Assignment # 05 EE 324: Control Systems Semester: Spring: 2013 Due Date: 18.05.2013 at 8a.m. The following concepts have been tested in this homework • Lead, Lag and Lead-Lag Compensators Note: Some Problems are provided with solutions to give you some insight. Lead Compensator Design Lag Compensator 1 Homework-5 Solution
34
Embed
Control Systems Compensator DESIGN via Root Locus.pdf
Solved Examples of Compensator DESIGN via Root Locus.pdf
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Assignment # 05 EE 324: Control Systems Semester: Spring: 2013
Due Date: 18.05.2013 at 8a.m.
The following concepts have been tested in this homework
• Lead, Lag and Lead-Lag Compensators
Note: Some Problems are provided with solutions to give you some insight.
Lead Compensator Design
Lag Compensator
1 Homework-5 Solution
2 Homework-5 Solution
Lead-Lag Compensator Design3 Homework-5 Solution
A-'7-6. Consider a system with an unstable plant as shown in Figure 7-34(a). Using the root-locus approach, design a proportional-plus-derivative controller (that is, determine the values of K, and T ~ ) such that the damping ratio of the closed-loop system is 0.7 and the undamped natural frequency w , is 0.5 radlsec.
Solution. Note that the open-loop transfer function involves two poles at s = 1.085 and s = -1.085 .. and one zero at s = -l/Td, which is unknown at this point.
Since the desired closed-loop poles must have w, = 0.5 rad/sec and f = 0.7, they must be located at
s = 0.5/180° f. 45.573"
(6 = 0.7 corresponds to a line having an angle of 45.573" with the negative real axis.) Hence, the desired closed-loop poles are at
The open-loop poles and the desired closed-loop pole in the upper half-plane are located in the diagram shown in Figure 7-34(b). The angle deficiency at point s = -0.35 f j0.357 is
Figure 7-34 (a) PD control of an unstable plant; (b) root-locus diagram for the system.
Example Problems and Solutions
4 Homework-5 Solution
JAMEEL AHMAD
Highlight
JAMEEL AHMAD
Text Box
Some Solved Problems on Compensators
Figure 7-35 Control system.
This means that the zero at s = - l /Td must contribute 11.939", which, in turn, determines the location of the zero as follows:
Hence, we have
The value of T, is
The value of gain Kp can be determined from the magnitude condition as follows:
or
Hence,
By substituting the numerical values of Td and Kp into Equation (7-12), we obtain
which gives the transfer function of the desired proportional-plus-derivative controller.
A-7-7. Consider the control system shown in Figure 7-35. Design a lag compensator G,(s) such that the static velocity error constant K, is 50 sec-' without appreciably changing the location of the orig- inal closed-loop poles, which are at s = -2 d~ j f i .
462 Chapter 7 / Control Systems Design by the Root-Locus Method
5 Homework-5 Solution
JAMEEL AHMAD
Highlight
Solution. Assume that the transfer function of the lag compensator is
Since K , is specified as 50 sec-', we have
Thus
Now choose K, = 1. Then
Choose T = 10. Then the lag compensator can be given by
The angle contribution of the lag compensator at the closed-loop pole s = -2 + jV6' is
v'6 v% /Gc(s) I = tan-' - - tan-' -
s=-2+jG -1.9 -1.995
which is small.Thus the change in the location of the dominant closed-loop poles is very small. The open-loop transfer function of the system becomes
The closed-loop transfer function is
To compare the transient-response characteristics before and after the compensation, the unit-step and unit-ramp responses of the compensated and uncompensated systems are shown in Figures 7-36(a) and (b), respectively.The steady-state error in the unit-ramp response is shown in Figure 7-36(~).
Example Problems and Solutions 463
6 Homework-5 Solution
JAMEEL AHMAD
Highlight
Figure 7-36 (a) Unit-step responses of the compensated and uncompensated systems; (b) unit- ramp responses of both systems; (c) unit-ramp responses showing steady-state errors.
