首页资源分类DSP > Digital PID Controller Design

Digital PID Controller Design

已有 445176个资源

下载专区

上传者其他资源

    文档信息举报收藏

    标    签:digitalPID

    分    享:

    文档简介

    这是一篇较为详尽的关于数字PID设计的参考资料,希望对大家有所帮助。

    文档预览

    Digital PID Controller Design 1 r[n] e[n] C(z) u[n] Digital PID Controller Design »[n] y[n] P (z) ² Plant and Controller N (z) G(z) = ; D(z) C(z) = NC(z) : DC (z) ² The characteristic polynomial of the closed loop system ¦(z) := DC(z)D(z) + NC(z)N (z) 2 Digital PID Controller Design TCHEBYSHEV REPRESENTATION AND ROOT CLUSTERING Tchebyshev representation of real polynomials ² Consider a real polynomial P (z) = anzn + an¡1zn¡1 + ¢ ¢ ¢ + a1z + a0 ² The image of P (z) evaluated on the circle C½ of radius ½, centered at the origin is: © ª P (z) : z = ½ejμ; 0 · μ · 2¼ : ¡¢ ¡¢ ² As the coe±cients ai are real P ½ejμ and P ½e¡jμ are conjugate complex numbers, and so it su±ces to determine the image of the upper half of the circle: © ª P (z) : z = ½ejμ; 0 · μ · ¼ : 3 Digital PID Controller Design ² Since z k ¯¯ z=½ejμ = ½k(cos kμ + j sin kμ); ¡¢ P ½ejμ = (|an½n cos nμ + ¢ ¢{¢z+ a1½ cos μ + a0}) +j |(an½n sin nμ +{¢z¢ ¢ + a1½ sin μ}) R¹(½;μ) I¹(½;μ) = R¹(½; μ) + jI¹(½; μ): ² Consider (½ejμ)k = ½k cos kμ + j½k sin kμ ² Write u = ¡ cos μ and de¯ne the generalized Tchebyshev polynomials as follows: ck(u; ½) = ½kck(u); sk(u; ½) = ½ksk(u); k = 0; 1; 2 ¢ ¢ ¢ and note that sk(u; ½) = ¡1 ¢ d [ck(u; ½)] ; k du ¡ k = 1; 2; ¢ ¢ ¢ ¢ ck+1(u; ½) = ¡½uck(u; ½) ¡ 1 ¡ u2 ½sk(u; ½); k = 1; 2; ¢ ¢ ¢ 4 Digital PID Controller Design ² The generalized Tchebyshev polynomials are for k = 1; ¢ ¢ ¢ 5: k ck(u; ½) sk(u; ½) 1 ¡½u ½ 2 ½2 (2u2 ¡ 1) ¡2½2u 3 ½3 (¡4u3 + 3u) ½3 (4u2 ¡ 1) 4 ½4 (8u4 ¡ 8u2 + 1) ½4 (¡8u3 + 4u) 5 ½5 (¡16u5 + 20u3 ¡ 5u) ½5 (16u4 ¡ 12u2 + 1) ... ... ... ¡¢ p ² With this notation, P ½ejμ = R(u; ½) + j 1 ¡ u2T (u; ½) =: Pc(u; ½) where R(u; ½) = ancn(u; ½) + an¡1cn¡1(u; ½) + ¢ ¢ ¢ + a1c1(u; ½) + a0 T (u; ½) = ansn(u; ½) + an¡1sn¡1(u; ½) + ¢ ¢ ¢ + a1s1(u; ½): ² R(u; ½) and T (u; ½) are polynomials in u and ½. ² The complex plane image of P (z) as z traverses the upper half of the circle C½ can be obtained by evaluating Pc(u; ½) as u runs from ¡1 to +1. 5 Digital PID Controller Design LEMMA For a ¯xed ½ > 0, (a) if P (z) has no roots on the circle of radius ½ > 0, (R (u; ½) ; T (u; ½)) have no common roots for u 2 [¡1; 1] and R (§1; ½) 6= 0. (b) if P (z) has 2m roots at z = ¡½ (z = +½), then R (u; ½) and T (u; ½) have m roots each at u = +1 (u = ¡1). (c) if P (z) has 2m ¡ 1 roots at z = ¡½ (z = +½), then R (u; ½) and T (u; ½) have m and m ¡ 1 