ECE5530: Multivariable Control Systems II. 3–1
LINEAR QUADRATIC REGULATOR
3.1: Cost functions; deterministic LQR problem
Cost functions
■ The engineering tradeoff in control-system design is
Fast response Slower response
Large intermediate states versus Smaller intermediate states
Large control effort Smaller control effort
■ Optimizing a tradeoff like this can be turned into a mathematical
optimization by defining a cost function that must be minimized.
■ The cost function used to compare controllers should include a
measure of size of output errors and size of control
J DZ tf
0
´T .t/W.t/´.t/ dt D k´.t/k2W.t/;Œ0;tf ! ;
where ´.t/ is assumed to include both output errors and control
inputs and W.t/ is psd at all times.
■ The cost function may be expanded to give more detailed description
J DZ tf
0
h
eT .t/ uT .t/i
2
4W1.t/ 0
0 W2.t/
3
5
2
4e.t/
u.t/
3
5 dt
DZ tf
0
eT .t/W1.t/e.t/ C uT .t/W2.t/u.t/ dt
where W1.t/ and W2.t/ are both psd.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–2
■ This cost function is not as general, but often sufficient.
■ Note we can write e.t/ D r.t/ ! Cyx.t/.
■ Then,
J DZ tf
0
.r.t/ ! Cyx.t//T W1.t/.r.t/ ! Cyx.t// C uT .t/W2.t/u.t/ dt:
■ Often, we can formulate a goal such that we want to drive x.t/ ! 0 or
some linear combination of states to zero. This is called a regulator.
■ Then,
J DZ tf
0
xT .t/Q.t/x.t/ C uT .t/R.t/u.t/ dt;
where Q.t/ D C Ty W1.t/Cy and R.t/ D W2.t/, for example.
■ Sometimes, final output error is more critical than intermediate state
errors. e.g., flight of a missile
J D!
r.tf / ! Cyx.tf /"T
V!
r.tf / ! Cyx.tf /"
CZ tf
0
uT .t/Ru.t/ dt:
■ Value of cost function depends on reference input applied,
disturbance applied, initial conditions, final conditions, constraints on
state or control.
" Collectively, these are known as test conditions. e.g., nominal or
worst-case.
" For valid comparison between controllers, same test conditions
must be used.
The deterministic LQR problem
■ In our case, the deterministic linear quadratic regulator (LQR)
problem assumes that we have full state information available, and
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–3
that we desire to minimize the cost function
J.xk; uk/ D xTN HdxN C
N !1X
kD0
ŒxTk Qdxk C uT
k Rduk!;
subject to xkC1 D Axk C Buk (if the system dynamics are discrete), or
J.x.t/; u.t// D xT .tf /Hx.tf / CZ tf
0
xT .t/Qx.t/ C uT .t/Ru.t/ dt
subject to Px.t/ D Ax.t/ C Bu.t/ (if system dynamics are continuous).
" xTN HdxN or xT .tf /Hx.tf / is the penalty for “missing” the desired
final state.
" xTk Qdxk or xT .t/Qx.t/ is the penalty on excessive state size.
" uTk Rduk or uT .t/Ru.t/ is the penalty on excessive control effort.
■ Matrices H , Hd , Q, Qd and R, Rd are symmetric matrices that put
more (or less) cost on each term in the cost function.
" We require H; Hd # 0, Q; Qd # 0 and R; Rd > 0.
" The weighting of these matrices is relative to each other; if we
multiply everything by 2, the solution does not change (but the
minimum cost is doubled).
■ The solution is an input trajectory, either u.t/ or uk.
■ There are two approaches to optimization problems such as this:
" Calculus of variations [Burl];
" Dynamic programming [Bay].
■ We will look at both, since both are common in the literature.
" We’ll use calculus of variations to derive the continuous-time result.
" We will review the dynamic-programming results from ECE5520
for the discrete-time problem.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–4
Lagrange multipliers
■ The LQR optimization is subject to the constraint imposed by the
system dynamics: e.g., Px.t/ D Ax.t/ C Bu.t/.
■ Without the constraint, we might consider optimizing the cost function
by using its gradient, rJ .
" The gradient at any location points in the direction of the steepest
increase in the function.
" Following the negative gradient will lead us down to a minimum of
the function.
" Whenever the gradient is zero, we are either at a minimum, a
maximum, or a saddle point.
" Second-derivative tests can tell us what is the case at any
particular location.
■ Now, suppose that we must minimize a function J.x/ subject to the
constraint c.x/ D 0. Plotting the constraint itself:
c.x/ D 0.
The directions you can
move are perpendicular to
the gradient.
The gradient of c.x/ at any
point is orthogonal to the
curve c.x/ D 0.
■ If rJ has any component aligned with the constraint, then the
constrained cost can be reduced by moving along the constraint in
the negative direction of that component.
■ However, when rJ is perpendicular to the constraint, that point is a
local minimum, maximum, or saddle point.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–5
■ Since the gradient of the constraint is perpendicular to the constraint,
we have that rJ D !"rc, and the Lagrange multiplier " is the
proportionality factor.
■ So, we satisfy a constrained optimization when r.J C "c/ D 0.
" We do so by making an augmented cost Ja.x; "/ D J.x/ C "c.x/,
" Taking partial derivatives of Ja.x; "/, and
" Setting these derivatives to zero.
■ We will use this approach to solve the LQR problem.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–6
3.2: Optimization via calculus of variations
■ Differentiation is primary tool for optimization w.r.t. a (scalar) variable.
