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If the polar plot for Y 1 j o approaches the critical point (À1Y 0), the system is at the limit of stability, the logarithmic magnitude is 0 dB, and the phase angle is p on the Bode diagram. Let us now consider the translational mechanism with electrical drive discussed in the preceding section. From Eq. (9.29) we have Y 1 j o k H jojot 1 1j ot 2 1 X 9X48 We assume k H 1, t 1 0X2, t 2 0X1. The Bode characteristics are presented in Fig 9.7. The critical frequency is o c 1. The gain margin is estimated by the segment AB, d % 20 dB, and the phase margin is evaluated by the segment CD, w %À80 . 10. Design of Closed-Loop Control Systems by Pole-Zero Methods In the preceding sections we analyzed the design and adjustment of the system parameters in order to provide a desirable quality of the control Figure 9.7 Bode characteristics for the translational mechanism with electrical drive. 10. Design of Closed-Loop Control Systems by Pole-Zero Methods 649 Control system. But often it is not simple to adjust a parameter of a technological process that has a complex con®guration. It is preferable to reconsider the structure of the control system and to introduce new structure components that allow a better selection and adjustment of the parameters for the overall system. These new structure components are called controllers. 10.1 Standard Controllers In Fig. 10.1 we present a feedback control system in which a controller ensures the quality of the control system. The adjustment of the controller parameters in order to provide suitable performance is called compensation. The transfer function of the controller is designated as Y c s X c s E s Y 10X1 where Es is the system error and X c s de®nes the output of the controller. This variable acts as the input for the second component, the driving system, which represents an interface between the controller and the mechanical process: Y D s X p s X c s X 10X2 If we suppose that the transfer function of the mechanical system (process) is Y p s, the open-loop system has the transfer function Y 1 sY c sÁY D sÁY p s and the closed-loop control system Y s Y c sÁY D sÁY p s 1 Y c sÁY D sÁY p sÁY T s Y 10X3 where Y T s represents the transfer function of the transducer on the feed- back path. In order to facilitate the selection of the best control structure, several types of standard controllers are used. (a) The P controller (proportional controller) is de®ned by the equation x c tK p Á etX 10X4 Figure 10.1 A feedback control system with controller. 650 Control Control This controller provides a proportional output as a function of the error Y c sK p X 10X5 (b) I controller (integration controller): x c t 1 T i etdt 10X6 Y c s 1 T i s X 10X7 (c) PI controller (proportional-integration controller): x c tK p et 1 T i etdt 10X8 Y c sK p 1 1 T i s X 10X9 (d) PD controller (proportional-derivative controller): x c tK p etT d det dt 10X10 Y c sK p 1 T d sX 10X11 (e) PDD 2 controller (proportional-derivative-derivative controller): x c tK p T d 1 T d 2 d 2 et dt 2 T d 1 T d 2 det dt et 10X12 Y c sK p 1 T d 1 s1 T d 2 sX 10X13 (f) PID controller (proportional-integration-derivative controller): x c tK p et 1 T i tdt T d det dt 10X14 Y c sK p 1 1 T i s T d s X 10X15 The design of a control system requires the selection of a type of controller, the arrangement of the system structure, and then the selection and adjustment of the controller parameters in order to obtain a set of desired performances. If the theoretical design of the controller requires a transfer function more complex than those of PID or PDD 2 controllers, it is preferable to connect several types of standard controllers that can achieve the desired performance. 