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STABILIZATION OF SECOND-ORDER UNSTABLE DELAY PROCESSES BY PID CONTROLLERS
STABILIZATION OF SECOND-ORDER UNSTABLE DELAY PROCESSES BY PID CONTROLLERS Qing-Guo Wang ∗,1 Xiang Cheng ∗ Xiang Lu ∗ ∗
ECE Dept., National University of Singapore, Singapore
Abstract: Based on the well known Nyquist stability criterion, the stabilization of second-order unstable time-delay processes by PID controllers is investigated. Explicit stabilizability results in terms of the upper limit of time-delay size is established. The analysis not only provides both theoretical understanding on such stabilization issue, but also meaningful for the teaching of Nyquist stability criterion in control eduction. Keywords: Nyquist plot, stabilization, unstable time-delay Processes, PID controller
1. INTRODUCTION Time-delay is commonly encountered in chemical, biological, mechanical and electronic systems. There are some unstable processes in industry such as chemical reactors and their stabilization is essential for successful operations. Especially, unstable processes coupled with time delay makes control system design a difficult task, which has attracted increased attention from control community (Chidambaram, 1997). Recently, many techniques (Shafiei and Shenton, 1994; Wang et al., 1999) have been reported to improve PID tuning for unstable delay processes. However, these works do not provide a clear scenario on what kind of process could be stabilized by PID controllers. Silva et al. (2004) investigated the complete set of stabilizing PID controllers based on the HermiteBiehler theorem for quasi-polynomials. Hwang and Hwang (2004) applied D-partition method to characterize the stability domain in the space of system and controller parameters. However, these approaches are mathematically involved, and the 1
Author to whom all correspondence should be addressed, Email: [email protected], Tel: (+65) 6874 2282, Fax: (+65) 6779 1103
maximal stabilizable time delay for some typical yet simple processes still remains obscure. In this paper, we focus on the second-order unstable time-delay process and provide a thorough study for its stabilization based on the well-known Nyquist stability criterion. The maximum stabilizable delay is derived and the solution for corresponding stabilizing controller is given. The paper is organized as follows. After the problem statement in Section 2, some preliminaries are introduced in Section 3. Then the stabilization analysis is addressed in Section 4. Finally, Section 5 concludes the paper. 2. PROBLEM STATEMENT Suppose that the process of interest is controlled in the unity feedback system (Figure 1) with PID controller or its special cases, namely, P, PI, PD. Normalized process is adopted throughout the paper for uniform formulation, which could be derived by scaling down the time-delay and all time constants by T1 and absorbing the process gain ¯ into the controller. The normalized transfer K functions of process and controller are expressed as
G(s) =
1 e−Ls , (s − 1)(T s + 1)
(1)
and
KI ) s respectively. In the rest of the paper, stabilization C(s) = KP (1 + KD s +
R (s)
+
E (s)
−
C (s)
G (s)
Y (s)
Fig. 1. Unity output feedback system of (1) is addressed with one of the following four controllers:C1 (s) = KP , C2 (s) = KP (1 + KsI ), C3 (s) = KP (1 + KD s), and C4 (s) = KP (1 + KD s+ KsI ). The corresponding open-loop transferfunction, Ql (s) = G(s)Cl (s), l ∈ {1, 2, 3, 4}, is re-written as Ql (s) = Gi (s)Cl (s) =
KN (s) −Ls e , L > 0, (2) sv D(s)
where K is the gain, v a non-negative integer representing type of the loop, N (s) and D(s) both rational polynomials of s with N (0) = D(0) = 1. If the loop has a pole at the origin, then the Nyquist contour needs to be modified by replacing the origin with a infinitely small semicircle of s = rejφ with r → 0 and −π/2 < φ < π/2. The Nyquist stability theorem is now applied loop (2). Theorem 1. Given the loop transfer function Ql (s) in (2) with P + poles inside the Nyquist contour, the closed-loop system in Figure 1 is stable if and only if the Nyquist plot of Ql encircles the critical point, (−1, 0), P + times anticlockwise.
Due to the delay element in the open-loop Ql (s) (2), the phase of Ql (jω), ΦQl (ω), approaches −∞ when frequency ω → ∞. Consequently, if limω→∞ |Ql (jω)| ≥ 1, the Nyquist curve of Ql (s) will encircle/pass the critical point infinite times clockwise, which violates Theorem 1 and the closed-loop is unstable. Hence, the following lemma follows. Lemma 1. For the open-loop Ql (s) in (2), lim |Ql (jω)| < 1
In contrast, if K < −1, Ql (0) is now left to the critical point. This causes Ql (jω) + 1 to have the total phase change of (2m + 1)π with an integer m for the frequency from zero to positive infinity, and thus an odd number of times for the entire frequency range, making an unstable closed-loop for P + = 0. Therefore, K > −1 is necessary for stability if P + = 0. The case that the loop has one integrator (v = 1) can be analyzed similarly. And K > 0 is necessary for stability if P + = 0, while K < 0 is necessary for stability if P + = 1. Lemma 2. Consider the open-loop Ql (s) in (2), the necessary condition for closed-loop stability is (i). For v = 0, K > −1 if P + = 0; and K < −1 if P + = 1. (ii). For v = 1, K > 0 if P + = 0; and K < 0 if P+ = 1 . Lemma 3. Given the open-loop transfer function Ql (s) described in (2), a necessary condition for the closed-loop stability is: polynomial H(s) =
dm+1 v Ls dsm+1 [s D(s)e ] eLs
(4)
has all its zeros lie in the open left half plane, where m is the degree of N (s).
3. PRELIMINARY
ω→∞
If K > −1, Ql (0) is also right to the critical point. This causes Ql (jω) + 1 to have the total phase change of an integral multiple (zero included) of 2π for entire positive frequency range. Thus the Nyquist curve encircles the critical point an even number of times for the entire frequency range, making an unstable closed-loop for P + = 1. Therefore, K < −1 is necessary for stability if P + = 1.
(3)
is necessary for the closed-loop stability. Suppose first that the loop has no integrator (v = 0). Then Ql (0) = K is finite. The Nyquist curve starts at Ql (0) = K and, |Ql (j∞)| < 1 due to (3), should end right to the critical point, (−1, 0), to meet Theorem 1 for stability.