■ Our objective requires differentiation of a real scalar cost function
w.r.t. the control input (a function of time).
■ Can be accomplished by using a generalization of the differential
called the variation.
■ Consider: The real scalar function of a scalar J.x/ has a local
minimum at x$ iff
J.x$ C ıx/ # J.x$/
for all ıx sufficiently small. Equivalently,
#J.x$; ıx/ D J.x$ C ıx/ ! J.x$/ # 0:
■ #J.x$; ıx/ is called the increment of J . Expand via Taylor series
#J.x$; ıx/ D J.x$ C ıx/ ! J.x$/
DdJ.x$/
dxıx C
d2J.x$/
dx2ıx2 C h:o:t: # 0:
■ Above, ıx is called the differential of x, and the term linear in ıx
(which is ŒdJ.x$/=dx!ıx in this case) is called the differential of J .
■ When dealing with a functional (a real scalar function of functions), ıx
is called the variation of x and the term linear in ıx is called the
variation of J and is denoted ıJ.x$; ıx/. So,
#J.x$; ıx/ D ıJ.x$; ıx/ C h:o:t: # 0:
■ The variation of J is a generalization of the differential and can be
applied to the optimization of a functional.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–7
NECESSARY CONDITION OF OPTIMALITY: The variation of J is zero at x$
for all ıx.
EXAMPLE: Consider
J.x/ D x2 C 6x C 8:
1. Find the increment #J .
2. Simplify, and extract the variation ıJ .
3. Set ıJ D 0 and solve.
■ The increment of J is
#J.x; ıx/ D .x C ıx/2 C 6.x C ıx/ C 8 ! .x2 C 6x C 8/
D x2 C 2xıx C ıx2 C 6x C 6ıx C 8 ! x2 ! 6x ! 8
D .2x C 6/ıx C ıx2:
■ For optimality, the variation of J must be zero
ıJ.x; ıx/ D .2x C 6/ıx D 0
for all ıx ➠ Therefore, x D !3:
■ Using standard calculus, we would havedJ
dxD 2x C 6 D 0 and would
get the answer more quickly. However, the calculus of variations can
be directly extended to the optimization of a functional.
Lagrange multipliers (again)
■ The optimal control problem is a constrained minimization problem.
■ Want minimum cost subject to dynamics of plant Px D Ax C Bu.
■ The calculus of variations applies to unconstrained minimization
problems.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–8
■ Lagrange multipliers convert a constrained minimization problem to a
higher-order unconstrained minimization problem. Can then use
calculus of variations.
■ Consider minimizing J.x/, x 2 Rn, subject to the constraint c.x/ D 0.
" The differential in J must be parallel to the differential in c for an
optimal solution.
■ That is, generalizing our prior solution (and dividing both sides by ıx),
we must haveıJ.x$/
ıxC "
ıc.x$/
ıxD 0
for some scalar ". Can also be written as the solution to an
unconstrained augmented cost function
Ja.x; "/ D J.x/ C "c.x/:
■ Note that we are just adding zero to our cost function in a clever way
that enforces the constraint. When c.x/ is a vector, we use
Ja.x; "/ D J.x/ C "T c.x/:
EXAMPLE: We wish to minimize J.x/ D x21 C x2
2 subject to the constraint
that c.x/ D 2x1 C x2 C 4 D 0.
■ The augmented cost function is
Ja.x; "/ D x21 C x2
2 C ".2x1 C x2 C 4/:
■ The increment of the augmented cost function is
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–9
#Ja.x; "/ D Ja.x C ıx; " C ı"/ ! Ja.x; "/
D .x1 C ıx1/2 C .x2 C ıx2/
2 C ." C ı"/
#
2.x1 C ıx1/
C.x2 C ıx2/ C 4
$
! x21 ! x2
2 ! "Œ2x1 C x2 C 4!
D .2x1 C 2"/ıx1 C .2x2 C "/ıx2 C .2x1 C x2 C 4/ı"
Cıx21 C ıx2
2 C 2ı"ıx1 C ı"ıx2:
■ Then,
ıJa.x; "; ıx; ı"/ D .2x1 C 2"/ıx1 C .2x2 C "/ıx2 C .2x1 C x2 C 4/ı" D 0:
■ Therefore,
ıJa
ıx1D 2x1 C 2" D 0
ıJa
ıx2D 2x2 C " D 0
ıJa
ı"D 2x1 C x2 C 4 D 0:
■ Notice that the unconstrained minimization problem simply has
2x1 D 2x2 D 0. Adding the constraint is seen by the added “"” terms
plus the ıJa=ı" term.
■ Solving these three equations and three unknowns givesh
x1 x2 "i
Dh
!1:6 !0:8 1:6i
:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–10
3.3: The LQR problem solved via calculus of variations
■ For the continuous-time LQR problem, we wish to minimize
J.x.t/; u.t// D1
2xT .tf /Hx.tf / C
1
2
Z tf
0
xT .t/Qx.t/ C uT .t/Ru.t/ dt
subject to the constraint that Px.t/ D Ax.t/ C Buu.t/.
■ We add the constraint to the integral term to get
Ja.x.t/; u.t/; ".t// D J.x.t/; u.t// CZ tf
0
"T .t/ŒAx.t/ C Buu.t/ ! Px.t/! dt:
D1
2xT .tf /Hx.tf / C
Z tf
0
1
2xT .t/Qx.t/ C
1
2uT .t/Ru.t/
C"T .t/ŒAx.t/ C Buu.t/ ! Px.t/!
