Control Theory:Feedback and the Pole-Shifting Theorem

Joseph Downs

September 3, 2013

Introduction (+ examples)What is Control Theory?Examples

State-SpaceWhat is a State?Notation

(State) FeedbackFeedback Mechanisms

LTI SystemsLinearity and StationarityHow to Apply the PST

The Pole-Shifting TheoremStatement of the Pole-Shifting TheoremConsequences of the Pole-Shifting Theorem

Conclusion

What is Control Theory?

I steer physical quantities to desired values

I mathematical description of engineering process

I application of dynamical systems theory

I introduce a control, u

What is Control Theory?

I steer physical quantities to desired values

I mathematical description of engineering process

I application of dynamical systems theory

I introduce a control, u

What is Control Theory?

I steer physical quantities to desired values

I mathematical description of engineering process

I application of dynamical systems theory

I introduce a control, u

What is Control Theory?

I steer physical quantities to desired values

I mathematical description of engineering process

I application of dynamical systems theory

I introduce a control, u

Examples

I cruise control

I precision amplification (lasers + circuits)

I biological motor control systems

Examples

I cruise control

I precision amplification (lasers + circuits)

I biological motor control systems

Examples

I cruise control

I precision amplification (lasers + circuits)

I biological motor control systems

The Handstand Problem: Setup

I Iα =∑τi

I mL2θ = mgL sin θ − u

I θ = g sin θL − u

mL2

The Handstand Problem: Setup

I Iα =∑τi

I mL2θ = mgL sin θ − u

I θ = g sin θL − u

mL2

The Handstand Problem: Setup

I Iα =∑τi

I mL2θ = mgL sin θ − u

I θ = g sin θL − u

mL2

What is a State?

I contains ”sufficient information”

I describes system past

I useful for deterministic systems

I state variable, x

I allows us to write x = φ(t, x , u)

What is a State?

I contains ”sufficient information”

I describes system past

I useful for deterministic systems

I state variable, x

I allows us to write x = φ(t, x , u)

What is a State?

I contains ”sufficient information”

I describes system past

I useful for deterministic systems

I state variable, x

I allows us to write x = φ(t, x , u)

What is a State?

I contains ”sufficient information”

I describes system past

I useful for deterministic systems

I state variable, x

I allows us to write x = φ(t, x , u)

What is a State?

I contains ”sufficient information”

I describes system past

I useful for deterministic systems

I state variable, x

I allows us to write x = φ(t, x , u)

Notation

I convenient to ”vectorize”

I x1 = f1(x1, x2, ..., xn)x2 = f2(x1, x2, ..., xn)...xn = fn(x1, x2, ..., xn)

I →I x = f(x)

Notation

I convenient to ”vectorize”

I x1 = f1(x1, x2, ..., xn)x2 = f2(x1, x2, ..., xn)...xn = fn(x1, x2, ..., xn)

I →I x = f(x)

Notation

I convenient to ”vectorize”

I x1 = f1(x1, x2, ..., xn)x2 = f2(x1, x2, ..., xn)...xn = fn(x1, x2, ..., xn)

I →

I x = f(x)

Notation

I convenient to ”vectorize”

I x1 = f1(x1, x2, ..., xn)x2 = f2(x1, x2, ..., xn)...xn = fn(x1, x2, ..., xn)

I →I x = f(x)

The Handstand Problem: Notation

I x1 = θ

I x2 = θ

I x :=

(x1x2

)=

(x2

g sin x1L − u

mL2

)=: f(x,u)

The Handstand Problem: Notation

I x1 = θ

I x2 = θ

I x :=

(x1x2

)=

(x2

g sin x1L − u

mL2

)=: f(x,u)

The Handstand Problem: Notation

I x1 = θ

I x2 = θ

I x :=

(x1x2

)=

(x2

g sin x1L − u

mL2

)=: f(x,u)

Feedback Mechanisms

I plug calculated values in as control

I u = ψ(t, x)

I achieved by measuring physical quantities

I state vs. output feedback

Feedback Mechanisms

I plug calculated values in as control

I u = ψ(t, x)

I achieved by measuring physical quantities

I state vs. output feedback

Feedback Mechanisms

I plug calculated values in as control

I u = ψ(t, x)

I achieved by measuring physical quantities

I state vs. output feedback

Feedback Mechanisms

I plug calculated values in as control

I u = ψ(t, x)

I achieved by measuring physical quantities

I state vs. output feedback

Linearity and Time-Invariance

I linear: x = A(t)x + B(t)u

I time-invariant: x = φ(t, x,u) = f (x,u)

I much simpler mathematically

Linearity and Time-Invariance

I linear: x = A(t)x + B(t)u

I time-invariant: x = φ(t, x,u) = f (x,u)

I much simpler mathematically

Linearity and Time-Invariance

I linear: x = A(t)x + B(t)u

I time-invariant: x = φ(t, x,u) = f (x,u)

I much simpler mathematically

How to Apply the Pole-Shifting Theorem

I assume time-invariance: ∂φ(t,:,:)∂t � 1

I linearize about an equilibrium point (x = 0)

I A =

(∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

)

I B =

(∂f1∂u∂f2∂u

)

How to Apply the Pole-Shifting Theorem

I assume time-invariance: ∂φ(t,:,:)∂t � 1

I linearize about an equilibrium point (x = 0)

I A =

(∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

)

I B =

(∂f1∂u∂f2∂u

)

How to Apply the Pole-Shifting Theorem

I assume time-invariance: ∂φ(t,:,:)∂t � 1

I linearize about an equilibrium point (x = 0)

I A =

(∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

)

I B =

(∂f1∂u∂f2∂u

)