Proof: The stability of characteristic function F0 (s) = sv D(s) + KN (s)e−Ls is equivalent to that of F1 (s) = sv D(s)eLs + KN (s). It follows from the results in Kharitonov et al. (2005) that H(s)eLs , the m + 1 derivative of F1 (s), is also stable. Therefore H(s) is stable, or its zeros lie in the open left half plane. Lemma 4. Given the open-loop transfer function Ql (s) with P + > 0, and any integer k, if (i). ΦQl (ω) < −2kπ + 3π, and dΦQl (ω) < 0 for ΦQl (ω) ≤ −2kπ − π (ii). dω hold for ω ≥ 0, then the closed-loop system is stable only if ¡ ¢ max ΦQl (ω)|ω>0 > −2kπ + π. (5) Proof: Anti-clockwise encirclement is not obtainable for the portion of the loop Nyquist curve
corresponding either to s = rejφ with r → 0 since possible poles of Ql (s) would cause the curve to rotate clockwise only, or to s = jw which meet (ii) as its phase keeps decreasing. Taking into account (i), it can occur only if the curve has the phase increase in the phase range of −2kπ − π < ΦQl (ω) < −2kπ + 3π, and traverses the negative real axis from the third quadrant to the second quadrant therein, that is, there holds (15). The proof is complete. In the following section, the stabilization analysis is provided using these Lemmas. 4. SECOND-ORDER UNSTABLE PROCESS WITH A STABLE POLE 4.1 P/PI controller For P controller, C1 (s) = KP , the open-loop frequency response is given by Q1 (jω) =
KP e−jLω , (jω − 1)(jT ω + 1)
with P + = 1 and v = 0. By Lemma 2(i), K = −KP < −1 is necessary. Then the magnitude is p MQ1 (ω) = KP / (1 + ω 2 )(1 + T 2 ω 2 ), which decreases from KP to 0. The phase is ΦQ1 (ω) = −Lω+arctan(ω)−arctan (T ω)−π, (6) with its first and second order derivatives as d 1 T ΦQ1 (ω) = −L + − , 2 dω 1+ω 1 + T 2 ω2 2ω 2T 3 ω d2 ΦQ1 (ω) = − + . 2 2 2 dω (1 + ω ) (1 + T 2 ω 2 )2 For ω > 0, there holds ΦQ1 (0) = −π and ΦQ1 < −π/2. Moreover, for ΦQ1 ≤ −3π, dΦQ1 /dω is always negative. It follows from Lemma 4 that Φ51 (ω) > −π for some ω > 0 is necessary for closed-loop stability. And this requires 2 2 dΦQ1 (ω)/dω > 0 for some ω. Let = √ √ d ΦQ1 (ω)/dω 2 2 0 yields ω1 = 0, and ω2 = (T T + T + T )/T 2 . d ΦP (ω) becomes Then the maximum value for dω 1 − L − T, ¶ µ ω = ω1 ,√0 < T < 1; d ΦQ1 (ω) = max T T −1 √ √ dω − L, T T +T + T +1 ω = ω2 , T ≥ 1 If T ≥ 1, it is easy to check from (6) that ΦQ1 (ω) < −π for ω > 0, and thus the closed-loop is always unstable. If 0 < T < 1, the stabilization d ΦQ1 (ω) turns to be requirement for dω d ΦQ1 (ω)|ω=ω1 = 1 − L − T > 0, or L < 1 − T. dω (7) In this case, the phase will increase from −π first and then decrease, and there is one and
only one intersection with the negative real axis with ΦQ1 = −π. In order for the anticlockwise encirclement of critical point to occur, this intersection should lie between −1 and 0, that is MQ1 (ωc1 ) < 1, ΦQ1 (ωc1 ) = −π, or equivalently q 2 )(1 + T 2 ω 2 ). (8) 1 < KP < (1 + ωc1 c1 And for ω > ωc1 , MQ1 is always less than 1 so that there is no encirclement around the critical point thereafter. Consequently, there is exactly one anticlockwise encirclement when L < 1 − T and (8) are all true. As for PI controller, C2 (s) = KP (1 + open-loop frequency response is Q2 (jω) = KP
KI s ),
1 − j KωI . (jω − 1)(jT ω + 1)
the (9)
It follows from Lemma 2(ii) that KP KI > 0 is necessary for closed-loop stability. Assume KP > 0 and KI > 0 first, then the loop has its magnitude as s ¡ ¢2 1 + KωI MQ2 = KP , (1 + ω 2 )(1 + T 2 ω 2 ) which always decreases from ∞ to 0. The phase of (9) is ¶ µ KI −arctan(T ω)−π, ΦQ2 = −Lω+arctan(ω)−arctan ω (10) with its derivative being KI T 1 dΦQ2 ω2 = −L + + . ¡ KI ¢2 − dω 1 + ω2 1 + (T ω)2 1+ ω
It follows that ΦQ2 < −π/2. Moreover, when ΦQ2 ≤ −3π, the derivative of phase is always negative. In consequence, anticlockwise encirclement is possible only when there exists some ω > 0 such that ΦQ2 (ω) > −π by invoking Lemma 4. In case of L ≥ 1 − T , it is readily seen from the previous P-control discussion that ΦQ2 = ΦQ1 − arctan(KI /ω) ≤ ΦQ1 , ΦQ1 and then ΦQ2 are always less than −π. In consequence, the Nyquist curve has no anticlockwise encirclement around the critical point and the closed-loop is unstable when KP > 0, KI > 0 and L ≥ 1. In case of L < 1 − T , ΦQ1 > −π holds from some small ω, and it is always possible to find ΦQ2 (ω) > −π by reducing KI and in turn arctan(KI /ω). Thus KI should be chosen to ensure max(ΦQ2 (ω)|ω>0 ) > −π, which is not empty. Then the Nyquist curve will have two crossings with the negative real axis with phase angle −π. In order to have anticlockwise encirclement around the critical point, KP should be chosen such that MQ2 (ωc2 ) < 1 < MQ2 (ωc1 ),
(11)
where 0 < ωc1 < ωc2 are the two phase crossover frequencies satisfying ΦQ32 (ω) = −π. Inequality
(11) is always feasible since MQ2 is monotonically decreasing.
Nyquist Diagram
Nyquist Diagram
0.2
0.25
0.2
0.15
0.15 0.1 0.1
Imaginary axis
0.05 Imaginary axis
Assume KP < 0 and KI < 0 then, the phase is ¶ µ KI −arctan(T ω), ΦQ2 = −Lω+arctan(ω)−arctan ω
0
−0.05
1 Theorem 2. The process, G5 (s) = (T s+1)(s−1) e−Ls , is stabilizable by P controller (C1 (s) = KP ) or PI controller (C2 (s) = KP (1 + KsI )) if and only if 0 < L < 1 − T . If 0 < L < 1 − T , the stabilizing gain for P controller is bounded by q 2 )(1 + T 2 ω 2 ). (12) 1 < KP < (1 + ωc1 c1
with −Lωc1 + arctan(ωc1 ) − arctan (T ωc1 ) = 0. And the stabilizing parameters for PI controller satisfy KP > 0, KI > 0. (13) KI is chosen such that max(ΦQ2 (ω)|ω>0 ) > −π,
(14)
and the range of KP is given by v u (1 + ω 2 )(1 + T 2 ω 2 ) u c2 c2 < KP t ³ ´2 KI 1 + ωc2 v u (1 + ω 2 )(1 + T 2 ω 2 ) c1 c1 .
