Multivariable Control
Systems
• Ali Karimpour
• Assistant Professor
• Ferdowsi University of Mashhad
Chapter 8
Multivariable Control System Design: LQG Method
Topics to be covered include:
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
2
LQG Control
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
3
LQG Control
In traditional LQG Control, it is assumed that the plant dynamics are linear and known
and that the measurement noise and disturbance signals (process noise) are stochastic
with known statistical properties.
x = Ax + Bu + w
y = Cx + v
That is, wd and wn are white noise processes with covariances
 
Evv  = V  0
E wwT = W  0
and
T
Ewv T = 0
EvwT = 0
The problem is then to devise a feedback-control law which minimizes the ‘cost’
(
)
1 T T


J = E  lim  z Qz + u T Ru dt 
T → T 0

where z = Mx, Q = QT  0 and R = RT  0
4
LQG Control
The solution to the LQG problem is prescribed by the separation theorem, which
states that the optimal result is achieved by adopting the following procedure.
• First, obtain an optimal estimate of the state x


Optimal in the sense that E ( x − xˆ )T ( x − xˆ )
is minimized
• Then use this estimate as if it were an exact measurement of the state to solve
the deterministic linear quadratic control problem.
5
LQG Control: Optimal state feedback

(
)
J r =  z T Qz + u T Ru dt
0
where z = Mx, Q = QT  0 and R = RT  0
The optimal solution for any initial state is
u(t ) = − K r x(t )
where
K r = R −1 B T X
Where X=XT ≥ 0 is the unique positive-semidefinite solution of the
algebraic Riccati equation
AT X + XA − XBR−1 BT X + M T QM = 0
6
LQG Control: Kalman filter
The Kalman filter has the structure of an ordinary
state-estimator or observer, as
xˆ = Axˆ + Bu + K f ( y − cxˆ )


The optimal choice of K f which minimizes E x − xˆ  x − xˆ  is :
T
K f = YC T V −1
Where Y=YT ≥ 0 is the unique positive-semidefinite solution of the
algebraic Riccati equation
YAT + AY − YC T V −1CY + WT = 0
7
LQG Control: Combined optimal state estimation and optimal state feedback
KLQG(s)
 A − BK r − K f C
K LQG ( s)  
− Kr

K f   A − BR −1 B T X − YC T V −1C YC T V −1 
=


0  
− R −1 B T X
0 
Exercise 1: Proof the relation of KLQG(s) according to above figure.
8
Robustness Properties
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
9
Robustness Properties
For an LQR-controlled system (i.e. assuming all the states are available and no
stochastic inputs) it is well known (Kalman, 1994; Safonov and Athans, 1997) that,
if the weight R is chosen to be diagonal, the sensitivity function
(
S = I + K r (sI − A) B
−1
)
−1
satisfies
 (S ( j))  1, 
Nyquist plot in MIMO case
From this it can be shown that the system will
have a gain margin equal to infinity, a gain
reduction margin (lower gain margin) equal
to 0.5 and a (minimum) phase margin of 60˚
-1
-1
in each plant input control channel.
So K r (sI − A) B will lie outside the
−1
unit circle center at - 1
10
Robustness Properties
 i = 0 and 0.5  ki   , i = 1, 2 , ..., m
k i = 1 and  i  60  , i = 1, 2 , ... , m
11
Robustness Properties
Example 8-1: LQR design of a first order process. Consider a first order process
x = ax + u
y=x
For a non-zero initial state the cost function to be minimized is
G( s) =
1
s−a
Jr = 

0
(x
2
)
+ Ru 2 dt
The algebraic Riccati equation becomes
aX + Xa − XR −1 X + 1 = 0
X = aR +
(aR )2 + R

