[Preprint] The paper "A Study of Disturbance Observers with Unknown Relative Degree" [1] by the authors could not include the proofs for Theorem 5 and Theorem 6 due to the page limit. We provide them in this supplementary document, and an example is included with simulation results

Jo, Nam Hoon

Joo, Youngjun

Shim, Hyungbo

Automatica

SNU Open Repository and Archive

Addendum for“AStudyofDisturbanceObserverswithUnknownRelativeDegree of thePlant”NamH. Jo a, Youngjun Joo b, and Hyungbo Shim baDepartment of Electrical Engineering, Soongsil University, Seoul, KoreabASRI, Department of Electrical and Computer Engineering, Seoul National University, Seoul, KoreaAbstractThe paper “A Study of Disturbance Observers with Unknown Relative Degree of the Plant” [1] by the authors could notinclude the proofs for Theorem 5 and Theorem 6 due to the page limit. We provide them in this supplementary document,and an example is included with simulation results.Proof of Theorem 5: We now consider the case wheremβ − mα < 0. From the Newton diagram of this case(Fig. 1), it is seen that there are two groups of vanish-ing roots of δ̄(s; τ). The first group consists of k roots ofthe form s∗(τ) = γaτ1 + o(τ1) where γa is the roots ofφa(γ) = bk + · · ·+ bk̄+1γk−k̄−1 + ak̄γk−k̄ + · · ·+ a0γk =bk + · · ·+ b0γk = γkNQ(1/γ; 1). The condition (ii) guar-antees that φa is Hurwitz. On the other hand, it is seenfrom Fig. 1 that the second group has the roots of theform s∗(τ) = γbτ(l−k)/(l+mβ−mα−k), with γb being theroots of φb(γ) = βmβ + bkαmαγl+mβ−mα−k. Note that1 ≤ l+mβ−mα−k = r.deg(Q)+r.deg(P )−r.deg(Pn) ≤2 by (i) and by r.deg(Q) ≥ r.deg(Pn). If its value isFig. 1. Newton diagram for the case r.deg(P ) < r.deg(Pn).1, then (iii) guarantees that φb is Hurwitz (of first or-der). If its value is 2 (so that l − k > 2), then two rootsof the second group are s∗(τ) = (±iγ̄ + ŝ(τ))τ (l−k)/2where γ̄ =√βmβ/(bkαmα) and ŝ is a continuous func-tion to be found such that ŝ(0) = 0. Let τ̂ = τ1/2 andA(τ̂) = iγ̄ + ŝ(τ̂2) so that s∗(τ̂2) = Aτ̂ l−k. From (4)of [1], it is seen that the power of τ̂ in each τ̂2j q̄j(Aτ̂l−k)begins with (l−k)(l+mβ−mα−j)+2j if 0 ≤ j ≤ k, andwith (l− k)(l− j) + 2j if k+ 1 ≤ j ≤ l, and increases by(l−k) in both cases. Since l−k > 2, the term of the sec-ond lowest power in the polynomial δ̄(Aτ̂ l−k, τ̂2) comesfrom the lowest power term of τ̂2(k−1)q̄k−1(Aτ̂l−k) andof τ̂2(l−1)q̄l−1(Aτ̂l−k) but not from others. Writing δ̄ inascending power of τ̂ , we haveδ̄(s∗; τ̂2) = [bkαmαA2 + βmβ ]τ̂2l+ [bk−1αmαA3 + al−1βmβA]τ̂2l−mβ+mα + · · · =[2ibkαmα γ̄ŝ+ (· · · )ŝ2]τ̂2l + [iγ̄(al−1βmβ − bk−1αmα γ̄2)+ (· · · )ŝ+ (· · · )ŝ2 + (· · · )ŝ3]τ̂2l−mβ+mα + · · ·=: δ̂(ŝ; τ̂).The corresponding Newton diagram (Fig. 2) suggeststhat it has one root ŝ∗(τ̂) of the form γ̂τ̂ c + o(τ̂ c) wherec = mα −mβ and γ̂ is the root ofφ̂(γ̂) = 2bkαmα γ̂ + (al−1βmβ − bk−1αmα γ̄2)= 2bkαmα[γ̂ +βmβ2b2kαmα(al−1bk − bk−1)].Preprint submitted to Automatica 8 April 2014Fig. 2. Newton diagram for δ̂(ŝ; τ̂)/τ̂2l under the conditionr.deg(P ) < r.deg(Pn).The condition (iv) implies that φ̂ is Hurwitz. The resultis the same with A(τ̂) = −iγ̄ + ŝ(τ̂2).Proof of Theorem 6: Conclusions of Theorem 6 areeasily derived from the proof of Theorem 5. Indeed, thecondition (d) implies that φa(γ) is not Hurwitz. On theother hand, if the condition (a) holds, then φb(γ) isnot Hurwitz because l + mβ −mα − k ≥ 3. Regardingthe condition (e), it implies that φb(γ) has at least oneRHP root. Finally, suppose that βmβ/αmα > 0 whiler.deg(Q) = r.deg(Pn) − r.deg(P ) + 2. Then, s∗(τ) is ofthe form s∗(τ) = γ̂τmα−mβ + o(τmα−mβ ). But, γ̂ is pos-itive because of the condition (f).An Illustrative Example: A numerical example isgiven to illustrate the method presented in Section 3.1.Let h.gain(P ) denote the high-frequency gain of P (s)and define sets of transfer functions (having finite coef-0 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4[sec]Output, y  P1aP1bP1c0 1 2 3 4 5 6 7 8 9 10−50510152025[sec]Control, u  P1aP1bP1cFig. 3. Simulation results for P1,a, P1,b, P1,c (plants hav-ing relative degree 1) in the presence of