2-1 State Space

State‐Space Design

N.H. Jo, H. Shim, Y.I. Son

자동화 및 DSP 연구실

Lecture Topics

State space representation

Stability

Controllability & Observability

RHP pole/zero cancellation

State feedback

Observer

Integral controller

Internal model principle


State Variables

Electrical circuit example

Differential equations in terms of energy storage variables

Inductor currents

capacitor voltages


State Variables

Inverted pendulum

State variables



SS Representation is NOT Unique

State space representation is NOT unique

State variables x1, x2 x1: inductor currentx2: capacitor voltage


SS Representation is NOT Unique

State space representation is NOT unique (continued)

State variables: inductor current

: capacitor voltage


SS Representation from Transfer Function

How to write a state space representation from given transfer function ?

Control canonical form

Observer canonical form


Control/Observer Canonical Form

Control canonical form

Observer canonical form


Phase Variable Form

Phase variable form


Transfer Function from SS Representation

How to obtain the transfer function from state space representation ?

Matlab commands[nu,de]=ss2tf(a,b,c,d)[A,B,C,D]=tf2ss(num,den)

Example: Single-input single-output system


Transfer Function from SS Representation

Example: Multi-input multi-output system


State Transformations

State TransformationHow can we go from one state space realization to another ?New state vector z :

P : nonsingular matrix

Transformed equation

Two important propertiesTransfer function is NOT changed by state transformation

Eigenvalues are NOT changed by state transformation



State Transformation into Modal Canonical FormState transformation

λi , vi : eigenvalue & eigenvector of A, i.e.,

New state space representation

Eigenvalues and Eigenvectors Eigenvalue λ

For x ≠ 0, λ x = A xCharacteristic equation : det (λI – A) = 0

Eigenvector xiFor each eigenvalue λi , λi xi = A xi



State Transformation into Control canonical FormState transformation

New state space representation

Controllability is NOT affected by the state transformation


State Transition Matrix

Time domain response Linear system

Laplace transform

State transition matrix

Time domain response


State Transition Matrix

ExampleThe linear system

The state transition matrix


Stability

BIBO StabilityBased on transfer function representation

A system is BIBO stable if system output is bounded for all bounded inputs

Necessary and Sufficient Condition All the transfer function poles must be in the LHP

Asymptotic Stability (A.S.)Based on the state space representation

is asymptotically stable if all the states approach zero with time, i.e.,

Necessary and Sufficient Condition All eigenvalues must be in the LHP


Stability

What is the difference between BIBO stability and A.S. ? Does one imply the other ?

Transfer function

In the absence of pole-zero cancellations, BIBO stability and Asymptotic stability are equivalent

Two stabilities differ only when RHP pole-zero cancellationsoccur (including imaginary axis pole-zero cancellation)

Example

Asymptotic stabilityEigenvalues = 1, -2 ⇒ unstable

BIBO stability

Pole= -2 ⇒ stable


Stability

Two stabilities are different for the system

BIBO stable but not asymptotically stable

Why does this happen ? Diagonalization by state transformation

The unstable mode at +1 is not connected to the outputSo, even if the first state may “blow up,” you will not be aware of it


Stability - Lyapunov Theorem

Positive definitenessA symmetric matrix M is said to be positive definite, denoted by M >0, if x’Mx > 0 for every nonzero x

Lyapunov theoremAll eigenvalues of A have negative real parts ⇔for any positive definite matrix N, the Lyapunov equation

has a unique solution M and M is positive definite The solution can be expressed as

The sketch of the proof


Internal Stability

Motivating exam.

Is it O.K. to cancel RHP plant poles by compensator zero, for example,

BIBO stability of TYR(s)

How about BIBO stability of TYD(s) ?

The slightest disturbances in the system will grow unbounded

A design lessonWe must NEVER cancel the RHP plant poles by RHP compensator zeros, for this will render the closed-loop system internally unstable A controller example


Internal Stability

Block diagram of a feedback systemPlant, controller, sensor dynamicsReference input, sensor noise, disturbance inputs are included

Internal stabilityAll signals within the feedback system should remain bounded forall bounded inputsAll possible Transfer function between all inputs and outputs should be stable

Only nine Transfer function between (R,D,N) and (U,Y,W) are sufficient

How to check this ?

Necessary and sufficient conditionThe product of KGH has NO pole-zero cancellations in the RHP(including jw-axis) Transfer function (1+KGH) has NO zeros in the RHP(including jw-axis)

Previous example revisited : KGH has RHP pole-zero cancellation at +1 ⇒ Unstable !!