X 2 − 2aRX − R = 0
K r = X / R = a + a 2 + 1/ R
x = ax + u = − a 2 + 1 / R x
12
Robustness Properties
Example 8-1: LQR design of a first order process. Consider a first order process
G( s) =
x = ax + u
y=x
1
s−a
)
(
u = Kr x = a + a2 +1/ R x
Let R = 
Let R → 0
Closed loop pole is : p = − a
x = ax + u = − a 2 + 1 / R x
2a
u = K r x = (a + a )x = 
0
if a  0
if a  0
Closed loop pole is : p = − a ....... → - 
13
Robustness Properties
So, an LQR-controlled system has good stability margins at the plant inputs,
So K r (sI − A) B will lie outside the
−1
unit circle center at - 1
Arguments dual to those employed for the LQR-controlled system can then be used to
show that, if the power spectral density matrix V is chosen to be diagonal, then at the
input to the Kalman gain matrix Kf there will be an infinite gain margin, a gain
reduction margin of 0.5 and a minimum-phase margin of 60˚.
So c(sI − A) K f will lie outside the
−1
unit circle center at - 1
14
Robustness Properties
So, an LQR-controlled system has good stability margins at the plant inputs,
And Kalman filter has good stability margins at the inputs to Kf
For an LQG-controlled system with a combined Kalman filter and LQR control law
are there any guaranteed stability margins?
Unfortunately there are no guaranteed stability margins.
This was brought starkly to the attention of the control community by Doyle (1978 )
(in a paper entitled “Guaranteed Margins for LQR Regulators” with a very compact
abstract which simply states “There are none”).
Doyle showed, by an example, that there exist LQG combinations with
arbitrarily small gain margins.
15
Robustness Properties
Why there are no guaranteed stability margins in LQG controller.
L3 (s) = K r (s) B
(Regulator transfer function)
guaranteed stability margins
L 4 ( s ) = C ( s ) K f
L2
L3
(Kalman Filter transfer function)
L4
guaranteed stability margins
L2 ( s ) = −G ( s ) K LQG ( s )

