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Advanced and Multivariable Control
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Formulas
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Stability 𝑥̇
Equilibrium pair (𝑥̅ , 𝑢̅ ) → 𝑓(𝑥̅ , 𝑢̅ ) = 0
The equilibrium 𝑥̅ is
- Isolated if there exists 𝛿 (region of attraction) such that there does not exist any other
equilibrium in 𝐵
𝛿
( 𝑥̅ , 𝛿
) ≔ {𝑥: |
| 𝑥 − 𝑥̅
| | ≤ 𝛿}
- Stable if for any 𝜀 > 0 there exists 𝛿 > 0 such that, for all initial
states satisfying |
| 𝑥
0
− 𝑥̅
| | ≤ 𝛿, it holds that |
| 𝑥(𝑡) − 𝑥̅
| | ≤ 𝜀
If 𝑥̅ is stable and lim
𝑡→∞
|
| 𝑥
( 𝑡
) − 𝑥̅
| | = 0 , then 𝑥̅ is asymptotically stable
𝑥̅ is globally asymptotically stable if it is asymptotically stable for any 𝑥
0
𝑛
In linear systems stability is a property of the system, while in nonlinear ones it is a property
of the equilibrium
Nonlinear systems :
𝑥̇
( 𝑡
) = 𝑓(𝑥
( 𝑡
) , 𝑢
( 𝑡
) )
𝑙𝑖𝑛
→ 𝛿𝑥̇
( 𝑡
) = 𝐴𝛿𝑥
( 𝑡
)
𝐴 =
𝜕𝑓
𝜕𝑥
|
( 𝑥
̅ ,𝑢
̅ )
𝐵 =
𝜕𝑓
𝜕𝑢
|
( 𝑥̅ ,𝑢̅
)
- If all eigenvalues of 𝐴 have 𝑅𝑒 < 0 , then the equilibrium (𝑥̅ , 𝑢̅ ) is asymptotically
stable
- If at least one eigenvalue of 𝐴 has 𝑅𝑒 > 0 , then the equilibrium
is unstable
- If all eigenvalues of 𝐴 have 𝑅𝑒 < 0 or 𝑅𝑒 = 0 , no conclusion can be drawn on the
stability of the equilibrium
Phase plane → allows one to have an idea of the region of attraction
Sometimes functions cannot be linearized → analysis of energy functions
Lyapunov function : 𝑉(𝑥) = 𝑇(𝑥) + 𝑈(𝑥) kynetic + potential energy
If there exists a function 𝑉(𝑥), continuous with its derivative, positive definite in 𝑥̅ and
such that along the state trajectories 𝑉
̇ (𝑥) is semidefinite negative in 𝑥̅ , then 𝑥̅ is a stable
equilibrium
(𝑥) < 0 in 𝑥̅ → asymptotically stable equilibrium
< 0 in 𝑥̅ along the state trajectories → globally asymptotically stable
equilibrium
(𝑥) > 0 in 𝑥̅ → unstable equilibrium
What if 𝑉
(𝑥) ≤ 0? → Krasowski-La Salle theorem (sufficient condition)
If there exists a function 𝑉(𝑥) continuous with its derivative, positive definite in 𝑥̅ such
that 𝑉
(𝑥) ≤ 0 in 𝑥̅ and the set 𝑆 ≔ {𝑥: 𝑉
(𝑥) = 0 } does not contain perturbed
trajectories compatible with the system, then 𝑥̅ is asymptotically stable
Lyapunov stability for linear systems → given any matrix 𝑄 = 𝑄
′
> 0 , there exists a
matrix 𝑃 = 𝑃
′
> 0 verifying the following Lyapunov equation: 𝐴
′
𝑃 + 𝑃𝐴 = −𝑄 (necessary
and sufficient condition for as. stability)
Proof of sufficiency → if we have a system 𝑥̇ (𝑡) = 𝐴𝑥(𝑡) with the Lyapunov function
𝑇
𝑃𝑥, we have 𝑉
𝑇
𝑇
𝑇
𝑇
𝑇
𝑇
For nonlinear systems linearizable at the equilibrium:
- Linearize the system
- Solve the Lyapunov equation for the linearized system with 𝑄 > 0 → compute the
matrix 𝑃 > 0
- Use the Lyapunov function 𝑉(𝑥) = 𝑥′𝑃𝑥 for the analysis of the equilibrium of the
original nonlinear system
Design of digital
regulators
Stategy 1 – discretise a continuous-time regulator
Strategy 2 – design a discrete time regulator for a discrete time system
𝐺
( 𝑧
) = 𝐶
( 𝑧𝐼 − 𝐴
)
− 1
𝐵 + 𝐷 {
𝑥(𝑘 + 1 ) = 𝐴𝑥(𝑘) + 𝐵𝑢(𝑘)
𝑦
( 𝑘
) = 𝐶𝑥
( 𝑘
)
( 𝑘
)
𝑥 is constant at the equilibrium in DT → 𝑥̅ = (𝐼 − 𝐴)
