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���BASIC CONTROL THEORY�LECTURE 5�� �

TOPICS

  • The concepts of controllability and observability
  • Determining controllability and observability from the canonical form of the state equation
  • Determining controllability and observability from the Kalman conditions
  • The Kalman decomposition
  • State feedback (Ackermann formula)
  • Compensation of the static error
  • State feedback based on the state equation enhanced with an integrator
  • State estimation
  • State feedback with state estimation

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CONTROLLABILITY

The key issue in control is whether the input signal can influence all state variables independently.

The system is state-controllable if its state vector can be moved from the initial state to the arbitrarily specified end state by a control signal u over a finite time.

The individual components of the state variable must be independent of each other.

If this is true for the y output signal only, the system is output controllable.

Determination of controllability from the canonical form of the state equation:

The eigenvalues of matrix A are different.

None of the rows (elements) of matrix b (column vector) are zero.

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

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DETERMINING CONTROLLABILITY FROM THE GENERAL FORM OF THE STATE EQUATION (KALMAN CONDITION)

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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Zero initial conditions are supposed.

Solution of the state equation:

As

(Cayley-Hamilton theorem)

The solution is:

The values of the integrals are constant.

The rank of the controllability hypermatrix should be n.

The rank of the output controllability matrix

should be equal to the number of the output signals.

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OBSERVABILITY

Observability gives an answer to the question, whether the initial values of the state variables at the starting point of the measurements can be reconstructed by measuring the input and output signals during a certain time.

It is enough to perform the investigation only for u(t)=0, i.e. for the motion generated by the initial values of the state variables.

Determination of observability from the canonical form:

The eigenvalues of matrix A should be different.

None of the column vectors of matrix c is zero.

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

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DETERMINING OBSERVABILITY FROM THE GENERAL FORM OF THE STATE EQUATION (KALMAN CONDITION)�THE KALMAN DECOMPOSITION�

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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The rank of the observability hypermatrix should be n.

THE KALMAN DECOMPOSITION

A system can be decomposed into four sub-systems, as:

controllable and observable

controllable and non-observable

non-controllable and observable

non-controllable and non-observable sub-systems.

The transfer function contains information only

on the controllable and observable sub-system.

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EXAMPLES FOR DETERMINING CONTROLLABILITY AND OBSERVABILITY

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

The canonical state equation of a system is:

Is the system state controllable? Is it output controllable? Is it observable? no yes no

Determine the transfer function!

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EXAMPLES FOR DETERMINING CONTROLLABILITY AND OBSERVABILITY

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

Given the block diagram of a control system

Equivalent scheme

The state equation with the assigned state variables:

Is the system state controllable, is it output controllable,

is it observable?

The rank is 1<2, not state controllable

The rank is 1, output controllable

The rank is 2, observable.

Explanation: pole cancellation

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JORDAN FORM

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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The transfer function of a system is:

Its block scheme is:

The state equation:

As the three poles are identical, the state variables

are not independent, so the system is not state controllable

and not observable.

Matrix A is not diagonal, this is the JORDAN form.

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STATE FEEDBACK

The state equation:

The transfer function:

Generally d=0.

The poles are roots of the characteristic equation.

The control signal can be created by feeding back the state variables by constants.

or

The effect is acceleration.

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

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STATE FEEDBACK

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

The characteristic equation of the system with state feedback:

The vector k is determined so that the poles are in the required position.

(Pole placement)

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STATE FEEDBACK WITH POLE PLACEMENT

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

ACKERMANN FORMULA:

Another principle: LQ controller, when the state feedback vector k

is calculated by minimising an integral criterion.

(It can be calculated easily from the observable canonical form.)

(command acker in matlab.)

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EXAMPLE FOR STATE FEEDBACK

The parameter matrices of the state equation of a continuous system:

Solution

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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The system is controlled with state feedback. Determine the state feedback vector ensuring that the poles of the closed loop system with state feedback be located at p1=-5, p2=-6.

or

Comparing the coefficients:

whence

és

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EXAMPLE FOR STATE FEEDBACK

The transfer function of an unstable continuous system is:

Give the state equation in controllable canonical form.

Determine the stabilising state feedback vector. (Prescribing the poles of the closed loop system mirror the unstable pole to the imaginary axis, and let the stable pole at its location.)

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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Solution

The state equation in controllable canonical form:

The prescribed poles of the feedback system: -6 and -3.

whence

és

Or:

An unstable system can be stabilised easily.

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COMPENSATION OF THE STATIC ERROR

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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The gain of the resulting transfer function is not 1.

The resulting transfer function of closed loop:

The callibration coefficient kr is:

Another solution compensating the static error is enhancing the control system

with an integrator.

be

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STATE FEEDBACK BASED ON THE STATE EQUATION ENHANCED WITH AN INTEGRATOR

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

The number of the state variables is increased by 1.

The state feedback vector has to be determined

considering the extended system.

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STATE FEEDBACK BASED ON THE STATE EQUATION ENHANCED WITH AN INTEGRATOR

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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Let us suppose d=0

The control system eliminates the static error

In case of step reference input.

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STATE ESTIMATION

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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If the state variables are not measurable, their values have to be estimated.

Another form of the state estimation circuit:

Its poles can be prescribed by the choice of L.

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STATE FEEDBACK WITH STATE ESTIMATION

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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The output signal with

feedback from the original

and from the estimated

state variables

The first state variable

and its estimation

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REMARK ABOUT THE SPEED OF THE SYSTEM,� THE ESTIMATION AND THE CONTROL CIRCUIT

  • The state estimation circuit should be the fastest
  • Then the state feedback circuit should be faster than the dynamics of the process.

BARS RUTH*, KEVICZKY LÁSZLÓ**, HETTHÉSSY JENŐ*, MAX GYULA*, VÁMOS TIBOR**, *BME AAIT, **SZTAKI

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2019

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THANK YOU FOR YOUR ATTENTION����