L1 ( s ) = − K LQG ( s )G ( s ) = K r ( s ) −1 + BK r + K f C

−1
K f C ( s ) B
The most important loop but no guaranteed stability margins
16
Loop Transfer Recovery
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
17
Loop transfer recovery (LTR) procedures
Assume that the plant model G(s) is minimum-phase and that it has
at least as many inputs as outputs.
The LQG loop transfer function
L2 ( s ) = −G ( s ) K LQG ( s )
→
L4 ( s ) = C ( s ) K f
Guaranteed stability margins
If Kr in the LQR problem is designed to be large using the sensitivity recovery
procedure of Kwakernaak (1969).
The LQG loop transfer function
L1 ( s ) = − K LQG ( s )G ( s )
→
L3 (s) = K r (s) B
Guaranteed stability margins
If Kf in the Kalman filter to be large using the robustness recovery procedure of
Doyle and Stein (1979).
18
Loop transfer recovery (LTR) procedures
Assume that the plant model G(s) is minimum-phase
and that it has at least as many inputs as outputs.
The LQG loop transfer function
L2 ( s ) = −G ( s ) K LQG ( s )
→
L2
L4 ( s ) = C ( s ) K f
Recovering robustness at the plant output
The LQG loop transfer function
L1 ( s ) = − K LQG ( s )G ( s )
→
L3 ( s ) = K r  ( s ) B
L1
Recovering robustness at the plant input
19
Loop transfer recovery (LTR) procedures
Recovering robustness at the plant input
Assume that the plant model G(s) is minimum-phase
and that it has at least as many inputs as outputs.
L1 ( s ) = − K LQG ( s )G ( s )
→
L1
L3 ( s ) = K r  ( s ) B
Step I: First, design the linear quadratic problem whose transfer function KrΦ(s)B
is desirable.
This is done, in an iterative fashion, by manipulate the matrices Q and R, emphasis of
The design is on aspects such as gains, possibly ‘balancing’ the principal gains, and
adjusting the low frequency behavior.
Step II: When the singular values of KrΦ(s)B are thought to be satisfactory, LTR is
achieved by designing Kf in the Kalman filter by setting Г=B, W=I and V= ρI ,where
ρ is a scalar. As ρ tends to zero
L1 ( s ) = − K LQG ( s )G ( s )
→
L3 ( s ) = K r  ( s ) B
20
Loop transfer recovery (LTR) procedures
Recovering robustness at the plant output
Assume that the plant model G(s) is minimum-phase
and that it has at least as many inputs as outputs.
L2 ( s ) = −G ( s ) K LQG ( s )
→
L2
L4 ( s ) = C ( s ) K f
Step I: First, we design a Kalman filter whose transfer function CΦ(s)Kf is desirable.
By choosing the power spectral density matrices W and V so that the minimum singular
value of CΦ(s)Kf is large enough at low frequencies for good performance and its
maximum singular value is small enough at high frequencies for robust stability.
Step II: When the singular values of CΦ(s)Kf are thought to be satisfactory, loop
transfer recovery is achieved by designing Kr in an LQR problem with M=C, Q=I
and R= ρI, where ρ is a scalar. As ρ tends to zero
L2 ( s ) = −G ( s ) K LQG ( s )
→
L4 ( s ) = C ( s ) K f
21
Loop transfer recovery (LTR) procedures
L2
L1
LTR procedure guaranteed to work only with minimum-phase plants.
• Since it relies on the ‘cancellation’ of some of the plant dynamics
by the filter Dynamics)
• If RHP zeros exist in the plant the procedure may still work,
particularly if these zeros lie beyond the operation bandwidth of the
22
system as finally designed.
Loop transfer recovery (LTR) procedures
Proof: Recovering robustness at the plant input
Г=B, W=I and V= ρI ,As ρ tends to zero
L1
?
The LQG loop transfer function at the plant input is:
L1 ( s ) = K LQG ( s )G ( s ) = − K r (sI − A + BK r + K f C ) K f C (sI − A) B
−1
  ( s) = (sI − A)−1
Let 
−1
 ( s) = ( sI − A + BK r )
−1
(
L1 ( s ) = K LQG ( s )G ( s ) = − K r  −1 + K f C
)
−1
K f CB
By matrix-inversion lemma we have
(
L1 ( s ) = K LQG ( s)G ( s ) = − K r  −1 + K f C
)
−1
K f CB = − K rK f (I + CK f ) CB
−1
(I )
Now the algebraic Riccati equation is:
YAT + AY − YC T V −1CY + WT = 0
Exercise: Derive equation I .
Let W = W0 + q 
q?→ 
23
Loop transfer recovery (LTR) procedures
Proof: Recovering robustness at the plant input
L1
Г=B, W=I and V= ρI ,As ρ tends to zero
(
L1 ( s ) = K LQG ( s)G ( s ) = − K r  −1 + K f C
)
−1
K f CB = − K rK f (I + CK f ) CB
−1
(I )
Now the algebraic Riccati equation is:
YAT + AY − YC T V −1CY + WT = 0
Let W = W0 + q 
YAT AY YC TV −1CY W0 T
+
−
+
+   T = 0
q
q
q
q
YCTV −1CY = q  T
It can be shown (Kwakernaak and Sivan, 1973) that, if
• C(sI-A)-1ГW1/2 has no RHP zero
• and if it has at least as many outputs as rank(Σ), then
Y
=0
q → q
lim
q →
(
lim qT
q →
)
1/ 2
K f = YC T V −1
= YC TV −1/ 2
V −1/ 2V −1/ 2
K f → q1/ 2 1/ 2V −1/ 2 as q24→ 
Loop transfer recovery (LTR) procedures
Proof: Recovering robustness at the plant input
L1
Г=B, W=I and V= ρI ,As ρ tends to zero
YAT AY YC TV −1CY W0 T
+
−
+
+   T = 0
q
q
q
q
It can be shown (Kwakernaak and Sivan, 1973) that, if
K f → q1/ 2 1/ 2V −1/ 2 as q → 
• C(sI-A)-1ГW1/2 has no RHP zero
• and if it has at least as many outputs as rank(Σ), then
In particular if we choose
and provided
C(sI-A)-1B
 = B,
=I
K f → q1/ 2 BV −1/ 2
has no zeros in LHP, then
as q → 
Substituting this in the LQG loop transfer function at the plant input leads to:
(
L1 ( s) = K LQG ( s )G ( s ) = − K r  −1 + K f C
)
−1
K f CB = − K rK f (I + CK f ) CB ( I )
(
= −q1/ 2 K rBV −1/ 2 I + q1/ 2CBV −1/ 2
−1
)
−1
CB
as q → 
25
Loop transfer recovery (LTR) procedures
Proof: Recovering robustness at the plant input
L1
Г=B, W=I and V= ρI ,As ρ tends to zero
(
L1 ( s ) = K LQG ( s )G ( s) = −q1/ 2 K rBV −1/ 2 I + q1/ 2CBV −1/ 2
(
L1 ( s ) = K LQG ( s )G ( s ) = − K rBV −1/ 2 CBV −1/ 2
)
−1
CB
= − K rB(CB ) CB
−1
(
)
−1
CB
as q → 
as q → 
as q → 
 =  I + BK r −1
)
as q → 
L1 ( s ) = K LQG ( s)G ( s ) = − K r I + BK r  B C I + BK r  B CB
−1
By push-through rule:
(
L1 ( s) = K LQG ( s )G ( s ) = − K rBI + K rB  CBI + K rB 
−1
Finally we have:
L1 ( s ) = K LQG ( s )G ( s ) = − K rB = L3 ( s )
−1
−1
)
−1 −1
as q → 
CB
as q → 
So : L1 (s) → L3 (s)
as q26→ 
Shaping the Principal Gains (Singular Values)
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
27
Shaping the principal gains (singular values)
In order to exploit LTR
technique, we must to know:
• How to modify W and V in order to bring about
desirable changes in C(sI-A)-1Kf
• How to modify Q and R in order to bring about
desirable changes in Kr(sI-A)-1B
In order to obtain an intuitive grasp of this, consider the Kalman filter. (let u=0)
x = Ax + Bu + w
y = Cx + v
xˆ = Axˆ + K f ( y − cxˆ )
This is now looks like a
feedback system which is to:
track (in a sense) the
‘reference input’ z, while
rejecting the measurement
errors v.
28
Shaping the principal gains (singular values)
To shape the principal-gain plots we can do one of two things:
• Modify the plant model by augmenting it with additional dynamics
For example adding integrator in each loop.
• Modify the matrices ГWГT and V in a more sophisticated way,
We can use ГWГT to increase the smallest principal
gain of the sensitivity matrix, or decrease the
largest one near some particular frequency.
29
Shaping the principal gains (singular values)
Modify the matrices W and V
Let Ff as the return difference of Kalman filter,
F f ( s ) = I + C (sI − A) K f
−1
G f ( s ) = C (sI − A) 
−1
and define
Then we can show that
F f ( s )VF fT (− s ) = V + G f ( s )WG Tf (− s )
(I)
Suppose we choose V=I. Then
F f ( j ) F fH ( j ) = I + G f ( j )WG Hf ( j )
from which it follows that
 i (F f ( j ) ) =  1 +  i2 (G f ( j )W 1/ 2 ) 
1/ 2
( II )
Exercise I : Derive equation I . (Hint Maciejowski 1989 pp. 227-231)
Exercise II: Derive equation II .
30
Shaping the principal gains (singular values)
Modify the matrices W and V