Unit-Ramp Responses of Compensated and Uncompensated Systems
9 -
Uncompensated system has steady-state error of 0.4
Un~t-Step Responses of Compensated and Uncompensated Systems 1 2 . , , , , , , , , , -
J Compensated system
7----------- Uncompensated system
t Sec
(b)
0
0 4
0 2
Unit-Rarno Resoonse (35 < t < 40)
-
-
. . 40
39 5
39
3 3 8 5 Compensdted system 4 8 38 w
2 3 7 5
5 37 5 5 3 6 5 - Uncompensated system
36
35 5
3535 35 5 36 36 5 37 37 5 38 38 5 39 3 9 5 40
~ / i i i i i i i i ; , o t Sec
(a)
t Sec
( c )
464 Chapter 7 / Control Systems Design by the Root-Locus Method
7 Homework-5 Solution
A-7-44. Consider a unity-feedback control system whose feedforward transfer function is given by
Design a compensator such that the dominant closed-loop poles are located at s = -2 f @V"3 and the static velocity error constant K, is equal to 80 sec-'.
Solution. The static velocity error constant of the uncompensated system is K, = # = 0.625. Since K, = 80 is required, we need to increase the open-loop gain by 128. (This implies that we need a lag compensator.) The root-locus plot of the uncompensated system reveals that it is not possible to bring the dominant closed-loop poles to -2 -+ j 2 s by just a gain adjustment alone. See Figure 7-37. (This means that we also need a lead compensator.) Therefore, we shall employ a lag-lead compensator.
Let us assume that the transfer function of the lag-lead compensator to be
where Kc = 128. This is because
Root-Locus Plot of G(s) = 10/[s(s+2)(s+8)]
Figure 7-37 Root-locus plot of G ( s ) = 101 [ S ( S + 2 ) ( s + 8)). Real Axis
10
8
6
4
v, 2 -
4 Do 0 z - -2
-4
-6
-8
Example Problems and Solutions
-
- Des~red closed-loop
- '----la
-
-
-
-
- lo l0 -5 0 5 10
8 Homework-5 Solution
JAMEEL AHMAD
Highlight
JAMEEL AHMAD
Highlight
Figure 7-38 Graphical determination of the zero and pole of the lead portion of the compensator.
and we obtain Kc = 128. The angle deficiency at the desired closed-loop pole s = -2 + j 2 f l is
Angle deficiency = 120" + 90" + 30" - 180" = 60"
The lead portion of the lag-lead compensator must contribute this angle.To choose Tl we may use the graphical method presented in Section 7-5.
The lead portion must satisfy the following conditions:
and
The first condition can be simplified as
By using the same approach as used in Section 7-5, the zero ( s = 1 1 ~ ~ ) and pole ( s = PIT,) can be determined as follows:
See Figure 7-38. The value of P is thus determined as
/3 = 14.419
466 Chapter 7 / Control Systems Design by the Root-Locus Method
9 Homework-5 Solution
JAMEEL AHMAD
Highlight
For the lag portion of the compensator, we may choose
Figure 7-39 (a) Root-locus plot of compensated system; (b) root- locus plot near the origin.
Then
Noting that
the angle contribution of the lag portion is -1.697" and the magnitude contribution is 0.9837.This means that the dominant closed-loop poles lie close to the desired location s = -2 & j 2 G . Thus the compensator designed,
is acceptable. The feedforward transfer function of the compensated system becomes
A root-locus plot of the compensated system is shown in Figure 7-39(a).An enlarged root-locus plot near the origin is shown in Figure 7-39(b).
q o x ; i . . . . . . . . . . . . . . . . . . . . . . - .:.
-. . . . .:. . . . . . . . .;, . . . . . . . .
-40 -20 0 20 40 60
10 Homework-5 Solution
Figure 7-39 (Continued)
Figure 7-40 (a) Unit step responses of compensated and uncompensated systems; (b) unit- ramp responses of both systems.
To verify the improved system performance of the compensated system, see the unit-step responses and unit-ramp responses of the compensated and uncompensated systems shown in Figures 7-40 (a) and (b), respectively.
Root-Locus Plot of Cqmpensated System near the Origin
Unit-Step Responses of Compensated and Uncompensated Systems
10
8
4
2 a ? , z " -2
-4
-6
-8
t Sec
(4
Chapter 7 / Control Systems Design by the Root-Locus Method
-5 0 5 10 Real Axis
(b)
-
- I - Desired closed-Io ,p -
-
-
-
-
-
poks
11 Homework-5 Solution
Figure 7-40 (Continued)
Un~t-Ramp Responses of Compensated and Uncompensated Systems
9 -
8 -
7 -
6 -
Uncompensated system -
t Sec
('J)
Figure 7-41 Control system.
