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Journal of Control Science and Engineering
Volume 2011, Article ID 671703, 5 pages
http://dx.doi.org/10.1155/2011/671703
Research Article

Relay Feedback Analysis for Double Integral Plants

1Department of Electrical and Computer Engineering, National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260
2Institute of Complexity Science College of Automation Engineering, Qingdao University, Qingdao 266071, China
3Faculty of Mathematics and Mechanics, St. Petersburg State University, Universitetskij pr. 28, Petrodvoretz, St. Petersburg 198504, Russia

Received 18 October 2010; Revised 10 January 2011; Accepted 21 February 2011

Academic Editor: Bijoy K. Ghosh

Copyright © 2011 Zhen Ye et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Double integral plants under relay feedback are studied. Complete results on the uniqueness of solutions, existence, and stability of the limit cycles are established using the point transformation method. Analytical expressions are also given for determining the amplitude and period of a limit cycle from the plant parameters.

1. Introduction

Relay feedback forms one important class of nonlinear systems and can cause complex nonlinear behaviors. Its early analysis can be traced to 1950–1960s, and afterwards, two basic approaches, the time domain approach [1, 2] and the frequency domain approach [3], emerged. Although they are almost identical, the frequency approach is more popular because of its ease of manipulation. However, as a general method for relay analysis, such a frequency approach also has some limits in itself. Firstly, only necessary conditions on the existence and stability of possible limit cycles are obtained. Secondly, the conditions are usually expressed as a summation of infinite items. It is certainly desirable to find both sufficient and necessary conditions of the existence and stability of limit cycles for relay feedback systems, as well as to give such conditions explicitly in terms of system parameters without any requirement on numerical computation.

In our recently published paper [4], we presented a complete relay analysis for a class of servo plants, 𝐺(𝑠)=𝐾/[𝑠(𝑠+𝑎)], (𝑎>0), including the uniqueness of solutions, existence, and stability of limit cycles, and its amplitude and period, using a similar analysis to [5]. Naturally, it is desirable to investigate whether such an analysis can be extended to double integral plants, that is, the case of 𝑎=0. Most people believe that double integral plants have no limit cycle under relay feedback [1, 3], but our analysis shows that this is not true. Actually, a limit cycle for double integral case is observable, which completely depends on the system initial conditions. Moreover, we also find that the analysis for the case of 𝑎>0 in [4] is not applicable to the case of 𝑎=0. This is because for 𝑎>0, the derivative of system output is stable and bounded as 𝐺(𝑠)=𝐾/(𝑠+𝑎). While for 𝑎=0, 𝐺(𝑠)=𝐾/𝑠 and the derivative of system output becomes unstable and not bounded anymore. Hence, the extension is not straightforward and the results for double integral plants are quite different from our previous ones.

This paper is organized as follows. Section 2 gives the main result followed by the proofs and a few remarks, where more details and explanations are given. Conclusions are drawn in Section 3.

2. The Result

Consider a double integral plant 𝐺(𝑠)=𝐾𝑠2,(𝐾0),(1) whose state-space representation in the controllable canonical form is given by ̇𝑥(𝑡)=0100𝐴𝑥(𝑡)+01𝐵𝑢(𝑡),(2)𝑦(𝑡)=𝐾0𝐶𝑥(𝑡),(3) where 𝑥(𝑡)=[𝑥1(𝑡),𝑥2(𝑡)]𝑇2, 𝑦(𝑡),𝑢(𝑡) are the state, output, and input of the system, respectively. The plant is under the relay feedback control: 𝑢(𝑡)=𝑢if𝑒(𝑡)<𝜀,or𝑒(𝑡)𝜀+,𝑢𝑡=𝑢,𝑢+if𝑒(𝑡)>𝜀+,or𝑒(𝑡)𝜀,𝑢𝑡=𝑢+,(4) where 𝑢+ and 𝑢 are the relay amplitudes, 𝑒(𝑡)=𝑦(𝑡), 𝜀+ and 𝜀 are the relay hysteresis with 𝜀𝜀+. We assume 𝑢+𝑢 since otherwise (4) becomes a constant but no longer a relay control. The relay control is depicted in Figure 1. The initial function for 𝑡=𝑡0 is 𝑢𝑡0𝑢+if𝑒𝑡0>𝜀+,𝑢if𝑒𝑡0<𝜀,𝑢0𝒰if𝜀𝑒𝑡0𝜀+,(5) where 𝑡0 is the initial time and 𝒰={𝑢,𝑢+}. We call (1)–(5) a relay feedback system (RFS) which is depicted in Figure 2, where 𝐶(𝑠) represents the controller.

