problem 4.9

problem 4.9 - 'Figure 441: Paper machine response data for...

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Unformatted text preview: 'Figure 441: Paper machine response data for Problem 4.9 9. ¥The unit-step response of apaper machine is shown‘in Fig. 4.41(a) where the input into the system is stock flow onto the wire and the output is basis ' I ' weight (thickness). The time delay and slope of the transient response may V) be determined from the figure. (a) Find the proportional, PI, and PID—controller parameters Using the Zeigler—Nichols transient—response method. (b) Using proportional feedback control, control designers have obtained a closed—loop system with the‘unit impulse response shown in Fig. 4.4l(b). . When the gain K“ = 8.556, the system is on the verge of ' instability. Determine the proportional-, PL, and FED-controller pa- rameters according to the Zeigler—Nichols ultimate sensitivity method. Solution: (a) From step response: L = 7d 2 0.65 sec '_ 1 N 0.2. _ _, R" 1- * 1.25 —0.65 " 0‘33 se“ From Table 4.1: Controller Gain P K = fi—L = 4.62 L PI K=°-9 =4. K 5 182 CHAPTER 4. BASIC PROPERTIES OF FEEDBACK Figure 4.42: Unit impulse response for paper machine in Problem 4.10 (b) From the impulse response: .P z 2.33 sec. and from Table 4.2: -‘ Controller Gain P K = 0.5m. = 4.28 ' 1 . PI K = 0.45Ku = 3.85 T, = ——P = 1.86 1.2 u ' 1 PID K = 0.6Ku = 5.13 T1 = —12-Pu = 1.12TD = g 10. ¥A paper machine has the transfer function e—Za 3s+1’ Where the input is stock flow onto the wire and the output is basis weight or thickness. G(s) = (3.) Find the PID-controller parameters'using the Zeigler—Nichols tuning rules. (b) The system becomes marginally stable for a proportional gain of Ku = 3.044 as shown by the unit impulse response in Fig. 4.42. Find the optimal FED-controller parameters according to the Zeigler— Nichols tuning rules. Solution: (a) From the transfer function: L = rd 2 2 sec R = g z 0.33 sec‘1 P9 = 0.28 183 From Table 4.1: 1 Controller Gain P . K = EELS PI ‘K=%=1.35 TI=E=6.66 I PID K=112—-§3=1.8 TI=2L=4TD=0.5L=1.0 (b) From the impulse response: P r: 7 sec From Table 4.2: Controller Gain P. K PI K 0.5Ku = 1.52 I 0.45Ku e 1.37 T1 = T153, = 5.83 PID K ' 0.6Ku = 1.82 T1 = éPu = 3.5TD = éPu = 0.875 11. Consider the system with the plant transfer function 1 32+1“ We would lilre to use PID control on this system. It is known that the system’s actuator is a saturation non-linearity with a. slope of .1 and S 10. 0(3) = (a) Design the values of hp, 193, and k1 so that the closed loop charac- teristic equation has roots at s = ——1, —1 :lsz. Connect the derivative term to the output, not to the error with the other terms, and use the . —-kD3 modified form 013 + 1 pure derivative term. ' . (b) Using SIMULINK‘as in Fig.4.23 add an anti—windup‘system using the techniques discussed in this chapter. Experiment with different values for the anti-windup feedback gain Ka, and select a value that ' gives good response to large-steps. ' (0) Plot both the step tracking response and the control efiort for steps that cause the actuator to saturate. Qualitativer describe the effect of the anti-windup on both the output response and the control effort. Solution: since MATLAB cannot realize the unrealistic - (a) We start with an ideal PID controller of the form, D(s)=kp+£sI-+kps and determine the values from the characteristic equation. 