Post on 06-Oct-2020
Topic #17
16.31 Feedback Control
State-Space Systems
• Closed-loop control using estimators and regulators.
• Dynamics output feedback
• “Back to reality”
Copyright 2001 by Jonathan How. All Rights reserved
1
Fall 2001 16.31 17—1
Combined Estimators and Regulators
• Can now evaluate the stability and/or performance of a controller
when we design K assuming that u = −Kx, but we implementu = −Kx
• Assume that we have designed a closed-loop estimator with gain L
˙x(t) = Ax(t) +Bu(t) + L(y − y)y(t) = Cx(t)
• Then we have that the closed-loop system dynamics are given by:
x(t) = Ax(t) +Bu(t)˙x(t) = Ax(t) +Bu(t) + L(y − y)y(t) = Cx(t)
y(t) = Cx(t)
u = −Kx
• Which can be compactly written as:∙x˙x
¸=
∙A −BKLC A−BK − LC
¸ ∙x
x
¸⇒ xcl = Aclxcl
• This does not look too good at this point — not even obvious that
the closed-system is stable.
λi(Acl) =??
Fall 2001 16.31 17—2
• Can fix this problem by introducing a new variable x = x− x andthen converting the closed-loop system dynamics using the
similarity transformation T
xcl ,∙x
x
¸=
∙I 0
I −I¸ ∙
x
x
¸= Txcl
— Note that T = T−1
• Now rewrite the system dynamics in terms of the state xcl
Acl ⇒ TAclT−1 , Acl
— Note that similarity transformations preserve the eigenvalues, so
we are guaranteed that
λi(Acl) ≡ λi(Acl)
• Work through the math:
Acl =
∙I 0
I −I¸ ∙
A −BKLC A−BK − LC
¸ ∙I 0
I −I¸
=
∙A −BK
A− LC −A + LC¸ ∙
I 0
I −I¸
=
∙A−BK BK
0 A− LC¸
• Because Acl is block upper triangular, we know that the closed-loop
poles of the system are given by
det(sI − Acl) , det(sI − (A−BK)) · det(sI − (A− LC)) = 0
Fall 2001 16.31 17—3
• Observation: The closed-loop poles for this system con-
sist of the union of the regulator poles and estimator poles.
• So we can just design the estimator/regulator separately and com-
bine them at the end.
— Called the Separation Principle.
— Just keep in mind that the pole locations you are picking for these
two sub-problems will also be the closed-loop pole locations.
• Note: the separation principle means that there will be no ambi-
guity or uncertainty about the stability and/or performance of the
closed-loop system.
— The closed-loop poles will be exactly where you put them!!
— And we have not even said what compensator does this amazing
accomplishment!!!
Fall 2001 16.31 17—4
The Compensator
• Dynamic Output Feedback Compensator is the combina-
tion of the regulator and estimator using u = −Kx˙x(t) = Ax(t) +Bu(t) + L(y − y)
= Ax(t)−BKx + L(y − Cx)
⇒ ˙x(t) = (A− BK − LC)x(t) + Lyu = −Kx
• Rewrite with new state xc ≡ xxc = Acxc +Bcy
u = −Ccxcwhere the compensator dynamics are given by:
Ac , A−BK − LC , Bc , L , Cc , K— Note that the compensator maps sensor measurements to ac-
tuator commands, as expected.
• Closed-loop system stable if regulator/estimator poles placed in the
LHP, but compensator dynamics do not need to be stable.
λi(A−BK − LC) =??
Fall 2001 16.31 17—5
• For consistency in the implementation with the classical approaches,
define the compensator transfer function so that
u = −Gc(s)y— From the state-space model of the compensator:
U(s)
Y (s), −Gc(s)= −Cc(sI −Ac)−1Bc= −K(sI − (A− BK − LC))−1L
⇒ Gc(s) = Cc(sI−Ac)−1Bc
• Note that it is often very easy to provide classical interpretations
(such as lead/lag) for the compensator Gc(s).
• One way to implement this compensator with a reference command
r(t) is to change the feedback to be on e(t) = r(t) − y(t) ratherthan just −y(t)
Gc(s) G(s)- -6
—
r e yu
⇒ u = Gc(s)e = Gc(s)(r − y)— So we still have u = −Gc(s)y if r = 0.— Intuitively appealing because it is the same approach used for
the classical control, but it turns out not to be the best approach.
Fall 2001 16.31 17—6
Mechanics
• Basics:
e = r − y, u = Gce, y = Gu
Gc(s) : xc = Acxc +Bce , u = Ccxc
G(s) : x = Ax +Bu , y = Cx
• Loop dynamics L = Gc(s)G(s) ⇒ y = L(s)e
x = Ax +BCc xcxc = +Ac xc +Bce
L(s)
∙x
xc
¸=
∙A BCc0 Ac
¸ ∙x
xc
¸+
∙0
Bc
¸e
y =£C 0
¤ ∙ xxc
¸
• To “close the loop”, note that e = r − y, then∙x
xc
¸=
∙A BCc0 Ac
¸ ∙x
xc
¸+
∙0
Bc
¸µr − £ C 0
¤ ∙ xxc
¸¶=
∙A BCc−BcC Ac
¸ ∙x
xc
¸+
∙0
Bc
¸r
y =£C 0
¤ ∙ xxc
¸— Acl is not exactly the same as on page 17-1 because we have re-
arranged where the negative sign enters into the problem. Same
result though.