!
dt:
■ The Lagrange multiplier ".t/ is often referred to as the costate for
reasons that will become clear later.
■ We find the optimal control by first forming the increment of Ja.
#Ja.x; u; "; ıx; ıu; ı"/
D Ja.x C ıx; u C ıu; " C ı"/ ! Ja.x; u; "/
D1
2
!
x.tf / C ıx.tf /"T
H!
x.tf / C ıx.tf /"
CZ tf
0
1
2.x C ıx/T Q.x C ıx/ C
1
2.u C ıu/T R.u C ıu/
C." C ı"/T ŒA.x C ıx/ C Bu.u C ıu/ ! . Px C ı Px/! dt
!1
2xT .tf /Hx.tf / !
Z tf
0
1
2xT Qx C
1
2uT Ru C "T .Ax C Buu ! Px/ dt;
where time dependence has been omitted to simplify the expression.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–11
■ Expand, noting that xT Qıx D .xT Qıx/T D ıxT QT x D ıxT Qx,
#Ja D1
2ıxT .tf /Hıx.tf /
CZ tf
0
1
2ıxT Qıx C
1
2ıuT Rıu C ı"T .Aıx C Buıu ! ı Px/ dt
CxT .tf /Hıx.tf / CZ tf
0
xT Qıx C uT Rıu C ı"T .Ax C Buu ! Px/
C"T .Aıx C Buıu ! ı Px/ dt:
■ The variation of Ja must equal zero
ıJa.x; u; "; ıx; ıu; ı"/ D xT .tf /Hıx.tf /
CZ tf
0
.xT Q C "T A/ıx C .uT R C "T Bu/ıu
Cı"T .Ax C Buu ! Px/ ! "T ı Px dt D 0:
■ The last term, "T ı Px, deserves special attention. It is a function of ıx
so is not an independent variable. We can eliminate it from the
equation via integration-by-partsZ tf
0
"T .t/ı Px.t/ dt D "T .t/ıx.t/ˇˇtf0
!Z tf
0
P"T .t/ıx.t/ dt:
■ Note that the initial state is fixed, so its partial is zero soZ tf
0
"T .t/ı Px.t/ dt D "T .tf /ıx.tf / !Z tf
0
P"T .t/ıx.t/ dt:
■ Substituting,
ıJa.x; u; "; ıx; ıu; ı"/ D!
xT .tf /H ! "T .tf /"
ıx.tf /
CZ tf
0
.xT Q C "T A C P"T /ıx C .uT R C "T Bu/ıu
Cı"T .Ax C Buu ! Px/ dt D 0:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–12
■ Since the variations ıx, ıu, ı" are arbitrary, this expression is zero iff
"T .tf / D xT .tf /H; 0 D uT .t/R C "T .t/Bu;
P"T .t/ D !xT .t/Q ! "T .t/A; Px.t/ D Ax.t/ C Buu.t/:
■ From the third result, the optimal control is u.t/ D !R!1BTu ".t/, where
the inverse exists since R > 0. Still need to solve for ".t/ . . .
■ Combining two remaining equations (substitute u.t/ as above)
"
Px.t/
P".t/
#
D
"
A !BuR!1BTu
!Q !AT
#
„ ƒ‚ …
Hamiltonian matrix
"
x.t/
".t/
#
D Z
"
x.t/
".t/
#
■ We solve this system of equations noting that x.0/ D x0 and
".tf / D Hx.tf /.
"
x.tf /
".tf /
#
D eZ.tf !t /
"
x.t/
".t/
#
D
"
ˆ11.tf ! t/ ˆ12.tf ! t/
ˆ21.tf ! t/ ˆ22.tf ! t/
#"
x.t/
".t/
#
■ Substitute ".tf /
"
x.tf /
Hx.tf /
#
D
"
ˆ11.tf ! t/ ˆ12.tf ! t/
ˆ21.tf ! t/ ˆ22.tf ! t/
#"
x.t/
".t/
#
■ Eliminating x.tf / by substituting first equation into second
H!
ˆ11.tf ! t/x.t/ C ˆ12.tf ! t/".t/"
D ˆ21.tf !t/x.t/Cˆ22.tf !t/".t/:
■ Solve for ".t/
".t/ D!
ˆ22.tf ! t/ ! Hˆ12.tf ! t/"!1 !
Hˆ11.tf ! t/ ! ˆ21.tf ! t/"
x.t/
D P.t/x.t/;
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–13
so
u.t/ D !R!1BTu P.t/x.t/ D !K.t/x.t/:
Solving for P.t/ via the Hamiltonian System
■ From above, ".t/ D P.t/x.t/. Differentiating (via product rule) we get
P".t/ D PP .t/x.t/ C P.t/ Px.t/:
■ Substitute for P".t/ and Px.t/ to get
!Qx.t/ ! AT ".t/ D PP .t/x.t/ C P.t/ŒAx.t/ ! BuR!1BTu ".t/!:
■ Substitute ".t/ D P.t/x.t/ to get
Œ PP .t/ C P.t/A C AT P.t/ C Q ! P.t/BuR!1BTu P.t/!x.t/ D 0:
■ Since this is valid for arbitrary x.t/ we get
PP .t/ D !P.t/A ! AT P.t/ ! Q C P.t/BuR!1BTu P.t/:
■ This is called the differential (matrix) Riccati equation.