roots, respectively at u = +1 (u = ¡1). p (d) if P (z) has qi pairs of complex roots at z = ¡½ui § j½ 1 ¡ u2i , for ui 6= §1, then R (u; ½) and T (u; ½) each have qi real roots at u = ui. ² When the circle of interest is the unit circle, that is ½ = 1, we will write Pc(u; 1) = Pc(u) and also R(u; 1) =: R(u); T (u; 1) =: T (u): 6 Digital PID Controller Design Interlacing Conditions for Root Clustering and Schur Stability THEOREM P (z) has all its zeros strictly within C½ if and only if (a) R(u; ½) has n real distinct zeros ri, i = 1; 2; ¢ ¢ ¢ ; n in (¡1; 1). (b) T (u; ½) has n ¡ 1 real distinct zeros tj, j = 1; 2; ¢ ¢ ¢ ; n ¡ 1 in (¡1; 1). (c) The zeros ri and tj interlace: ¡1 < r1 < t1 < r2 < t2 < ¢ ¢ ¢ < tn¡1 < rn < +1: The three conditions given in the above Theorem may be referred to as interlacing conditions on R(u; ½) and T (u; ½). By setting ½ = 1 in the above Theorem we obtain conditions for Schur stability in terms of interlacing of the zeros of R(u) and T (u). 7 Digital PID Controller Design Tchebyshev Representation of Rational Functions ² Let p Pi(z)jz=¡½u+j½p1¡u2 =Ri(u;½)+j 1¡u2Ti(u;½);i =1;2: ² Q(z)jz=¡½u+j½p1¡u2 = P1(z)P2 P2(z)P2 (z¡1) (z¡1) ¯¯¯¯ p z=¡½u+j½ 1¡u2 z¡ ¡R(}u|;½) ¢ ¢{ = R1(u; ½)R2(u; ½) + 1 ¡ u2 T1(u; ½)T2(u; ½) R22(u; ½) + (1 ¡ u2) T22(u; ½) p z T (}u|;½) { +j 1 ¡ u2 (T1(u; ½)R2(u; ½) ¡ R22(u; ½) + (1 ¡ u2) R1(u; T22(h; ½)T2(u; ½) ½)) : ² R(u; ½), T (u; ½) are rational functions of the real variable u which runs from -1 to +1. 8 Digital PID Controller Design ROOT COUNTING FORMULAS LEMMA Let the real polynomial P (z) have i roots in the interior of the circle C½ and no roots on the circle. Then: ¢¼0 [ÁP (μ)] = ¼i LEMMA Let Q(z) = P1(z) P2(z) where the real polynomials P1(z) and P2(z) have i1 and i2 roots, respectively in the interior of the circle C½ and no roots on the circle. Then ¢¼0 [ÁQ(μ)] = ¼ (i1 ¡ i2) = ¢+¡11[ÁQC (u)]: 9 Digital PID Controller Design ² Let t1; ¢ ¢ ¢ ; tk denote the real distinct zeros of T (u; ½) of odd multiplicity, for u 2 (¡1; 1), ordered as follows: ¡1 < t1 < t2 < ¢ ¢ ¢ < tk < +1. Suppose also that T (u; ½) has p zeros at u = ¡1 and let f i(x0) denote the i-th derivative to f (x) evaluated at x = x0. THEOREM Let P (z) be a real polynomial with no roots on the circle C½ and suppose that T (u; ½) has p zeros at u = ¡1. Then the number of roots i of P (z) in the interior of the circle C½ is given by i = 1 Sgn £ T (p)(¡1; ¤ ½) μ Sgn [R(¡1; ½)] 2 + Xk 2 (¡1)jSgn [R (tj ; ½)] + ¶ (¡1)k+1Sgn [R(+1; ½)] : j=1 10 Digital PID Controller Design ² The result derived above can now be extended