 (F ( j ) ) = 1 +  (G f ( j )W
−1
f
2
1/ 2
)
 (F f−1 ( j ) ) =  1 +  2 (G f ( j )W 1/ 2 ) 
−1 / 2
−1 / 2
So we can reduce  (F f−1 ( j ) ) by increasing (G f ( j )W
1/ 2
)
, etc.
But the point is not merely to reduce all the singular values of F f−1 ( j ) , but to reduce
the largest one, relative to smallest.
One way is: Suppose we need adjustment at ω1
G f ( j1 )W
1/ 2
= YU
m
H
=  yi  i uiH
i =1
Now let
W 1/ 2 → W 1/ 2 ( I +  u j u Hj )
So
G f ( j1 )W
1/ 2
( I +  u j u ) = YU
H
j
m
H
=  yi  i uiH + (1 +  ) j y j u Hj
i j
so the jth singular value has been changed by a factor (1+α), while all the other
31
singular values have been left unchanged.
Shaping the principal gains (singular values)
Modify the matrices W and V
W
1/ 2
→ W
1/ 2
( I +  u j u ) G f ( j1 )W
Example: Let
H
j
1/ 2
1 3 0
G f = 6 1 0
0 0 1
( I +  u j u ) = YU
H
j
m
H
=  yi  i uiH + (1 +  ) j y j u Hj
i j
4 0 0 
W = 0 9 0 
0 0 16
Singular value of GfW1/2 is:
G f W 1/ 2
0
0 − 0.86 0.51 0
− 0.47 − 0.88 0 13.43
=  − 0.88 0.47 0  0
7.60 0  − 0.51 − 0.86 0
 0
0
1  0
0
4  0
0
1
We want to change 7.6 to 3*7.6 so we change W1/2 by
W 1/ 2 = W 1/ 2 ( I + 2 u2 u2H ) G f W 1/ 2
0
0 − 0.51 − 0.86 0
 0.88 − 0.47 0 22.80
= − 0.47 − 0.88 0  0
13.43 0  0.86 − 0.51 0
 0
0
1  0
0
4  0
0 32 1
Shaping the principal gains (singular values)
Modify the matrices W and V
Now let
So
W 1/ 2 → W 1/ 2 ( I +  u j u Hj )
G f ( j1 )W
1/ 2
( I +  u j u ) = YU
H
j
m
H
=  yi  i uiH + (1 +  ) j y j u Hj
i j
so the jth singular value has been changed by a factor (1+α), while all the other
singular values have been left unchanged.
The problem with this approach is that uj is usually a complex vector, whereas we
wish to keep W1/2 real.
Once again we are faced with the problem of approximating a complex matrix by a
real matrix, and as before we can employ the align algorithm.
In this case other algorithms may be more appropriate, however, since we really
want to approximate uj rather than align it.
In particular, Re{uj} is sometimes an adequate approximation.
A further possibility is to approximate uj by the output direction of the matrix
33
[Re{uj} Im{uj}] which corresponds to its largest singular value.
Some Practical Consideration
• LQG Control
• Robustness Properties
• Loop transfer recovery (LTR) procedures
- Recovering robustness at the plant output
- Recovering robustness at the plant input
- Shaping the principal gains (singular values)
• Some practical consideration
34
Some practical consideration
LTR procedures are limited in their applicability.
Their main limitation is to minimum phase plants.
This is because the recovery procedures work by canceling the plant
zeros, and a cancelled non-minimum phase zero would lead to instability.
The cancellation of lightly damped zeros is also of concern because of
undesirable oscillations at these modes during transients.
A further disadvantage is that the limiting process
For full recovery
(  → 0)
Introduces high gains which may cause problems with unmodelled
dynamics.
The recovery procedures are not usually taken to their limits.
The result is a somewhat ad-hoc design procedure.
35
Loop transfer recovery (LTR) procedures
Design example
Consider the aircraft model AIRC described in the following state-space model.
x = Ax + Bu
y = Cx + Du
0
1.1320
0
− 1.0000 
0
0
0
 0