A-7-9. Consider the system shown in Figure 7-41. Design a lag-lead compensator such that the static velocity error constant K, is 50 sec-' and the damping ratio 5 of the dominant closed-loop poles is 0.5. (Choose the zero of the lead portion of the lag-lead compensator to cancel the pole at s = -1 of the plant.) Determine all closed-loop poles of the compensated system.
Solution. Let us employ the lag-lead compensator given by
where p > 1. Then
K, = lim sG,(s)G(s) s-0
K,(T,S + 1)(T2s + 1) = lims
1
'-' (5, + 1)(pT2s + 1) S ( S + 1) ( s + 5)
- Kc - - 5
The specification that K, = 50 sec-' determines the value of Kc, or
Kc = 250
Example Problems and Solutions
12 Homework-5 Solution
JAMEEL AHMAD
Highlight
We now choose Tl = 1 so that s + ( 1 1 ~ ~ ) will cancel the (s + 1 ) term of the plant. The lead portion then becomes
For the lag portion of the lag-lead compensator we require
where s = s1 is one of the dominant closed-loop poles. For s = sl , the open-loop transfer func- tion becomes
Noting that at s = sl the magnitude and angle conditions are satisfied, we have
where k = 0,1,2,. . . . In Equations (7-13) and (7-14), P and s, are unknowns. Since the damping ratio 5 of the dominant closed-loop poles is specified as 0.5, the closed-loop pole s = s1 can be writ- ten as
where x is as yet undetermined. Notice that the magnitude condition, Equation (7-13), can be rewritten as
Noting that Kc = 250, we have
The angle condition, Equation (7-14), can be rewritten as
Chapter 7 / Control Systems Design by the Root-Locus Method
13 Homework-5 Solution
We need to solve Equations (7-15) and (7-16) for P and x. By several trial-and-error calculations, it can be found that
Thus
The lag portion of the lag-lead compensator can be determined as follows: Noting that the pole and zero of the lag portion of the compensator must be located near the origin, we may choose
Figure 7-43 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.
t Sec
(a)
Unit-Ramp Response of Compensated System
t Sec
(b)
Notice that the dominant closed-loop poles s = -1.8308 f j3.2359 differ from the dominant closed-loop poles s = hs, assumed in the computation of P and T,. Small deviations of the dom- inant closed-loop poles s = -1.8308 h j3.2359 from s = f s, = -1.9054 + j3.3002 are due to the approximations involved in determining the lag portion of the compensator [See Equations (7-17) and (7-18)].
Figures 7-43(a) and (b) show the unit-step response and unit-ramp response of the de- signed system, respectively. Note that the closed-loop pole at s = -0.1684 almost cancels the zero at s = -0.16025. However, this pair of closed-loop pole and zero located near the origin produces a long tail of small amplitude. Since the closed-loop pole at s = -17.205 is located
Example Problems and Solutions 47 3
16 Homework-5 Solution
very much farther to the left compared to the closed-loop poles at s = -1.8308 + j3.2359, the effect of this real pole on the system response is very small. Therefore, the closed-loop poles at s = -1.8308 f j3.2359 are indeed dominant closed-loop poles that determine the response characteristics of the closed-loop system. In the unit-ramp response, the steady-state error in following the unit-ramp input eventually becomes 1/K, = $ = 0.02.
A-7-10. Figure 7-44(a) is a block diagram of a model for an attitude-rate control system.The closed-loop transfer function for this system is
The unit-step response of this system is shown in Figure 7-44(b). The response shows high- frequency oscillations at the beginning of the response due to the poles at s = -0.0417 & j2.4489. The response is dominated by the pole at s = -0.0167. The settling time is approximately 240 sec.
It is desired to speed up the response and also eliminate the oscillatory mode at the beginning of the response. Design a suitable compensator such that the dominant closed-loop poles are at s = -2 + j 2 a .
t Hydraulic servo Aircraft I -
Rate gyro
(a)
Unit-Steo Res~onse of Uncom~ensated System
Figure 7-44 (a) Attitude-rate control system; (b) unit-step response.
b I t
50 100 150 200 250 300 Time (sec)
(b)
Chapter 7 / Control Systems Design by the Root-Locus Method
17 Homework-5 Solution
. .., .• • •--
--
•,1
• " !