to the case of rational functions. Let Q(z) = P1(z) P2(z) where Pi(z); i = 1; 2 are real rational functions. ² Tchebyshev representation of Q(z) on the circle C½. Let R(u; ½); T (u; ½) be de¯ned by: R(u; ½) = R1(u; ½)R2(u; ½) + (1 ¡ u2)T1(u; ½)T2(u; ½) T (u; ½) = T1(u; ½)R2(u; ½) ¡ R1(u; ½)T(u; ½) ² Suppose that T (u; ½) has p zeros at u = ¡1 and let t1 ¢ ¢ ¢ tk denote the real distinct zeros of T (u; ½) of odd multiplicity ordered as ¡1 < t1 < t2 < ¢ ¢ ¢ < tk < +1. THEOREM Let Q(z) = P1 (z) P2 (z) where Pi(z); i = 1; 2 are real polynomials with i1 and i2 zeros respectively inside the circle C½ and no zeros on it. Then i1¡i2 = 1 2 Sgn £ T (p)(¡1; ¤ ½) μ Sgn [R(¡1; ½)]+2 Xk (¡1)j Sgn [R (tj ; ½)]+(¡1)k+1 Sgn [R(+1; ¶ ½)] : j=1 11 Digital PID Controller Design DIGITAL PI, PD AND PID CONTROLLERS ² For PI controllers, ³ ´ C(z) = KP + KIT ¢ z z ¡ 1 = (KP + KIT ) z ¡ z¡1 KP KI T +KP = K1 (z ¡ K2) z¡1 where KP = K1K2; KI = K1 ¡ K1K2 : T ² For PD controllers, C(z) = KP + KD T ¢ z ¡ z 1 = ¡ KP + ¢μ KD T z z ¡ KD T KP + KD T ¶ =: K1 (z ¡ K2) z where KP = K1 ¡ K1K2; KD = K1K2T: ² The general formula of a discrete PID controller, using backward di®erences to preserve causality, C(z) = KP + KIT ¢ z z ¡1 + KD T ¢ z ¡1 z =: K2z2 + K1z + z(z ¡ 1) K0 KP = ¡K1 ¡ 2K0; KI = K0 + K1 T + K2 ; KD = K0T: where 12 Digital PID Controller Design COMPUTATION OF THE STABILIZING SET Constant Gain Stabilization N (z) ² Plant G(z) = D(z) ² The closed-loop characteristic polynomial is ±(z) = D(z) + KN(z): ² Tchebyshev representations of D(z) and N(z) ¡¢ p D ejμ ¡¢ = RD (u) + j 1 p ¡ u2TD(u) N ejμ = RN (u) + j 1 ¡ u2TN (u); ² Note also that ¡¢ p N e¡jμ = RD(u) ¡ j 1 ¡ u2TD(u) and N (z¡1) = Nr (z) zl where Nr(z) is the reverse polynomial and l is the degree of N(z). 13 Digital PID Controller Design ² ±(z)N (z¡1) = D(z)N (z¡1) + KN(z)N (z¡1) • ±(z )Nr zl (z ) ¯¯¯¯ z =ejμ = ³ p ´³ p ´ RD(u) + j 1 ¡ u2TD(u) RN (u) ¡ j 1 ¡ u2TN (u) £ ¡ ¢ ¤ +K RN2 (u)¡+ 1 ¡¢u2 TN2 (u) £ ¡ ¢ ¤ = |RD(u)RN (u) + 1 ¡ u2 TD(u)TN{(zu) + K RN2 (u) + 1 ¡ u2 TN2 (u) } p R(K;u) +j 1 ¡ u2 |[TD(u)RN (u) {¡z RD(u)TN (u)}] p T (u) = R(K; u) + j 1 ¡ u2T (u): ² The imaginary part of the above expression has been rendered independent of K as a result of multiplying ±(z) by N(z¡1). parameter separation Ready to apply the root counting formulas 14 Digital PID Controller Design Constant Gain Stabilization Algorithm ² Let ti; i = 1; 2 ¢ ¢ ¢ ; k denote the real zeros of odd multiplicity of the ¯xed T (u), for u in (¡1; +1) and set t0 = ¡1; tk+1 = +1. ² Write Sgn [R(K; tj)] = xj; j = 0; 1; ¢ ¢ ¢ ; k + 1 ² Let