0 − 0.0538 − 0.1712

− 0.1200 1.0000

0
0
.
0705
0




,B= 0

A = 0
0
0
1.0000
0
0
0




0
0
.
0485
0
−
0
.
8556
−
1
.
0130
4
.
4190
0
−
1
.
6650




0 − 0.2909
 1.5750
0
1.0532 − 0.6859
0
− 0.0732
1 0 0 0 0
0 0 0 
C = 0 1 0 0 0 , D = 0 0 0
0 0 1 0 0
0 0 0
the model has three inputs, three outputs and five states.
36
Loop transfer recovery (LTR) procedures
Design example
THE SPECIFICATION
• We shall attempt to achieve a bandwidth of about l0 rad/sec for each loop.
• Integral action in each loop, little interaction between outputs.
• Good damping of step responses and zero steady-state error in the face of step
demands or disturbances.
PROPERTIES OF THE PLANT
• The time responses of the plant to unit step signals on inputs 1 and 2 exhibit very
severe interaction between outputs.
• The poles of the plant (eigenvalues of A) are
0 , − 0.78  1.03 j , − 0.0176  0.1826 j
so the system is stable (but not asymptotically stable).
• Thus this plant has no finite zeros, and we do not expect any limitations on
performance to be imposed by zeros.
37
Loop transfer recovery (LTR) procedures
Design example
Recovering robustness at the plant output
The LQG loop transfer function
L2 ( s ) = −G ( s ) K LQG ( s )
→
L4 ( s ) = C ( s ) K f
Recovering robustness at the plant input
The LQG loop transfer function
L1 ( s ) = − K LQG ( s )G ( s )
→
L3 ( s ) = K r  ( s ) B
Here we use Recovering robustness at the plant output
38
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
We need to choose the matrices Г, W, V, which appear in
YAT + AY − YC T V −1CY + WT = 0
and obtain the Kalman-filter gain Kf from
K f = YC T V −1
We shall write
K f = LQE ( A , , C , W , V )
It is generally advisable to start with simple choices of Г, W, V, inspect L4
Then adjust Г, W, V accordingly, and so gradually improve L4
One of the simplest possible choices is Г=B, W=I3 and V=I3.
39
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
So try with
0.0732 − 0.2507
 0.9897
 0.0732