4.2 PD/PID controller For PD controller, C3 (s) = KP (1 + KD s), the open-loop frequency response is 1 + jKD ω , (15) Q3 (jω) = KP (jω − 1)(jT ω + 1)
0
−0.05
−0.1 −0.1 −0.15 −0.15
which is always less than π/2. Moreover, for ΦQ2 ≤ −π, its derivative is always negative. It follows from Lemma 4 that Q2 (jω) has no anticlockwise encirclement, and that the closedloop is unstable when KP < 0 and KI < 0.
0.05
−0.2 −1.6
−0.2
−1.4
−1.2
(a) G5 = C1 = 1.5
−1
−0.8 −0.6 Real Axis
−0.4
−0.2
e−0.3s (0.5s+1)(s−1)
0
0.2
−0.25 −1.4
−1.2
−1
−0.8
−0.6 Real Axis
−0.4
−0.2
e−0.3s (0.5s+1)(s−1)
and (b) G5 = C2 = 1.3 + 0.026/s
0
0.2
and
Fig. 2. Nyquist plots of G5 with P and PI controllers with P + = 1 and v = 0. It follows from Lemma 2(i) that K = KP > 0 is necessary for stabilization. Then the magnitude and phase are s 2 ω2 1 + KD , (16) MQ3 = KP (1 + ω 2 )(1 + T 2 ω 2 ) and ΦQ3 = −Lω+arctan(ω)+arctan(KD ω)−arctan(T ω)−π, (17) respectively, with ΦQ3 (0) = −π and ΦQ3 < 0 for ω > 0. Notice that µ 2 ¶ M d KQ23 2 4 2 P 2ω(T 2 KD ω + 2T 2 ω 2 + 1 + T 2 − KD ) = , 2 2 2 2 2 dω (1 + ω ) (1 + T ω ) (18) 2 2 it follows that if (1+T −K ) > 0, or equivalently D √ KD < 1 + T 2 , then dMQ3 /dω < 0 always holds, and MQ3 decreases from KP to 0 when ω √ increases 1 + T 2, from 0 to ∞. Otherwise, if KD > then dMQ3 /dω is positive when ω is small and turns negative when ω increases, so that MQ3 increases from KP first and then decreases to 0 as ω increases. As for the phase, one sees that ΦQ3 (0) = −π and 1 KD T dΦQ3 = −L + + 2 ω2 − 1 + T 2 ω2 dω 1 + ω2 1 + KD (19) with ¯ dΦQ3 ¯¯ = −L + 1 + KD − T. (20) dω ¯ω=0 When ΦQ3 ≤ −3π, the derivative of phase is always negative. It follows from Lemma 4 the closed-loop is stable only if max (ΦQ3 (ω)) > −π. According to the combinations of signs of (1+T 2 − 2 KD ) and (−L+1+KD −T ), the stabilization issue is addressed by four cases correspondingly. Case A In this case, p ½ KD < 1 + T 2 , 1 + KD − T − L > 0 and this leads to L<
p 1 + T 2 − T + 1.
(21)
(22)
Given arbitrary L satisfies (22), KD is chosen within the range
L + T − 1 < KD <
p 1 + T 2,
(23)
which is not empty. Since dΦQ3 (0)/dω > 0, the stabilization is possible. In order for the correct encirclement to occur, the first intersection of Nyquist curve with the real axis for positive frequency should between −1 and 0. It follows that MQ3 (ωc1 ) < 1, which leads to s 2 )(1 + T 2 ω 2 ) (1 + ωc1 c1 , (24) 1 < KP < 2 ω2 1 + KD c1 combined with the requirement KP > 1. (24) is also not empty since dMQ3 /dω < 0. Moreover, when (24) is true, MQ3 (ω) is always less than 1 for ω > ωc1 , and Q3 does not encircle the critical point for ω > ωc1 . Consequently, PD controller could always stabilize process (1) under case A when (22), (23) and (24) are all true. In the rest of this subsection, it is demonstrated that PD controller could not stabilize process (1) if √ L ≥ 1 + T 2 − T + 1. Case B In this case, p ½ KD > 1 + T 2 , (25) 1 + KD − T − L > 0 √ 1 + T 2 − T + 1 is assumed. For and L ≥ convenience of analysis, let p δ KD = 1 + T 2 + δ 2 , δ > 0 and ω0 = T also denote by ωcp the smallest positive frequency that ΦQ3 (ωcp ) = −π. Let dΦQ3 /dω = 0, it follows from (17) that a3 x3 + a2 x2 + a1 x + a0 = 0,
(26)
2 2 with x = ω 2 , a3 = −LKD T , a 2 = KD T 2 + 2 2 2 2 2 2 KD T − T KD − LKD T − LT 2 − LT 2 − LKD , 2 2 2 2 2 a1 = −LT − LKD − L − T KD − T + T + KD + KD T 2 + KD , a0 = 1 + KD − L − T . Since a3 < 0, a2 < 0, and a0 > 0, the roots satisfy x1 + x2 + x3 = − aa12 < 0 and x1 x2 x3 = − aa30 > 0, and there only exists one positive root. In other words, dΦQ3 /dω = 0 has only one positive solution, such that ΦQ3 (ω) increases first at small frequency and then decreases. Thus ΦQ3 (ω) > −π when 0 < ω < ωcp and ΦQ3 (ω) < −π when ω > ωcp . On the other hand, let µ ¶ δ δ Ψ(δ) = ΦQ3 (ω0 ) = −L + arctan T T ¶ µ p δ 1 + T 2 + δ 2 − arctan(δ) − π, + arctan T
< 0, then ΦQ3 (ω0 ) < It easily verifies that dΨ(δ) dδ Ψ(δ)|δ=0 = −π, and in turn ω0 > ωcp . Thus 2
(MQ3 (ωcp )) = KP2
2 2 1 + KD ωcp 2 (1 + ωcp )(1 + T 2 ωc p2 ) 2
> KP2 = (MQ3 (0)) ,
