Pole Placement
Pole Placement
A majority of the design techniques in modern
control theory is based on the state-feedback
configuration. That is, instead of using
controllers with fixed configurations in the
forward or feedback path, control is achieved by
feeding back the state variables through real
constant gains.
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Definition of Pole Placement
• (Pole Assignment, Pole Allocation)
Placing the poles or eigenvalues of the
closed-loop system at specified locations.
• Poles can be arbitrarily placed if and only if
the system is controllable.
• Pole placement is easier if the system is
given in controllable form.
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Pole Placement
4
Pole Placement
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State Feedback
State Feedback: involves the use of the state vector to
compute the control action for specified system
dynamics.
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State Feedback
X(k+1)=Ax(k)+Bu(k)
Y(k)=C x(k)
U(k)=-K x(k)+v(k)
X(k+1)=[A-BK]x(k)+B v(k)
Let Ad= A-BK The closed-loop system state
space: X(k+1)= Ad x(k)+B v(k)
Y(k)=C x(k)
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Output Feedback
Output Feedback
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Output Feedback
U(k)=-K y(k)+v(k)= -K C x(k)+ v(k)
X(k+1)=[A-BKC]x(k)+B v(k)
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Pole Placement
Theorem 9.1: State Feedback. If the pair (A,B) is
controllable, then there exists a feedback gain
matrix K that arbitrarily assigns the system poles
to any set [λi i=1,….n]. Furthermore, if the pair
(A,B) is stabilizable, then the controllable modes
can all be arbitrarily assigned.
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Procedure 9.1: Pole Placement by
Equating Coefficients
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Example 9.1
Assign the eigenvalues [0.3±j0.21] to the pair
0
A 
3
1
4
0 
B 
1
(λ-0.3-j0.2)(λ-0.3+j0.2)= λ2-0.6λ+0.13
The closed loop matrix is
𝐴−
𝑏𝑘 𝑇
0 1
0 𝑘
=
−
3 4
1 1
𝑘2
0
=
3 − 𝑘1
1
4 − 𝑘2
The closed loop characteristic polynomial is
Det{λI-(A-𝑏𝑘 𝑇 )}=λ2-(4-𝑘2 )λ-(3-𝑘1 )
4-𝑘2 =0.6 → 𝑘2 =3.4
3-𝑘1 =0.13 → 𝑘1 = 3.13
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MATLAB
>> A = [0, 1; 3, 4];
>>B = [0; 1];
>> poles = [0.3 + j*.2, 0.3 – j*0.2];
>> K = place(A, B, poles)
K  3.1300 3.4000
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Pole Placement by Transformation to Controllable Form
Any controllable single-input-single-output (SISO) system can be
transformed into controllable form using the transformation
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Procedure 9.2
1. Obtain the characteristic polynomial of the pair
(A, B) using the Leverrier algorithm described in
Section 7.4.1.
2. Obtain the transformation matrix 𝑇 −1 using the
coefficients of the polynomial from step 1.
3. Obtain the desired characteristic polynomial
coefficients from the given eigenvalues using
(9.10).
4. Compute the state feedback matrix using (9.15).
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Example 9.2
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Solution
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Servo Problem
The schemes shown in Figures 9.1 and 9.2
are regulators that drive the system state to
zero starting from any initial condition
capable of rejecting impulse disturbances. In
practice, it is often necessary to track a
constant reference input r with zero steadystate error.
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Servo Problem
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Servo Problem
The reference input of (9.2) becomes v(k)  F r(k),
and the control law is chosen as
with r(k) the reference input to be tracked. The
corresponding closed-loop system equations are
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where the closed-loop state matrix is
The z-transform of the corresponding output is
given by
The steady-state tracking error for a unit step input
is given by
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Servo Problem
For zero steady-state error, we require the
condition
If the system is square (m  l) and Acl is stable (no
unity eigenvalues), we solve for the reference gain
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Example 9.5
Design a state–space controller for the
discretized state–space model of the DC
motor speed control system described in
Example 6.8 (with T  0.02) to obtain
(1) zero steady-state error due to a unit step,
(2) a damping ratio of 0.7, and
(3) a settling time of about 1 s.