83+352+43+2=33+kpsz+(1+kp)s+k1 4.2. 184 CHAPTER 4. BASIC PROPERTIES OF FEEDBACK from which k, = 3, k1 = 2, and k3 = 3. The step response of the sys— tem with the imperfect derivative is shown below. The two methods ‘in the text can yield the same response. . The step response for an input of size 9 is shown below with and without anti-windup. Note that with an input limited to 10, the output cannot be maintained at any values greater than this. The system with. anti-windup has a. much better step-response and lower control effort.The SIMULINK block diagram for the solution is shown below. With no antiwindup the gain is set to K = 0 and for antiwindup, K = 30. Simulink diagram for Problem 4.11 . Problems and solutions for Section 4.3 12. Consider the second-order plant 1 G“) = (s + 1)(5s + 1)‘ 185 (a) Determine the system type and error constant with respect to track'- ing polynomial reference inputs of the system for P, PD, anE‘PID controllers (as configured in Fig.4.2(b)) Let kp = 19, k; = 9.5, and [CD = 4. ‘ (b) Determine the system type and error constant of the system with respect to disturbance inputs for each of the three regulators in part (a) with respect to rejecting polynomial disturbances w(t) at the input to the plant. (c) Is this system better at tracking references or rejecting disturbances? Explain your response briefly. - (d) Verifyyour results for parts (a) and (b) using MATLAB by plot- ting unit step and ramp responses for both tracking and disturbance rejection, Solution: ‘ (a) o P: Y(s) _ kpG(s) _ 19 R(s) - 1 + kpG(s) — 552 + 63 + 20 Using the FVT, yss = é—g- and the steady-stateerror is e” = Type0,Kp=.kp=19 ‘ ' 0 PD: ' ,Y(s) _ D(3)G(s) _ 19+4s R(s) _ 1 + D(s)G(s) _ 532 + 105 + 20 ' Using the FVT, y” = g) and the steady-state error is e” = Type 0, K1, = 1:1, = 19 O PID: _v ' ' Y(s) _ ' D(s)G(s) _ 852 4 383 + 19 3(5) _ 1 + D(3)G(s) ‘ 1033 + 2052 + 405 + 19 Using the FVT, ys‘a = land the steady-state error is zero. Type 1: K1) = [91 = v o P: 2 Y(s) __ G(s) _ 1 W(s) ” 1 + kpG(-s) " 552 + 63 + 20 Using the FVT, yss = Type 0,Kp = 19. 0 PD: Y(s) __ G(s) _ 1 W(s) ’ '1 + D(s)G(s) ’ 532 + 103 + 20 Using the FVT, y” = Type 0, K1, = 19 ° I E53530 Test-(II) SF.20oq« Answer shena- i -_ fin";wa _ ____ ,;+__-——_____ - . .. ._ a .fl4»___Ji.E__QL.6J§u__—‘é_ .e_ _ _.;I;_2_§_fiL_E_Q.L£__ID_;__3flL_BJiQ_P_&SQ : m”) (g+ ? 22—61%: ~ g”) ' (gnu-H) CSféfmo) ,. st:_fl.______-_ _ ,— 4°°°CS‘+5*I) _ _3"'+5+/ .. __ . ______.E‘_+s_:§: _,__:LMMW_QS—w_ __ H K Wmsib _.4___._______ y LN ‘ L. ‘ ' _ LW- _ ____‘______n__; 1, 9,_V_(t)_;._(_/_*I_)_l¢)_____ ____L________‘____'________ I A. .1 .1 - ____—_..;____..___E_C$)_ LEFCSQ%LS.)___. ..__._-_._ ._.—___-. a )1 c s ) ' ' . . .— ‘ E— _ 6 _._;__ ______ _ _ —_— "1:3? U—Ifg—(TFCW r: m R _ —"‘_——_"___ L.___J_€? _——_--if_-- . '_-_E_=_E :t gaging __________ - =_E:_Ll+_C6t)_i-Dé_.__..___.. 9% Q ___;__;L_.~_L _ 8-H) (C6 I2 4'- 6D 5L;— 3 Fig—— _.__- ’ l m ._.._._. __G I _———————____4.— I —€ E ‘ .. o . .. -* _¢ - _ 1.209% _L_°v __ V _%C:T\___O—t g, 2 5m?!" _ n 717) d Lug 1’41“! f”) ll a” .r f . .. ...
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problem 4.9 - 'Figure 441: Paper machine response data for...

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