− 1
- 𝑥̅ is stable if for any 𝜀 > 0 there exists 𝛿 > 0 such that, for all initial states satisfying
|
| 𝑥
0
− 𝑥̅
| | ≤ 𝛿, it holds that |
| 𝑥(𝑘) − 𝑥̅
| | ≤ 𝜀
𝑘→∞
|
| 𝑥
( 𝑘
) − 𝑥̅
| | = 0 , then 𝑥̅ is asymptotically stable
Necessary and sufficient confìdition for the asymptotic stability is that all the eigenvalues
of 𝑨 have modulus < 𝟏
For nonlinear system, after linearization:
- If all eigenvalues of 𝐴 have | ∙ | < 1 , then the equilibrium (𝑥̅ , 𝑢̅ ) is asymptotically
stable
- If at least one eigenvalue of 𝐴 has | ∙ | > 1 , then the equilibrium (𝑥̅ , 𝑢̅ ) is unstable
- If all eigenvalues of 𝐴 have
< 1 or
= 1 , no conclusion can be drawn on the
stability of the equilibrium
Lyapunov method
If there exists a function 𝑉(𝑥), continuous and positive definite in 𝑥̅ such that Δ𝑉(𝑥) =
− 𝑉(𝑥) ≤ 0 in a neighbor of 𝑥̅ , then 𝑥̅ is a stable equilibrium. Moreover if
Δ𝑉(𝑥) < 0 in a neighbor of 𝑥̅ , then 𝑥̅ is an asymptotically stable equilibrium
Krasowski-La Salle theorem (sufficient condition)
If there exists a function 𝑉(𝑥) positive definite in 𝑥̅ , with Δ𝑉(𝑥) ≤ 0 in 𝑥̅ and the set
𝑆 ≔ {𝑥: Δ𝑉(𝑥) = 0 } does not contain perturbed trajectories compatible with the system,
then 𝑥̅ is asymptotically stable
Necessary and sufficient condition for the asymptotic stability of the linear system
𝑥(𝑘 + 1 ) = 𝐴𝑥(𝑘) is that for any matrix 𝑄 = 𝑄
′
> 0 there exists a matrix 𝑃 = 𝑃
′
solving the Lyapunov equation 𝐴
′
Induced p-norm of a matrix ||𝐴||
𝑖𝑝
𝑑≠ 0
|
| 𝐴𝑑
| |
𝑝
|
| 𝑑
| |
𝑝
Induced 2 - norm ||𝐴||
𝑖 2
𝑑≠ 0
||𝐴𝑑||
2
|
| 𝑑
| |
2
Norm of a “map” A 𝑠𝑢𝑝
𝑢≠ 0
||𝐴𝑢||
2
||𝑢||
2
𝑚𝑎𝑥
′
Norm of a system |
| 𝑢
| |
2
=
√ ∫ (𝑢
′
( 𝜏
) 𝑢
( 𝜏
) )𝑑𝜏
+∞
0
< +∞
Gain of a system 𝛾 = |
| 𝑆
| |
∞
= 𝑠𝑢𝑝
𝑢∈𝐿 2
|
| 𝑦
| |
2
|
| 𝑢
| |
2
= 𝑠𝑢𝑝
𝑢∈𝐿 2
|
| 𝑆(𝑢)
| |
2
|
| 𝑢
| |
2
↔ |
| 𝑦
| |
2
≤ |
| 𝑆
| |
∞
|
| 𝑢
| |
2
The gain is the supremum of the modulus of the frequency response 𝐺(𝑠). While in the
SISO case we just have to study one diagram, in the MIMO case amplification depends
both on sv and inputs
A system 𝑦 = 𝑆(𝑢) is I/O stable if it has finite gain
- Infinite gain norm 𝛾 = ||𝐺||
∞
- Gain at a given frequency 𝜎(𝐺(𝑗𝜔)) ≤
||𝐺(𝑗𝜔)𝑈(𝑗𝜔)||
2
|
|𝑈(𝑗𝜔)| |
2
Small gain theorem
Assume that 𝑆
1
and 𝑆
2
are I/O stable systems. Then the
feedback system is I/O stable if |
1
∞
2
∞
If the systems are linear ||𝑆
1
2
∞
< 1 (only sufficient conditions)
Stability of feedback systems with static sector nonlinearity
In view of the small gain theorem, I/O stability of the
feedback is guaranteed if 𝑘
2
𝜔
The closed-loop system is I/O stable if the Nyquist
diagram of 𝐺(𝑠) does not encompass, intersect or touch
the circle with diameter given by the segment [−
1
𝑘
1
1
𝑘
2
]
The sufficient condition to guarantee I/O stability is 𝑘
2
|𝐺(𝑗𝜔)|
∞
< 1 (small gain theorem)
Control design
SISO
𝑜
𝑦
𝑢
𝑜
𝑦
𝑢
1
1 +𝐿(𝑠)
1
𝐿(𝑗𝜔)
𝑠
𝜔
∞
- Complementary sensitivity 𝑇(𝑠) =
𝐿(𝑠)
1 +𝐿(𝑠)
1
𝐿(𝑗𝜔)
𝑇
𝜔
∞
- Control sensitivity 𝐾(𝑠) =
𝑅
( 𝑠
)
1 +𝐿(𝑠)
All four transfer functions must be studied to check the presence of hidden and forbidden
cancellations
Stability:
- Nyquist criterion
- Bode criterion
- Root locus