0
.
9436
−
0
.
0642


K f 1 = lqe( A , B, C ,W ,V ) = − 0.2507 − 0.0642 1.7807 


−
0
.
4934
−
0
.
0485
1
.
6190


− 0.8076 − 0.2483 2.2215 
The loop transfer function L4 ( s) = C (sI − A)−1 K f 1 is:
40
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
L4 ( s ) = C (sI − A) K f 1
−1
Singular Values
100
BW around 1 rad/sec
80
Constant gain at low
frequencies
Decreasing with 20 db/dec
at low frequencies
Singular Values (dB)
60
40
20
0
-20
-40
-60
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
2
10
10
41
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The first thing to do is to insert integral action, by augmenting the plant model.
Placing poles of the augmented model at the origin leads to problems in the
recovery step later, so in this case we place them at -0.001, which is virtually at
the origin, when compared to the required bandwidth 10 rad/sec
Aw = −0.001 I 3
Bw = I 3
Cw = I 3
Dw = 03,3
We could also have chosen Cw more carefully, with the aim of adjusting the
low frequency gains. The augmented model is
 A Cw 
Aa = 

 0 Aw 
 B
Ba =  
0
Ca = C 0
Da = 03,3
D 
a =  w 
 Bw 
42
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
 A Cw 
Aa = 

 0 Aw 
 B
Ba =  
0
Ca = C 0
Da = 03,3
D 
a =  w 
 Bw 
Now we have
Kf2
0.0909 − 0.1648
 1.4129
 0.0909

1
.
3501
−
0
.
0539


− 0.1648 − 0.0539 2.0429 


−
0
.
4556
0
.
0062
2
.
1018

= lqe( Aa , a , Ca , W , V ) = 
 − 1.2024 − 0.2823 2.4314 


− 0.6484 − 0.0329 0.7587 
 0.0225
0.9964
0.0624 


 − 0.7573 0.0576 − 0.6463
43
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The loop transfer function L4 ( s) = Ca (sI − Aa )K f 2 is:
Singular Values
140
The gain was added around
60 db at low frequency
according to integrator
60 db
By tuning W one can
manipulate gains.
100
80
Singular Values (dB)
We want to increase the
gain at low frequency.
Kf1
Kf2
120
60
40
20
0
-20
-40
-60
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
44
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
By tuning W one can manipulate gains. But how?
YAT + AY − YC T V −1CY + WT = 0
In a case of diagonal system every diagonal element of W corresponds to a
singular value.
But in non-diagonal system we must use singular value decomposition.
Ca ( j 0.001 I − Aa ) aW 1/ 2 = YU H
−1
Which gives
0.509


u3 = 0.158 + 0.0005 j 
0.846 − 0.0005 j 
4.7942 106

=
0

0


0
0

4637
0

0
651.2119
45
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
0.509
u3  0.158
0.846
By tuning W one can manipulate gains. But how?
Now let α=9 (α+1=10) so we have (better value for α+1 is 4637/651=7.12)
(
W31/ 2 = I 3 + 9 Re u3u3
so
Kf3
T
)
W3 = W
1/ 2
3
(W )
1/ 2 T
3
 2.77 − 0.37 − 0.50
− 0.37 1.54