Conwhich prevents anticlockwise encirclement. √ sequently, stabilization for L ≥ 1 + T 2 − T + 1 is impossible for case B. Case C In this case, p ½ KD < 1 + T 2 , (27) 1 + KD − T − L < 0 √ and assume L ≥ 1 + T 2 + 1 − T . It follows that p ΦQ3 (ω) ≤ arctan(ω) + arctan( 1 + T 2 ω) p − arctan(T ω) − ( 1 + T 2 + 1 − T )ω − π ,Θ(ω). Letting dΘ(ω)/dω = 0 leads to (b2 x2 + b1 x + 2 2 b0 )x √ = 0, with x = (1 + T √ )T 2 > 0, √ ω , b2 = 4 2 2 2 2 b1 = 1 + T T +2 1√+ T T +2T √+ 1 + T 2 + 1−T 5 −2T 3 > 0, b0 = 1 + T 2 T 2 + 1 + T 2 +1− T 3 > 0. And there is no positive solution for x or ω. Combined with the fact that dΘ(ω)/dω|ω=∞ < 0, it is seen that dΘ(ω)/dω < 0 for ω > 0 and Θ(ω) < Θ(0) = −π. Thus, ΦQ3 (ω) < −π holds for ω > 0, and√there is no anticlockwise encirclement when L ≥ 1 + T 2 − T + 1. Case D In this case, p ½ KD > 1 + T 2 , (28) 1 + KD − T − L < 0 √ 1 + T 2 + 1 − T . Still let and assume L > dΦQ3 /dω = 0, it follows from (26) that ai < 0 for i = 0, 1, 2, 3. And there is no positive root for x or ω. Thus dΦQ3 /dω keeps the negative sign when ω > 0. Consequently, ΦQ3 (ω) < −π for ω > 0, and once again there is no anticlockwise encirclement. Summarizing the cases A-D yields the theorem. 1 Theorem 3. The process, G5 (s) = (T s+1)(s−1) e−Ls , is stabilizable by PD controller √ C3 (s) = KP (1 + KD s)√if and only if L < 1 + T 2 − T + 1. If L < 1 + T 2 − T + 1, the stabilizing controller parameters can be found from p (29) L + T − 1 < KD < 1 + T 2 ,
and
s 1 < KP <
2 )(1 + T 2 ω 2 ) (1 + ωc1 c1 2 ω2 1 + KD c1
(30)
with −Lωc1 + arctan(ωc1 ) + arctan(KD ωc1 ) − arctan(T ωc1 ) = 0. As for PID controller, C4 (s) = KP (1+KD s+ KsI ), The open-loop transfer function is Q4 (s) = KP
KD s2 + s + KI −Ls e . s(s − 1)(T s + 1)
It follows from Lemma 3 that the closed-loop stability requires H(s) = T L3 s3 + (9T L2 + L3 − T L3 )s2 + (18T L + 6L2 − 6T L2 − L3 )s + 6T + 6L−6T L−3L2 be stable. Then the constant term
It follows from Lemma 2 that KP KI > 0 is necessary for closed-loop stability. In case that KP > 0 and KI > 0, the open-loop has its magnitude s 1 + (KD ω − KωI )2 MQ4 (ω) = KP . (1 + ω 2 )(1 + T 2 ω 2 )
Nyquist Diagram
Nyquist Diagram
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2 Imaginary axis
Imaginary axis
6T + 6L√ − 6T L − 3L2 > 0 is necessary, which leads to L < 1 + T 2 − T + 1. Also noting that PD is a special case of PID controller, it is concluded that PID √ controller could stabilize G5 if and only if L < 1 + T 2 − T + 1.
0
0
−0.2
−0.2
−0.4
−0.4
−0.6
−0.8 −1.2
−0.6
−1
−0.8
−0.6
−0.4 −0.2 Real Axis
0
0.2
−1.2s
e (0.5s+1)(s−1)
(a) G5 = C3 = 1.04s + 1.04
0.4
0.6
−0.8 −1.2
−1
−0.8
−0.6
−0.4 −0.2 Real Axis
0
0.2
0.4
0.6
e−1.2s (0.5s+1)(s−1)
and (b) G5 = and C4 = 1.166s + 1.06 + 0.053/s
Fig. 3. Nyquist plots of G5 with PD and PID controllers It is manifested from the investigations that: for And the phase is ΦQ¡4 (ω) = −Lω¢ + arctan(ω) − the process under consideration, the maximum KI arctan(T ω) + arctan KD ω − ω − π. Then the stabilizable time delay with PD/PID controller stabilizing controller parameters could be found is larger than that with P/PI controller. At the from same time, the maximum stabilizable time delay max (ΦQ4 ) > −π, (31) with P controller is equivalent to that with PI cons troller, and the maximum stabilizable time delay 2 2 (1 + ωc1 )(1 + T 2 ωc1 ) with PD is equivalent to that with PID controller. < K P 1 + (KD ωc1 − ωKc1I )2 Hence when only stabilization of these processes s 2 2 is needed, P or PD controller is sufficient. This 2 (1 + ωc2 )(1 + T ωc2 ) , (32) < proposed stabilization analysis approach is also KI 2 1 + (KD ωc2 − ωc2 ) applicable to several other low-order processes, which in not included here. with ωc1 < ωc2 the two positive phase crossover frequencies solved from −Lω+arctan(ω)−arctan(T ω)+ ¢ ¡ To deal with practical process with time delay, arctan KD ω − KωI = 0, and if the time delay is within the stabilizing range s given in this paper, then the corresponding PID 2 2 2 (1 + ωc )(1 + T ωc ) , (33) KP < parameters could be determined to stabilize the I 2 1 + (KD ωc − K ωc ) plant. And then the problem is reduced to confor subsequent phase crossover frequencies ΦQ3 (ωc ) = troller design for stable process with time delay, where there are many techniques available −2kπ − π with ωc > ωc2 . 1 e−1.2s , since Example 2. Let G5 = (0.5s+1)(s−1) √ L = 1.2 < 1 + 0.52 − 0.5 + 1, it follows from Theorem 3 that the process is stabilizable by PD controller. According to (29), a stabilizing gain of KD could be found from the range (0.7, 1.118). Let KD = 1, then ωc1 = 0.821, and in turn 1 < KP < 1.081. Choose KP = 1.04, then (1.04+1.04s) −1.2s e . C3 = 1.03 + 0.824s, and Q3 = (0.5s+1)(s−1) The Nyquist plot is shown in Fig 3(a). When PID controller is considered, Choose KD = 1.1 and KI = 0.05, (31) is met, then KP is bounded by (1.047, 1.0732) from (32). Choose KP = 1.06, the PID controller is C4 = 1.166s + 1.06 + 0.053/s and the open-loop transfer function turns to be 2 +1.06s+0.053 −1.2s e . The Nyquist plot Q4 = 1.166s s(0.5s+1)(s−1) is given in Figure 3(b).
5. CONCLUSION In this paper, the stabilization of second-order unstable time-delay process is addressed. The maximum allowable time-delay for different controllers is derived, and the stabilizing gain range is given.