671703.fig.001
Figure 1: Relay function.
671703.fig.002
Figure 2: Relay feedback system.

If the RFS generates a limit cycle, let 𝑇+ and 𝐴+ be the half period and the extreme value corresponding to 𝑢(𝑡)=𝑢+, respectively, and 𝑇 and 𝐴 be the half period and the extreme value corresponding to 𝑢(𝑡)=𝑢, respectively, as shown in Figure 3. 𝑇𝑟 (resp., 𝑇𝑑) is the rising (resp., decreasing) period between the switching time, 𝑡𝑖 (resp., 𝑡𝑖+1), when the limit cycle trajectory traverses 𝒮+ (resp., 𝒮) and the peak (resp., valley) time.

671703.fig.003
Figure 3: Limit cycles.

We are now in a position to state the sufficient and necessary conditions for the existence of solutions, the existence and stability of limit cycles, and the amplitudes and periods of limit cycles.

Theorem 1. Consider the RFS for double integral plant 𝐺(𝑠) in (1). (i)A unique solution exists for any initial condition.(ii)A limit cycle exists if and only if 𝐾𝑢+>0>𝐾𝑢, 𝜀+=𝜀=𝜀 and any of the following holds: (a)𝑥1(𝑡0)𝜀/𝐾;(b)𝑥1(𝑡0)=𝜀/𝐾 but 𝑥(𝑡0)0. If this is the case, the limit cycle is formed after the first switch, which is unique with two switchings per period.(iii)If a limit cycle exists, it is globally marginally stable. (iv)If a limit cycle exists, its amplitude and period are described by 𝐴+=𝐾𝑥22𝑡02𝑢𝑡0𝐾𝑥1𝑡0+𝜀2𝑢+,𝐴=𝐾𝑥22𝑡02𝑢𝑡0𝐾𝑥1𝑡0+𝜀2𝑢,𝑇+=2𝐾2𝑥22𝑡02𝐾𝑢𝑡0𝐾𝑥1𝑡0+𝜀𝐾𝑢+,𝑇=2𝐾2𝑥22𝑡02𝐾𝑢𝑡0𝐾𝑥1𝑡0+𝜀𝐾𝑢.(6)