I r-,1
II
•/
1/
)00
..
...
•
8-7-7. 'lb9 801\21;ion to such a problem is not unique. "We"imzill _t tlI<>801~lCl11B to the problem in wI1at follows. Nots that frem the requ1remsntstated in the problem. the daD1nant clossd-100p po1BB mat; have ~ • 0.5 and Wn:= 3, or
S=-I.S"±j 2. Nel
Notice that the angle deficiency is . '. . .Angle deficiency = 180 - 120 - 100.894 • - 40.894
Method 1, If 1I'B c:booes the zero ~" the lead """I"""sator at a • -1 80 that itwill cancel the~ pole at a • -1, tIlen the ..............tor pole _ be l"""teclat 8 • -3, or
or
(j. (s) = I<
5+/S+3
)_.Dl..-'Tz
5+1S-f-~
'lb9 value ~ It can be cIetem1ne<1 by Use ~ the magnitude condition.
K = IS(S+.I)30
or
31< 5+/S+ .3
10
S(S+I) =J
$+15i-3
'lb9 open-100p trlInafer _ion is
- 104 -
18 Homework-5 Solution
jameel
Highlight
jameel
Highlight
jameel
Highlight
,(j,,{r) t7/f) = s (s+ 3)
The closec!-loop transfer fun6tion C(s)!R(s) teoromI8 as follows:
crs-J _ --=",,,,,,,,,,,P__K(s) - S I. + 3 SO +.f
Method 2: Referring to the figure shown below, if we bisect angle OPA and' take20.447· each side, thep the locations of the zero and pole are found as follows:
zero at s • -1.9432pole at s • -4.6458
Thus, Ge(S) can be given as
(iris) = I< 7iS-+/ =k I!... s+/.fP-.1Z =2. 'II K S+f.,,..1Z ,72 S T I T2. S T ,. *~rr .7 S'rf4/N"f
.P
j3
jz
o I % ()-
-it
The value of J( can be determined by 1188 at ~ EgnitueSe condiUCIl.
It is interssting to ccmpare the ststic velocity error constants for thetwo systems designed above.
For the system designed by MethDd 1.
K~ =.Ii- s f = 3S ..p S(St3)
For the system designee! by IIet!lod 2.
o.$"/ fZp.s+ I
P·2/S-ZS+/
It)S(S+I)
'DIe system designed by MethDd 2 gives a 1s<ger value ~ the static velocityerror ccnstant. '!his means that the. system designed by _ 2 will giveSIIIS1ler steady-state errors in following ramp inputs t:ban the system designedby_l.
In what follows., we c:ompare the unit-step respcnses Of the three systems:the original unc:mpensated system, the system designed by _ 1, and thesystem designed by _ 2. 'DIe HATLl\Il _am used to obtain the unit-stepresponse curves is given below. 'DIe resulting unit-step response curves areshown on the next page.
" ........ Comparison of unlt-step responses for three systems ........
num - [0 0 10):den - [1 1 10/:num1 - (0 0 9]:denl - [1 3 91:num2 - (0 0 2.644 5.1381:den2 - [0.2152 1.2152 3.644 5.138);I - 0:0.02:8:c - steplnum,den,tJ:c1 - atep(num1.den1,t);c2 - step(num2,den2,t);plotlt,c, ',',t,c1,'.' ,t,c2,'-.')gridtille('Comparlaon of Unlt-Slep Responses for Three Svotemo')xleball'l sec',ylabel('Outputo') .textn.5,1.6,'Uncompensated system')leXlI1.1.0.5,'CompensetedOVOlem with K - 0.5138. Tl - 0.5148, T2 - 0.2152'),exl(1.1,O.3,'Compenseted ovotem with K - 0;3, T1 - 1, T2 - 0.3333')
Un~t-Ramp Responses of Compensated and Uncompensated Systems
9 -
8 -
7 -
6 -
Uncompensated system -
t Sec
('J)
Figure 7-41 Control system.