Controllability

Motivating example 1Can the input u control x2 ?

Open circuit across y ⇒ cannot control x2

Motivating example 2Can the input u transfer x1 and x2 to any values ?If, x1(0)=x2(0)=0

⇒ x1(t)=x2(t), ∀ t ≥ 0x1 and x2 cannot be controlled independently


Controllability

DefinitionThe system is controllable if there exists a control u(t) that will take the state of the system from any initial state x0 to any desired final state xf in a finite time interval

Theorem: ControllabilityThe system (A,B) is controllable iff the controllability matrix Chas full row rank

PBH rank test: The system (A,B) is controllable iff the matrix has full row rank at every eigenvalue λ of A

The system (A,B) is controllable iffK can be chosen s.t. λ (A-BK) are arbitrary

How to find an uncontrollable mode ?The modes associated with λ are uncontrollable ⇔


Controllability

Theorem: ControllabilityThe system (A,B) is controllable ⇔ the controllability Gramian

is nonsingular for any t>0

How to transfer x0 to x1 at time t1 ?A control input, that transfers x0 to x1 at time t1, is given by

Why ?


Observability

Motivating example 1Can we estimate the behavior of x1 by measuring y ?

The current passing through 2-Ω resistor always equals the current source u⇒ The response excited by the initial state x1 will not appear in y⇒ The initial state x1 cannot be observed from the output

Motivating example 2Can we estimate the behavior of xby measuring y ?If, u(t) = 0, ∀ t ≥ 0 ⇒ y(t) = 0, ∀ t ≥ 0

No matter what x(t) is

x(t) cannot be estimated by measuring y


Observability

DefinitionThe system

is observable if, for any x(0), there is a finite time τ such that x(0) can be determined from u(t) and y(t) for 0 · t · τRoughly speaking, observability condition is required in order to design an observer

Theorem: ObservatilityThe system (A,C) is observable iff the observability matrix O has full column rank

PBH rank test: The system (A,C) is observable iff the matrixhas full column rank at every eigenvalue λ of A

The system (A,C) is observable iffL can be chosen s.t. λ (A-LC) are arbitrary


Controllability, Observability

Theorem of dualityThe pair (A,B) is controllable ⇔

The pair (AT,BT) is observable

Canonical decompositionKalman decomposition theorem

Controllable and observableControllable and unobservableUncontrollable and observableUncontrollable and unobservable

Controllability vs. ObservabilityControllability: whether or not the state can be controlled from the inputObservability: whether or not the initial state can be observed from the output


Lack of Controllability or Observability

Example

Transfer function

If b1=0 The mode at 1: uncontrollable, pole 1: canceled out

If c1=0The mode at 1: unobservable, pole 1: canceled out

If b2=0 The mode at 2: uncontrollable, pole -2: cancelled out

If c2=0 The mode at 2: unobservable, pole -2: cancelled out

Lack of either controllability or observability ⇒ pole-zero cancellation in the transfer function


Unstable Pole-Zero Cancellation

When an unstable pole is canceled by a zero,

it does not really disappear,

it simply becomes either uncontrollable or unobservable

If uncontrollable, you will observe the state blow up, but you can do nothing about it

If unobservable, you will not even be aware that something is wrong because the unstable state does not appear at the output

In either case, the results are disastrous


Inverted Pendulum Example

Transfer function

Measuring θ aloneUnstable pole/zero cancellation (at 0) A.S.: unstable ⇒ stabilization is NOT possible by measuring θ alone

Measuring x aloneNo pole/zero cancellationA.S.: stable ⇒ Stabilization is possible by measuring cart position x alone


Inverted Pendulum Example


Controllable ?

Full row rank ⇒ controllable


Inverted Pendulum

Stabilizable by measuring pendulum angle θ only ?Observability check

Sensing pendulum angle ⇒ C=[1 0 0 0]

Observability matrix

: singular ⇒ unobservable

Unobservability can be expected by pole/zero cancellation

Stabilizable by measuring cart positioin x only ?Sensing cart position : C=[0 0 1 0]

: nonsingular ⇒ observable

Sensor location problemIt makes a big difference which state variable we measure !!


State Feedback Control Law

Control-law design for Reference input is set to zero at this time

Choose K so that eigenvalues of (A-BK)

are in desirable locations, e.g., LHP

MATLABK=acker(a,b,p)

Calculation of K for u = -K xUseful for small( · 10 ) number of state variables

K=place(a,b,p)Numerically more reliable than ‘acker’Restriction : NONE of the desired poles may be repeated


Pole Placement Theorem

Example 1

Desired pole location : -4, -4, -5 ⇒ K=[75 49 10]

Example 2

Desired pole location : -2, -3 No such controller exist !