Approximation of the complementary sensitivity 𝑇(𝑠) ≅
𝜔
𝑛
2
𝑠
2
𝑛
+𝜔
𝑛
2
𝜑
𝑚
100
Design requirements for 𝐿(𝑠) are on 𝜑 𝑚
and 𝜔
𝑐
−𝜏𝑠
introduced a negative 𝜑
𝑚
- Systems with zeros with positive real part → negative 𝜑
𝑚
- Systems with unstable poles → sufficiently high gain
It’s proven that, given the gain margin 𝑔
𝑚
𝑚
1
𝑀 𝑇
and 𝑔
𝑚
𝑀 𝑆
𝑀 𝑆
− 1
𝑚
≥ 2 arcsin (
1
2 𝑀
𝑇
1
𝑀
𝑇
𝑚
≥ 2 arcsin (
1
2 𝑀
𝑆
1
𝑀
𝑆
Additive uncertainty → 𝐺
𝑎
- Asymptotically stable closed-loop system
𝑎
(𝑠) asymptotically stable → #𝑝𝑜𝑙𝑒𝑠(𝐺) = #𝑝𝑜𝑙𝑒𝑠(𝐺
̅
)
𝑎
∞
< 1 small 𝑀 𝑆
Multiplicative uncertainty → 𝐺
𝑚
- Asymptotically stable closed-loop system
𝑚
(𝑠) asymptotically stable → #𝑝𝑜𝑙𝑒𝑠(𝐺) = #𝑝𝑜𝑙𝑒𝑠(𝐺
̅
)
𝑚
∞
< 1 small 𝑀
𝑇
Crossover frequency
Define by 𝜔
𝐵
the frequency where 𝑆(𝑠) crosses − 3 𝑑𝐵
from below and 𝜔
𝐵𝑇
the frequency where 𝑇(𝑠) crosses
− 3 𝑑𝐵 from above
𝐵
𝑐
𝐵𝑇
Sensitivity function designed considering shape, minimum 𝜔 𝐵
, small asymptotic error for
constant reference signals and 𝑀 𝑆
𝑆
𝑆
𝑑𝑒𝑠𝑖𝑟𝑒𝑑
− 1
1
|𝑊(𝑠)|
𝑆
∞
[
] [
] = [
]
𝑃(𝑠) = [
] → the normal rank of 𝑃(𝑠) is its rank ∀𝑠 (except at most a finite
number of singularities)
Invariant zeros are the values of 𝑠 such that the rank of 𝑃(𝑠) is lower than the normal
rank → they have the blocking property
The polynomial 𝑍(𝑠) is the greatest common divisor of all the numerators of all the
minors of order 𝑟 of 𝐺(𝑠), where 𝑟 is its normal rank
For DT systems the formulation is the same, the only difference lies in the derivative
action in 𝑧 = 1
MIMO system
analysis
Series → 𝑌
2
2
1
1
Feedback → {
1
𝑝
1
2
− 1
1
1
1
𝑚
2
1
− 1
Frequency response
|
| 𝑌
( 𝑗𝜔
)| |
2
||𝑈(𝑗𝜔)||
2
|
| 𝐺(𝑗𝜔)𝑈
( 𝑗𝜔
)| |
2
||𝑈(𝑗𝜔)||
2
|
| 𝐺
( 𝑗𝜔
) 𝑈
( 𝑗𝜔
)| |
2
||𝑈(𝑗𝜔)||
2
The minimum and maximum sv are called “principal gains”
Condition number: 𝛾(𝐺(𝑗𝜔)) =
𝜎(𝐺
( 𝑗𝜔
) )
𝜎(𝐺
( 𝑗𝜔
) )
A feedback system is asymptotically stable if:
- There are no hidden cancellations of unstable modes
- Bounded inputs applied at any point of the system produce bounded outputs at any
point of the system
Nyquist theorem for MIMO systems
→ The closed loop system with loop TF 𝐿(𝑠) and negative feedback is asymptotically
stable iff the Nyquist plot of det(𝐼 + 𝐿(𝑠)) does not pass through the origin and
the number of encirclements around the origin in 𝑃
𝑜𝑙
(number of poles of 𝐿(𝑠) with
positive real part)
Stability from the closed loop transfer functions: the feedback system is internally stable
iff the following transfer functions are asymptotically stable
1
− 1
2
− 1
3
− 1
4
𝑢
− 1
5
𝑢
− 1
If 𝐺
1
(𝑠) and 𝐺
2
(𝑠) have a common pole/zero the cancellation could not occur if the
poles of 𝐺
1
(𝑠) and/or 𝐺
2
(𝑠) are not included in the poles of 𝐺
1
2
Small gain theorem → a closed loop system made by asymptotically stable linear systems
and with loop transfer function 𝐿(𝑠) is asymptotically stable if ||𝐿(𝑠)||
∞
Static performance for MIMO systems – integrators
{
0 = 𝐴𝑥̅ + 𝐵𝑢̅ + 𝑀𝑑
𝑦
𝑜
= 𝐶𝑥̅ + 𝑁𝑑
→ [
𝐴 𝐵
𝐶 0
] [
𝑥̅
𝑢̅
] = [
0 −𝑀
𝐼 −𝑁
] [
𝑦
𝑜
𝑑
] [
𝐴 𝐵
𝐶 0
] = Σ
If we want that, for constant 𝑑, the output reaches a constant reference value it must hold
that at steady state {