i±, iNr denote the number of zeros of ±(z) and Nr(z) inside the unit circle. For simplicity assume that N (z) has no unit circle zeros and therefore neither does Nr(z). i± + iNr ¡ l = 1 Sgn £ T ¤ (p)(¡1) 2 μ Xk ¶ ¢ Sgn [R(K; ¡1)] + 2 (¡1)jSgn [R (K; tj)] + (¡1)k+1Sgn [R(K; +1)] : j=1 15 Digital PID Controller Design Example z4 + 1:93z3 + 2:2692z2 + 0:1443z ¡ 0:7047 G(z) = : z5 ¡ 0:2z4 ¡ 3:005z3 ¡ 3:9608z2 ¡ 0:0985z + 1:2311 ² Then RD(u) = ¡16u5 ¡ 1:6u4 + 32:02u3 ¡ 6:3216u2 ¡ 13:9165u + 4:9919 TD(u) = 16u4 + 1:6u3 ¡ 24:02u3 + 7:1216u + 3:9065 RN (u) = 8u4 ¡ 7:72u3 ¡ 3:4616u2 + 5:6457u ¡ 1:9739 TN (u) = ¡8u3 + 7:72u2 ¡ 0:5384u ¡ 1:7857 ² and T (u) = TD(u)RN (u) ¡ RD(u)TN (u) = ¡11:2752u4 + 7:5669u3 + 16:7782u2 ¡ 14:1655u + 1:203: ² The roots of T (u) of odd multiplicity and lying in (¡1; 1) are 0:0963 and 0:8358. 16 Digital PID Controller Design R(K; u) = 11:2752u5 + 12:1307u4 ¡ 40:6359u3 ¡ 7:1779u2 + 40:8322u ¡¡16:8293 ¡ 19:6615u ¡ 5:4727 ¢ +K ¡11:2752u4 + 9:7262u3 + 15:0696u2 ¡ 20:3653u + 7:085 : ² Since i± = 5 for stability, and iNr = 2 and l = 4, we must have: £ ¤μ ¶ Sgn T (p)(¡1) Sgn[R(K; ¡1)]¡2Sgn[R(K; 0:0963)]+2Sgn[R(K; 0:8358)]¡Sgn[R(K; 1)] = 6 £ ¤ ² Since Sgn T (p)(¡1) = +1, we have the only feasible string given by: Sgn[R(K,-1)] Sgn[R(K, 0.0963)] Sgn[R(K, 0.8358)] Sgn[R(K, 1)] 1 -1 1 -1 17 Digital PID Controller Design ² This translates into the following set of inequalities: R(K; ¡1) = ¡23:348 + 21:5185K > 0 ) K > 1:085 R(K; 0:0963) = ¡12:998 + 5:2709K < 0 ) K < 2:466 R(K; 0:8358) = ¡0:9232 + 0:7673K > 0 ) K > 1:2032 R(K; 1) = ¡0:4050 + 0:2403K < 0 ) K < 1:6854: ² The closed loop system is stable for 1:2032 < K < 1:6854. ² In this example, we have xj; j = 0; 1; 2; 3. Each xj may assume the value +1 or ¡1 since 0 is excluded as we are testing for stability. This leads to 24 = 16 possible strings which may satisfy the signature requirement. In this example, only one string of the possible 16 satis¯es the signature requirement. 18 Digital PID Controller Design Stabilization with PI Controllers N (z) ² Plant and Controller: P (z) = , D(z) C(z) = K1 (z ¡ K2) z¡1 ² The characteristic polynomial: ±(z) =(z¡1)D(z)+K1(z¡K2)N(z) ² Writing the Tchebyshev representations of D(z), N (z) and N (z¡1) ² Then to achieve parameter separation, we calculate ¡¢ ³ p ´³ p ´ ±(z)N z¡1 ju=¡ cos μ = ¡u ¡ 1 + j 1 ¡ u2 P1(u) + j 1 ¡ u2P2(u) p +jK1 1 ¡ u2P3(u) ¡ K1 (u + K2) P3(u) where ¡ ¢ P1(u) = RD(u)RN (u) + 1 ¡ u2 TD(u)TN (u) P2(u) = RN (u)TD¡(u) ¡ TN¢ (u)RD(u) P3(u) = RN2 (u) + 1 ¡ u2 TN2 (u): 19 Digital PID Controller Design ±(z)N ¡z¡1¢¯¯z=ejμ;u=¡ cos μ = ± (z )Nr zl (z ) ¯¯¯¯ z=ejμ ;pu=¡ cos μ = R (u; K1; K2) + 1 ¡ u2T (u; K1) ¡ ¢ where R (u; K1; K2) = ¡(u + 1)P1(u) ¡ 1 ¡ u2 P2(u) ¡ K1 (u + K2) P3(u) T (u; K1) = P1(u) ¡ (u + 1)P2(u) + K1P3(u): ² For a ¯xed value of K1, we calculate the real distinct zeros ti of T (u; K1) of odd multiplicity for u 2 (¡1; 1): ¡1 < t1 < ¢ ¢ ¢ < tk < +1: ² Let i±, iNr be the number of zeros of ±(z) and Nr(z) inside the unit circle, respectively, then we have £ ¤μ i± + iNr ¡ l = 1 2 Sgn T (p)(¡1) Sgn [R (¡1; K1; K2)] +2 Pk j=1 (¡1)j Sgn [R (tj ; K1; K2)] + (¡1)k+1Sgn [R (+1; K1; ¶ K2)] : 20 Digital PID Controller Design Stabilization with PD Controllers N (z) ² Plant and Controller: P (z) = , D(z) C(z) = K1 (z ¡ K2) z ² The characteristic polynomial: ±(z) = zD(z) + K1 (z ¡ K2) N(z) ² Consider ±(z)N ¡ z ¡1¢¯¯ z =ejμ ;u=¡ cos μ = R (u; K1; K2) + p j1 ¡ u2T (u; K1) where ¡ ¢ R (u; K1; K2) = ¡uP1(u) ¡ 1 ¡ u2 P2(u) ¡ K1 (u + K2) P3(u) T (u; K1) = K1P3(u) + P1(u) ¡ uP2(u): ² Parameter separation has again been achieved, that is, K1 appears only in the imaginary part and for ¯xed K1 the real part is linear in K2. ² Thus the application of the root counting formulas will yield linear inequalities in K2, for ¯xed K1. 21 Digital PID Controller Design STABILIZATION WITH PID CONTROLLERS ² PID Controller: C(z) = K2z2 + K1z + K0 z(z ¡ 1) ² The characteristic polynomial becomes ¡ ¢ ±(z) = z(z ¡ 1)D(z) + K2z2 + K1z + K0 N (z) ² Multiplying the characteristic polynomial by z¡1N (z¡1), ¡¢ ¡ ¢¡ ¢ ¡¢ z¡1±(z)N z¡1 = (z ¡ 1)D(z)N z¡1 + K2z + K1 + K0z¡1 N (z)N z¡1 : ² Using the Tchebyshev representations, we have ¡¢ ¡ ¢ z¡1±(z)N z¡1 = ¡(u + p 1)P1(u) ¡ 1 ¡ u2 P2(u) ¡ [(K0 + K2) u ¡ K1] P3(u) +j 1 ¡ u2 [¡(u + 1)pP2(u) + P1(u) + (K2 ¡ K0) P3(u)] = R (u; K0; K1; K2) + j 1 ¡ u2T (u; K0; K2) : 22 ² Let K3 := K2 ¡ K0: Digital PID Controller Design ² Then KP = ¡K1 ¡ 2K0; KI = ; K0 +K1 +K2 T and KD = K0T: ² Hence we rewrite R (u; K0; K1; K2) and T (u; K0; K2) as follows. ¡ ¢ R (u; K1; K2; K3) = ¡(u + 1)P1(u) ¡ 1 ¡ u2 P2(u) ¡ [(2K2 ¡ K3) u ¡ K1] P3(u) T (u; K3) = P1(u) ¡ (u + 1)P2(u) + K3P3(u) ² The parameter separation achieved : K3 appears only in the imaginary part and K1; K2; K3 appear linearly in the real part. ² Thus by applying root counting formulas to the rational function on the left, and imposing the stability requirement yields linear inequalities in the parameters for ¯xed K3. ² The solution is completed by sweeping over the range of K3 for which an adequate number of real roots tk exist. 