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State Space Modelling and Control

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x(0) = -[B : AB : ... : An-1B] x(0) = -QU ... The exercises can be calculated using paper and pencil. Use Matlab to check the result. ... – PowerPoint PPT presentation

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Title: State Space Modelling and Control


1
State Space Modelling and Control
  • Morten Kristiansen
  • Bjørn Langeland

2
Lesson 3
3
Modelling
  • Modelling and analysis of dynamic systems with
    regard to design of a control for these systems
  • A systems dynamical behaviour can be described by
    differential equations and be expressed at state
    space form formalism by the state space equation
    and output equation
  • These systems are modelled graphically as shown
    below (D 0)

4
Analysis of system response
  • It applies that the response of the system is
    given by
  • I.e. Given an initial state, x(0) and an input,
    u(t), can the response characteristic of the
    system be drawn

5
Todays agenda
  • Design of controllers
  • Open and closed loop control systems
  • Regulator and servo controllers
  • Determination of the input
  • System characteristics
  • Is it possible to control the system?
    (controllability)?
  • Ackermannss formula

6
Control of systems
  • We want to control the output of a system by
    giving it a certain input, such that
  • the system follows a given reference or that
  • the systems stays i a certain state (e.g. a
    certain position or velosity)
  • Furthermore we want the system to be stable and
    to behave by some given performance measures.
  • It is up to the designer of the system to specify
    the performance measures
  • Examples of performance measures
  • the system reacts fast on a given input
  • it is not acceptable that the system has an
    overshoot of more that xx
  • steady state must be reached after xx sec.
  • Hence, a system can have several performance
    measures

7
Open loop system
  • If the input of the system can be completely
    planned and without any disturbances or changes
    of the system. Then an open loop control can be
    used.
  • Open loop system illustrated graphically
  • F is called the reference matrix. It described
    the relation between the reference and the input.
  • Does the system have any disturbances, it is not
    possible to compensate for these, because the
    changes of the systems states are not recorded.

Model
Control (Gain)
8
Closed loop system
  • To compensate for disturbances or if the input is
    dependent on the response of the system. Then a
    closed loop control is used.
  • L is called the feedback gain matrix.
  • The states of the system are feed back and is
    used to determine the input u.
  • u is a function of the system state and a whished
    reference
  • u(t) -Lx(t) Fr(t) called the control law

Model
Control
9
Servo and regulator controllers
  • If we wish that the systems response follows a
    change of the reference then we call it a servo
    system.
  • If there are no reference (r(t) 0) then the
    system is called a regulator. Here the wish is to
    control the system so a certain parameter is kept
    constant.
  • Graphically it look like
  • The control then law looks like
  • u(t) -Lx(t)

10
Determination of L substitution
  • The control law, u(t) -Lx(t), is inserted in
    the state equation. That gives the state equation
    for a closed loop system
  • To control the system is u(t) -Lx(t) used to
    decide the input. By this the size of L can be
    found. It is made from the state equation for the
    closed loop equation
  • Stability and the systems response characteristic
    is decided from the eigenvalues of the A matrix
    of the system.
  • For a closed loop system applies that à (A-BL)

11
Pole-placement
  • Eigenvalues are found from the characteristic
    polynomial
  • lI - (A-BL) 0
  • Above is described
  • A and B system, also known from modelling of the
    system
  • l is the characteristic of the system, also know
    from the designers specification of the system.
  • L is the state feedback gain matrix, and is not
    known.
  • Eigenvalues also called poles.
  • We design and control by choosing L, so the
    placement of the poles are made most appropriate
    - Pole placement.