■ It is a nonlinear differential equation with boundary condition
P.tf / D H , solved backwards in time.
Steady-State Solution
■ As the differential equation for P.t/ is simulated backward in time
from the terminal point, it tends toward steady-state values as t ! 0.
It is much simpler to approximate the optimal control gains as a
constant set of gains calculated using Pss.
0 D PssBR!1BT Pss ! PssA ! AT Pss ! Q:
■ This is called the (continuous-time) algebraic Riccati equation (ARE).
In MATLAB, care.m
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–14
3.4: Solving the Riccati equation
Solving the differential Riccati equation via simulation
■ The differential Riccati equation may be solved numerically by
integrating the matrix differential equation
PP .t/ D P.t/BR!1BT P.t/ ! P.t/A ! AT P.t/ ! Q
backward in time.
■ An easy way to do this is to use Simulink with matrix signals.
EXAMPLE: Consider the continuous-time system
Px.t/ D
"
1 0
2 0
#
x.t/ C
"
1
0
#
u.t/
y.t/ Dh
0 1i
x.t/:
■ Solve the differential matrix Riccati equation that results in the control
signal that minimizes the cost function
J D1
2xT .5/
"
2 0
0 2
#
x.5/ C1
2
Z 5
0
ŒyT .t/y.t/ C uT .t/u.t/! dt:
■ First, note that the open-loop system is unstable, with poles at 0
and 1. It is controllable and observable.
■ The cost function is written in terms of y.t/ but not x.t/. However,
since there is no feedthrough term, we can also write it as
J D1
2xT .5/
"
2 0
0 2
#
x.5/ C1
2
Z 5
0
!
xT .t/C T Cx.t/ C uT .t/u.t/"
dt:
This is a common “trick”.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–15
■ Therefore, the penalty matrices are Q D C T C and R D $ D 1.
■ We can simulate the finite-horizon case to find P.t/.
P=P.signals.values;t=squeeze(t.signals.values);P2=reshape(P,[min(size(P))^2 1 length(P)]);plot(t,squeeze(P2)’)
Integrator has "initial condition" Ptf.Final time.
To plot:
tP
MatrixMultiply
Product
K*u
MatrixGain A’
u*K
MatrixGain A
1/s−1
Gain
B*inv(R)*B’
tfQ
Clock
P(t)P(t)
0 1 2 3 4 50
1
2
3
4Solving for P
Time (s)
Valu
eSolving the algebraic Riccati equation manually
■ We can also solve the infinite-horizon case (analytically, for this
example). Consider the ARE
0 D AT P C PA C C T C ! PBR!1BT P"
0 0
0 0
#
D
"
1 2
0 0
#"
p11 p12
p21 p22
#
C
"
p11 p12
p21 p22
#"
1 0
2 0
#
C
"
0 0
0 1
#
!
"
p11 p12
p21 p22
#"
1 0
0 0
#"
p11 p12
p21 p22
#
D
"
p11 C 2p12 p12 C 2p22
0 0
#
C
"
p11 C 2p12 0
p12 C 2p22 0
#
C
"
0 0
0 1
#
!
"
p211 p11p12
p11p12 p212
#
:
■ This matrix equality represents a set of three simultaneous equations
(because P is symmetric). They are:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–16
2p11 ! p211 C 4p12 D 0 1 ! p2
12 D0:
p12 C 2p22 ! p11p12 D 0
■ The final equation gives us p12 D ˙1. If we select p12 D !1 then the
first equation will have complex roots (bad). So, p12 D 1.
■ Then, p11 D 1 ˙p
5. If p11 D 1 !p
5 then P cannot be positive
definite. Therefore, p11 D 1 Cp
5 D 3:236.
■ Finally, we get p22 Dp
5=2 D 1:118.
■ These are the same values as the steady-state solution found by
integrating the differential Riccati equation.
■ The static feedback control signal is
u.t/ D !R!1BT Pssx.t/ D !h
3:236 1i
x.t/:
For this feedback, the closed-loop poles are at !p
5
2˙
p3
2j (stable).
Solving the algebraic Riccati equation generally
■ Can sometimes solve ARE manually, as above.
■ Or, can solve ARE by substituting P.tf / D H and solving differential
equation backwards in time until steady-state achieved.
" Usually a bad idea due to computation involved and propagation of
numeric errors.
■ Instead, recall
"
Px.t/
P".t/
#
D
"
A !BuR!1BTu
!Q !AT
#"
x.t/
".t/
#
D Z
"
x.t/
".t/
#
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–17
■ Diagonalize the Hamiltonian matrix. This can be done if all
eigenvalues distinct. If not, perturb Q or R.
"
P1.t/
P2.t/
#
D
"
!ƒ 0
0 ƒ
#"
´1.t/
´2.t/
#
■ The matrix ƒ is diagonal, and contains RHP poles of Hamiltonian
matrix.
■ We achieved the transformation via the transformation matrix ‰
"
x.t/
".t/
#
D
"
‰11 ‰12
‰21 ‰22
#"
´1.t/
´2.t/
#
■ The reverse relationship is then
"
´1.t/
´2.t/
#
D
"
.‰!1/11 .‰!1/12
.‰!1/21 .‰!1/22
#"
x.t/
".t/
#
■ Solving for ´2.t/ we get
´2.t/ D eƒt´2.0/ D .‰!1/21x.t/ C .‰!1/22".t/
D!