23 Digital PID Controller Design Example 1 ² Plant: G(z) = z2 ¡ 0:25 ² Then RD(u) = 2u2 ¡ 1:25, TD(u) = ¡2u, RN (u) = 1, TN (u) = 0 P1(u) = 2u2 ¡ 1:25; P2(u) = ¡2u; P3(u) = 1 ² Since G(z) is of order 2 and C(z), the PID controller, is of order 2, the number of roots of ±(z) inside the unit circle is required to be 4 for stability. ² From Theorem (Root counting for a real polynomial), ii ¡ i2 = |(i± +{ziNr}) ¡ |(l +{z1}) i1 i2 where i± and iNr are the numbers of roots of ±(z) and the reverse polynomial of N (z) inside the unit circle, respectively and l is the degree of N (z). 24 Digital PID Controller Design ² Since the required i± is 4, iNr = 0, and l = 0, i1 ¡ i2 is required to be 3. ² To illustrate the example in detail, we ¯rst ¯x K3 = 1:3. ² Then the real roots of T (u; K3) in (¡1; 1) are ¡0:4736 and ¡0:0264. ² Furthermore, Sgn[T (¡1)] = 1, i1 ¡ i2 = 3 requires that: μ ¶ 1 Sgn[T (¡1)] Sgn[R(¡1)] ¡ 2Sgn[R(¡0:4736)] + 2Sgn[R(¡0:0264)] ¡ Sgn[R(1)] = 3 2 ² We have only one valid sequence satisfying the above equation, Sgn[R(-1)] Sgn[R(-0.4736)] Sgn[R(-0.0264)] Sgn[R(1)] 2 (i1 ¡ i2) 1 -1 1 -1 6 ² From this valid sequence, we have the following set of linear inequalities. ¡1:3 + K1 + 2K2 > 0 ¡0:9286 + K1 + 0:9472 < 0 1:1286 + K1 + 0:0528K2 > 0 ¡0:2 + K1 ¡ 2K2 < 0: 25 Digital PID Controller Design 23 2 32 3 2 32 32 3 KP ¡2 ¡1 0 K0 ¡2 ¡1 0 0 1 ¡1 K1 4 KI 5 = 4 1 T 1 T 1 T 54 K1 5=4 1 T 1 T 1 T 54 1 0 0 5 4 K2 5 KD T 2 0 0 3 2K2 3 T 00 01 0 K3 ¡1 ¡2 2 K1 = 4 1 T 2 T ¡ 1 T 5 4 K2 5: 0 T ¡T K3 Stability regions in (K1, K2, K3) space (left) and (KP , KI, KD) space (right) 26 Digital PID Controller Design Maximally Deadbeat Control ² The design scheme attempts to place the closed loop poles in a circle of minimum radius ½. Let S½ denote the set of PID controllers achieving such a closed loop root cluster. ² We show below how S½ can be computed for ¯xed ½. The minimum value of ½ can be found by determining the value ½¤ for which S½¤ = Á but S½ 6= Á; ½ > ½¤. ² PID Controller: C(z) = K2z2 + K1z + K0 z(z ¡ 1) ² The characteristic equation ¡ ¢ ±(z) = z(z ¡ 1)D(z) + K2z2 + K1z + K0 N (z): ² Note that p D(z)jz=¡½u+j½p1¡u2 = RD (u; ½) + j 1 p ¡ u2TD(u; ½) N (z)jz=¡½u+j½p1¡u2 = RN (u; ½) + j 1 ¡ u2TN (u; ½) 27 Digital PID Controller Design N ¡ ½2z ¡1 ¢¯¯ z=¡½u+j p ½ 1¡u2 = N (z)jz=¡½u¡jp½p1¡u2 = RN (u; ½) ¡ j 1 ¡ u2TN (u; ½): ² We evaluate ¡¢ £ ¡ ¢ ¤¡ ¢ ½2z¡1±(z)N ½2z¡1 = ½2z¡1 z(z ¡ 1)D(z) + | K2{zz2 + K1z + K0 N (z) N } ½2z¡1 ±(z) over the circle C½ ½2z¡1±(z)N ¡ ½2 z¡1 ¢¯¯ p z=¡½u¡+j½ 1¡u¢2 £¡ ¢ ¤ = ¡½2(½u p + 1)£P1(u; ½) ¡ ½3 1 ¡ u2 P2(u; ½) ¡ ¡K0 + K2½2 ½u ¢ ¡ K1 ½2 ¤ P3(u; ½) +j 1 ¡ u2 ½3P1(u; ½) ¡ ½2(½u + 