Proof. For a RFS, the input 𝑢(𝑡) is a piecewise constant function. Without loss of generality, suppose 𝑢(𝑡)=𝑢𝟏𝑡𝑡0+𝑖=1(1)𝑖1𝑢+𝑢𝟏𝑡𝑡𝑖,𝑡𝑡0,,𝑡0<𝑡1<,(7) as shown in Figure 4, where 𝟏(𝑡)=1,𝑡0,0,𝑡<0.(8)Define the switching planes: 𝒮+=𝜉2𝐶𝜉=𝜀+,𝒮=𝜉2𝐶𝜉=𝜀.(9) If the trajectory of 𝑥(𝑡) traverses 𝒮+ (resp., 𝒮), that is, 𝐶𝑥(𝑡𝑖)=𝜀+ (resp., 𝐶𝑥(𝑡𝑖)=𝜀) at some instant 𝑡=𝑡𝑖>𝑡0 with 𝐶𝑥(𝑡𝑖)<𝜀+ (resp., 𝐶𝑥(𝑡𝑖)>𝜀) and 𝐶𝑥(𝑡+𝑖)>𝜀+ (resp., 𝐶𝑥(𝑡+𝑖)<𝜀), then the instant 𝑡=𝑡𝑖 is called a switching time. In particular, 𝑡𝑖 denotes the switching time when the 𝑖th switching takes place.
The state response of (2) to 𝑢(𝑡) in (7) is given by 𝑥(𝑡)=e𝐴(𝑡𝑡0)𝑥𝑡0+𝑡𝑡0e𝐴(𝑡𝜏)𝐵𝑢(𝜏)d𝜏=1𝑡𝑡001𝑥𝑡0+𝑡𝑡022𝑡𝑡0𝑢𝟏𝑡𝑡0+𝑖=1(1)𝑖1𝑡𝑡𝑖22𝑡𝑡𝑖𝑢+𝑢𝟏𝑡𝑡𝑖,(10) where 𝑡[𝑡0,), 𝑡0<𝑡1<. It is easy to show that both 𝑥1(𝑡) and 𝑥2(𝑡) are continuous at switching time 𝑡𝑖,𝑖=1,2,, as 𝑡𝑡𝑖=0 for 𝑡=𝑡𝑖. However, ̇𝑥2(𝑡1)=lim𝑡𝑡1̇𝑥2(𝑡)=𝑢, while ̇𝑥2(𝑡+1)=lim𝑡𝑡+1̇𝑥2(𝑡)=̇𝑥2(𝑡1)=𝑢+(𝑢+𝑢)=𝑢+ indicating ̇𝑥2(𝑡) is not continuous at 𝑡=𝑡1 due to 𝑢+𝑢 by our assumption. Straightforwardly, this discontinuity persists for the general case of 𝑡=𝑡𝑖.
If considering the time between two consecutive switchings only, that is, 𝑡[𝑡𝑖,𝑡𝑖+1), (10) is simplified as 𝑥1(𝑡)=𝑥1𝑡𝑖+𝑡𝑡𝑖𝑥2𝑡𝑖+𝑡𝑡𝑖22𝜇,𝑥2(𝑡)=𝑥2𝑡𝑖+𝑡𝑡𝑖𝜇,(11) where 𝑖=0,1,2,, and 𝜇=𝑢+if𝑖isodd,𝑢if𝑖iseven.(12) Since 𝑒(𝑡)=𝑦(𝑡)=𝐾𝑥1(𝑡) and 𝑒(𝑡)=𝑦(𝑡)=𝐾𝑥2(𝑡), it yields 𝑒(𝑡)=𝑒𝑡𝑖+𝑡𝑡1𝑒𝑡𝑖𝑡𝑡𝑖22𝐾𝜇,(13)𝑒(𝑡)=𝑒𝑡𝑖𝑡𝑡𝑖𝐾𝜇.(14)
It can be seen from (13) that 𝑒(𝑡𝑖), 𝑒(𝑡𝑖) and 𝐾𝜇 will affect behavior of 𝑒(𝑡) and thus determine whether or not 𝑒(𝑡) will reach the switching level of 𝜀+ or 𝜀. Without loss of generality, suppose the initial condition 𝑒(𝑡0)𝜀+, thus 𝑢(𝑡0)=𝑢. To see how 𝑒(𝑡) evolves with time, the following four cases are considered which are mutually exclusive and cover all possible cases.
Case 1. 𝑒(𝑡0)0 and 𝐾𝑢0. For 𝑡[𝑡0,𝑡1), 𝑒(𝑡)=𝑒(𝑡0)(𝑡𝑡0)𝐾𝑢0, which implies 𝑒(𝑡)𝑒(𝑡0)𝜀+ so that 𝑥(𝑡) never traverses 𝒮+ for all 𝑡>𝑡0. The trajectory of 𝑥(𝑡) is governed by (11) with 𝑡𝑖=𝑡0 and 𝜇=𝑢.Case 2. 