Pole Placement Approach - DSpace@MIT Home

Topic #14

16.31 Feedback Control Systems

State-Space Systems

• Full-state Feedback Control

• How do we change the poles of the state-space system?

• Or, even if we can change the pole locations.

• Where do we put the poles?

– Heuristics

– Linear Quadratic Regulator

• How well does this approach work?

Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare(http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

Fall 2007 16.31 14–1

Pole Placement Approach

• So far we have looked at how to pick K to get the dynamics to

have some nice properties (i.e. stabilize A)

λi(A) � λi(A−BK)

• Classic Question: where should we put these closed-loop poles?

• Approach #1: use time-domain specifications to locate domi-

nant poles – roots of:

s2 + 2ζωns + ω2n = 0

– Then place rest of the poles so they are “much faster” than the

dominant behavior.

� Example: could keep the same damped frequency wd and

then move the real part to be 2–3 times faster than real part

of dominant poles ζωn

� Just be careful moving the poles too far to the left because

it takes a lot of control effort

• Rules of thumb for 2nd order response:

10-90% rise time tr =1 + 1.1ζ + 1.4ζ2

ωn

Settling time (5%) ts = 3ζωn

Time to peak amplitude tp = πωn�

1− ζ2

Peak overshoot Mp = e−ζωntp

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Fall 2007 16.31 14–2

• Approach #2: Could also choose closed-loop poles to mimic a

system that has performance similar to what you want

– Just set pole locations equal to those of the prototype system.

– Various options exist

• Bessel Polynomial Systems of order k → Gp(s) = 1

Bk(s)

Figure 1: Bessel root locations.

• All scaled to give settling times of 1 second, which you can change

to ts by dividing the poles by ts.

October 20, 2007

k = 1 2

1.2

1

0.8

0.6

0.4

0.2

0

Time (sec)

Step response ofGp for various k

Step

resp

onse

G

0 0.5 1 1.5 2 2.5 3

10

Pole locations of B1(s)k-4.6200-4.0530 j2.3400±-5.0093, - 3.9668 j3.7845±-4.0156 j5.0723, - 5.5281 j1.6553±±-6.4480, - 4.1104 j6.3142 - 5.9268 j3.0813±±-4.2169 j7.5300, - 6.2613 j4.4018, - 7.1205 j1.4540±±±-8.0271, - 4.3361 j8.7519, - 6.5714 j5.6786, - 7.6824 j2.8081±±±-4.4554 j9.9715, - 6.8554 j6.9278, - 8.1682 j4.1057, - 8.7693 j1.3616±±±±9.6585, - 4.5696 j11.1838, - 7.1145 j8.1557, - 8.5962 j5.3655, - 9.4013 j2.6655±±±±

j1.3071±-4.6835 j-12.4022, - 7.3609 j9.3777, - 8.9898 j6.6057, - 9.9657 j3.9342, - 10.4278 ± ± ± ±

123456789

10

Roots of normalized bessel polynomials corresponding to a settling time of 1 second

Figure by MIT OpenCourseWare.


Fall 2007 16.31 14–3

• Procedure for an nth order system:

– Determine the desired settling time ts

– Find the k = n polynomial from the table.

– Divide pole locations by ts

– Form desired characteristic polynomial Φd(s) and use acker/place

to determine the feedback gains.

– Simulate to check performance and control effort.

• Example:

G(s) =1

s(s + 4)(s + 1)

with

A =

⎡⎣ −5 −4 0

1 0 0

0 1 0

⎤⎦ B =

⎡⎣ 1

0

0

⎤⎦so that n = k = 3.

– Want ts = 2 sec. So there are 3 poles at:

−5.0093/2 = −2.5047 and

(−3.9668± 3.7845i)/2 = −1.9834± 1.8922i

– Use these to form Φd(s) and find the gains using acker

• The Bessel approach is fine, but the step response is a bit slow

(little overshoot allowed)

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Fall 2007 16.31 14–4

• Approach #3: Select the poles to match the nth order polyno-

mial that was designed to minimize the ITAE “integral of the time

multiplied by the absolute value of the error”

JITAE =

� ∞

0

t |e(t)| dt

in response to a step function.

• Both Bessel and ITAE are tabulated in FPE-508.

– Comparison for k = 3 (Given for ω0 = 1 rad/sec, so slightly

different than numbers given on previous page)

φBd = (s + 0.9420)(s + 0.7465± 0.7112i)

φITAEd = (s + 0.7081)(s + 0.5210± 1.068i)

– So the ITAE poles are not as heavily damped.

� Some overshoot

� Faster rise-times.