Fall 2001 16.31 17—7
Simple Example
• Let G(s) = 1/s2 with state-space model given by:
A =
∙0 1
0 0
¸, B =
∙0
1
¸, C =
£1 0
¤, D = 0
• Design the regulator to place the poles at s = −4± 4jλi(A−BK) = −4± 4j ⇒ K =
£32 8
¤— Time constant of regulator poles τc = 1/ζωn ≈ 1/4 = 0.25 sec
• Put estimator poles so that the time constant is faster τe ≈ 1/10— Use real poles, so Φe(s) = (s + 10)
2
L = Φe(A)
∙C
CA
¸−1 ∙0
1
¸=
Ã∙0 1
0 0
¸2+ 20
∙0 1
0 0
¸+
∙100 0
0 100
¸! ∙1 0
0 1
¸−1 ∙0
1
¸=
∙100 20
0 100
¸ ∙0
1
¸=
∙20
100
¸
Fall 2001 16.31 17—8
• Compensator:
Ac = A− BK − LC=
∙0 1
0 0
¸−∙0
1
¸ £32 8
¤− ∙ 20100
¸ £1 0
¤=
∙ −20 1
−132 −8¸
Bc = L =
∙20
100
¸Cc = K =
£32 8
¤
• Compensator transfer function:
Gc(s) = Cc(sI −Ac)−1Bc , UE
= 1440s + 2.222
s2 + 28s + 292
• Note that the compensator has a low frequency real zero and two
higher frequency poles.
— Thus it looks like a “lead” compensator.
Fall 2001 16.31 17—9
10−1
100
101
102
103
100
101
102
Freq (rad/sec)
Mag
Plant G Compensator Gc
10−1
100
101
102
103
−200
−150
−100
−50
0
50
Freq (rad/sec)
Pha
se (
deg)
Plant G Compensator Gc
Figure 1: Plant is pretty simple and the compensator looks like a lead
2—10 rads/sec.
10−1
100
101
102
103
100
101
102
Freq (rad/sec)
Mag
Loop L
10−1
100
101
102
103
−280
−260
−240
−220
−200
−180
−160
−140
−120
Freq (rad/sec)
Pha
se (
deg)
Figure 2: Loop transfer function L(s) shows the slope change near
ωc = 5 rad/sec. Note that we have a large PM and GM.
Fall 2001 16.31 17—10
−15 −10 −5 0 5−15
−10
−5
0
5
10
15
Real Axis
Imag
Axi
s
Figure 3: Freeze the compensator poles and zeros and look at the root
locus of closed-loop poles versus an additional loop gain α (nominally
α = 1.)
10−1
100
101
102
103
10−2
10−1
100
101
102
Freq (rad/sec)
Mag
Plant G closed−loop Gcl
Figure 4: Closed-loop transfer function.
Fall 2001 16.31 17—11
Figure 5: Example #1: G(s) = 8·14·20(s+8)(s+14)(s+20)
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Plant G Compensator Gc
10−1
100
101
102
103
−200
−150
−100
−50
0
50
Freq (rad/sec)
Pha
se (
deg)
Plant G Compensator Gc
10−1
100
101
102
103
10−2
10−1
100
101
102
Freq (rad/sec)
Mag
Plant G closed−loop Gcl
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Loop L
10−1
100
101
102
103
−250
−200
−150
−100
−50
0
Freq (rad/sec)
Pha
se (
deg)
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−150
−100
−50
0
50Gm=10.978 dB (at 40.456 rad/sec), Pm=69.724 deg. (at 15.063 rad/sec)
100
101
102
103
−400
−350
−300
−250
−200
−150
−100
−50
0
Fall 2001 16.31 17—12
Figure 6: Example #1: G(s) = 8·14·20(s+8)(s+14)(s+20)
−80 −70 −60 −50 −40 −30 −20 −10 0 10 20−50
−40
−30
−20
−10
0
10
20
30
40
50
Real Axis
Imag
Axi
s
−50 −45 −40 −35 −30 −25 −20 −15 −10 −5 0−25
−20
−15
−10
−5
0
5
10
15
20
25
Real Axis
Imag
Axi
s
3 — closed-loop poles, 5 — open-loop poles, 2 — Compensator poles, ◦ — Compensator zeros
Fall 2001 16.31 17—13
• Two compensator zeros at -21.54±6.63j draw the two lower fre-quency plant poles further into the LHP.
• Compensator poles are at much higher frequency.
• Looks like a lead compensator.