0
.
15


− 0.50 0.15
2.51 


−
1
.
11
0
.
54
3
.
27

= lqe( Aa , a , Ca , W3 , V ) = 
 − 4.61 1.21
4.42 


2.89 
 − 3.94 1.64
 − 0.88 1.48
0.71 


46
3.05 
− 7.24 3.16
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The loop transfer function L4 ( s) = Ca (sI − Aa )K f 3 is:
Singular Values
140
100
Kf1
Kf2
80
Kf3
120
We need at least 7 rad/sec.
Singular Values (dB)
Band width problem?
60
40
20
0
W3 must increase.
-20
-40
-60
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
47
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
W3 must increase.
Maximum singular value of L4 ( s ) = Ca (sI − Aa )K f 3, x , 4
Singular Values
W3 and 10W3 and 100W3 are
considered
160
K f 3 = lqe( Aa , a , Ca , W3 , V )
K f 4 = lqe( Aa , a , Ca ,100W3 , V )
Kf3
120
Kfx
Kf4
100
Singular Values (dB)
K fx = lqe( Aa , a , Ca ,10W3 , V )
140
80
60
40
20
Which one is ok?
0
-20
-40
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
2
10
10
48
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The loop transfer function L4 ( s) = Ca (sI − Aa )K f 4 is:
K f 4 = lqe( Aa , a , Ca ,100W3 , V )
Singular Values
160
140
We need to find closed loop
transfer functions
T f ( s) = I − S f ( s)
−1
100
)
−1
Singular Values (dB)
(
S f ( s) = (I + L4 ) = I + Ca ( sI − Aa ) K f 4
−1
120
80
60
40
20
0
-20
-40
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
2
10
10
49
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
We need to find closed loop transfer functions
(
S f ( s) = (I + L4 ) = I + Ca ( sI − Aa ) −1 K f 4
−1
)
−1
T f ( s) = I − S f ( s)
Singular Values
5
BT is between 6.5 till 12 rad/sec
0
B is between 2.5 till 5.5 rad/sec

= 1.6(4dB)
Singular Values (dB)
Tf
-3
We could therefore terminate the
Kalman filter design and move on to
the recovery step.
However we shall suppose that we
wish to improve the sensitivity
further.
-5
-10
-15
-20
0
10
1
10
Frequency (rad/sec)
50
2
10
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
However we shall suppose that we wish to improve the sensitivity further.
Ca ( j 5.5 I − Aa ) a (100W )1/ 2 = YU H
−1
− 0.71
− 0.49
− 0.4979 

U =  − 0.32 + 0.03 j 0.85 − 0.10 j − 0.39 + 0.06 j 
− 0.62 + 0.04 j 0.12 + 0.003 j 0.77 − 0.05 j 
0
0 
0.8076
 =  0
0.3382
0 
 0
0
0.1020
W51/2 can be shape as follows:
(

W51/ 2 = 10W31/ 2 I + 0.2382 Re u1u1
1
−1
0.8076
H
)(I + 1.9567 Reu u )(I + 8.8039 Reu u )
H
2 2
1
−1
0.3382
(
W5 = W51/ 2 W51/ 2
H
3 3
1
−1
0.1020
)
T
51
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
(

W51/ 2 = 10W31/ 2 I + 0.069 Re u1u1
H
)(I + 1.639 Reu u )(I + 7.409 Reu u )
H
2 2
W5 = W
1/ 2
5
H
3 3
(W )
1/ 2 T
5
So we have
Kf5
 10.31
 − 0.19