We place one integrator per error and 𝑅′(𝑠) stabilizes the plant
In the design of 𝑅′(𝑠) we must consider
{
[
𝑥̇ (𝑡)
𝑣̇ (𝑡)
] = [
𝐴 0
−𝐶 0
] [
𝑥(𝑡)
𝑣(𝑡)
] + [
𝐵
0
] 𝑢
( 𝑡
)
0 𝑀
𝐼 −𝑁
] [
𝑦
𝑜
𝑑
] = 𝐴
̅
[
𝑥(𝑡)
𝑣(𝑡)
] + 𝐵
̅
𝑢(𝑡) + [
0 𝑀
𝐼 −𝑁
] [
𝑦
𝑜
𝑑
]
𝑣
( 𝑡
)
[ 0 1
] [
𝑥(𝑡)
𝑣(𝑡)
] = 𝐶
̅
[
𝑥(𝑡)
𝑣(𝑡)
]
) must be reachable → iff the original pair is reachable and there are no
derivative actions
) must be observable → iff the original pair is observable
Dynamic performance for MIMO systems
Requirements on 𝑇(𝑠) and 𝑆(𝑠) can be transformed into
requirements on minimum and maximum singular values
The problem is that there is not a unique loop transfer
function 𝐿(𝑠), the Bode criterion does not exist and
multivariable root locus is extremely complex → good motivations to use automatic
synthesis procedure
Integrator of DT systems →
1
𝑧− 1
{
[
𝑥(𝑘 + 1 )
𝑣(𝑘 + 1 )
] = [
𝐴 0
−𝐶 0
] [
𝑥(𝑘)
𝑣(𝑘)
] + [
𝐵
0
] 𝑢
( 𝑘
)
0 𝑀
𝐼 −𝑁
] [
𝑦
𝑜
𝑑
]
𝑣
( 𝑘
)
[ 0 1
] [
𝑥(𝑘)
𝑣(𝑘)
]
The system is reachable and observable iff the original system is reachable and observable
and does not have invariant zeros in 𝑧 = 1
Pole placement 𝑥̇
𝑛
𝑚
𝑚,𝑛
𝑚
Necessary and sufficient condition for the design of an asymptotically stable observer
is that (𝐴, 𝐶) is observable
- Estimated 𝑥̂ is used in the pole assignement control law
Systems with disturbances
(𝑡) = 𝐴𝑥̂ (𝑡) + 𝐵𝑢(𝑡) + 𝑀𝑑(𝑡) + 𝐿[𝑦(𝑡) − 𝐶𝑥̂ (𝑡) − 𝐷𝑢(𝑡) − 𝑁𝑑(𝑡)]
- Nonmeasurable disturbance 𝑥̂ ↛ 𝑥
(𝑡) = 𝐴𝑥̂ (𝑡) + 𝐵𝑢(𝑡) + 𝐿[𝑦(𝑡) − 𝐶𝑥̂ (𝑡) − 𝐷𝑢(𝑡)]
Estimation of constant disturbances → we treat 𝑑(𝑡) like an additional unknown state
𝑟
{
[
𝑥̇ (𝑡)
𝑑
̇
(𝑡)
] = [
𝐴 𝑀
0 0
] [
𝑥(𝑡)
𝑑(𝑡)
] + [
𝐵
0
] 𝑢(𝑡) = 𝐴
̅
[
𝑥(𝑡)
𝑑(𝑡)
] + 𝐵
̅
𝑢(𝑡)
𝑦(𝑡) = [ 𝐶 𝑁
] [
𝑥(𝑡)
𝑑(𝑡)
] + 𝐷𝑢(𝑡) = 𝐶
̅
[
𝑥(𝑡)
𝑑(𝑡)
] + 𝐷𝑢(𝑡)
enlarged system
To use an observer on the enlarged system the pair (𝐴
) must be observable: it’s
observable iff (𝐴, 𝐶) is observable and 𝑟 ≤ 𝑝
We can add a compensator that forces the transfer
function from the disturbance to the output to be
null
The eigenvalues of the closed loop system with pole placement + observer are those of
𝐴 − 𝐵𝐾 and 𝐴 − 𝐿𝐶
Separation principle → we can independently design state feedback control law and the
observer
Regulator transfer function
[
)]
{
𝐴
̅
= 𝐴 − 𝐿𝐶 − 𝐵𝐾 + 𝐿𝐷𝐾
𝐵
̅ = 𝐵 − 𝐿𝐷
By applying again the state feedback control law
− 1
[
− 1
]
{
𝑅
( 𝑠
) = 𝐾
( 𝑠𝐼 − 𝐴
̅ )
− 1
𝐿
Δ
( 𝑠
) = −𝐾
( 𝑠𝐼 − 𝐴
̅ )
− 1
𝐵
̅
Reduced order observer → we want to apply a state transformation so that the 𝑝 outputs
coincide with 𝑝 new states and we then estimate the 𝑛 − 𝑝 remaining states
𝑥̃ = 𝑇𝑥 = [
1
] 𝑥 = [
𝑟
] 𝑥̃
𝑟
𝑛−𝑝
1
is any non singular matrix (usually chosen as the identity matrix)
{
𝑥̃
̇
(𝑡) = 𝐴
̃
𝑥̃ (𝑡) + 𝐵
̃
𝑢(𝑡)
𝑦(𝑡) = 𝐶
̃
𝑥̃ (𝑡)
→ 𝐴
̃
= 𝑇𝐴𝑇
− 1
= [
𝐴
̃
11
𝐴
̃
12
𝐴
̃
21
𝐴
̃
22
] 𝐵
̃
= 𝑇𝐵 = [
𝐵
̃
1
𝐵
̃
2
] 𝐶
̃
= 𝐶𝑇
− 1
=
[𝐼
𝑝
0 ]
{
𝑦̇ (𝑡) = 𝐴
̃
11
𝑦(𝑡) + 𝐴
̃
12
𝑥̃
𝑟
(𝑡) + 𝐵
̃
1
𝑢(𝑡)
𝑥̃
𝑟
̇ ( 𝑡
) = 𝐴
̃
21
𝑦
( 𝑡
)
̃
22
𝑥̃
𝑟
( 𝑡
)
̃
2
𝑢(𝑡)
We define 𝜂(𝑡) = 𝑦̇ (𝑡) − 𝐴
11
1
𝑢(𝑡) and 𝜁(𝑡) = 𝐴
̃
21
𝑦(𝑡) + 𝐵
̃
2
𝑢(𝑡) and we design
an observer for this new system
{
𝑥̃
̇
𝑟
(𝑡) = 𝐴
̃
22
𝑥̃
𝑟
(𝑡) + 𝜁(𝑡)
𝜂(𝑡) = 𝐴
̃
12
𝑥̃
𝑟
(𝑡)
the problem is that 𝜂 contains the derivative of 𝑦
