✫ ✪ Chapter 3 State Feedback - Pole Placement Motivation Whereas classical control theory is based on output feed- back, this course mainly deals with control system design by state feedback. This model-based control strategy con- sists of Step 1. State feedback control-law design. Step 2. Estimator design to estimate the state vector. Step 3. Compensation design by combining the control law and the estimator. Step 4. Reference input design to determine the zeros. ESAT–SCD–SISTA CACSD pag. 93
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Chapter 3
State Feedback - Pole Placement
Motivation
Whereas classical control theory is based on output feed-
back, this course mainly deals with control system design
by state feedback. This model-based control strategy con-
sists of
Step 1. State feedback control-law design.
Step 2. Estimator design to estimate the state vector.
Step 3. Compensation design by combining the control
law and the estimator.
Step 4. Reference input design to determine the zeros.
ESAT–SCD–SISTA CACSD pag. 93
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Schematic diagram of a state-space design
example
ux = Ax+Bu
x
C
r˙x = Ax+Bu
+L(y − r − Cx)estimatorcontrol-law
−K
plant
compensator
x y
−
ESAT–SCD–SISTA CACSD pag. 94
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Control-law design by state feedback : a motiva-tionExample: Boeing 747 aircraft control
10s+10
ss+0.3333aircraft
washout circuitactuatoru y
The complete lateral model of a Boeing 747 (see also page
22), including the rudder actuator (an hydraulic servo) and
washout circuit (a lead compensator), is
x = Ax +Bu,
y = Cx +Du.
ESAT–SCD–SISTA CACSD pag. 95
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where
A =
−10 0 0 0 0 0
0.0729 −0.0558 −0.997 0.0802 0.0415 0
−4.75 0.598 −0.115 −0.0318 0 0
1.53 −3.05 0.388 −0.465 0 0
0 0 0.0805 1 0 0
0 0 1 0 0 −0.3333
B =
1
0
0
0
0
0
, C =[
0 0 1 0 0 −0.3333]
, D = 0.
The system poles are
−0.0329± 0.9467i, −0.5627, −0.0073, −0.3333, −10.
ESAT–SCD–SISTA CACSD pag. 96
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The poles at −0.0329 ± 0.9467i have a damping ratio
ζ = 0.03 which is far from the desired value ζ = 0.5. The
following figure illustrates the consequences of this small
damping ratio.
Initial condition response with β = 1◦.
0 50 100 150−0.01
−0.005
0
0.005
0.01
0.015
Time (secs)
Am
plitu
de
To improve this behavior, we want to design a control law
such that the closed loop system has a pair of poles with a
damping ratio close to 0.5.
ESAT–SCD–SISTA CACSD pag. 97
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General Format of State Feedback
Control law
u = −Kx, K : constant matrix.
For single input systems (SI):
K =[
K1 K2 · · · Kn
]
For multi input systems (MI):
K =
K11 K12 · · · K1n
K21 K22 · · · K2n
... ... . . . ...
Kp1 Kp2 · · · Kpn
Note: one sensor is needed for each state ⇒ disadvantage.
We’ll see later how to deal with this problem (estimator
design).
ESAT–SCD–SISTA CACSD pag. 98
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Structure of state feedback control
uC
Plant
x yx = Ax+Bu
u = −Kx
Control-law
ESAT–SCD–SISTA CACSD pag. 99
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Pole Placement
Closed-loop system:
x = Ax +Bu, u = −Kx. ⇒ x = (A−BK)x
poles of the closed loop system
mroots of det (sI − (A−BK))
Pole-placement:
Choose the gain K such that the poles of
the closed loop systems are in specified
positions.
ESAT–SCD–SISTA CACSD pag. 100
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More precisely, suppose that the desired locations are given
by
s = s1, s2, · · · , snwhere si, i = 1, · · · , n are either real or complex conjugated
pairs, choose K such that the characteristic equation
αc(s)∆= det (sI − (A− BK))
equals
(s− s1)(s− s2) . . . (s− sn).
ESAT–SCD–SISTA CACSD pag. 101
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Pole-placement - direct methodFind K by directly solving
det (sI − (A− BK)) = (s− s1)(s− s2) · · · (s− sn)
and matching coefficients in both sides.
Disadvantage:
• Solve nonlinear algebraic equations, difficult for n>3.
Example Let n = 3, m = 1. Then the following
3rd order equations have to be solved to find K =[
K1 K2 K3
]
:
∑
1≤i≤3
(aii − biKi) =∑
1≤i≤3
si,
∑
1≤i<j≤3
∣∣∣∣∣
aii − biKi aij − biKj
aji − bjKi ajj − bjKj
∣∣∣∣∣
=∑
1≤i<j≤3
sisj,
∣∣∣∣∣∣∣
a11 − b1K1 a12 − b1K2 a13 − b1K3
a21 − b2K1 a22 − b2K2 a23 − b2K3
a31 − b3K1 a32 − b3K2 a33 − b3K3
∣∣∣∣∣∣∣
= s1s2s3.
• You never know whether there IS a solution K. (But
THERE IS one if (A,B) is controllable!)
ESAT–SCD–SISTA CACSD pag. 102
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Pole-placement for SI: Ackermann’s methodLet Ac, Bc be in a control canonical form, then