𝑒(𝑡0)0 and 𝐾𝑢<0. For 𝑡[𝑡0,𝑡1), it follows from (14) that 𝑒(𝑡)=𝐾𝑢>0, which implies there always exists some 𝑡1>𝑡0 such that 𝑒(𝑡1)=𝜀+ and 𝑒(𝑡1)>0, that is, 𝑥(𝑡) traverses 𝒮+ at 𝑡=𝑡1. For 𝑡[𝑡1,𝑡2), it follows from (14) that 𝑒(𝑡)=𝑒(𝑡1)(𝑡𝑡𝑖)𝐾𝑢+>0 if 𝐾𝑢+0, which implies 𝑒(𝑡)>𝑒(𝑡1)=𝜀+𝜀 so that 𝑥(𝑡) never traverses 𝒮 for all 𝑡>𝑡1. The trajectory of 𝑥(𝑡) is governed by (13) and (14) with 𝑡𝑖=𝑡1 and 𝜇=𝑢+. If 𝐾𝑢+>0, it follows from (14) that 𝑒(𝑡)=𝐾𝑢+<0, which implies there always exists some 𝑡2>𝑡1 with 𝑒(𝑡2)=𝜀 and 𝑒(𝑡2)<0 so that 𝑥(𝑡) traverses 𝒮 at 𝑡=𝑡2. Afterwards, with the same analysis as above, it is straightforward to verify that 𝑥(𝑡) will traverse 𝒮+ and 𝒮 alternatively and consecutively.Case 3. 𝑒(𝑡0)>0 and 𝐾𝑢<0. For 𝑡[𝑡0,𝑡1), it follows from (14) that 𝑒(𝑡)=𝑒(𝑡𝑖)(𝑡𝑡𝑖)𝐾𝑢>0, which implies there always exists 𝑡1>𝑡0 such that 𝑒(𝑡1)=𝜀+ and 𝑒(𝑡1)>0, that is, 𝑥(𝑡) will traverse 𝒮+ at 𝑡=𝑡1. From 𝑡=𝑡1 onwards, following the same analysis in Case 2, if 𝐾𝑢+0, 𝑥(𝑡) never traverses 𝒮 for all 𝑡>𝑡1, the trajectory of 𝑥(𝑡) is governed by (13) and (14) with 𝑡𝑖=𝑡1 and 𝜇=𝑢+. Otherwise, if 𝐾𝑢+>0, 𝑥(𝑡) will traverse 𝒮 and 𝒮+ alternatively and consecutively.Case 4. 𝑒(𝑡0)>0 and 𝐾𝑢0. For 𝑡[𝑡0,𝑡1), it follows from (14) that 𝑒(𝑡)=𝐾𝑢0. If 𝑒(𝑡)<0, there always exists 𝑡0𝑚 such that 𝑒(𝑡)𝑒(𝑡0𝑚) for all 𝑡𝑡0. If 𝑒(𝑡0𝑚)𝜀+, then 𝑒(𝑡)𝑒(𝑡𝑚)𝜀+ so that 𝑥(𝑡) never traverses 𝒮+ for 𝑡𝑡0. The trajectory of 𝑥(𝑡) is governed by (13) and (14) with 𝑡𝑖=𝑡0 and 𝜇=𝑢. On the contrary, if 𝑒(𝑡𝑚)>𝜀+, there always exists 𝑡1 satisfying 𝑡0<𝑡1<𝑡0𝑚 such that 𝑒(𝑡1)=𝜀+ and 𝑒(𝑡1)>0, that is, 𝑥(𝑡) traverses 𝒮+ at 𝑡=𝑡1. If 𝑒(𝑡)=0, then 𝑒(𝑡)=𝑒(𝑡0)>0, which also implies 𝑥(𝑡) traverses 𝒮+ at some 𝑡=𝑡1. For 𝑡[𝑡1,𝑡2), it follows from (14) that 𝑒(𝑡)=𝑒(𝑡1)(𝑡𝑡𝑖)𝐾𝑢+>0 if 𝐾𝑢+0, which implies 𝑒(𝑡)>𝑒(𝑡1)=𝜀+𝜀 so that 𝑥(𝑡) never traverses 𝒮 for all 𝑡>𝑡1. The trajectory of 𝑥(𝑡) is governed by (13) and (14) with 𝑡𝑖=𝑡1 and 𝜇=𝑢+. If 𝐾𝑢+>0, it follows from (14) that 𝑒(𝑡)=𝐾𝑢+<0, which implies there always exists some 𝑡2>𝑡1 such that 𝑒(𝑡2)=𝜀 and 𝑒(𝑡2)<0, that is, 𝑥(𝑡) traverses 𝒮 at 𝑡=𝑡2. For 𝑡[𝑡2,𝑡3), it follows from (14) that 𝑒(𝑡)=𝑒(𝑡2)𝐾𝑢<0, which implies 𝑥(𝑡) never traverses 𝒮+ again for all 𝑡>𝑡2. The trajectory of 𝑥(𝑡) is governed by (14) with 𝑡𝑖=𝑡2 and 𝜇=𝑢.
In view of the above analysis, the solution to (2) always exists. This completes (i).