• Problem with both of these approaches is that they completely

ignore the control effort required – the designer must iterate.

October 20, 2007


Fall 2007 16.31 14–5

Linear Quadratic Regulator

• An alternative approach is to place the pole locations so that the

closed-loop system optimizes the cost function:

JLQR =

� ∞

0

�y(t)TQy(t) + u(t)TRu(t)

�dt

where:

– yTQy = xT (CTQC)x {assume D = 0} is the State Cost

– uTRu is called the Control Cost with penalty R

– Basic form of Linear Quadratic Regulator problem.

• State cost written using output yTQy, but we could define a new

system output z = Czx that is not based on a physical sensor

measurement.

⇒ JLQR =

� ∞

0

�xT (t)(CT

z Q̃Cz)x(t) + ρ u(t)Tu(t)�dt

– Selection of z used to isolate the system states you are most

concerned about, and thus would like to be regulated to “zero”.

– R = ρI effectively sets the controller bandwidth

• MIMO Optimal control is a time invariant linear state feedback

u(t) = −Klqrx(t)

where Klqr found by solving an Algebraic Riccati Equation(ARE)

0 = ATP + PA + CTQC − PBR−1BTP

Klqr = R−1BTP

– Details to follow.

October 20, 2007


Fall 2007 16.31 14–6

Figure 2: Example #1: G(s) = 8·14·20(s+8)(s+14)(s+20) with control penalty ρ and 10ρ

October 20, 2007


Fall 2007 16.31 14–7

Figure 3: Example #2: G(s) = 0.94s2−0.0297 with control penalty ρ and 10ρ

October 20, 2007


Fall 2007 16.31 14–8

Figure 4: Example #3: G(s) = 8·14·20(s−8)(s−14)(s−20) with control penalty ρ and 10ρ

October 20, 2007


Fall 2007 16.31 14–9

Figure 5: Example #4: G(s) = (s−1)(s+1)(s−3) with control penalty ρ and 10ρ

October 20, 2007


Fall 2007 16.31 14–10

Figure 6: Example #5: G(s) = (s−2)(s−4)(s−1)(s−3)(s2+0.8s+4)s2 with control penalty ρ and

10ρ

October 20, 2007


Fall 2007 16.31 14–11

Regulator Summary

• Dominant second and prototype design approaches (Bessel and

ITAE) place the closed-loop pole locations with no regard tothe amount of control effort required.

– Designer must iterate on the selected bandwidth (ωn) to ensure

that the control effort is reasonable.

• LQR selects closed-loop poles that balance between state errors

and control effort.

– Easy design iteration using R

– Sometimes difficult to relate the desired transient response to

the LQR cost function.

– Key thing is that the designer is focused on system performance

issues rather than the mechanics of the design process

October 20, 2007


Fall 2007 16.31 14–12

LQR Examples - (step4.m)

1 % LQR examples for 16.31

2 % Fall 2007

3 % Jonathan How, MIT

4 set(0,’DefaultLineLineWidth’,2)

5

6 if 1

7 clear all;fig=1;

8 % system

9 G=tf(8*14*20,conv([1 8],conv([1 14],[1 20])));

10 [a,b,c,d]=ssdata(G);

11 R=1e-3;

12 k=lqr(a,b,c’*c,R);

13 k2=lqr(a,b,c’*c,10*R);

14

15 figure(fig);clf;fig=fig+1;

16 % find the feedforward gains

17 Nbar=inv(-c*inv(a-b*k)*b);

18 Nbar2=inv(-c*inv(a-b*k2)*b);

19

20 sys1=ss(a-b*k,b*Nbar,c,d);

21 sys2=ss(a-b*k2,b*Nbar2,c,d);

22

23 t=[0:.005:1];

24 [y,t,x]=step(sys1,t);

25 [y2,t2,x2]=step(sys2,t);

26


28 plot(t,y,’--’,t2,y2,’LineWidth’,2);axis([0 1 0 1.2])

29 grid;

30 legend([’\rho= ’,num2str(R)],[’\rho= ’,num2str(10*R)],’Location’,’SouthEast’)

31 xlabel(’time (sec)’,’FontSize’,14);ylabel(’Y output’,’FontSize’,14);

32 title(’Step Response’,’FontSize’,14)

33 hold on

34 plot(t2([1 end]),[.1 .1]*y2(end),’r--’);

35 plot(t2([1 end]),[.1 .1]*9*y2(end),’r--’);

36 hold off

37

38 text(.4,.6,[’K=’,num2str(k)],’FontSize’,14)

39 text(.4,.8,[’Nbar=’,num2str(Nbar)],’FontSize’,14)