12
Pole-placement
  • On one hand we have eigenvalues which gives the
    system the decided characteristic m1, m2, m3, ..
    mn
  • From these can a wished characteristic polynomial
    be written
  • f(l) (l-m1) (l-m2) (l-m3)... (l-mn) ln
    a1ln-1 an-1l an
  • On the other hand are the real eigenvalues for
    the system given by the roots in
  • lI - (A-BL) 0
  • from this the following polynomial is calculated
    ln b1ln-1 bn-1l bn
  • where b-coefficients are the function of l1, l2,
    ln
  • By comparing the coefficients, a og b, in the two
    polynomials can
  • L l1 l2 ln be decided.

13
Design approach
  • We can by this way decide L by
  • Selecting a row of m1, m2, m3, .. mn which
    describe the systems decided characteristic.
  • calculate L
  • simulate y(t) for u -Lx(t)
  • If y(t) is not as expected/wished then a row of
    new mis are chosen.
  • It is an iterative process to decide about L.

14
System characteristics
  • The three important parameters used to
    characterise the response of a system
  • Risetime
  • required time for the response to go from 10 of
    steady state to 90 of steady state.
  • Peak overshoot
  • the response maximum percentage
  • exceeding of steady state.
  • dependent on the systems damping.
  • Settling time
  • the time before the response is within
  • 1, 2 or 5 of steady state.

15
System characteristics
  • If it is wished to have a fast reacting system
    (small rise time) then a small damping should be
    chosen. That causes however a very large
    overshoot and a long settling time.
  • A compromise between a relative small damping for
    a short rising time and not to big overshoot and
    by then a not too long settling time.
  • It is a requirement that the system is stable,
    and settles towards steady state.
  • The characteristics of the system is
    given/decided by the systems eigenvalues/poles,
    and they should be in the negative complex half
    part.
  • See also chapter 5

16
Pole placement vs. System response
2. order system
1. order system
17
Can a system be controlled?
  • We can also create a control input by deciding L.
    That can be done by placing the eigenvalues
    appropriate.
  • BUT, it is not always possible to find a L for a
    given system?
  • That a system is controllable means
  • the system can be brought from one arbitrary
    condition to another arbitrary condition within
    finite time.
  • E.g.
  • Can an input signal, u, be found, which can
    influence the system so the system can be
    transferred from one arbitrary start condition,
    x(t0), to an arbitrary final condition, x(t1),
    within finite time.
  • If any x(t0) can be transferred to x(t1) is the
    system fully controllable.

18
Conditions for system controllability
  • If
  • start time t0 0 final time t1
  • start state x(t0) x(0) final state x(t1)
    0
  • For full controllability are applies x(t0) ?
    x(t1)
  • consequently x(0) ? 0
  • From earlier we have that
  • For full controllability are also required

19
Controllability matrix
  • To prove that a system is fully controllable
    should we be able to find x(0). (Solve the
    integral from the previous slide).
  • From the proof in the book
  • x(0) -B AB An-1B
  • x(0) -QU
  • To decide a U, which can affect the system to
    change from x(0) to x(t1) should the following
    equation be solved U x(0)Q-1
  • Q should be invert able. (Q-1 should exists).
  • The condition for Q-1 is found, is that Q should
    be non-singular which means have a rank n
    (complete rank).
  • Q is called the controllable matrix.

20
Determination of L - Ackermanns formula
  • A general method to determine L is Ackermanns
    formula.
  • Given a SISO system
  • The wished eigenvalues for the system are m1,
    m2, m3, .. mn
  • Therefore f(l) ln a1ln-1 an-1l an
  • Ackermanns formula says that L 0 0 0 0
    1 Q-1 f(A)
  • where Q B AB A2B An-1B
  • and f(A) An a1An-1 an-1A an

21
SimuLink example model regulator
22
SimuLink example - responses
Eigenvalues 0.1 0.2
Eigenvalues -44i -4-4i
Eigenvalues -1i -1-i
23
Todays exercises
  • Controllability
  • B-11-13 (controllable?)
  • Regulator design
  • B-12-3
  • B-12-5
  • B-12-6
  • The exercises can be calculated using paper and
    pencil.
  • Use Matlab to check the result.
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