.‰!1/21 C .‰!1/22P.t/"
x.t/:
■ As t ! 1, x.t/ ! 0 for the cost function to remain finite. Therefore
limt!1
´2.t/ D limt!1
eƒt´2.0/ D limt!1
!
.‰!1/21 C .‰!1/22P.t/"
x.t/ D 0
which forces ´2.0/ D 0 and therefore ´2.t/ D 0. Then
x.t/ D ‰11´1.t/I
".t/ D ‰21´1.t/:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–18
■ Since ".t/ D P x.t/ D ‰21.‰11/!1x.t/ then
P D ‰21.‰11/!1
and the steady-state optimal feedback gain matrix is
K D R!1BTu P D R!1BT
u ‰21.‰11/!1:
■ Summary:
1. Find the eigenvalues and eigenvectors of the Hamiltonian matrix.
2. Select the eigenvectors associated with stable eigenvalues and
write as
"
‰11
‰21
#
:
3. Compute the steady-state P D ‰21.‰11/!1.
4. Compute the steady-state control gain as K D R!1BTu ‰21.‰11/
!1.
■ “Schur decomposition” algorithm also exists—numerically more
stable.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–19
3.5: Symmetric root locus
■ We have found that the closed-loop dynamics are governed by"
Px.t/
P".t/
#
D
"
A !BuR!1BTu
!Q !AT
#"
x.t/
".t/
#
D Z
"
x.t/
".t/
#
■ We have seen how to eliminate ".t/ using the solution to a Riccati
equation (more later), but the Hamiltonian is still useful since it can
tell us where the closed-loop poles are.
" Closed-loop poles are given by det.sI ! Z / D 0.
" Can evaluate this using the block-matrix determinant identity:
det
"
A B
C D
#
D det.A/ det.D ! CA!1B/:
■ So, substituting terms from the Hamiltonian,
det.sI ! Z /
D jsI ! Aj % detŒ.sI C AT / ! Q.sI ! A/!1BuR!1BTu !
D jsI ! Aj % jsI C AT j % detŒI ! .sI C AT /!1Q.sI ! A/!1BuR!1BTu ! D 0:
■ Note: det.I C ABC / D det.I C CAB/ so,
det.sI ! Z /
D jsI ! Aj % jsI C AT j % detŒI C R!1BTu .!sI ! AT /!1Q.sI ! A/!1Bu!:
■ Let Q D C Ty QyCy. Also
Gyu.s/ D Cy.sI ! A/!1Bu
GTyu.!s/ D BT
u .!sI ! AT /!1C Ty
D.s/ D det.sI ! A/
D.!s/ D det.sI C A/T .!1/n
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–20
■ Therefore
det.sI ! Z / D .!1/nD.s/D.!s/ detŒI C R!1GTyu.!s/QyGyu.s/!
D .!1/nD.s/D.!s/ŒI C R!1QyGyu.s/Gyu.!s/! if SISO:
■ This is called the “symmetric root locus” for reasons that will become
clear.
" In SISO case, there will be easy drawing rules.
" In MIMO case, don’t have easy drawing rules, but still have
closed-loop pole symmetry such that if p is a pole of the system,
then !p is also a pole of the system.
■ The stable poles are roots of the regulator. The “unstable” poles when
solving the differential equation in the forward direction become stable
poles when solving in the backward direction, so are poles of ".t/.
Symmetric root locus in MATLAB
■ We want to plot the root locus
1 C1
$GT .!s/G.s/ D 0:
■ We need to find a way to represent GT .!s/G.s/ as a state-space
system in MATLAB.
G.s/ D C.sI ! A/!1B C D
and
GT .!s/ D BT%
!sI ! AT&!1
C T C DT :
■ This can be represented in block-diagram form as:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–21
u.t/ y.t/x.t/Px.t/ ".t/P".t/
R R
A
B C
D
!AT
BT!C T
DT
■ The overall system has state
"
Px.t/
P".t/
#
D
"
A 0
!C T C !AT
#"
x.t/
".t/
#
C
"
B
!C T D
#
u.t/
y.t/ Dh
DT C BT
i"
x.t/
".t/
#
C DT Du.t/:
function srl(sys)
[A,B,C,D]=ssdata(sys);
bigA=[A zeros(size(A)); -C'*C -A'];
bigB=[B; -C'*D];
bigC=[D'*C B'];
bigD=D'*D;
srlsys=ss(bigA,bigB,bigC,bigD);
rlocus(srlsys);
EXAMPLE: Let
G.s/ D1
.s ! 1:5/.s2 C 2s C 2/:
Note that G.s/ is unstable.
−3 −2 −1 0 1 2 3−3
−2
−1
0
1
2
3
Imag
Axi
s
Real Axis
Symmetric Root Locus
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–22
3.6: Stochastic LQR
■ An intermediate step to developing controllers for systems with
incomplete, noisy measurements.
■ We consider full state available, noisy system.
Px.t/ D Ax.t/ C Buu.t/ C Bww.t/
and w.t/ & N .0; Sw/ and white; x.0/ stochastic too: EŒx.0/! D mo and
EŒx.0/x.0/T ! D †x.0/.
Cost functions involving random signals
■ Need to use expectation to come up with real measure of
performance
J D E
#Z tf
0
´T .t/W.t/´.t/ dt
$
DZ tf
0
E!
´T .t/W.t/´.t/"
dt:
■ For a finite final time, constant W and stationary random inputs, this
is proportional to
J D E!