 0.57

6.79

= lqe( Aa , a , Ca , W5 ,V ) =
 − 52.64

 − 119.65
 − 12.72

− 312.55
− 0.19
0.57 
7.77
− 0.12 
− 0.12 10.08 

− 1.00
51.00 
1.27
6.31 

4.28
− 12.75 
31.02
− 1.50 

13.69 − 133.37
52
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The loop transfer function L4 ( s) = Ca (sI − Aa )K f 5 is:
Singular Values
200
The principle gains have clearly
been “squeezed together” near 10
rad/sec, and at higher frequency.
Singular Values (dB)
150
100
50
0
-50
-3
10
-2
10
-1
10
0
10
Frequency (rad/sec)
1
2
10
10
53
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
We need to find closed loop transfer functions
(
S f ( s) = (I + L4 ) = I + Ca ( sI − Aa ) −1 K f 5
)
−1
T f ( s) = I − S f ( s)
Singular Values
5
0
Singular Values (dB)
−1
-5
-10
-15
-20
0
10
1
10
Frequency (rad/sec)
54 102
Loop transfer recovery (LTR) procedures
Design example
Kalman filter design
The characteristic loci of
L4 ( s ) = Ca (sI − Aa )K f 5 is:
5
All the loci remain outside or on the
boundary of the unit circle as
predicted by Kalman filter theory,
from which we know that:
4
3
2
Im
1
 i (S f )  1
0
-1
-2
-3
-4
-5
-10
-9
-8
-7
-6
-5
Re
-4
-3
-2
-1
55
0
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
Since the system has no transmission zero in the RHP so arbitrary good recovery
should be possible.
To obtain LTR, we solve the following Riccati equation
AT X + XA − XBR−1 BT X + M T QM = 0
With M=Ca, Q=I and R=ρI
We shall write
K r = lqr ( Aa , Ba , Ca Ca , I 3 )
T
When Kr has been find the controller realization is:
ALTR = Aa − Ba K r − K f 5Ca
BLTR = K f 5
C LTR = − K r
DLTR = 03,3
Now we need
L2 ( s ) = −G ( s ) K LQG ( s )
→
L4 ( s ) = C ( s ) K f
56
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
L2 ( s ) = −G ( s ) K LQG ( s )
Now we need
→
L4 ( s ) = C ( s ) K f 5
-2
ρ=10
=1e-2
-4
ρ=10
=1e-4
200
200
L4
150
L4
150
L2
100
Singular Values (dB)
Singular Values (dB)
100
50
0
-50
-100
-3
10
L2
50
0
-50
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
-100
-3
10
-2
10
-1
10
0
10
1
2
10
10
Frequency (rad/sec)
57
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
L2 ( s ) = −G ( s ) K LQG ( s )
Now we need
-6
=1e-6
ρ=10
200
→
L4 ( s ) = C ( s ) K f 5
-8
=1e-8
ρ=10
200
L4
150
L2
Singular Values (dB)
Singular Values (dB)
150
L4
100
50
0
-50
-3
10
L2
100
50
0
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
-50
-3
10
-2
10
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
58
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
We have for ρ=10-8
36.3 82 1 0 0
− 5960 − 1158 7853
T
K r = lqr ( Aa , Ba , Ca Ca , I 3 ) =  − 700 9933
910
4.3
9.8 0 1 0
− 8000 − 7.8 − 6205 − 79.4 180 0 0 1
59
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
Characteristic loci at the output of compensated plant, for ρ=10-6 and ρ=10-8
3
ρ=10-6
ρ=10-8
2
1
Im
0
-1
-2
-3
-4
-7
-6
-5
-4
-3
Re
-2
-1
0
60
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
Principle gains of S and T, for ρ=10-6 and ρ=10-8
10
-6
=1e-6
ρ=10
10
T
S
5
0
Singular Values (dB)
Singular Values (dB)
T
S
5
0
-5
-5
-10
-10
-15
-15
-20
0
10
=1e-8-8
ρ=10
1
10
Frequency (rad/sec)
2
10
-20
0
10
1
10
2
10
Frequency (rad/sec)
61
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
Closed-loop step responses to step responses to different inputs.
Step Response
From: In(2)
From: In(1)
From: In(3)
To: Out(1)
1.5
1
0.5
0
To: Out(2)
1
0.5
0
1.5
To: Out(3)
Amplitude
1.5
1
0.5
0
0
1
2 0
1
Time (sec)
2 0
1
2
62
Loop transfer recovery (LTR) procedures
Design example
Recovery at the plant output
The closed-loop poles are located at
− 156.93  156.93 j
− 5.48  0.16 j
− 2.74  4.62 j
− 10072
− 48.14  48.14 j
− 2.74  4.92 j
− 3.90  3.90 j
63