disturbanced(t) = sin(2πt).ficients and of minimum phase)P12 = {P (s)| 1 ≤ r.deg(P ) ≤ 2, 0.1 ≤ h.gain ≤ 8}P3 = {P (s)| r.deg(P ) = 3, 0.1 ≤ h.gain ≤ 8}P4 = {P (s)| r.deg(P ) = 4, 0.1 ≤ h.gain ≤ 8, µ(P ) ≥ 8}.It is assumed that the primary control goal is to achievezero steady-state error (to step response) with overshootless than 15% and settling time less than 6 seconds. Wewill show that, for any plant P (s) ∈ P := P12∪P3∪P4, arobust controller can be designed in order to achieve thecontrol goal. As discussed in Section 3.1, we first choosePn(s) =1s(s+ 2)(s+ 3),Q(s) =1(τs)3 + 3(τs)2 + 3(τs) + 1,(1)which guarantees that, for P ∈ P4,µ(P )− µ(Pn) ≥ 8− 5 >83=K̄pKna0a2a21. (2)Next, selectC(s) = 5 (3)so that the unity feedback control system composed ofPn(s) and C(s) achieves the primary control goal. Then,according to Theorems 3 and 5, the DOB control system(with small τ) will be stable for any P ∈ P.To verify the stability as well as the performance, thecomputer simulations are carried out using the DOBcontroller with (1), (3), and τ = 0.01. In addition, thedisturbance and the reference inputs are chosen as d(t) =sin(2πt) and r(t) = 1. For simulation purpose, we con-sider the following plants of variation:P1,a =2s+ 6, P1,b =0.2s+ 4, P1,c =5s− 1,P2,a =2(s+ 2)(s+ 4), P2,b =0.2(s+ 1)(s+ 3),P2,c =5(s+ 2)(s− 1),P3,a =1sP2,a, P3,b =1sP2,b, P3,c =1sP2,c,P4,a =1sP3,a, P4,b =1sP3,b, P4,c =s+ 1s(s+ 8)P3,c.It is observed that (a) all the plants except P4,b belong toP, (b) all the plants have different high-frequency gainsfrom Pn(s), (c) (2) is satisfied by P1,a, P2,a, P3,a, P4,a andP4,c but not by the others, and (d) P1,c, P2,c and P3,c areunstable.Fig. 3 and Fig. 4 show the simulation results for P1,a,P1,b, P1,c and P2,a, P2,b, P2,c, respectively. Although20 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4[sec]Output, y  0 1 2 3 4 5 6 7 8 9 10−505101520[sec]Control, u  P2aP2bP2cP2aP2bP2cFig. 4. Simulation results for P2,a, P2,b, P2,c (plants hav-ing relative degree 2) in the presence of disturbanced(t) = sin(2πt).there is the disturbance signal d(t), it seems that plantoutputs are not affected by d(t). In addition, it is seenthat the performance of each plant can be recovered tothat of nominal one so that the primary control goalis achieved for any plant belonging to P12. The simu-lation results for P3,a, P3,b, P3,c are depicted in Fig. 5.It is also seen that the recovery of the nominal closed-loop system performance is achieved. From Figs. 3–5,it is verified that, when r.deg(P ) = r.deg(Pn) or 1 ≤r.deg(P ) ≤ 2, the control system can be stabilized re-gardless of whether or not the condition (2) (i.e., (iii) ofTheorem 3) is satisfied.Fig. 6 shows the simulation results for P4,a, P4,b, P4,c. Itis seen that P4,a and P4,c can by stabilized by the DOBcontroller and the nominal performance is recovered. Onthe other hand, the instability occurs for P4,b /∈ P, whichindicates that the condition (2) is very critical when0 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4[sec]Output, y  0 1 2 3 4 5 6 7 8 9 10−505101520[sec]Control, u  P3aP3bP3cP3aP3bP3cFig. 5. Simulation results for P3,a, P3,b, P3,c (plants hav-ing relative degree 3) in the presence of disturbanced(t) = sin(2πt).0 1 2 3 4 5 6 7 8 9 1000.511.5[sec]Output, y  0 1 2 3 4 5 6 7 8 9 10−20−1001020[sec]Control, u  P4aP4bP4cP4aP4bP4cFig. 6. Simulation results for P4,a, P4,b, P4,c (plants hav-ing relative degree 4) in the presence of disturbanced(t) = sin(2πt).r.deg(P ) = r.deg(Pn) + 1.Finally, it should be remarked that, with the help ofthe DOB controller, plant output of any P (s) ∈ P isalmost indistinguishable from that of nominal plant inthe absence of disturbance input.References[1] N.H. Jo, Y. Joo, and H. Shim, “A Study of DisturbanceObservers with Unknown Relative Degree of the Plant,” toappear at Automatica.3

Addendum for "A Study of Disturbance Observers with Unknown Relative Degree"

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Addendum for "A Study of Disturbance Observers with Unknown Relative Degree"

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