´T .t/W ´.t/"
:
■ We can compute the cost by noticing that J is a scalar and that
J D tracefJ g.J D trace
˚
E!
´T .t/W ´.t/"'
:
■ The trace operator is invariant under cyclic permutation
J D trace˚
W E!
´.t/´T .t/"'
D trace fW †´g :
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–23
■ We can compute
†´ DZ 1
0
Z 1
0
gcl.%1/Rw.%1 ! %2/gTcl.%2/ d%1 d%2
where gcl is the impulse response of the closed-loop system from
input w to output ´.
■ If w is white with spectral density Sw, this simplifies
†´ DZ 1
0
gcl.%/SwgTcl.%/ d%:
■ We can also compute †´ from †x if ´.t/ D C´x.t/
†´ D C´E!
x.t/xT .t/"
C T´ D C´†xC T
´ :
Then J D tracefW C´†xC T´ g, where
Acl†x C †xATcl C BclSwBT
cl D 0
may be solved for †x using lyap.
Finite horizon, white process noise
■ Moving from cost functions in general to specific LQR case: cost is
now of the form
Js D E
#
xT .tf /Hx.tf / CZ tf
0
xT .t/Qx.t/ C uT .t/Ru.t/ dt
$
:
■ The controller must be some kind of feedback system (state or
output) since x.0/ is not known a priori. Linearity in the feedback is
assumed (and is optimal if the disturbance inputs are Gaussian).
■ The system state contains all the information about past inputs and
past states that contribute to the future behavior of the system.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–24
■ Also, since w.t/ is white, noise inputs are uncorrelated with past
inputs and cannot be predicted from past inputs or states.
■ Therefore, the current state contains all the available information on
the future behavior of the plant.
■ Therefore, control at any time instant same as deterministic LQR with
unknown initial state.
Finite horizon, colored process noise
■ If we know that the noise is colored, we should augment the shaping
filter dynamics.
■ Design stochastic regulator for augmented system. Same design
procedure as deterministic since input is white Gaussian.
■ Solution includes feedback on disturbance dynamics “state.” Non
physical—we cannot measure xh.t/.
■ Therefore will need to estimate xh.t/. . . more on this later with LQG.
Infinite horizon, white process noise
■ Extend analysis of stochastic LQR to case of infinite time horizon
tf ! 1.
■ Already did this for deterministic LQR. Had no particular problems
because
" System asymptotically stable,
" Only disturbance is initial condition x.0/ which goes to zero as
tf ! 1. Therefore JLQR finite.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–25
■ Cannot make similar claims for the stochastic case. Have continuing
driving noise so even optimized cost ! 1.
■ Solution: Use a time-averaged cost
J D limtf !1
E
#
1
tf
Z tf
0
xT .t/Qx.t/ C uT .t/Ru.t/ dt:
$
■ Then, same optimality conditions as for finite horizon case since both
are fixed end-time problems.
■ Consider: A time-invariant system.
■ If control properly posed, optimum closed-loop system stable. x.t/
and u.t/ are stationary random processes, and we can write
J D limt!1
E!
x.t/T Qx.t/ C u.t/T Ru.t/"
:
■ Provides steady-state mean-square response.
■ Can use this form to analyze controller performance. Consider control
u D !Kx. Then
Px.t/ D .A ! BuK/x.t/ C Bww.t/
y.t/ D Cyx.t/:
■ Then,
J D limt!1
E!
xT .t/Qx.t/ C xT .t/KT RKx.t/"
D limt!1
E!
xT .t/˚
Q C KT RK'
x.t/"
D limt!1
trace˚
.Q C KT RK/E!
x.t/x.t/T"'
D trace˚
.Q C KT RK/†x;ss
'
;
where †x;ss solves
.A ! BuK/†x;ss C †x;ss.A ! BK/T C BwSwBTw D 0:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–26
■ We will later use this cost when comparing the performance of LQR
versus LQG.
H2 optimal control
■ A final observation is that we can pose the stochastic LQR problem
as an H2 optimization.
■ The H refers to the Hardy space of all stable, LTI systems.
■ The subscript “2” denotes the applicable system norm.
■ The H2 problem is to find the LTI controller for the plant
Px.t/ D Ax.t/ Ch
Bu Ii"
u.t/
w.t/
#
I
2
64
m.t/
y1.t/
u1.t/
3
75 D
2
64
I
Q1=2
0
3
75x.t/ C
2
64
0 0
0 0
R1=2 0
3
75
"
u.t/
w.t/
#
;
that stabilizes the closed-loop system and minimizes the system
2-norm
J2 D#Z 1
0
tracefgTcl.t/gcl.t/ dt
$1=2
D kGclk2 ;
where gcl.t/ is the impulse response from w.t/ to the reference
output (combination of u1.t/ and y1.t/).
■ Equivalent to steady-state stochastic regulator with Sw D I . Then,
steady-state mean square value of reference output is
E
(h
yT1 .1/ uT
1 .1/i"
y1.1/
u1.1/
#)
D EŒxT .1/Qx.1/ C uT .1/Ru.1/!;
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–27
which is also the square of the closed-loop 2-norm kGclk22.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–28
3.7: The LQR problem solved via dynamic programming
■ We now consider the discrete-time LQR problem, and a different
method of solution known as dynamic programming.
" Review from final chapter of ECE5520 materials.