A-7-9. Consider the system shown in Figure 7-41. Design a lag-lead compensator such that the static velocity error constant K, is 50 sec-' and the damping ratio 5 of the dominant closed-loop poles is 0.5. (Choose the zero of the lead portion of the lag-lead compensator to cancel the pole at s = -1 of the plant.) Determine all closed-loop poles of the compensated system.
Solution. Let us employ the lag-lead compensator given by
where p > 1. Then
K, = lim sG,(s)G(s) s-0
K,(T,S + 1)(T2s + 1) = lims
1
'-' (5, + 1)(pT2s + 1) S ( S + 1) ( s + 5)
- Kc - - 5
The specification that K, = 50 sec-' determines the value of Kc, or
Kc = 250
Example Problems and Solutions
29 Homework-5 Solution
jameel
Highlight
jameel
Highlight
We now choose Tl = 1 so that s + ( 1 1 ~ ~ ) will cancel the (s + 1 ) term of the plant. The lead portion then becomes
For the lag portion of the lag-lead compensator we require
where s = s1 is one of the dominant closed-loop poles. For s = sl , the open-loop transfer func- tion becomes
Noting that at s = sl the magnitude and angle conditions are satisfied, we have
where k = 0,1,2,. . . . In Equations (7-13) and (7-14), P and s, are unknowns. Since the damping ratio 5 of the dominant closed-loop poles is specified as 0.5, the closed-loop pole s = s1 can be writ- ten as
where x is as yet undetermined. Notice that the magnitude condition, Equation (7-13), can be rewritten as
Noting that Kc = 250, we have
The angle condition, Equation (7-14), can be rewritten as
Chapter 7 / Control Systems Design by the Root-Locus Method
30 Homework-5 Solution
We need to solve Equations (7-15) and (7-16) for P and x. By several trial-and-error calculations, it can be found that
Thus
The lag portion of the lag-lead compensator can be determined as follows: Noting that the pole and zero of the lag portion of the compensator must be located near the origin, we may choose
Figure 7-43 (a) Unit-step response of the compensated system; (b) unit-ramp response of the compensated system.
t Sec
(a)
Unit-Ramp Response of Compensated System
t Sec
(b)
Notice that the dominant closed-loop poles s = -1.8308 f j3.2359 differ from the dominant closed-loop poles s = hs, assumed in the computation of P and T,. Small deviations of the dom- inant closed-loop poles s = -1.8308 h j3.2359 from s = f s, = -1.9054 + j3.3002 are due to the approximations involved in determining the lag portion of the compensator [See Equations (7-17) and (7-18)].
Figures 7-43(a) and (b) show the unit-step response and unit-ramp response of the de- signed system, respectively. Note that the closed-loop pole at s = -0.1684 almost cancels the zero at s = -0.16025. However, this pair of closed-loop pole and zero located near the origin produces a long tail of small amplitude. Since the closed-loop pole at s = -17.205 is located
Example Problems and Solutions 47 3
33 Homework-5 Solution
very much farther to the left compared to the closed-loop poles at s = -1.8308 + j3.2359, the effect of this real pole on the system response is very small. Therefore, the closed-loop poles at s = -1.8308 f j3.2359 are indeed dominant closed-loop poles that determine the response characteristics of the closed-loop system. In the unit-ramp response, the steady-state error in following the unit-ramp input eventually becomes 1/K, = $ = 0.02.
A-7-10. Figure 7-44(a) is a block diagram of a model for an attitude-rate control system.The closed-loop transfer function for this system is
The unit-step response of this system is shown in Figure 7-44(b). The response shows high- frequency oscillations at the beginning of the response due to the poles at s = -0.0417 & j2.4489. The response is dominated by the pole at s = -0.0167. The settling time is approximately 240 sec.
It is desired to speed up the response and also eliminate the oscillatory mode at the beginning of the response. Design a suitable compensator such that the dominant closed-loop poles are at s = -2 + j 2 a .
t Hydraulic servo Aircraft I -
Rate gyro
(a)
Unit-Steo Res~onse of Uncom~ensated System
Figure 7-44 (a) Attitude-rate control system; (b) unit-step response.
b I t
50 100 150 200 250 300 Time (sec)
(b)
Chapter 7 / Control Systems Design by the Root-Locus Method