We can also see from the analysis in (i) that consecutive switchings between 𝒮+ and 𝒮 take place if and only if 𝐾𝑢+>0>𝐾𝑢. Under this condition, let 𝑡𝑖 be the switching instant of 𝑥(𝑡) on the plane 𝒮+, that is, 𝑒(𝑡𝑖)=𝜀+ and 𝑒(𝑡)<𝜀+, 𝑢(𝑡)=𝑢 for sufficiently close 𝑡<𝑡𝑖. This implies 𝑒(𝑡𝑖)0. The case 𝑒(𝑡𝑖)=0 can occur only when 𝜀=𝜀+ and then (𝜀+,0) is a fixed point of 𝑥(𝑡). But a fixed point is unique. Hence, no limit cycle exists in this case.Assume 𝑒(𝑡𝑖)>0, then 𝑢(𝑡) switches from 𝑢 to 𝑢+ at 𝑡=𝑡𝑖. In Case 2, we have shown that 𝑒(𝑡) decreases and becomes negative until 𝑥(𝑡) traverses 𝒮 at 𝑡=𝑡𝑖+1. Consider the Poincare mapping 𝜒=𝑒𝑡𝑖𝜙+(𝜒)=𝑒𝑡𝑖+1.(15) The function 𝜙+ is defined on the ray [0,+) and takes values in the ray (,0]. Let 𝜒=𝑒(𝑡𝑖)>0 and 𝜏=𝑡𝑖+1𝑡𝑖. By definition, 𝑒(𝑡𝑖)=𝜀+ and 𝑒(𝑡𝑖+1)=𝜀. Since both 𝑒(𝑡) and 𝑒(𝑡) are continuous at 𝑡=𝑡𝑖+1, it follows from (13) and (14), respectively, by taking 𝑡=𝑡𝑖+1 that 𝜀=𝜀++𝜏𝜒𝜏22𝐾𝑢+,(16)𝜙+(𝜒)=𝜒𝜏𝐾𝑢+.(17) It follows from (17) that 𝐾𝑢+=𝜒𝜙+(𝜒)𝜏.(18) Substituting (18) into (16) yields 𝜙+(𝜒)=𝜒+2𝜀𝜀+𝜏𝜒<0,(19) which implies ||||𝜙+(𝜒)𝜒||||1.(20) If 𝜀+>𝜀, “>” holds in (20) so that the Poincare mapping (15) is not a contraction and no limit cycles exist. If 𝜀=𝜀+=𝜀, “=” holds in (20), which implies a limit cycle is formed after the first switching. As 𝑒(𝑡𝑖)>0, 𝑥(𝑡1)0 must be satisfied so that (a) and (b) can be easily derived with the help of (11). This completes (ii).
Once a limit cycle is formed, if 𝑒(𝑡𝑖) is changed due to some disturbance, 𝑥(𝑡) cannot remain at the original limit cycle, but form a new one with different amplitude and period, which means the limit cycle is marginally stable. Since the initial condition 𝑥(𝑡0) is arbitrarily chosen, the limit cycle is globally marginally stable. This completes (iii).
It has been shown in (ii) that for 𝐾𝑢+>0>𝐾𝑢, a limit cycle always exists after the first switching. For 𝑡[𝑡1,𝑡2), it follows from (14) and with the help of (20) that 𝑒𝑡2=𝑒𝑡1𝑇+𝐾𝑢+=𝑒𝑡1.(21) Thus, 𝑇+=2𝑒𝑡1𝐾𝑢+,𝑒𝑡1>0.(22) Let 𝑒(𝑡)=𝑒(𝑡1)(𝑡𝑡1)𝐾𝑢+=0, then 𝑡𝑡1=𝑒(𝑡1)/(𝐾𝑢+). Substituting it into (13) yields 𝐴+=𝑒(𝑡)𝜀=𝑒𝑡122𝐾𝑢+.(23) In the similar way, we have 𝑇=2𝑒𝑡1𝐾𝑢,(24)𝐴=𝑒𝑡122𝐾𝑢.(25) For 𝑡[𝑡0,𝑡1), it follows from (13) that 𝑒𝑡1=𝑒𝑡0+𝑡𝑡0𝑒𝑡0𝑡𝑡022𝐾𝑢𝑡0=𝜀,(26) which gives 𝑡𝑡0=𝑒𝑡0𝑒𝑡022𝐾𝑢𝑡0𝜀𝑒𝑡0𝐾𝑢𝑡0,(27) where negative root is ignored since 𝑒(𝑡1)>0. Substituting (27) into (14) yields 𝑒𝑡1=𝑒𝑡022𝐾𝑢𝑡0𝜀𝑒𝑡0=𝐾2𝑥22𝑡02𝐾𝑢𝑡0𝐾𝑥1𝑡0+𝜀.(28) Substituting (28) into (22)–(24), respectively, yields 𝐴+=𝐾𝑥22𝑡02𝑢𝑡0𝐾𝑥1𝑡0+𝜀2𝑢+,𝐴=𝐾𝑥22𝑡02𝑢𝑡0𝐾𝑥1𝑡0+𝜀2𝑢,𝑇+=2𝐾2𝑥22𝑡02𝐾𝑢𝑡0𝐾𝑥1𝑡0+𝜀𝐾𝑢+,𝑇=2𝐾2𝑥22𝑡02𝐾𝑢𝑡0𝐾𝑥1𝑡0+𝜀𝐾𝑢,(29) which completes (iv).