1)P2(u; ½) + K2½2 ¡ K0 ½P3(u; ½) where ¡ ¢ P1(u; ½) = RD(u; ½)RN (u; ½) + 1 ¡ u2 TD(u; ½)TN (u; ½) P2(u; ½) = RN (u; ½)TD¡(u; ½) ¡¢TN (u; ½)RD(u; ½) P3(u; ½) = RN2 (u; ½) + 1 ¡ u2 TN2 (u; ½): 28 Digital PID Controller Design ² By letting K3 := K2½2 ¡ K0, ² we have ½2z¡1±(z)N ¡ ½2 z ¡1¢¯¯ p z=¡½u¡+j½ 1¡u¢2 £¡ ¢ ¤ = ¡½2(½u p + 1)£P1(u; ½) ¡ ½3 1 ¡ u2 P2(u; ½) ¡ 2K2½2 ¡ K3¤ ½u ¡ K1½2 P3(u; ½) +j 1 ¡ u2 ½3P1(u; ½) ¡ ½2(½u + 1)P2(u; ½) + K3½P3(u; ½) : ² Fix K3, use the root counting formulas, develop linear inequalities in K2; K3 and sweep over the requisite range of K3. This procedure is then performed as ½ decreases until the set of stabilizing PID parameters just disappears. 29 Digital PID Controller Design Example ² We consider the same plant used in the previous example. ² Left ¯gure shows the stabilizing set in the PID gain space at ½ = 0:275. 30 Digital PID Controller Design ² For a smaller value of ½, the stabilizing region in PID parameter space disappears. This means that there is no PID controller available to push all closed loop poles inside a circle of radius smaller than 0:275. ² From this we select a point inside the region that is K0 = 0:0048; K1 = ¡0:3195; K2 = 0:6390; K3 = 0:0435: ² From the relationship between parameters, we have 2 4 KP KI 32 5 = 64 ¡1 1 KD T 0 ¡2½2 ½2 1 + TT ½2T 2 1 ¡ 32 3 2 75 4 K1 K2 5 = 4 3 0:3099 0:3243 5 T ¡T K3 0:0048 ² Right ¯gure shows the closed loop poles that lie inside the circle of radius ½ = 0:275. The roots are: 0:2500 § j0:1118 and 0:2500 § j0:0387: 31 Digital PID Controller Design ² We select several sets of stabilizing PID parameters from the set obtained in the previous example (i.e., ½ = 1) and compare the step responses between them. 32 Maximum Delay Tolerance Design Digital PID Controller Design ² Finding the maximum values of L¤ such that the stabilizing PID gain set that simultaneously stabilizes the set of plants z¡LG(z) = N (z) ; zLD(z) is not empty. for L = 0; 1; ¢ ¢ ¢ ; L¤ ² Let Si be the set of PID gains that stabilizes the plant z¡iG(z). Then \Li=0Si stabilizes ziG(z) for all i = 0; 1; ¢ ¢ ¢ ; L: 33 Example Digital PID Controller Design ² The right ¯gure shows the stabilizing PID gains when L = 0; 1. As seen in the ¯gure, the size of the set is reduced as the delay increases. 34 Digital PID Controller Design ² In many systems, the set disappears for a large value of L¤. This is the maximum delay that can be stabilized by any PID controllers. 35

    Top_arrow
    回到顶部
    EEWORLD下载中心所有资源均来自网友分享,如有侵权,请发送举报邮件到客服邮箱bbs_service@eeworld.com.cn 或通过站内短信息或QQ:273568022联系管理员 高进,我们会尽快处理。