REFERENCES Chidambaram, M. (1997). Control of unstable systems: a review. J. Energy, Heat Mass Transfer 19, 49–56. Hwang, C. and J. H. Hwang (2004). Stabilisation of first-order plus dead-time unstable processes using PID controllers. IEE Proc.Control Theory Appl. 151(1), 89–94. Kharitonov, V.L., S.-I. Niculescu, J. Moreno and W. Michiels (2005). Static output feedback stabilization: Necessary conditions for multiple delay controllers. IEEE Trans. Automatic Control 50(1), 82–86. Shafiei, Z. and A. T. Shenton (1994). Tuning of PID-type controllers for stable and unstable systems with time-delay. Automatica 30(10), 1609–1615. Silva, G.J., A. Datta and S.P. Bhattacharyya (2004). PID controllers for time-delay sytems. Birkh¨auser. Boston. Wang, Q.G., H.W. Fung and Y. Zhang (1999). PID tuning with exact gain and phase margins. ISA Transactions 38, 243–249.
ECE Dept., National University of Singapore, Singapore
Abstract: Based on the well known Nyquist stability criterion, the stabilization of second-order unstable time-delay processes by PID controllers is investigated. Explicit stabilizability results in terms of the upper limit of time-delay size is established. The analysis not only provides both theoretical understanding on such stabilization issue, but also meaningful for the teaching of Nyquist stability criterion in control eduction. Keywords: Nyquist plot, stabilization, unstable time-delay Processes, PID controller
1. INTRODUCTION Time-delay is commonly encountered in chemical, biological, mechanical and electronic systems. There are some unstable processes in industry such as chemical reactors and their stabilization is essential for successful operations. Especially, unstable processes coupled with time delay makes control system design a difficult task, which has attracted increased attention from control community (Chidambaram, 1997). Recently, many techniques (Shafiei and Shenton, 1994; Wang et al., 1999) have been reported to improve PID tuning for unstable delay processes. However, these works do not provide a clear scenario on what kind of process could be stabilized by PID controllers. Silva et al. (2004) investigated the complete set of stabilizing PID controllers based on the HermiteBiehler theorem for quasi-polynomials. Hwang and Hwang (2004) applied D-partition method to characterize the stability domain in the space of system and controller parameters. However, these approaches are mathematically involved, and the 1
Author to whom all correspondence should be addressed, Email: [email protected], Tel: (+65) 6874 2282, Fax: (+65) 6779 1103
maximal stabilizable time delay for some typical yet simple processes still remains obscure. In this paper, we focus on the second-order unstable time-delay process and provide a thorough study for its stabilization based on the well-known Nyquist stability criterion. The maximum stabilizable delay is derived and the solution for corresponding stabilizing controller is given. The paper is organized as follows. After the problem statement in Section 2, some preliminaries are introduced in Section 3. Then the stabilization analysis is addressed in Section 4. Finally, Section 5 concludes the paper. 2. PROBLEM STATEMENT Suppose that the process of interest is controlled in the unity feedback system (Figure 1) with PID controller or its special cases, namely, P, PI, PD. Normalized process is adopted throughout the paper for uniform formulation, which could be derived by scaling down the time-delay and all time constants by T1 and absorbing the process gain ¯ into the controller. The normalized transfer K functions of process and controller are expressed as
G(s) =
1 e−Ls , (s − 1)(T s + 1)
(1)
and
KI ) s respectively. In the rest of the paper, stabilization C(s) = KP (1 + KD s +
R (s)
+
E (s)
−
C (s)
G (s)
Y (s)
Fig. 1. Unity output feedback system of (1) is addressed with one of the following four controllers:C1 (s) = KP , C2 (s) = KP (1 + KsI ), C3 (s) = KP (1 + KD s), and C4 (s) = KP (1 + KD s+ KsI ). The corresponding open-loop transferfunction, Ql (s) = G(s)Cl (s), l ∈ {1, 2, 3, 4}, is re-written as Ql (s) = Gi (s)Cl (s) =
KN (s) −Ls e , L > 0, (2) sv D(s)
where K is the gain, v a non-negative integer representing type of the loop, N (s) and D(s) both rational polynomials of s with N (0) = D(0) = 1. If the loop has a pole at the origin, then the Nyquist contour needs to be modified by replacing the origin with a infinitely small semicircle of s = rejφ with r → 0 and −π/2 < φ < π/2. The Nyquist stability theorem is now applied loop (2). Theorem 1. Given the loop transfer function Ql (s) in (2) with P + poles inside the Nyquist contour, the closed-loop system in Figure 1 is stable if and only if the Nyquist plot of Ql encircles the critical point, (−1, 0), P + times anticlockwise.
Due to the delay element in the open-loop Ql (s) (2), the phase of Ql (jω), ΦQl (ω), approaches −∞ when frequency ω → ∞. Consequently, if limω→∞ |Ql (jω)| ≥ 1, the Nyquist curve of Ql (s) will encircle/pass the critical point infinite times clockwise, which violates Theorem 1 and the closed-loop is unstable. Hence, the following lemma follows. Lemma 1. For the open-loop Ql (s) in (2), lim |Ql (jω)| < 1
In contrast, if K < −1, Ql (0) is now left to the critical point. This causes Ql (jω) + 1 to have the total phase change of (2m + 1)π with an integer m for the frequency from zero to positive infinity, and thus an odd number of times for the entire frequency range, making an unstable closed-loop for P + = 0. Therefore, K > −1 is necessary for stability if P + = 0. The case that the loop has one integrator (v = 1) can be analyzed similarly. And K > 0 is necessary for stability if P + = 0, while K < 0 is necessary for stability if P + = 1. Lemma 2. Consider the open-loop Ql (s) in (2), the necessary condition for closed-loop stability is (i). For v = 0, K > −1 if P + = 0; and K < −1 if P + = 1. (ii). For v = 1, K > 0 if P + = 0; and K < 0 if P+ = 1 . Lemma 3. Given the open-loop transfer function Ql (s) described in (2), a necessary condition for the closed-loop stability is: polynomial H(s) =
dm+1 v Ls dsm+1 [s D(s)e ] eLs
(4)
has all its zeros lie in the open left half plane, where m is the degree of N (s).
3. PRELIMINARY
ω→∞
If K > −1, Ql (0) is also right to the critical point. This causes Ql (jω) + 1 to have the total phase change of an integral multiple (zero included) of 2π for entire positive frequency range. Thus the Nyquist curve encircles the critical point an even number of times for the entire frequency range, making an unstable closed-loop for P + = 1. Therefore, K < −1 is necessary for stability if P + = 1.
(3)
is necessary for the closed-loop stability. Suppose first that the loop has no integrator (v = 0). Then Ql (0) = K is finite. The Nyquist curve starts at Ql (0) = K and, |Ql (j∞)| < 1 due to (3), should end right to the critical point, (−1, 0), to meet Theorem 1 for stability.