𝑥̂
̇
𝑟
( 𝑡
) = 𝐴
̃
22
𝑥̂ 𝑟
( 𝑡
)
( 𝑡
)
( 𝑡
) − 𝐴
̃
12
𝑥̂ 𝑟
( 𝑡
) ] = (𝐴
̃
22
− 𝐿𝐴
̃
12
)𝑥̂ 𝑟
( 𝑡
)
̃
21
𝑦
( 𝑡
)
̃
2
𝑢(𝑡) + 𝐿𝜂(𝑡)
We sum and subtract (𝐴
̃
22
− 𝐿𝐴
̃
12
)𝑦(𝑡) and we define 𝜉(𝑡) = 𝑥̂
𝑟
(𝑡) − 𝐿𝑦(𝑡)
Discrete Time systems
Control law → 𝑢
Closed loop → 𝑥
The problem is exactly the same and the same algorithms can be used
Observers can be of two types:
𝑥̂ (𝑘 + 1 |𝑘) = 𝐴𝑥̂ (𝑘|𝑘) + 𝐵𝑢(𝑘) + 𝐿[𝑦(𝑘) − 𝐶𝑥̂ (𝑘|𝑘 − 1 ) − 𝐷𝑢(𝑘)]
𝑒̂ (𝑘|𝑘 − 1 ) = 𝑥(𝑘) − 𝑥̂ (𝑘|𝑘 − 1 ) estimation error
Deadbeat observers have all 𝑒𝑖𝑔(𝐴 − 𝐿𝐶) at the origin (error → 0 in at most 𝑛 steps)
𝑥̃ (𝑘 + 1 |𝑘 + 1 ) = 𝐴𝑥̃ (𝑘|𝑘) + 𝐵𝑢(𝑘) + 𝐿[𝑦(𝑘 + 1 ) − 𝐶(𝐴𝑥̃ (𝑘|𝑘) + 𝐵𝑢(𝑘)) − 𝐷𝑢(𝑘)]
𝑒̃ (𝑘|𝑘) = 𝑥(𝑘) − 𝑥̃ (𝑘|𝑘) estimation error
The gain 𝐿 must be designed to assign 𝑒𝑖𝑔(𝐴 − 𝐿𝐶𝐴) → (𝐴, 𝐶𝐴) must be observable
(𝐴, 𝐶𝐴) is observable iff (𝐴, 𝐶) is observable and 𝐴 is nonsingular (if it is singular
though, leave the poles in zero!)
Constant disturbances → as in the continuous time case, we must enlarge the system
treating the disturbance as a new unknown state with 𝑑(𝑘 + 1 ) = 𝑑(𝑘)
Reduced order observer → as in the continuous time case, we apply a state transformation,
however here we don’t have problems with derivatives and we can directly estimate 𝑥̂ 𝑟
- Regulator transfer function with state predictor
{
𝑥̂ (𝑘 + 1 |𝑘) = 𝐴𝑥̂ (𝑘|𝑘 − 1 ) + 𝐵𝑢(𝑘) + 𝐿[𝑦(𝑘) − 𝐶𝑥̂ (𝑘|𝑘 − 1 )]
𝑢(𝑘) = −𝐾𝑥̂ (𝑘|𝑘 − 1 )
𝜉
̇ ( 𝑡
) = (𝐴
̃
22
− 𝐿𝐴
̃
12
)𝜉
( 𝑡
)
̃
21
− 𝐿𝐴
̃
11
̃
22
𝐿 − 𝐿𝐴
̃
12
𝐴
̃
21
)𝑦
( 𝑡
)
̃
2
− 𝐿𝐵
̃
1
)𝑢(𝑡)
If we want to cancel out a pole in −𝑎 of 𝐺(𝑠) → 𝐴(𝑠) = (𝑠 + 𝑎)𝐴
′
′
′
The same procedure can be followed to cancel zeros of 𝐵(𝑠)
For DT system, nothing changes
Optimal control The control problem is transformed into an optimization one (allows to consider
nonlinear systems)
Optimal control is the precursor of MPC
Generic stabilization problem: min
𝑢
𝐽 = ∫
( 𝑥
′
( 𝜏
) 𝑄𝑥
( 𝜏
)
′
( 𝜏
) 𝑅𝑢(𝜏)
) 𝑑𝜏
𝑇
𝑡 0
′
( 𝑇
) 𝑆𝑥(𝑇)
- 𝑄 ≥ 0 weights the deviation of the state from zero
- 𝑅 > 0 weights the input
- 𝑆 ≥ 0 weights the deviation of the final state from zero
These matrices are design parameters
𝑖
𝑖
I am mainly interested in bringing the state quickly back to zero
𝑖
𝑖
I don’t want to use too much the control variables
1
2
I want the state 𝑥
1
to go to zero faster than 𝑥
2
𝑥̇ (𝑡) = 𝑓(𝑥(𝑡), 𝑢(𝑡)) 𝑓, 𝑙, 𝑚 continuously differentiable functions
We want to compute an optimal control 𝑢
𝑜
(𝑡) 𝑡 ∈ [𝑡
0
, 𝑇] minimizing
0
0
𝑇
𝑡 0
- 𝑚(𝑥(𝑇)) subject to the system’s dynamics and
state and input costraints
Denote by 𝑢
[𝑎,𝑏]
the control functions 𝑢(∙) in the interval [𝑎, 𝑏] and define
𝑜
(𝑥(𝑡), 𝑡) = min
𝑢[𝑡,𝑇]
𝑇
𝑡
+ 𝑚(𝑥(𝑇)) 𝑡 ∈ [𝑡
𝑜
, 𝑇]
𝑜
and 𝐽 depend on the current state 𝑥(𝑡), but not on its evolution up to time 𝑡
Bellman’s principle of optimality
From any point of an optimal trajectory, the remaining trajectory is
optimal for the corresponding problem over the remaining number
of stages, or time interval, initiated at that point
Hamilton Jacobi Bellman equation → {
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑡
= − min
𝑢
𝑥, 𝑢
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
𝑜