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Solution
The discretized transfer function of the system with
digital-to-analog converter (DAC) and analog-todigital converter (ADC) is
The corresponding state–space model, computed
with MATLAB, is
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Solution
The desired eigenvalues of the closed-loop system
are selected as {0.9  j0.09} (see Example 6.8).
This yields the feedback gain vector
and the closed-loop state matrix
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Solution
The feedforward gain is
The response of the system to a step reference
input r is shown in Figure 9.8. The system has
a settling time of about 0.84 s and percentage
overshoot of about 4% with a peak time of
about 1 s. All design specifications are met.
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Solution
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Integral Control
The control law (9.21) is equivalent to a
feedforward action determined by F to yield
zero steady-state error for a constant reference
input r. Because the forward action does not
include any form of feedback, this approach is
not robust to modeling uncertainties. Thus,
modeling errors (which always occur in
practice) will result in nonzero steady-state
error. To eliminate such errors, we introduce the
integral control shown in Figure 9.9, with a
new state added for each control error
integrated.
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Integral Control
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The resulting state–space equations are
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Integral Control
The eigenvalues of the closed-loop system state matrix
can be arbitrarily assigned by computing the gain matrix K using any of
the procedures for the regulator problem as described in Sections 9.1 and
9.2.
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Example 9.6
Solve the design problem presented in
Example 9.5 using integral control.
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Solution
The state–space matrices of the system are
In Example 9.5, the eigenvalues were selected as {0.9  j0.09}. Using
integral control increases the order of the system by one, and an
additional eigenvalue must be selected. The desired eigenvalues are
selected as {0.9  j0.09, 0.2}, and the additional eigenvalue at 0.2 is
chosen for its negligible effect on the overall dynamics. This yields the
feedback gain vector
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Solution
The response of the system to a unit step reference signal r is shown in
Figure 9.10. The figure shows that the control specifications are satisfied.
The settling time of 0.87 is well below the specified value of 1 s, and the
percentage overshoot is about 4.2%, which is less than the value
corresponding to ζ  0.7 for the dominant pair.
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Solution
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State Estimation
In most applications, measuring the entire
state vector is impossible or prohibitively
expensive. To implement state feedback
control, an estimate xˆ (k) of the state vector
can be used. The state vector can be
estimated from the input and output
measurements by using a state estimator
or observer.
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Estimators
In the previous section, we have designed a
state feedback controller, assuming that all
states are measured. We also know, however,
that not all states can be measured, for a
detailed discussion on this topic. Does this
mean that the state feedback controllers cannot
be implemented? The answer is that we can
obtain a reasonably equivalent state feedback
controller
through
estimated
state
measurements. A state estimator, or simply an
estimator, is a mathematical construct that
helps estimate the states of a system.
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Full-Order Observer
To estimate all the states of the system, one could in theory
use a system with the same state equation as the plant to
be observed. In other words, one could use the open-loop
system
xˆ (k +1)  Axˆ (k) + Bu(k)
However, this open-loop estimator assumes perfect
knowledge of the system dynamics and lacks the feedback
needed to correct the errors that are inevitable in any
implementation. The limitations of this observer become
obvious on examining its error dynamics. We obtain the
error dynamics by subtracting the open-loop observer
dynamics from the system dynamics (9.1).
Where
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Full-Order Observer
The error dynamics are determined by the state
matrix of the system and cannot be chosen
arbitrarily. For an unstable system, the observer will
be unstable and cannot track the state of the
system.
A practical alternative is to feed back the difference
between the measured and the estimated output of
the system, as shown in Figure 9.12. This yields to
the following observer:
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Full-Order Observer
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Full-Order Observer
Subtracting the observer state equation from the system dynamics yields
the estimation error dynamics
The error dynamics are governed by the eigenvalues of the observer matrix
.
We transpose the matrix to obtain
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Full-Order Observer
Theorem 9.2: State Estimation. If the pair
(A, C) is observable, then there exists a
feedback gain matrix L that arbitrarily
assigns the observer poles to any set {l , i 
1, . . . , n}. Furthermore, if the pair (A, C) is
detectable, then the observable modes can
all be arbitrarily assigned.
i
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Example 9.8
Determine the observer gain matrix L for the
discretized state–space model of the
armature controlled DC motor described in
Example 7.15 with the observer eigenvalues
selected as {0.1, 0.2  j0.2}.
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Solution
Recall that the system matrices are
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HW 7
9.7
9.11
9.15
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