40 set(gca,’FontSize’,14)

41 print -dpng -r300 srl12.png

42

43 if 1


45 f=logspace(-1,2,400);

46 gcl1=freqresp(sys1,f);


48 loglog(f,abs(squeeze(gcl1)),f,abs(squeeze(gcl2)),’LineWidth’,2);%grid

49 axis([.1 100 .01 2])

50 xlabel(’Freq (rad/sec)’,’FontSize’,14);ylabel(’G_{cl}’,’FontSize’,14)

51 title(’Closed-loop Freq Response’,’FontSize’,14)

52 legend([’\rho= ’,num2str(R)],[’\rho= ’,num2str(10*R)],’Location’,’SouthWest’)



55 end

56

57 end % if 0

58

59 %%%%%%%%%

60 if 1

61 clear all

62 fig=4;

63

64 G=tf(8*14*20,conv([1 -8],conv([1 -14],[1 -20])))


66 R=.1


68 k2=lqr(a,b,c’*c,10*R);

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Fall 2007 16.31 14–13

69




73



76

77 t=[0:.005:1];


79 [y2,t2,x2]=step(sys2,t);

80



83 grid;




87 hold on

88 plot(t2([1 end]),[.1 .1]*y2(end),’r--’);

89 plot(t2([1 end]),[.1 .1]*9*y2(end),’r--’);

90 hold off

91

92 text(.4,.6,[’K=’,num2str(k)],’FontSize’,14)

93 text(.4,.8,[’Nbar=’,num2str(Nbar)],’FontSize’,14)



96

97 if 1


99 f=logspace(-1,2,400);




103 axis([.1 100 .01 2])






109 end

110

111 end % if 0

112 %%%%%%%%%%%%%%

113

114 if 1

115 clear all

116 fig=7;

117

118 G=tf(.94,[1 0 -0.0297]);


120 R=.1


122 k2=lqr(a,b,c’*c,10*R);

123




127



130

131 t=[0:.01:30];


133 [y2,t2,x2]=step(sys2,t);

134



137 grid;




141 hold on

142 plot(t2([1 end]),[.1 .1]*y2(end),’r--’);

October 20, 2007


Fall 2007 16.31 14–14

143 plot(t2([1 end]),[.1 .1]*9*y2(end),’r--’);

144 hold off

145

146 text(15,.6,[’K=’,num2str(k)],’FontSize’,14)

147 text(15,.8,[’Nbar=’,num2str(Nbar)],’FontSize’,14)



150

151 if 1


153 f=logspace(-3,3,400);




157 axis([.1 100 .01 2])






163 end

164

165 end % if 0

166

167 %%%%%%%%%%%%

168 if 1

169 clear all

170 fig=10;

171

172 G=tf([1 -1],conv([1 1],[1 -3]));


174 R=.1


176 k2=lqr(a,b,c’*c,10*R);

177




181



184

185 t=[0:.01:10];


187 [y2,t2,x2]=step(sys2,t);

188


190 plot(t,y,’--’,t2,y2,’LineWidth’,2);axis([0 10 -1 1.2])

191 grid;



194 title(’Unstable, NMP system Step Response’,’FontSize’,14)

195 hold on

196 plot(t2([1 end]),[.1 .1]*y2(end),’r--’);

197 plot(t2([1 end]),[.1 .1]*9*y2(end),’r--’);

198 hold off

199





204

205 if 1


207 f=logspace(-2,2,400);




211 axis([.1 100 .01 2])






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Fall 2007 16.31 14–15

217 end

218

219 end % if 0

220

221 %%%%%%%%%%%%

222 clear all

223 fig=13;

224

225 G=tf(conv([1 -2],[1 -4]),conv(conv([1 -1],[1 -3]),[1 2*.2*2 2^2 0 0]));


227 R=.1


229 k2=lqr(a,b,c’*c,10*R);

230




234



237

238 t=[0:.01:10];


240 [y2,t2,x2]=step(sys2,t);

241


243 plot(t,y,’--’,t2,y2,’LineWidth’,2);axis([0 10 -1 1.2])

244 grid;



247 title(’Unstable, NMP system Step Response’,’FontSize’,14)

248 hold on

249 plot(t2([1 end]),[.1 .1]*y2(end),’r--’);

250 plot(t2([1 end]),[.1 .1]*9*y2(end),’r--’);

251 hold off

252





257

258 if 1


260 f=logspace(-2,2,400);




264 axis([.1 100 .01 2])

265 xlabel(’Freq (rad/sec)’,’FontSize’,14)

266 ylabel(’G_{cl}’,’FontSize’,14)





271 end

October 20, 2007


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