then sa3 sa1 iịsa1 ỵiị ẳ sa1 þiÞ þ ðsa1 ÀiÞ þ i i 1=2sa3 sa1 ðsa1 ỵiị ỵ F i sị ẳ 1=2sa3 : sa1 sa1 ÀiÞ Following the methodology used in the previous examples, we obtain that the solution yðtÞ of system (10) is given by 1 X X 2ka À1 2ka Àa À1 ytị ẳ 1ịk Ct2k1 a1 ị etị ỵ 1ịkỵ1 Ct2k1a1 3a3 ị etị; kẳ1 Ai 1=ai s ci ; 1=ai ReðsÞ > maxðReðci ÞÞ; and the fractional part is F f sị ẳ 2p i Z  FðeÀip uÞ À Fðeip uÞ Ã du; sỵu 13ị valid for Resị > The above steps led us to realize that: k¼1 which is a real solution Integer/fractional inversion of each partial fraction The solution supplied by the approach presented above does not show the underlying structure of a TF This limitation is revealed when we try to compute its inversion by using the Bromwich integral for inverting the LT We start by fixing a branch cut line on the left complex half-plane, since the TF must be analytic on the right half plane Let us choose the left half real axis for the cut and assume that each term of the TF is continuous from above on the branch cut line As seen, it verifies lims!1 Hsị ẳ 0; j argsịj < p We will assume that lims!0 sHsị ẳ so that there is a finite initial value [3,11] Consider (6) where we illustrate a general decomposition of a TF with two pseudo-zeroes As seen the decomposition involves terms having the form: Fsị ẳ sb Y ; ðsai À ci Þ nðcj sai À ci saj ị For ẳ 1, we have no fractional component For < 1, we may have two components depending on the location of ci in the complex plane – If j argðci Þj > pai , then we not have the integer order component; it is a purely fractional system – If j argðci Þj pai , then it is mixed character system in the sense that we have both components 11ị jẳ1 ji where b is such lims!0 sFsị ẳ that lims!1 Fsị ẳ 0; j argsịj < p, and Remark We remember that a given pseudo-pole p, corresponding to an order a, is a pole, if when s ¼ jsjeih and p ¼ jpjei/ , we have jsj ¼ jpj1=a and h ¼ /=a However, we have Àp < h p and, therefore, we only obtain a pole if Àap < / ap Fig Integration path 16 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 1117 When j argci ịj ẳ p2 , the integer order component is sinusoidal; however, the fractional component exists also The stability condition comes only from the integer order component In fact, and as it is straightforward to verify, the integer order component is stable if p2 < j argðci Þj < pai , and unstable if j argðci Þj < p2 The case j argci ịj ẳ p2 corresponds to a critically stable system Concerning to the fractional part we can verify that FðeÀip uÞ À Fðeip uÞ, is a bounded function Therefore, the integral in (13) is also bounded and decreases to zero as t goes to infinite, but slowly Applying the above considerations to the general system (1) we are led to conclude that we can decompose it in two parcells with integer and fractional behaviors, namely: Integer term: it has an impulse responses corresponding to linear combinations of exponentials that, in the stable case, go to zero very fast Fractional term: they are long memory systems that exist always even there are no poles as when arguments of the pseudo-polynomial roots have absolute values greater than pa, where a is the corresponding derivative order smaller then Fig Integer part argðc1 Þ ¼ 0:71 p2 Example Consider the basic element Fsị ẳ p s1= s0:2 p : c1 ị2 ỵ iịs1= ỵ c1 s0:51 ị The Figs 2–5 illustrate the behaviour of the integer and fractional solutions for poles in both sides of the stability threshold: argðc1 Þ ¼ 0:71 p2 and argðc1 Þ ¼ 0:69 p2 , with jc1 j ¼ As expected, the fractional part does not change its behaviour: it is always stable This is in agreement with the results in [11] The instability and oscillation comes from the integer part Theorem The result stated in (13) can be generalized for any TF as n o in (1) Let Cp ¼ cj : paj < argcj ị paj ; j ẳ 1; 2; Á Á Á , be the set of the poles of the TF (of course, subset of the pseudo-poles) Then htị ẳ X ci 2Cp 1=ai Ai eci t etị ỵ 2p i Z à HðeÀip uÞ À Hðeip uÞ eÀut du Á eðtÞ: Fig Integer part argc1 ị ẳ 0:69 p2 Fig Fractional part argc1 ị ẳ 0:69 p2 Fig Fractional part argc1 ị ẳ 0:71 p2 M.D Ortigueira, G Bengochea / Journal of Advanced Research 25 (2020) 11–17 17 Compliance with Ethics Requirements This article does not contain any studies with human or animal subjects Acknowledgments This work was funded by Portuguese National Funds through the FCT – Foundation for Science and Technology under the project UIDB/00066/2020 The second author was supported by Autonomous University of Mexico City (UACM) under the project PICCyT-2019-15 References a s Fig Fractional parts of the impulse responses of systems Hsị ẳ sa1 ặ1ịs a2 ặ2ị The proof is not very difficult to obtain from the above results (see [3]) In Fig we depict the fractional parts of the response of the system in Example and another one resulting from it with the substitutions ỵ1 for and ỵ2 for As seen, the behaviour is similar, at least for large values of t Conclusions In this paper a study of non-commensurate fractional linear systems was done proposing a methodology similar to the one followed in the commensurate case For it a partial fraction decomposition was obtained using a recursive procedure Each partial fraction was inverted in two different ways: a Mittag–Leffler like procedure and a integer/fractional decomposition Some examples were presented to illustrate the proposed approach Declaration of Competing Interest The authors have declared no conflict of interest [1] Bengochea G, Ortigueira M, Verde-Star L Operational calculus for the solution of fractional differential equations with noncommensurate orders Math Methods Appl Sci 2019 [2] Hcheichi K, Bouani F Comparison between commensurate and noncommensurate fractional systems Int J Adv Comput Sci Appl (IJACSA) 2018;9(11) [3] Henrici P Applied and computational complex analysis, vol WileyInterscience; 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Firstly, we present our results related to simple fraction decomposition with non-commensurate order Then, we resolve several examples of lineal fractional systems with non-commensurate order... Computing the impulse response of some fractional linear systems In this section, in order to illustrate how to use our decomposition, we solve several fractional linear systems using the simple fraction... integer and fractional behaviors, namely: Integer term: it has an impulse responses corresponding to linear combinations of exponentials that, in the stable case, go to zero very fast Fractional