How to Apply the Pole-Shifting Theorem

I assume time-invariance: ∂φ(t,:,:)∂t � 1

I linearize about an equilibrium point (x = 0)

I A =

(∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

)

I B =

(∂f1∂u∂f2∂u

)

The Handstand Problem: Theorem Preparation

I x ≈ Ax + Bu

I A =

(0 1gL 0

)I B =

(0− 1

mL2

)I AB =

(− 1

mL2

0

)

The Handstand Problem: Theorem Preparation

I x ≈ Ax + Bu

I A =

(0 1gL 0

)

I B =

(0− 1

mL2

)I AB =

(− 1

mL2

0

)

The Handstand Problem: Theorem Preparation

I x ≈ Ax + Bu

I A =

(0 1gL 0

)I B =

(0− 1

mL2

)

I AB =

(− 1

mL2

0

)

The Handstand Problem: Theorem Preparation

I x ≈ Ax + Bu

I A =

(0 1gL 0

)I B =

(0− 1

mL2

)I AB =

(− 1

mL2

0

)

Pole-Shifting Theorem Statement

I A(n × n) and B(n ×m) are s.t.rank

([ B AB ... An−1B ]

)= n

I →I ∃F (m × n) s.t. eigenvalues of A + BF are arbitrary

Pole-Shifting Theorem Statement

I A(n × n) and B(n ×m) are s.t.rank

([ B AB ... An−1B ]

)= n

I →

I ∃F (m × n) s.t. eigenvalues of A + BF are arbitrary

Pole-Shifting Theorem Statement

I A(n × n) and B(n ×m) are s.t.rank

([ B AB ... An−1B ]

)= n

I →I ∃F (m × n) s.t. eigenvalues of A + BF are arbitrary

Pole-Shifting Theorem Consequences

I we can make a feedback law!

I u = Fx

I x = (A + BF )x

I control system reduces to a dynamical system!

Pole-Shifting Theorem Consequences

I we can make a feedback law!

I u = Fx

I x = (A + BF )x

I control system reduces to a dynamical system!

Pole-Shifting Theorem Consequences

I we can make a feedback law!

I u = Fx

I x = (A + BF )x

I control system reduces to a dynamical system!

Pole-Shifting Theorem Consequences

I we can make a feedback law!

I u = Fx

I x = (A + BF )x

I control system reduces to a dynamical system!

The Handstand Problem: Theorem Application

I F =(f1 f2

)

I A + BF =

(0 1

gL −

f1mL2

− f2mL2

)I χA+BF = λ2 + λ f2

mL2+ ( f1

mL2− g

L ) = 0

I λ = − f22mL2

± 12

√f2

2

m2L4− 4 f1

mL2+ 4g

L

I

(f2 > 0)&(f1 >f2

2

4mL2+ mgL)

→

Re(λ±) < 0

The Handstand Problem: Theorem Application

I F =(f1 f2

)I A + BF =

(0 1

gL −

f1mL2

− f2mL2

)

I χA+BF = λ2 + λ f2mL2

+ ( f1mL2− g

L ) = 0

I λ = − f22mL2

± 12

√f2

2

m2L4− 4 f1

mL2+ 4g

L

I

(f2 > 0)&(f1 >f2

2

4mL2+ mgL)

→

Re(λ±) < 0

The Handstand Problem: Theorem Application

I F =(f1 f2

)I A + BF =

(0 1

gL −

f1mL2

− f2mL2

)I χA+BF = λ2 + λ f2

mL2+ ( f1

mL2− g

L ) = 0

I λ = − f22mL2

± 12

√f2

2

m2L4− 4 f1

mL2+ 4g

L

I

(f2 > 0)&(f1 >f2

2

4mL2+ mgL)

→

Re(λ±) < 0

The Handstand Problem: Theorem Application

I F =(f1 f2

)I A + BF =

(0 1

gL −

f1mL2

− f2mL2

)I χA+BF = λ2 + λ f2

mL2+ ( f1

mL2− g

L ) = 0

I λ = − f22mL2

± 12

√f2

2

m2L4− 4 f1

mL2+ 4g

L

I

(f2 > 0)&(f1 >f2

2

4mL2+ mgL)

→

Re(λ±) < 0

The Handstand Problem: Theorem Application

I F =(f1 f2

)I A + BF =

(0 1

gL −

f1mL2

− f2mL2

)I χA+BF = λ2 + λ f2

mL2+ ( f1

mL2− g

L ) = 0

I λ = − f22mL2

± 12

√f2

2

m2L4− 4 f1

mL2+ 4g

L

I

(f2 > 0)&(f1 >f2

2

4mL2+ mgL)

→

Re(λ±) < 0

Conclusion

I control theory provides framework

I linearization simplifies problem

I simple rank condition permits use of feedback

I further topics: PID control, feedback linearization, robotics,etc.

I Thanks for your time!

Conclusion

I control theory provides framework

I linearization simplifies problem

I simple rank condition permits use of feedback

I further topics: PID control, feedback linearization, robotics,etc.

I Thanks for your time!

Conclusion

I control theory provides framework

I linearization simplifies problem

I simple rank condition permits use of feedback

I further topics: PID control, feedback linearization, robotics,etc.

I Thanks for your time!

Conclusion

I control theory provides framework

I linearization simplifies problem

I simple rank condition permits use of feedback

I further topics: PID control, feedback linearization, robotics,etc.

I Thanks for your time!

Conclusion

I control theory provides framework

I linearization simplifies problem

I simple rank condition permits use of feedback

I further topics: PID control, feedback linearization, robotics,etc.

I Thanks for your time!