10.2 P-Controller Performance We consider the transfer function of an open-loop system (5.8) rewritten in the form Y 1 s K s l Á Qs Rs Y 10X16 10. Design of Closed-Loop Control Systems by Pole-Zero Methods 651 Control where l 0Y 1Y 2(l b 2 determines the instability of the system) and Qs, Rs are polynomials with coef®cients of s 0 equal to 1: Q0 R0 1X From Eq. (10.16) we obtain K lim s30 s l Á Y 1 sX 10X17 If we consider the transfer function for l 0 (a type- zero system) in Eq. (5.8), Y 1 s A Á m i1 s z i n j1 s p j Y 10X18 where Àz i , Àp j represent the zeros and poles of the open-loop system, respectively, we obtain K A Á m i1 z i n j1 p j X 10X19 For a unit step input, from Eq. (6.6) we get the steady error E s 1 1 A Á m i1 z i n j1 p j X 10X20 We will now analyze closed-loop systems. First, we consider the transfer function of a thermal heating system (9.12)±(9.14), Y 1 s k 1 t 1 s 1 Y with the closed-loop transfer function Y s X 0 s X i s Y 1 s 1 Y 1 s k 1 * s p 1 Y 10X21 where p 1 À 1 k 1 t 1 Y k 1 * k 1 t 1 Y V b b ` b b X 10X22 and k 1 , t 1 are de®ned in Eq. (9.14). 652 Control Control The transient response for a unit step input x i t will be X 0 sY sÁX i s X 0 s k 1 * s p 1 Á 1 s X 10X23 Expanding Eq. (10.23) in a partial fraction expansion, we obtain X 0 s c 0 s c 1 s p 1 Y 10X24 where c 0 s Á X 0 sj s0 k 1 * p 1 10X25 c 1 s p 1 ÁX 0 sj sÀp 1 Àk 1 * p 1 X 10X26 Then, the relation (10.24) becomes x 0 t k 1 * p 1 À k 1 * p 1 e Àp 1 Át X 10X27 The transient response is presented in Fig. 10.2. It is composed of the steady-state output k 1 *ap 1 and an exponential term k 1 *ap 1 e Àp 1 t . The steady- state error will be E s 1 À x 0 I 1 À k 1 * p 1 X 10X28 We remark that when p 1 approaches the origin (jp 1 j decreases), the time constant 1ap 1 and also the duration of the transient response increase. It is clear that a fast transient response requires a large p 1 that will determine the increase of the steady-state error. As a second case, we consider the closed- loop transfer function for a translational mechanism (Fig. 5.4). The open-loop transfer function is given by Eq. (9.16). The closed-loop transfer function will be Y s o 2 n s 2 2zo n s o 2 n Y 10X29 Figure 10.2 Transient response for closed-loop control of a thermal system. 10. Design of Closed-Loop Control Systems by Pole-Zero Methods 653 Control where the natural frequency o n and damping ratio z are o 2 n k 1 t 1 z 1 t 1 o n Y V b b ` b b X 10X30 and k 1 , t 1 are given in Eq. (9.17). The system poles are (Fig. 10.3) p 1Y2 Àzo n Æ 1 À z 2 q X 10X31 In Section 6, we obtained that the steady-state error is E s 0 10X32 for a unit step input (6.8), and E s 2z o n Y 10X33 for a ramp input (6.12). The overshoot of the transient response can be obtained by using the identity s 2 2zo n s o 2 n s zo n 2 o n 1 À z 2 q 2 X 10X34 From Eq. (10.29), X 0 sY sÁX i s 1 s À s 2zo n s 2 2zo n s o 2 n Y 10X35 or X 0 s 1 s À s zo n s zo n 2 o n 1 À z 2 p 2 23 z 1 À z 2 p Á o n 1 À z 2 p s zo n 2 o n 1 À z 2 p 2 23 X 10X36 Figure 10.3 Poles of a closed-loop for a translational mechanism. 654 Control Control The inverse Laplace transform of Eq. (10.36) will give x 0 t1 À e Àzo n t 1 À z 2 p Á sin o n 1 À z 2 q t tan À1 1 À z 2 p z 2323 X 10X37 The transient response is shown in Fig. 10.4. The maximum value of the time response is obtained for dx o t dt 0X 10X38 We obtain the values of time for which x 0 t achieves the extremes [4] t ex kp o n 1 À z 2 p Y k 0Y 1Y 2Y FFFX 10X39 For k 0 we obtain the absolute minimum value at k 0; for k 1we obtain the peak value time T p p o n 1 À z 2 p X 10X40 If we substitute T p in Eq. (10.37) we obtain the overshoot s e Àpza 1Àz 2 p X 10X41 We see that for z 0, the overshoot is 100 (the system is at the limit of stability) and for z b 0X85 the overshoot approaches zero. The settling time (T s ) is de®ned as the time required for the system to settle within a certain percentage d of the input amplitude. From Eq. (10.36) we obtain the condition [4] e Àzo n T s 1 À z 2 p dY 10X42 Figure 10.4 Transient response of a closed-loop control for a translational mechanism. 