Mode at -1 cannot be moved

Pole Placement TheoremFor arbitrary pole placement, system must be controllable


Stabilizability

For uncontrollable systems, which poles can be moved ? Previous example

Mode at 1 : cannot be moved ⇒ uncontrollable modeMode at 2 : can be placed anywhere ⇒ controllable mode

PBH rank testUncontrollable modes are fixed

Controllable modes can be shifted

Is the controllability necessary for a system to be stabilized by using state feedback ?

Stabilizability condition is sufficient !!

StabilizabilityA system is stabilizable if

The unstable modes are controllable, or The uncontrollable modes are stable


Observer Design

A weak point of state feedback controlNot all the states are measurable

Some sensors are very expensivePhysically impossible to measure all the states

How to reconstruct all the states from a few measurements ?

Observer Design

When doesObserver error go to zero ⇔ (A-LC) is a stable matrix

If (A,C) observable ⇒ L can be chosen s.t. λ (A-LC) are arbitrary


Observer Design


Observer Design Example

Example: a simple pendulum

An observer is given by

How to compute an observer gainDuality

Observer gain can be computed using state feedback design procedureλ [(A-LC)] = λ [(AT-CTLT)]⇔ LT = [state feedback gain for (AT,CT) system]


Separation Principle

Control using observersHow to stabilize the system when all the states are not available ?Combined control law

State feedback control law is combined with an observer

Separation principlePlant equation with feedback

The overall system dynamics (plant + observer error)

Characteristic Equation

⇒ Closed-loop poles = controller poles + observer poles

The designs of the control law & the observer can be carried outindependently, yet when they are used together in this way, the poles remain unchanged


Observer-Based Controller

Observer-based controllerObserver-based controller = Control using observers

(= Output feedback controller)Observer-based controller can be used when full state are not availableBut, No guaranteed stability margins

A simple Pendulum System

State feedback controlDesired control roots: -2, -2

Observer Desired observer roots: -10, -10

Observer-based controller






Robustness & Stability Margins

Robustness & Stability MarginsUsually, we have imperfect model of our systemsStability margins provide some protection against model uncertainties

System designed with low marginsare inherently sensitive to model errors and may become unstable in actual operation

High margins provide good robustness properties

Robustness of LQR designReturn-difference inequalityStability margins of open loop system

GM : ½ < GM < ∞LQR gain matrix K can be multiplied by a large scalar or reduced by half with guaranteed closed-loop systems stability

PM : PM ≥ 60


Integral Control

How to handle tracking problem ?Adding an integrator will increase the system type

Integral controlNew state variable

The augmented plant

The Integral controller

K0 & KI should be chosen s.t.

is stableSystem type is increased (+1) ⇒ Tracking error = 0 is guaranteed (step input)


Integral Control

The double integrator system example

Desired pole location : -1±j, -5Control gain

Integral controller

Tracking error verification SS output due to unit step reference input


Integral Control


Internal Model Principle



Disk-drive Servomechanism

Because the data on the disk is not exactly on a centered circle, the servo must track a sinusoid of radian frequency w0 determined by the spindle speed



Objective Tracking a non-decaying input (with zero SSE) such as

steprampsinusoidal input

Rejecting a non-decaying disturbance (with zero SSE)

The key idea including the equations satisfied by these external signals as part of the problem formulation and solving the problem of control in an error space

so we are assured that the error approaches zero even if the output is following a non-decaying command



System equation

AssumptionReference input & disturbance satisfy differential equations of

order 2

The extension to more complex signals are not difficult

Initial conditions on the 2nd order differential equations (of reference input and disturbance) are unknown

For example, the input could be a ramp whose slope and initial value are unknown



Tracking error

Error-spaceThe state in error space

The control in error space

Overall system



Control law in the error spaceAssumption : is controllable

In fact, it is controllable if (A,B) is controllable and (A,B) does not have a zero at any of the roots of the reference-signal characteristic equation

Control law

Robustness: the state z will tend to zero for all perturbations as long as is stable


Disk-Drive Servomechanism

A simple normalized model of a computer disk-drive servomechanism

Tracking reference input r (t) = A sin w0t with zero SSE

Controller DesignError space representation

Characteristic Equation of (A-BK)

Controller






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2-1 State Space