Fall 2001 16.31 17—14
Figure 7: Example #2: G(s) = 0.94s2−0.0297
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Plant G Compensator Gc
10−1
100
101
102
103
−200
−150
−100
−50
0
50
Freq (rad/sec)
Pha
se (
deg)
Plant G Compensator Gc
10−1
100
101
102
103
10−2
10−1
100
101
102
Freq (rad/sec)
Mag
Plant G closed−loop Gcl
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Loop L
10−1
100
101
102
103
−250
−200
−150
−100
−50
0
Freq (rad/sec)
Pha
se (
deg)
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−100
−50
0
50Gm=11.784 dB (at 7.4093 rad/sec), Pm=36.595 deg. (at 2.7612 rad/sec)
10−2
10−1
100
101
102
−300
−250
−200
−150
−100
Fall 2001 16.31 17—15
Figure 8: Example #2: G(s) = 0.94s2−0.0297
−8 −6 −4 −2 0 2−6
−4
−2
0
2
4
6
Real Axis
Imag
Axi
s
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Real Axis
Imag
Axi
s
3 — closed-loop poles, 5 — open-loop poles, 2 — Compensator poles, ◦ — Compensator zeros
Fall 2001 16.31 17—16
• Compensator zero at -1.21 draws the two lower frequency plant poles
further into the LHP.
• Compensator poles are at much higher frequency.
• Looks like a lead compensator.
Fall 2001 16.31 17—17
Figure 9: Example #3: G(s) = 8·14·20(s−8)(s−14)(s−20)
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Plant G Compensator Gc
10−1
100
101
102
103
−200
−150
−100
−50
0
50
Freq (rad/sec)
Pha
se (
deg)
Plant G Compensator Gc
10−1
100
101
102
103
10−2
10−1
100
101
102
Freq (rad/sec)
Mag
Plant G closed−loop Gcl
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Loop L
10−1
100
101
102
103
−250
−200
−150
−100
−50
0
Freq (rad/sec)
Pha
se (
deg)
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−100
−50
0
50Gm=−0.90042 dB (at 24.221 rad/sec), Pm=6.6704 deg. (at 35.813 rad/sec)
100
101
102
103
−350
−300
−250
−200
−150
Fall 2001 16.31 17—18
Figure 10: Example #3: G(s) = 8·14·20(s−8)(s−14)(s−20)
−140 −120 −100 −80 −60 −40 −20 0 20 40 60−100
−80
−60
−40
−20
0
20
40
60
80
100
Real Axis
Imag
Axi
s
−25 −20 −15 −10 −5 0 5 10 15 20 25−25
−20
−15
−10
−5
0
5
10
15
20
25
Real Axis
Imag
Axi
s
3 — closed-loop poles, 5 — open-loop poles, 2 — Compensator poles, ◦ — Compensator zeros
Fall 2001 16.31 17—19
• Compensator zeros at 3.72±8.03j draw the two higher frequencyplant poles further into the LHP. Lowest frequency one heads into
the LHP on its own.
• Compensator poles are at much higher frequency.
• Note sure what this looks like.
Fall 2001 16.31 17—20
Figure 11: Example #4: G(s) = (s−1)(s+1)(s−3)
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Plant G Compensator Gc
10−1
100
101
102
103
−200
−150
−100
−50
0
50
Freq (rad/sec)
Pha
se (
deg)
Plant G Compensator Gc
10−1
100
101
102
103
10−2
10−1
100
101
102
Freq (rad/sec)
Mag
Plant G closed−loop Gcl
10−1
100
101
102
103
10−2
10−1
100
101
Freq (rad/sec)
Mag
Loop L
10−1
100
101
102
103
−250
−200
−150
−100
−50
0
Freq (rad/sec)
Pha
se (
deg)
Frequency (rad/sec)
Pha
se (
deg)
; Mag
nitu
de (
dB)
Bode Diagrams
−80
−60
−40
−20
0
20Gm=−3.3976 dB (at 4.5695 rad/sec), Pm=−22.448 deg. (at 1.4064 rad/sec)
10−1
100
101
102
103
140
150
160
170
180
190
200
210
220
Fall 2001 16.31 17—21
Figure 12: Example #4: G(s) = (s−1)(s+1)(s−3)
−40 −30 −20 −10 0 10 20−40
−30
−20
−10
0
10
20
30
40
Real Axis
Imag
Axi
s
−10 −8 −6 −4 −2 0 2 4 6 8 10−10
−8
−6
−4
−2
0
2
4
6
8
10
Real Axis
Imag
Axi
s
3 — closed-loop poles, 5 — open-loop poles, 2 — Compensator poles, ◦ — Compensator zeros
Fall 2001 16.31 17—22
• Compensator zero at -1 cancels the plant pole. Note the very un-
stable compensator pole at s = 9!!
— Needed to get the RHP plant pole to branch off the real line and
head into the LHP.
• Other compensator pole is at much higher frequency.
• Note sure what this looks like.
• Separation principle gives a very powerful and simple way to develop
a dynamic output feedback controller
• Note that the designer now focuses on selecting the appropriate
regulator and estimator pole locations. Once those are set, the
closed-loop response is specified.
— Can almost consider the compensator to be a by-product.
• These examples show that the design process is extremely simple.