10. Design of Closed-Loop Control Systems by Pole-Zero Methods 655 Control and T s lnd 1 À z 2 p Àzo n X 10X43 The bandwidth (o B ) was discussed in Section 8. From Eqs. (8.35), (8.40), and (8.29) we obtain o 2 n o 2 n À o 2 B 2 2zo n o B 2 q 2 p 2 X 10X44 The bandwidth o B will be o B o n 1 À 2z 2 2 À 4z 2 4z 4 q r X 10X45 For example, for z 0X5, o B 1X27o n Y 10X46 and for z 0X7, o B o n X 10X47 10.3 Effects of the Supplementary Zero We consider a closed-loop control system as in Fig. 10.1 where the controller is de®ned by a PD transfer function (10.11). We assumed that the controlled process is represented by the translational mechanism (9.16). The closed- loop transfer function will be Y PD s o n z Ás z s 2 2zo n s o 2 n Y 10X48 where z is the zero introduced by the PD controller, z À 1 T D Y 10X49 and o 2 n k p t z 1 k p T D 2to n X 10X50 The transfer function of the open-loop control system from Fig. 10.5 represents a type-one system, so that the steady-state error will be E s 0Y 10X51 for a unit step input. For a ramp input signal, we obtain from Eq. (6.10) E s lim s30 1 s Á Y 1 PD s 45 Y 10X52 656 Control Control where Y 1 PD s Y PD s 1 À Y PD s Y Y 1 PD s o 2 n z Ás z ss 2zo n À o 2 n z X 10X53 Substituting Y 1 PD s in Eq. (10.52), we obtain E s PD 2z À o n z o n X 10X54 It is clear that the steady-state error decreases by the value l o n az .If we cancel the effect of the zero, z 3I, the PD steady-state error approaches the P steady-state error, E s PD 3 E s P 2z o n X From Eq. (10.54) we also have the condition 2z b o n z X 10X55 In order to analyze the transient response, we will rewrite (10.48) as Y PD s 1 s z Y P sY 10X56 where Y P s represents the closed-loop transfer function with a P controller discussed in the preceding section, Y P s o 2 n s 2 2zo n s o 2 n X 10X57 For a unit step input, the output x 0 t will be X 0 sY P sÁX i s 1 z sY P sX i sX 10X58 The inverse Laplace transformation of (10.58) will give x 0 PD tx 0 P t 1 z dx 0 P t dt Y 10X59 where x 0 PD , x 0 P denote the output signal for a PD controller or a P controller in the control system, respectively. It is clear that the overshoot of this system will be increased by the term 1az Ádx 0 P tadt (Fig. 10.6). Figure 10.5 A closed-loop control system with PD controller. 10. Design of Closed-Loop Control Systems by Pole-Zero Methods 657 Control From Eqs. (10.37) and (10.59) we obtain x 0 PD t1 À e Àzo n t l 2 À 2zl 1 p 1 À z 2 p sino n 1 À z 2 q t gY 10X60 where g tan À1 1 À z 2 p z À l 10X61 l o n z X 10X62 The maximum value is obtained by dx 0 PD t dt 0Y 10X63 which enables us to calculate the time [4]: T P PD p Àg À j o n 1 À z 2 p X 10X64 We remark that if z 3I, l 3 0, g j and the value of (10.64) is the same as that determined for the P-controller (10.40). The overshoot will be s PD l 2 À 2zl 1 q Á e ÀzÁpÀgÀja 1Àz 2 p X 10X65 The settling time T s PD can be determined by using the condition e Àzo n T s PD Á l 2 À 2zl 1 p 1 À z 2 p 0X05X 10X66 If we develop Eq. (10.66) and consider the settling time T s P de®ned by (10.42), (10.43), we obtain T s PD 1 zo n Á ln l 2 À 2zl 1 q T s P X 10X67 Figure 10.6 Transient response with PD controller. 658 Control Control [...]... obtain the pole phase angle j cosÀ1 z 50 30H X 10X126 The condition (10 .123 ) of the ramp steady-state error determines the natural frequency from Eq (10.33), 2z on 0X02X Therefore, we obtain on ! 63X5Y 10X127 oB 1X1oN Y 10X128 but the relation (10.45) requires and from the condition (10 .124 ), 91X oN 10X129 The inequalities (10 .127 ) and (10 .129 ) de®ne the natural frequency domain on , 6X5 oN... matrix form (12. 