■ Dynamic programming optimizes multi-step optimizations a single
step at a time, efficiently eliminating infeasible paths.
■ Consider the task of finding the lowest-cost route from point xo to xf ,
where there are many possible ways to get there.
➀
➁
➂
➃
➄
➅
➆
➇J12
J15 J58
J78J47
J24
J46
J68
J36
J23 J38
■ Then J $18 D min fJ15 C J58; J12 C J24 C J46 C J68; : : :g.
■ We need to make only one simple observation:
In general, if xi is an intermediate point between xo and xtf and
xi is on the optimal path, then J $of D Joi C J $
if .
■ This is called Bellman’s principle of optimality:
“An optimal path has the property that whatever the initial
conditions and control variables (choices) over some initial
period, the control (or decision variables) chosen over the
remaining period must be optimal for the remaining problem,
with the state resulting from the early decisions taken to be the
initial condition.”
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–29
Vector derivatives
■ In the derivation, we will need to take derivatives of vector/ matrix
quantities.
■ This small dictionary should help: x; y 2 Rn, A 2 R
n'n.
1.@
@x.xT Ay/ D Ay,
2.@
@x.yT Ax/ D AT y,
3.@
@x.xT Ax/ D .A C AT /x.
The discrete-time LQR problem
■ We want to choose uk such that we minimize
Ji;N D xTN HdxN C
N !1X
kDi
!
xTk Qdxk C uT
k Rduk
"
;
■ To find the optimum uk, we start at the last step and work backwards.
JN !1;N D xTN HdxN C xT
N !1QdxN !1 C uTN !1RduN !1:
■ We express xN as a function of xN !1 and uN !1 via the system
dynamics
JN !1;N D .AxN !1 C BuN !1/T Hd .AxN !1 C BuN !1/
CxTN !1Qd xN !1 C uT
N !1RduN !1
D xTN !1A
T HdAxN !1 C uTN !1B
T HdBuN !1
CxTN !1A
T HdBuN !1 C uTN !1B
T HdAxN !1
CxTN !1Qd xN !1 C uT
N !1RduN !1:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–30
■ We minimize over all possible inputs uN !1 by differentiation
@JN !1;N
@uN !1
D 2BT HdBuN !1 C 2BT HdAxN !1 C 2RduN !1 D 0
0 D 2%
Rd C BT HdB&
uN !1 C 2BT HdAxN !1:
■ Therefore,
u$N !1 D !
%
Rd C BT HdB&!1
BT HdAxN !1:
■ The exciting point is that the optimal uN !1, with no constraints on its
functional form, turns out to be a linear state feedback! To ease
notation, define
KN !1 D%
Rd C BT HdB&!1
BT HdA
such that
u$N !1 D !KN !1xN !1:
■ Now, we can express the value of J $N !1;N as
J $N !1;N D
"
AxN !1 ! BKN !1xN !1
!T
Hd
AxN !1 ! BKN !1xN !1
!
CxTN !1QdxN !1 C xT
N !1KTN !1RdKN !1xN !1
#
D xTN !1
"
.A ! BKN !1/T Hd.A ! BKN !1/ C Qd C KT
N !1RdKN !1
#
xN !1:
■ Simplify notation once again by defining PN D Hd and
PN !1 D .A ! BKN !1/T PN .A ! BKN !1/ C Qd C KT
N !1RdKN !1;
so that
J $N !1;N D xT
N !1PN !1xN !1:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–31
■ To see that this notation makes sense, notice that
JN;N D J $N;N D xT
N PN xN4D xT
N HdxN :
■ Now, we take another step backwards and compute the cost JN !2;N
JN !2;N D JN !2;N !1 C JN !1;N :
Therefore, the optimal policy (via dynamic programming) is
J $N !2;N D JN !2;N !1 C J $
N !1;N :
■ To minimize this, we realize that N ! 1 is now the goal state and
JN !2;N !1 D .AxN !2 C BuN !2/T PN !1 .AxN !2 C BuN !2/
CxTN !2Qd xN !2 C uT
N !2RduN !2:
■ We can find the best result just as before
u$N !2 D !KN !2xN !2
where
KN !2 D%
Rd C BT PN !1B&!1
BT PN !1A:
■ In general,
u$Œk! D !KkxŒk!
where
Kk D%
Rd C BT PkC1B&!1
BT PkC1A
and
Pk D .A ! BKk/T PkC1.A ! BKk/ C Qd C KTk Rd Kk;
■ This difference equation for Pk has a starting condition that occurs at
the final time, and is solved recursively backwards in time.
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–32
EXAMPLE: Simulate a feedback controller for the system
xkC1 D
"
2 1
!1 1
#
xk C
"
0
1
#
uk; x0 D
"
2
!3
#
such that the cost criterion
J D xT10
"
5 0
0 5
#
x10 C9X
kD1
xTk
"
2 0
0 0:1
#
xk C 2u2k
!
is minimized.
■ From the problem, we gather that
P10 D
"
5 0
0 5
#
; Qd D
"
2 0
0 0:1
#
; Rd D Œ2!:
■ Iteratively, solve for K9, P9, K8, P8 and so forth down to K1 and P1.
Then, uŒk! D !KkxŒk!.
