FEEDCON - Lesson 10 - Classical Design in the s Domain

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    Feedback ControlSystems Engineering

    Lesson 10: Classical Design in the s-domain

    Joshua NatividadFEEDCON

    6 July 2007

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    Objectives

    Learn how to determine stability ofdynamic systems

    Familiarize with Routh-Hurwitz stabilitycriterion

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    References

    R. S. Burns (2001), Advanced ControlEngineering , USA: Butterworth-Heinemann

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    Stability of DynamicSystems

    The response of a linear system to astimulus has two components:

    steady-state terms which are directly related

    to the input; and transient terms which are either exponential,

    or oscillatory with an envelope of exponentialform

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    Stability Defined

    If the exponential terms decay as timeincreases, then the system is said to bestable.

    If the exponential terms increase as timeincreases, then the system is said to beunstable.

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    Stability Examples

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    System Responses

    Whichvariabledoes thestability of

    the systemdependon?

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    System Responses

    Stability of adynamicsystemdepends uponthe sign of theexponential

    index in thetime responsefunction,which involvesfinding realroots of the

    characteristicequation.

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    Stability and Roots of theCharacteristic Equation

    The characteristic equation of asecond order system is given by

    whose rootsare found from

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    Response from Roots

    The roots determine the response of thesystem,

    Overdamping

    Critical Damping

    Underdamping

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    Response from Roots

    If the coefficient b were to be negative,then the roots would be

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    Stable Response

    Underdamping response, the timeresponse is given as

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    Unstable Response

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    Stability

    If any of the roots of thecharacteristic equation

    have positive real parts,then the system will be

    unstable.--The statement is true even for higher ordered systems.

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    Stability Criterion

    Routh (1905) and Hurwitz (1875) gives amethod of indicating the presence andnumber of unstable roots, but not their

    value.

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    Routh-Hurwitz Criterion

    If (a) is satisfied, then the necessary andsufficient condition for stability is either

    b) All the Hurwitz determinants of the

    polynomial are positive, or alternatively

    c) All coefficients of the first column ofRouthsarray have the same sign. The

    number of sign changes indicate thenumber of unstable roots.

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    Hurwitz Determinants

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    Routh Arrays

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    Example

    Check the stability of the system which hasthe following characteristic equation

    Test 1: All coefficients are present andhave the same sign. Proceed to Test 2.

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    Example

    4th

    order

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    Example

    3rd

    order

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    Example

    -4

    2

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    Example

    -4

    -4

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    Example

    +16

    --0

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    Example

    There are two sign changes in the columntherefore there are two roots with positivereal parts. Hence, the system is unstable.

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    Stability of a Closed Loop System

    The closed-loop transfer function wasdetermined to be

    The zero or roots is determined from thedenominator and equated to zero to form

    the characteristic equation

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    Stability of a Closed-Loop System

    Find the value of K1 such that the systemis unstable.

    What is the transfer function?

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    Stability of a Closed-Loop System

    What is the transfer function?

    The transfer function is

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    Stability of a Closed-Loop System

    The open loop gain constant is

    such that the transfer function becomes

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    Stability of a Closed-Loop System

    Since the roots can be found from

    the characteristic equation is therefore

    AlternativeMethod

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    Alternative Method

    Start with the definition of the transferfunction

    with H=1, the transfer function is writtenas

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    Alternative Method

    Multiply numerator and denominator with

    to get

    which simplifies into

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    Alternative Method

    Equate the denominator to zero in order toobtain the characteristic equation

    BACK

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    Stability of a Closed-Loop System

    TheRouth Arrayis

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    Stability of a Closed-Loop System

    To produce a sign change in the firstcolumn,K 2(!).

    SinceK= 8K1,

    to make the system

    just unstable,

    K1

    = 0.25.

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    Stability of a Closed-Loop System

    AtK= 2, the characteristic equation

    becomes

    Factorizing yields

    {

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    Stability of a Closed-Loop System

    And the transient response is

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    Special Cases of the Routh Array

    Case 2: All elements in a row arezero.

    If all the elements of a row is zero, replace

    that row with derivatives of an auxiliarypolynomial, formed from the elements ofthe previous row.

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