3), we will have P Q 0 1 AR k kf S À À M M A B X M 12X11 12X12 We assume that only the position is measurable, so that we have for the output y Cx Y 12X13 y x1 Y C 1 0X 12X14 where The second example is represented by a coupled spring±mass system shown in Fig 12. 2 The dynamic model is described by the differential equations @ m1 z1 k1 z1 À z2 F 12X15 m2... Equation (12. 2) can be rewritten in matrix form, x Ax BuY where P a11 Ta T 21 AT F T F R F an1 P 12X3 Q a1n a2n U U U U S a12 a22 ÁÁÁ ÁÁÁ an2 Á Á Á ann b11 T F BT F R F ÁÁÁ b1m bn1 ÁÁÁ 12X4 Q bnn U UY S and u u1 Y u2 Y F F F Y um T 12X5 de®nes the input vector of the system The initial state of the system is de®ned by the vector x0 x1 t0 Y x2 t0 Y F F F Y xn t0 T X 12X6 The... this system, M z kf z kz AP X 12X9 Control 673 12 State Variable Models 674 Control Control Figure 12. 1 The linear spring±mass± damper mechanical system The state variables that can de®ne this system rigorously are the position and the velocity We can write x1 z x2 z Y 12X10 the system state variables The input is pressure p u Equation (12. 10) can be rewritten as x1 x2 x2... be Y1 s 15kP X s s 95 10X120 The control system requires the following performance: Overshoot: s7 7X57X 10X121 Steady-state error: Es 0Y 10X122 for unit step input, and Es 7 27 10X123 for ramp input Bandwidth: 100X oB 10X124 The last condition of the bandwidth allows the estimation of the damping ratio From Eq (10.41) we obtain z 0X636X 10X125 If we use the pole representation... m1 U T U R 0 S 0 C 1 0 0 0 0 0 1 0 0 Q U U U U 1 U U kf S À m2 0 ! X 12X16 12X17 The mathematical model offered by the matrix equations (12. 3) and (12. 7) is called in the literature [9, 18] ``the input±state±output'' model In matrix form, the solution of Eq (12. 3) can be written as an exponential function [8, 9, 18]: t 12X18 x t exp At x 0 expA t À tBu td tX 0 The Laplace transform... we consider the input u 0, we obtain x t f t x 0Y 12X23 X s f sx 0X 12X24 or We can rewrite Eq (12. 23) by components: QP P Q P Q f11 t f12 t Á Á Á f1n t x1 0 x1 t T x2 t U T f21 t Á Á Á f2n t UT x2 0 U UT T U T U UT F U T F UT F SR F S R F S R F F F F xn t fn1 t fn2 t Á Á Á fnn t 12X25 xn 0 From this equation we see that the matrix coef®cient... system (Fig 12. 1) described by Eq (12. 11) The signal diagram ¯ow in Laplace variable is shown in Fig 12. 3 We note, therefore, that in order to determine the matrix coef®cients, it is necessary to evaluate the Xi s, changing the initial conditions xi 0 Thus, the coef®cient f11 s is obtained from the initial conditions x1 0 1; x2 0 0 677 Control 12 State Variable Models Figure 12. 3 The signal-¯ow... diagram for the linear spring± mass±damper mechanical system From Fig 12. 3 we can easily obtain 1 X1 1 X2 s kf 1 k À X1 À X2 Y X2 M s M then s f11 s X1 s s2 kf kf M k s M M X 12X28 If we repeat this procedure for all matrix coef®cients, we obtain f12 s 1 kf k s2 s M M s f21 s kf M kf k s M M s X f22 s kf k s2 s M M s2 12X29 The transition matrix f t is... m2 z2 kf z2 k2 z2 À k1 z1 À z2 0X Figure 12. 2 The coupled spring±mass system 675 12 State Variable Models Control We de®ne the state vector as x x1 Y x2 Y x3 Y x4 T Y where x 1 z1 x 2 z1 x 3 z2 x 4 z2 Y the input is u FY and the output variables are represented by positions z1 , z2 y y1 Y y2 T X Equations (12. 3) and (12. 7) will have the P 0 1 0 k1 T k1 TÀ 0 T m1 m1 AT . 10X127 but the relation (10.45) requires o B 1X1o N Y 10X128 and from the condition (10 .124 ), o N 91X 10X129 The inequalities (10 .127 ) and (10 .129 ) de®ne the natural frequency domain o n , 6X5 o N . 95 X 10X120 The control system requires the following performance: Overshoot: s7 7X57X 10X121 Steady-state error: E s 0Y 10X122 for unit step input, and E s 7 27 10X123 for ramp. 50 30 H X 10X126 The condition (10 .123 ) of the ramp steady-state error determines the natural frequency from Eq. (10.33), 2z o n 0X02X Therefore, we obtain o n ! 63X5Y 10X127 but the relation