- Compute the value 𝑢
𝑜
minimizing {𝑙(𝑥, 𝑢) +
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
𝑜
= 𝜅 (𝑥,
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
)
- Compute the function 𝐽
𝑜
(𝑥, 𝑡) satisfying the HJB equation
- Use
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑥
in the control law 𝑢
𝑜
= 𝜅 (𝑥,
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
𝑜
Computations must proceed backwards in time, we start from the final value 𝐽
𝑜
(𝑥, 𝑇) and
move on with reverse time
The resulting control law is a state feedback control law
Linear quadratic
control
0
0
, 𝑢(∙), 0 ) = ∫ (𝑥
′
(𝜏)𝑄𝑥(𝜏) + 𝑢
′
(𝜏)𝑅𝑢(𝜏))𝑑𝜏
𝑇
0
′
(𝑇)𝑆𝑥(𝑇)
′
≥ 0 𝑆 = 𝑆
′
≥ 0 𝑅 = 𝑅
′
> 0
Derivation of the LQ control law :
−
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑡
= min
𝑢
{𝑙
( 𝑥, 𝑢
)
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑥
𝑓(𝑥, 𝑢)} = min
𝑢
{𝑥
′
𝑄𝑥 + 𝑢
′
𝑅𝑢 +
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑥
(𝐴𝑥 + 𝐵𝑢)}
The derivative wrt 𝑢 is 2 𝑢
′
𝑅 +
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑥
𝑜
1
2
− 1
′
(
𝜕𝐽
𝑜
( 𝑥,𝑡
)
𝜕𝑥
)
′
We substitute 𝑢
𝑜
in the formula
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑡
= 𝑥
′
𝑄𝑥 −
1
4
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
− 1
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
′
𝜕𝐽
𝑜
(𝑥,𝑡)
𝜕𝑥
We know 𝐽
𝑜
′
𝑆𝑥 and we assume the optimal cost function to be quadratic over
the whole time interval → 𝐽
𝑜
′
By computing the derivatives of 𝐽
𝑜
(𝑥, 𝑡) wrt 𝑥 and 𝑡, and substituting them in the
equation, we obtain the differential Ricatti equation
{
𝑃
̇
(𝑡) + 𝑄 − 𝑃(𝑡)𝐵𝑅
− 1
𝐵
′
𝑃
′
(𝑡) + 𝑃(𝑡)𝐴 + 𝐴
′
𝑃(𝑡) = 0
𝑃
( 𝑇
) = 𝑆
Finite horizon optimal control law → 𝑢
𝑜
− 1
′
′
- 𝐾(𝑡) time varying and defined over a time interval
′
𝑛×𝑛
𝑜
′
- Difficult to solve (matrix differential equation) and not easy to use in standard
control problems
Infinite horizon optimal control law
0
′
′
∞
0
If (𝐴, 𝐵) is reachable the solution of the DRE for 𝑇 → ∞ and 𝑃(𝑇) = 0 tends to a constant
matrix 𝑃
̅
≥ 0 , solution of the ARE
The asymptotic control law is 𝑢(𝑡) = −𝑅
− 1
𝐵
′
𝑃
̅ 𝑥(𝑡) = −𝐾
̅ 𝑥(𝑡)
How to weight the states? → partition matrix 𝑄 = 𝐶
𝑞
′
𝑞
and define 𝑦̃ (𝜏) = 𝐶
𝑞
0
′
′
∞
0
we have to guarantee that the state is
observable from output 𝑦̃ ., i.e. that (𝐴, 𝐶
𝑞
) is fully observable
𝑞 1
′
𝑞 1
𝑞 2
′
𝑞 2
→ if (𝐴, 𝐶
𝑞 1
) is observable also (𝐴, 𝐶
𝑞 2
) is observable
𝑞
) is observable, 𝑃
is positive definite
′
′
∞
0
- Build the matrix 𝐴
- Compute the 𝐾
𝛼
matrix, solution of the LQ problem
- Implement the control law 𝑢
𝛼
Robustness of 𝑳𝑸
∞
for SISO systems
Kalman’s inequality:
2
The Nyquist curve cannot enter the red circle centred in
− 1. This guarantees that we can either have:
- Phase variations of ±60°
- Gain variations ( 0. 5 , ∞)
However at HF |𝑇(𝑗𝜔)| decreases with slope − 1 → the system has a small attenuation of
disturbances at HF (measurement disturbances, unmodeled dynamics)
In MIMO cases , if 𝑅 = 𝑑𝑖𝑎𝑔{𝑟
1
2
𝑚
} is diagonal, the closed loop system remains
stable in front of phase variations (±60°) and gain variations ( 0. 5 , ∞), but not at the
same time on the same channel!