Proof: The stability of characteristic function F0 (s) = sv D(s) + KN (s)e−Ls is equivalent to that of F1 (s) = sv D(s)eLs + KN (s). It follows from the results in Kharitonov et al. (2005) that H(s)eLs , the m + 1 derivative of F1 (s), is also stable. Therefore H(s) is stable, or its zeros lie in the open left half plane. Lemma 4. Given the open-loop transfer function Ql (s) with P + > 0, and any integer k, if (i). ΦQl (ω) < −2kπ + 3π, and dΦQl (ω) < 0 for ΦQl (ω) ≤ −2kπ − π (ii). dω hold for ω ≥ 0, then the closed-loop system is stable only if ¡ ¢ max ΦQl (ω)|ω>0 > −2kπ + π. (5) Proof: Anti-clockwise encirclement is not obtainable for the portion of the loop Nyquist curve
corresponding either to s = rejφ with r → 0 since possible poles of Ql (s) would cause the curve to rotate clockwise only, or to s = jw which meet (ii) as its phase keeps decreasing. Taking into account (i), it can occur only if the curve has the phase increase in the phase range of −2kπ − π < ΦQl (ω) < −2kπ + 3π, and traverses the negative real axis from the third quadrant to the second quadrant therein, that is, there holds (15). The proof is complete. In the following section, the stabilization analysis is provided using these Lemmas. 4. SECOND-ORDER UNSTABLE PROCESS WITH A STABLE POLE 4.1 P/PI controller For P controller, C1 (s) = KP , the open-loop frequency response is given by Q1 (jω) =
KP e−jLω , (jω − 1)(jT ω + 1)
with P + = 1 and v = 0. By Lemma 2(i), K = −KP < −1 is necessary. Then the magnitude is p MQ1 (ω) = KP / (1 + ω 2 )(1 + T 2 ω 2 ), which decreases from KP to 0. The phase is ΦQ1 (ω) = −Lω+arctan(ω)−arctan (T ω)−π, (6) with its first and second order derivatives as d 1 T ΦQ1 (ω) = −L + − , 2 dω 1+ω 1 + T 2 ω2 2ω 2T 3 ω d2 ΦQ1 (ω) = − + . 2 2 2 dω (1 + ω ) (1 + T 2 ω 2 )2 For ω > 0, there holds ΦQ1 (0) = −π and ΦQ1 < −π/2. Moreover, for ΦQ1 ≤ −3π, dΦQ1 /dω is always negative. It follows from Lemma 4 that Φ51 (ω) > −π for some ω > 0 is necessary for closed-loop stability. And this requires 2 2 dΦQ1 (ω)/dω > 0 for some ω. Let = √ √ d ΦQ1 (ω)/dω 2 2 0 yields ω1 = 0, and ω2 = (T T + T + T )/T 2 . d ΦP (ω) becomes Then the maximum value for dω 1 − L − T, ¶ µ ω = ω1 ,√0 < T < 1; d ΦQ1 (ω) = max T T −1 √ √ dω − L, T T +T + T +1 ω = ω2 , T ≥ 1 If T ≥ 1, it is easy to check from (6) that ΦQ1 (ω) < −π for ω > 0, and thus the closed-loop is always unstable. If 0 < T < 1, the stabilization d ΦQ1 (ω) turns to be requirement for dω d ΦQ1 (ω)|ω=ω1 = 1 − L − T > 0, or L < 1 − T. dω (7) In this case, the phase will increase from −π first and then decrease, and there is one and
only one intersection with the negative real axis with ΦQ1 = −π. In order for the anticlockwise encirclement of critical point to occur, this intersection should lie between −1 and 0, that is MQ1 (ωc1 ) < 1, ΦQ1 (ωc1 ) = −π, or equivalently q 2 )(1 + T 2 ω 2 ). (8) 1 < KP < (1 + ωc1 c1 And for ω > ωc1 , MQ1 is always less than 1 so that there is no encirclement around the critical point thereafter. Consequently, there is exactly one anticlockwise encirclement when L < 1 − T and (8) are all true. As for PI controller, C2 (s) = KP (1 + open-loop frequency response is Q2 (jω) = KP
KI s ),
1 − j KωI . (jω − 1)(jT ω + 1)
the (9)
It follows from Lemma 2(ii) that KP KI > 0 is necessary for closed-loop stability. Assume KP > 0 and KI > 0 first, then the loop has its magnitude as s ¡ ¢2 1 + KωI MQ2 = KP , (1 + ω 2 )(1 + T 2 ω 2 ) which always decreases from ∞ to 0. The phase of (9) is ¶ µ KI −arctan(T ω)−π, ΦQ2 = −Lω+arctan(ω)−arctan ω (10) with its derivative being KI T 1 dΦQ2 ω2 = −L + + . ¡ KI ¢2 − dω 1 + ω2 1 + (T ω)2 1+ ω
It follows that ΦQ2 < −π/2. Moreover, when ΦQ2 ≤ −3π, the derivative of phase is always negative. In consequence, anticlockwise encirclement is possible only when there exists some ω > 0 such that ΦQ2 (ω) > −π by invoking Lemma 4. In case of L ≥ 1 − T , it is readily seen from the previous P-control discussion that ΦQ2 = ΦQ1 − arctan(KI /ω) ≤ ΦQ1 , ΦQ1 and then ΦQ2 are always less than −π. In consequence, the Nyquist curve has no anticlockwise encirclement around the critical point and the closed-loop is unstable when KP > 0, KI > 0 and L ≥ 1. In case of L < 1 − T , ΦQ1 > −π holds from some small ω, and it is always possible to find ΦQ2 (ω) > −π by reducing KI and in turn arctan(KI /ω). Thus KI should be chosen to ensure max(ΦQ2 (ω)|ω>0 ) > −π, which is not empty. Then the Nyquist curve will have two crossings with the negative real axis with phase angle −π. In order to have anticlockwise encirclement around the critical point, KP should be chosen such that MQ2 (ωc2 ) < 1 < MQ2 (ωc1 ),
(11)
where 0 < ωc1 < ωc2 are the two phase crossover frequencies satisfying ΦQ32 (ω) = −π. Inequality
(11) is always feasible since MQ2 is monotonically decreasing.
Nyquist Diagram
Nyquist Diagram
0.2
0.25
0.2
0.15
0.15 0.1 0.1
Imaginary axis
0.05 Imaginary axis
Assume KP < 0 and KI < 0 then, the phase is ¶ µ KI −arctan(T ω), ΦQ2 = −Lω+arctan(ω)−arctan ω
0
−0.05
1 Theorem 2. The process, G5 (s) = (T s+1)(s−1) e−Ls , is stabilizable by P controller (C1 (s) = KP ) or PI controller (C2 (s) = KP (1 + KsI )) if and only if 0 < L < 1 − T . If 0 < L < 1 − T , the stabilizing gain for P controller is bounded by q 2 )(1 + T 2 ω 2 ). (12) 1 < KP < (1 + ωc1 c1
with −Lωc1 + arctan(ωc1 ) − arctan (T ωc1 ) = 0. And the stabilizing parameters for PI controller satisfy KP > 0, KI > 0. (13) KI is chosen such that max(ΦQ2 (ω)|ω>0 ) > −π,
(14)
and the range of KP is given by v u (1 + ω 2 )(1 + T 2 ω 2 ) u c2 c2 < KP t ³ ´2 KI 1 + ωc2 v u (1 + ω 2 )(1 + T 2 ω 2 ) c1 c1 .
