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Unformatted text preview: 'Figure 441: Paper machine response data for Problem 4.9 9. ¥The unitstep 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 FEDcontroller 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 = —12Pu = 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 PIDcontroller 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 FEDcontroller 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 nonlinearity 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 antiwindup feedback gain Ka, and select a value that '
gives good response to largesteps. ' (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 antiwindup 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 antiwindup. Note
that with an input limited to 10, the output cannot be maintained
at any values greater than this. The system with. antiwindup has a.
much better stepresponse 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 secondorder 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 steadystateerror 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 steadystate 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 steadystate 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
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 Fall '07
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