Kalman filter
𝑥
𝑦
𝑥
and 𝑣
𝑦
are gaussian white noises
[
𝑥
𝑦
]
→ 𝐸[𝑣(𝑡)] = 0 𝐸[𝑣(𝑡
1
2
′
] = 𝑉𝛿(𝑡
1
2
[
′
]
𝑛,𝑛
state noise
𝑝,𝑝
measurement noise
𝑛,𝑝
cross-covariance, usually = 0
We want to design an optimal state observer that considers the presence of the noises
We assume 𝑍 = 0 and 𝑥
0
= 𝑥( 0 ) with 𝐸[𝑥
0
] = 𝑥̅
0
𝐸[(𝑥
0
− 𝑥̅
0
)(𝑥
0
− 𝑥̅
0
)
′
] = 𝑃
̃
0
≥ 0
Also the initial state is uncorrelated from the noises 𝐸 [
0
[𝑣
𝑥
′
𝑦
′
]
]
(𝑡) = 𝐴𝑥̂ (𝑡) + 𝐵𝑢(𝑡) − 𝐿(𝑡)[𝑦(𝑡) − 𝐶𝑥̂ (𝑡)] filter/observer
𝑒(𝑡) = 𝑥(𝑡) − 𝑥̂ (𝑡) state estimation error
𝑒̇ (𝑡) = 𝐴𝑒(𝑡) − 𝐿(𝑡)[𝐶𝑒(𝑡) + 𝑣
𝑦
(𝑡)] + 𝑣
𝑥
(𝑡) = [𝐴 − 𝐿(𝑡)𝐶]𝑒(𝑡) + 𝐵
𝑐
𝑐
𝑐
If we set 𝑒̅ (𝑡) = 𝐸
[
]
𝑐
Choosing 𝑥̂ ( 0 ) = 𝑥̅
0
[
]
[
]
This guarantees that the state error has zero mean 𝑒̅ (𝑡) = 0
We want to minimize wrt the gain 𝐿(𝑡) the covariance of the estimation error
min
𝐿(𝑡)
′
(𝑡)𝛾 where 𝛾 ∈ ℛ
𝑛, 1
is a generic vector
(𝑡) is the state error covariance matrix ∈ ℛ
𝑛×𝑛
The solution to this problem is 𝐿
′
− 1
where 𝑃
is the solution of the
Ricatti equation 𝑃
′
′
− 1
(𝑡) with 𝑃
0
Extension of the result of 𝐿𝑄
∞
If (𝐴, 𝐵
𝑞
) is reachable, where 𝐵
𝑞
𝑇
𝑞
, and (𝐴, 𝐶) is observable
- The optimal estimator is 𝑥̂
̇ ( 𝑡
)
( 𝐴 − 𝐿
̅
𝐶
) 𝑥̂
( 𝑡
)
( 𝑡
)
̅
𝑦(𝑡)
with 𝐿
̅
= 𝑃
̃
̅
𝐶
′
𝑅
̃
− 1
where is the unique positive definite solution of
the stationary Ricatti equation 0 = 𝐴𝑃
̃
̅
̃
̅
𝐴
′
̃
− 𝑃
̃
̅
𝐶
′
𝑅
̃
− 1
𝐶𝑃
̃
̅
- The observer is asymptotically stable, i.e. all the eigenvalues of
𝐴 − 𝐿
̅
𝐶 have negative real part
What is 𝑍 ≠ 0? → new matrices {
∗
− 1
∗
− 1
′
and 𝐿
𝑡𝑜𝑡
∗
− 1
LQG control
𝑥
𝑦
𝑥
𝑦
, 𝑥( 0 ) satisfy all the assumptions for KF
The goal of 𝐿𝑄𝐺 is to minimize 𝐽 = lim
𝑇→∞
1
𝑇
𝐸 [
′
′
𝑇
0
]
- 𝑇 is required to have a finite cost function
- 𝐸 is required because 𝑥 and 𝑢 are stochastic processes
The optimal control law is given by the combination of the solution to the corresponding
deterministic LQ control problem and of the state estimated by the corresponding Kalman
filter
𝐾𝐹
[𝑦(𝑡) − 𝐶𝑥̂ (𝑡)]
𝐿𝑄
The structure is exactly equal to the one derived for pole placement control and the
separation principle holds as well
Regulator transfer function → 𝑈(𝑠) = −𝐾
− 1
𝐿𝑄𝐺 however does not inherit the robustness properties of 𝐿𝑄
∞
→ we can use 𝑳𝑻𝑹
(loop transfer recorvery) procedure
[𝑘
0
1
𝑛− 1
]
− 1
𝐵
𝐺
(𝑠)
𝐴
𝐺
(𝑠)
𝑏
𝑛− 1
𝑠
𝑛− 1
+⋯+𝑏
1
𝑠+𝑏
0
𝑠
𝑛
+𝑎
𝑛− 1
𝑠
𝑛− 1
+⋯+𝑎
0
The 𝐿𝑄 loop transfer function with robustness properties is 𝐿
𝑎
1
(𝑠) = 𝐾(𝑠𝐼 − 𝐴)
− 1
𝐵 =
𝜅(𝑠)
𝐴 𝐺
(𝑠)
Enlarged 2 × 2 plant of 𝐻 2
control – general formulation
{
𝑥̇ (𝑡) = 𝐴𝑥(𝑡) + 𝐵
1
𝑤(𝑡) + 𝐵
2
𝑢(𝑡)
𝑧 = 𝐶
1
𝑥(𝑡) + 𝐷
11
𝑤(𝑡) + 𝐷
12
𝑢(𝑡)
𝑣 = 𝐶
2
𝑥(𝑡) + 𝐷
21
𝑤(𝑡) + 𝐷
22
𝑢(𝑡)
→ 𝑃 = [
𝐴 𝐵
1
𝐵
2
𝐶
1
𝐷
11
𝐷
12
𝐶
2
𝐷
21
𝐷
22
]
By enlarging the system with shaping functions at the process
output , we are manipulating the performance variables 𝑧 with
dynamic filters. This forces the optimization algorithm to strictly confine the closed-loop
sensitivity functions within your specified deterministic frequency bounds
We could also enlarge the system with shaping
functions at the process input : in this case we color