4.2 PD/PID controller For PD controller, C3 (s) = KP (1 + KD s), the open-loop frequency response is 1 + jKD ω , (15) Q3 (jω) = KP (jω − 1)(jT ω + 1)
0
−0.05
−0.1 −0.1 −0.15 −0.15
which is always less than π/2. Moreover, for ΦQ2 ≤ −π, its derivative is always negative. It follows from Lemma 4 that Q2 (jω) has no anticlockwise encirclement, and that the closedloop is unstable when KP < 0 and KI < 0.
0.05
−0.2 −1.6
−0.2
−1.4
−1.2
(a) G5 = C1 = 1.5
−1
−0.8 −0.6 Real Axis
−0.4
−0.2
e−0.3s (0.5s+1)(s−1)
0
0.2
−0.25 −1.4
−1.2
−1
−0.8
−0.6 Real Axis
−0.4
−0.2
e−0.3s (0.5s+1)(s−1)
and (b) G5 = C2 = 1.3 + 0.026/s
0
0.2
and
Fig. 2. Nyquist plots of G5 with P and PI controllers with P + = 1 and v = 0. It follows from Lemma 2(i) that K = KP > 0 is necessary for stabilization. Then the magnitude and phase are s 2 ω2 1 + KD , (16) MQ3 = KP (1 + ω 2 )(1 + T 2 ω 2 ) and ΦQ3 = −Lω+arctan(ω)+arctan(KD ω)−arctan(T ω)−π, (17) respectively, with ΦQ3 (0) = −π and ΦQ3 < 0 for ω > 0. Notice that µ 2 ¶ M d KQ23 2 4 2 P 2ω(T 2 KD ω + 2T 2 ω 2 + 1 + T 2 − KD ) = , 2 2 2 2 2 dω (1 + ω ) (1 + T ω ) (18) 2 2 it follows that if (1+T −K ) > 0, or equivalently D √ KD < 1 + T 2 , then dMQ3 /dω < 0 always holds, and MQ3 decreases from KP to 0 when ω √ increases 1 + T 2, from 0 to ∞. Otherwise, if KD > then dMQ3 /dω is positive when ω is small and turns negative when ω increases, so that MQ3 increases from KP first and then decreases to 0 as ω increases. As for the phase, one sees that ΦQ3 (0) = −π and 1 KD T dΦQ3 = −L + + 2 ω2 − 1 + T 2 ω2 dω 1 + ω2 1 + KD (19) with ¯ dΦQ3 ¯¯ = −L + 1 + KD − T. (20) dω ¯ω=0 When ΦQ3 ≤ −3π, the derivative of phase is always negative. It follows from Lemma 4 the closed-loop is stable only if max (ΦQ3 (ω)) > −π. According to the combinations of signs of (1+T 2 − 2 KD ) and (−L+1+KD −T ), the stabilization issue is addressed by four cases correspondingly. Case A In this case, p ½ KD < 1 + T 2 , 1 + KD − T − L > 0 and this leads to L<
p 1 + T 2 − T + 1.
(21)
(22)
Given arbitrary L satisfies (22), KD is chosen within the range
L + T − 1 < KD <
p 1 + T 2,
(23)
which is not empty. Since dΦQ3 (0)/dω > 0, the stabilization is possible. In order for the correct encirclement to occur, the first intersection of Nyquist curve with the real axis for positive frequency should between −1 and 0. It follows that MQ3 (ωc1 ) < 1, which leads to s 2 )(1 + T 2 ω 2 ) (1 + ωc1 c1 , (24) 1 < KP < 2 ω2 1 + KD c1 combined with the requirement KP > 1. (24) is also not empty since dMQ3 /dω < 0. Moreover, when (24) is true, MQ3 (ω) is always less than 1 for ω > ωc1 , and Q3 does not encircle the critical point for ω > ωc1 . Consequently, PD controller could always stabilize process (1) under case A when (22), (23) and (24) are all true. In the rest of this subsection, it is demonstrated that PD controller could not stabilize process (1) if √ L ≥ 1 + T 2 − T + 1. Case B In this case, p ½ KD > 1 + T 2 , (25) 1 + KD − T − L > 0 √ 1 + T 2 − T + 1 is assumed. For and L ≥ convenience of analysis, let p δ KD = 1 + T 2 + δ 2 , δ > 0 and ω0 = T also denote by ωcp the smallest positive frequency that ΦQ3 (ωcp ) = −π. Let dΦQ3 /dω = 0, it follows from (17) that a3 x3 + a2 x2 + a1 x + a0 = 0,
(26)
2 2 with x = ω 2 , a3 = −LKD T , a 2 = KD T 2 + 2 2 2 2 2 2 KD T − T KD − LKD T − LT 2 − LT 2 − LKD , 2 2 2 2 2 a1 = −LT − LKD − L − T KD − T + T + KD + KD T 2 + KD , a0 = 1 + KD − L − T . Since a3 < 0, a2 < 0, and a0 > 0, the roots satisfy x1 + x2 + x3 = − aa12 < 0 and x1 x2 x3 = − aa30 > 0, and there only exists one positive root. In other words, dΦQ3 /dω = 0 has only one positive solution, such that ΦQ3 (ω) increases first at small frequency and then decreases. Thus ΦQ3 (ω) > −π when 0 < ω < ωcp and ΦQ3 (ω) < −π when ω > ωcp . On the other hand, let µ ¶ δ δ Ψ(δ) = ΦQ3 (ω0 ) = −L + arctan T T ¶ µ p δ 1 + T 2 + δ 2 − arctan(δ) − π, + arctan T
< 0, then ΦQ3 (ω0 ) < It easily verifies that dΨ(δ) dδ Ψ(δ)|δ=0 = −π, and in turn ω0 > ωcp . Thus 2
(MQ3 (ωcp )) = KP2
2 2 1 + KD ωcp 2 (1 + ωcp )(1 + T 2 ωc p2 ) 2
> KP2 = (MQ3 (0)) ,