671703.fig.004
Figure 4: A piecewise constant input.

Remark 1. In our previous relay analysis for a class of servo plant as 𝐺(𝑠)=𝐾/(𝑠(𝑠+𝑎)), (𝑎>0), we proved the existence of a limit cycle by showing the Poincare map 𝜙+ is a strict contraction, that is, |𝜙+(𝜒)|<1 for all 𝜒(0,+). Here for the case of 𝑎=0, differentiating both sides of (16) yields 𝜏(𝜒)=𝜏𝜏𝐾𝑢+𝜒.(30) Differentiating both sides of (16) and using the help of (20) yield ||𝜙+(𝜒)||=||1𝜏(𝜒)𝐾𝑢+||=||||𝜒𝜒𝜏𝐾𝑢+||||=||||𝜒𝜙+(𝜒)||||1.(31) If 𝜀+𝜀, |𝜙+(𝜒)|<1 for all 𝜒(0,+), but no limit cycle exists in this case as we show. It seems to be contradiction there and the reason is that for 𝑎=0, 𝜙+(𝜒) is not bounded any more. This also shows the previous results for 𝑎>0 cannot be simply extended to the double integral case here by just letting 𝑎0.

Remark 2. The limit cycle forms after the first switching but is marginally stable. This means the trajectory will not remain at the original limit cycle even if a very small disturbance is introduced. Nevertheless, it will form a new limit cycle with different amplitude and period immediately after the next switching.

Remark 3. The limit cycle is a parabolic curve with the symmetry of 𝑥(𝑡𝑖)=𝑥(𝑡𝑖+1), where 𝑡𝑖 is the switching time for 𝑖=1,2,. Its amplitude and period are only decided by the initial conditions, which is also consistent with the result from [6].

Example 1. Consider a double integral plant 𝐺(𝑠)=1/𝑠2 under the relay feedback with 𝜀=1, 𝑢+=1.0 and 𝑢=0.8. The initial conditions are assumed to be 𝑥1(𝑡0)=0.5 and 𝑥2(𝑡0)=1. The simulated response of the system is shown in Figure 5. By Theorem 1(iv), the amplitude and period of the limit cycle occurred are also calculated. The comparison of the results from the theory and the simulation as shown in Table 1, where one can see the error, is quite minor.

tab1
Table 1: Limit cycle characteristics of Example 1.
671703.fig.005
Figure 5: A piecewise constant input.

3. Conclusion

In this paper, a double integral plants under relay feedback is addressed. Complete results have been established on uniqueness of solutions, existence, and stability of limit cycles and its amplitude and period. Comparing with the plant 𝐺(𝑠)=𝐾/[𝑠(𝑠+𝑎)], 𝑎>0 in [4], analytical expressions for the amplitude and period of limit cycles with terms of system parameters are available, but the conditions on the existence of limit cycles are quite different. Reasons for such differences as well as some explanations are provided.

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