A=[2 1; -1 1]; B=[0; 1]; x0=[2; -3];
P=zeros(2,2,10); K=zeros(1,2,9);
x=zeros(2,1,11); x(:,:,1)=x0;
P(:,:,10)=[5 0; 0 5]; R=2; Q=[2 0; 0 0.1];
for i=9:-1:1,
K(:,:,i)=inv(R+B'*P(:,:,i+1)*B)*B'*P(:,:,i+1)*A;
P(:,:,i)=(A-B*K(:,:,i))'*P(:,:,i+1)*(A-B*K(:,:,i))+ ...
Q+K(:,:,i)'*R*K(:,:,i);
end
for i=1:9,
x(:,:,i+1)=A*x(:,:,i)-B*K(:,:,i)*x(:,:,i);
end
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–33
0 2 4 6 8 10−3
−2
−1
0
1
2State vector x[k]
Time sample, k
Valu
e
1 2 3 4 5 6 7 8 9−1
−0.5
0
0.5
1
1.5
2
2.5k2
k1
Feedback Gains K[k]
Time sample, k
Valu
e
2 4 6 8 100
10
20
30
40
50
60P11
P12=P21P22
Elements of the P matrix
Time sample, k
Valu
e
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–34
3.8: Infinite-horizon discrete-time LQR
■ If we let N ! 1, then Pk tends to a steady-state solution as k ! 0.
Therefore, Kk ! K. This is clearly a much easier control design, and
usually does just about as well.
■ To find the steady-state P and K, we let Pk D PkC1 D Pss in the above
equation.
Pss D .A ! BK/T Pss.A ! BK/ C Qd C KT RdK
and
K D%
Rd C BT PssB&!1
BT PssA
which may be combined to get
Pss D AT PssA ! AT PssB%
Rd C BT PssB&!1
BT PssA C Qd
which is called a (discrete-time) algebraic Riccati equation, and may
be solved in MATLAB using dare.m
EXAMPLE: For the previous example (with a finite end time), the solution
reached for P1 was
P1 D
"
49:5336 28:5208
28:5208 20:8434
#
:
In MATLAB, dare(A,B,Q,R) for the same system gives
Pss D
"
49:5352 28:5215
28:5215 20:8438
#
:
So, we see that the system settles very quickly to steady-state
behavior.
■ There are many ways to solve the the DARE, but when Qd has the
form C T C , and the system is SISO, there is a simple method which
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–35
yields the optimal closed-loop eigenvalues directly. (Note, when
Qd D C T C we are minimizing the output energy jykj2).
Chang–Letov method
■ The optimal eigenvalues are the roots of the equation
1 C1
$GT .´!1/G.´/ D 0
which are inside the unit circle, where
G.´/ D C.´I ! A/!1B C D:
(Proved earlier for the continuous-time version).
EXAMPLE: Consider G.´/ D1
´ ! 1so
1 C$!1
.´ ! 1/.´!1 ! 1/D 0
2 C $!1 ! ´ ! ´!1 D
´ D 1 C1
2$˙
s
1
4$2C
1
$:
■ The locus of optimal pole locations for all $ form a reciprocal root
locus.
Reciprocal root locus in MATLAB (SISO)
■ We want to plot the root locus
1 C1
$GT .´!1/G.´/ D 0;
where
G.´/ D C.´I ! A/!1B C D:
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–36
■ We know how to plot a root locus of the form
1 C KG 0.´/ D 0
so we need to find a way to convert GT .´!1/G.´/ into G 0.´/.
■ We know that
GT .´!1/ D BT%
´!1I ! AT&!1
C T C DT
D BT ´%
´I ! A!T&!1
.!A!T C T / C DT :
■ Combining G.´/ and GT .´!1/ in block-diagram form:
uŒk!
yŒk!
xŒk!xŒk C 1!"Œk!
"Œk C 1!
´!1 ´!1
A
B C
D
A!T
BT
!C T
DT
■ The overall system has state"
xŒk C 1!
"Œk C 1!
#
D
"
A 0
!A!T C T C A!T
#"
xŒk!
"Œk!
#
C
"
B
!A!T C T D
#
uŒk!
yŒk! Dh
!BT A!T C T C C DT C BT A!Ti"
xŒk!
"Œk!
#
C
!
DT D ! BT A!T C T D"
uŒk!:
function rrl(sys)
[A,B,C,D]=ssdata(sys);
bigA=[A zeros(size(A)); -inv(A)'*C'*C inv(A)'];
bigB=[B; -inv(A)'*C'*D];
bigC=[-B'*inv(A)'*C'*C+D'*C B'*inv(A)'];
bigD=-B'*inv(A)'*C'*D+D'*D;
rrlsys=ss(bigA,bigB,bigC,bigD,-1);
rlocus(rrlsys);
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett
ECE5530, LINEAR QUADRATIC REGULATOR 3–37
EXAMPLE: Let
G.´/ D.´ C 0:25/.´2 C ´ C 0:5/
.´ ! 0:2/.´2 ! 2´ C 2/:
Note that G.´/ is unstable.
−3 −2 −1 0 1 2 3−2
−1
0
1
2
Imag
Axi
s
Real Axis
Reciprocal root locus
OBSERVATIONS: For the “expensive cost of control” case, stable poles
remain where they are and unstable poles are mirrored into the unit
disc. (They are not moved to be just barely stable, as we might
expect!)
■ For the “cheap cost of control” case, poles migrate to the finite zeros
of the transfer function, and to the origin (deadbeat control).
Lecture notes prepared by Dr. Gregory L. Plett. Copyright © 2016, 2010, 2008, 2006, 2004, 2002, 2001, 1999, Gregory L. Plett