Conwhich prevents anticlockwise encirclement. √ sequently, stabilization for L ≥ 1 + T 2 − T + 1 is impossible for case B. Case C In this case, p ½ KD < 1 + T 2 , (27) 1 + KD − T − L < 0 √ and assume L ≥ 1 + T 2 + 1 − T . It follows that p ΦQ3 (ω) ≤ arctan(ω) + arctan( 1 + T 2 ω) p − arctan(T ω) − ( 1 + T 2 + 1 − T )ω − π ,Θ(ω). Letting dΘ(ω)/dω = 0 leads to (b2 x2 + b1 x + 2 2 b0 )x √ = 0, with x = (1 + T √ )T 2 > 0, √ ω , b2 = 4 2 2 2 2 b1 = 1 + T T +2 1√+ T T +2T √+ 1 + T 2 + 1−T 5 −2T 3 > 0, b0 = 1 + T 2 T 2 + 1 + T 2 +1− T 3 > 0. And there is no positive solution for x or ω. Combined with the fact that dΘ(ω)/dω|ω=∞ < 0, it is seen that dΘ(ω)/dω < 0 for ω > 0 and Θ(ω) < Θ(0) = −π. Thus, ΦQ3 (ω) < −π holds for ω > 0, and√there is no anticlockwise encirclement when L ≥ 1 + T 2 − T + 1. Case D In this case, p ½ KD > 1 + T 2 , (28) 1 + KD − T − L < 0 √ 1 + T 2 + 1 − T . Still let and assume L > dΦQ3 /dω = 0, it follows from (26) that ai < 0 for i = 0, 1, 2, 3. And there is no positive root for x or ω. Thus dΦQ3 /dω keeps the negative sign when ω > 0. Consequently, ΦQ3 (ω) < −π for ω > 0, and once again there is no anticlockwise encirclement. Summarizing the cases A-D yields the theorem. 1 Theorem 3. The process, G5 (s) = (T s+1)(s−1) e−Ls , is stabilizable by PD controller √ C3 (s) = KP (1 + KD s)√if and only if L < 1 + T 2 − T + 1. If L < 1 + T 2 − T + 1, the stabilizing controller parameters can be found from p (29) L + T − 1 < KD < 1 + T 2 ,
and
s 1 < KP <
2 )(1 + T 2 ω 2 ) (1 + ωc1 c1 2 ω2 1 + KD c1
(30)
with −Lωc1 + arctan(ωc1 ) + arctan(KD ωc1 ) − arctan(T ωc1 ) = 0. As for PID controller, C4 (s) = KP (1+KD s+ KsI ), The open-loop transfer function is Q4 (s) = KP
KD s2 + s + KI −Ls e . s(s − 1)(T s + 1)
It follows from Lemma 3 that the closed-loop stability requires H(s) = T L3 s3 + (9T L2 + L3 − T L3 )s2 + (18T L + 6L2 − 6T L2 − L3 )s + 6T + 6L−6T L−3L2 be stable. Then the constant term
It follows from Lemma 2 that KP KI > 0 is necessary for closed-loop stability. In case that KP > 0 and KI > 0, the open-loop has its magnitude s 1 + (KD ω − KωI )2 MQ4 (ω) = KP . (1 + ω 2 )(1 + T 2 ω 2 )
Nyquist Diagram
Nyquist Diagram
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2 Imaginary axis
Imaginary axis
6T + 6L√ − 6T L − 3L2 > 0 is necessary, which leads to L < 1 + T 2 − T + 1. Also noting that PD is a special case of PID controller, it is concluded that PID √ controller could stabilize G5 if and only if L < 1 + T 2 − T + 1.
0
0
−0.2
−0.2
−0.4
−0.4
−0.6
−0.8 −1.2
−0.6
−1
−0.8
−0.6
−0.4 −0.2 Real Axis
0
0.2
−1.2s
e (0.5s+1)(s−1)
(a) G5 = C3 = 1.04s + 1.04
0.4
0.6
−0.8 −1.2
−1
−0.8
−0.6
−0.4 −0.2 Real Axis
0
0.2
0.4
0.6
e−1.2s (0.5s+1)(s−1)
and (b) G5 = and C4 = 1.166s + 1.06 + 0.053/s
Fig. 3. Nyquist plots of G5 with PD and PID controllers It is manifested from the investigations that: for And the phase is ΦQ¡4 (ω) = −Lω¢ + arctan(ω) − the process under consideration, the maximum KI arctan(T ω) + arctan KD ω − ω − π. Then the stabilizable time delay with PD/PID controller stabilizing controller parameters could be found is larger than that with P/PI controller. At the from same time, the maximum stabilizable time delay max (ΦQ4 ) > −π, (31) with P controller is equivalent to that with PI cons troller, and the maximum stabilizable time delay 2 2 (1 + ωc1 )(1 + T 2 ωc1 ) with PD is equivalent to that with PID controller. < K P 1 + (KD ωc1 − ωKc1I )2 Hence when only stabilization of these processes s 2 2 is needed, P or PD controller is sufficient. This 2 (1 + ωc2 )(1 + T ωc2 ) , (32) < proposed stabilization analysis approach is also KI 2 1 + (KD ωc2 − ωc2 ) applicable to several other low-order processes, which in not included here. with ωc1 < ωc2 the two positive phase crossover frequencies solved from −Lω+arctan(ω)−arctan(T ω)+ ¢ ¡ To deal with practical process with time delay, arctan KD ω − KωI = 0, and if the time delay is within the stabilizing range s given in this paper, then the corresponding PID 2 2 2 (1 + ωc )(1 + T ωc ) , (33) KP < parameters could be determined to stabilize the I 2 1 + (KD ωc − K ωc ) plant. And then the problem is reduced to confor subsequent phase crossover frequencies ΦQ3 (ωc ) = troller design for stable process with time delay, where there are many techniques available −2kπ − π with ωc > ωc2 . 1 e−1.2s , since Example 2. Let G5 = (0.5s+1)(s−1) √ L = 1.2 < 1 + 0.52 − 0.5 + 1, it follows from Theorem 3 that the process is stabilizable by PD controller. According to (29), a stabilizing gain of KD could be found from the range (0.7, 1.118). Let KD = 1, then ωc1 = 0.821, and in turn 1 < KP < 1.081. Choose KP = 1.04, then (1.04+1.04s) −1.2s e . C3 = 1.03 + 0.824s, and Q3 = (0.5s+1)(s−1) The Nyquist plot is shown in Fig 3(a). When PID controller is considered, Choose KD = 1.1 and KI = 0.05, (31) is met, then KP is bounded by (1.047, 1.0732) from (32). Choose KP = 1.06, the PID controller is C4 = 1.166s + 1.06 + 0.053/s and the open-loop transfer function turns to be 2 +1.06s+0.053 −1.2s e . The Nyquist plot Q4 = 1.166s s(0.5s+1)(s−1) is given in Figure 3(b).
5. CONCLUSION In this paper, the stabilization of second-order unstable time-delay process is addressed. The maximum allowable time-delay for different controllers is derived, and the stabilizing gain range is given.
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