white noises with 𝑆
𝑑
and 𝑆
𝑛
to give them their true
spectral properties and allow the algorithm to focus
its rejection effort on the most critical frequency bands
Structure of the solution of the 𝐻
2
control problem
2
′
𝑃 where 𝑃 is the unique positive definite solution to
′
2
2
′
1
′
1
2
2
) with 𝐿 = 𝑃
2
′
where 𝑃
is the
unique positive definite solution to 𝐴𝑃
′
2
′
2
1
1
′
Structure of the solution of the H ∞
control problem
𝑧𝑤
∞
{
𝑥̂
̇
(𝑡) = 𝐴𝑥̂ (𝑡) + 𝛾
− 2
𝐵
1
𝐵
1
′
𝑃𝑥̂ (𝑡) + 𝐵
2
𝑢(𝑡) + 𝑍𝐿(𝑦(𝑡) − 𝐶
2
𝑥̂ (𝑡))
𝑢(𝑡) = −𝐾𝑥̂ (𝑡)
{
𝑍 =
( 𝐼 − 𝛾
− 2
𝑆𝑃
)
− 1
𝐿 = 𝑆𝐶
2
′
𝐾 = 𝐵
2
′
𝑃
Model reduction can be useful in various cases:
→ The original system to be controlled could be in nonminimal form
→ The model could contain pole/zero pairs very near to each other
→ The regulator computed with 𝐻
2
∞
synthesis could be of very high order
𝐴𝑡
′
𝐴
′
𝑡
∞
0
𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑙𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑔𝑟𝑎𝑚𝑖𝑎𝑛
𝐴𝑡
′
𝐴
′
𝑡
∞
0
𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑔𝑟𝑎𝑚𝑖𝑎𝑛
For asymptotically stable systems 𝑃 and 𝑄 can be computed as the positive definite
solutions of {
′
′
′
′
- Balanced realization → in general 𝑃 ≠ 𝑄, but with a proper state transformation
we can rewrite it in a balanced way
1
2
𝑛
1
2
𝑛
We want to find a reduced state space representation with only 𝑘 values
A big 𝜎 indicates a state highly controllable and observable, while a small 𝜎 suggests
that the state is difficult to manipulate and has little influence on the output
- Balanced truncation → one way to reduce the order of the system is to divide 𝑥
into 𝑥
1
(states to keep) and 𝑥
2
(states to discard)
The reduced model is described by (𝐴
11
1
1
, 𝐷) and 𝐺
𝑎
𝑘
It’s proven that ||𝐺(𝑠) − 𝐺
𝑎
𝑘
∞
𝑘+ 1
𝑘+ 2
𝑛
Another way to reduce the model is to neglect the dynamics of 𝑥
2
→ 𝑥̇
2
= 0
If 𝐴
22
is nonsingular, the reduced model is Σ: {
1
𝑟
1
𝑟
𝑟
1
𝑟
𝑟
11
12
22
− 1
21
𝑟
1
12
22
− 1
2
𝑟
1
2
22
− 1
21
𝑟
2
22
− 1
2
Also in this case ||
𝑎
𝑘
∞
𝑘+ 1
𝑘+ 2
𝑛
With this reduction method, the static gain is mantained → 𝐺
𝑎
𝑘
Optimal control
in DT
0
0
𝑛
𝑚
We want to minimize wrt the input 𝑢(𝑘
0
0
− 1 ) the cost function
0
0
𝑘
̅ − 1
𝑖=𝑘
0
0
subject the the system’s
dynamics and state/input constraints 𝑥
We define an optimal cost function and its terminal cost:
𝑜
(𝑥(𝑘), 𝑘) = min
𝑢(𝑘)
𝑜
𝑜
In optimal control we move from the end of the sequence backward
- Compute the optimal value of the input 𝑢
𝑜
= 𝜅(𝑥, 𝐽
𝑜
) from the minimization of
{𝑙(𝑥, 𝑢) + 𝐽
𝑜
(𝑓(𝑥, 𝑢), 𝑘 + 1 )} assuming that there exists a unique minimum
- Compute the function 𝐽
𝑜
(𝑥, 𝑘) satisfying the HJB equation
𝐽
𝑜
( 𝑥, 𝑘
) = 𝑙 (𝑥, 𝜅(𝑥, 𝐽
𝑜
( 𝑥, 𝑘
) )) + 𝐽
𝑜
(𝑓 (𝑥, 𝜅(𝑥, 𝐽
𝑜
( 𝑥, 𝑘
) )) , 𝑘 + 1 )
with boundary condition 𝐽
𝑜
(𝑥, 𝑘
̅
) = 𝑚(𝑥)
𝑳𝑸 control → 𝑥(𝑘 + 1 ) = 𝐴𝑥(𝑘) + 𝐵𝑢(𝑘)
0
0
) = ∑ [𝑥
′
′
(𝑖)𝑅𝑢(𝑖)]
𝑘
̅ − 1
𝑖=𝑘 0
′
)
We procede with the tentative solution and obtain 𝑢
′
′
′
− 1
′
𝐾(𝑘)
Infinite horizon 𝐿𝑄 → 𝐽 =
′
′
∞
𝑘= 0
If (𝐴, 𝐵) is reachable and (𝐴, 𝐶
𝑞
) is observable
