Chp3 Non Linear Stability_latest PDF

Embed Size (px)

Citation preview

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    1/37

    NON-LINEAR

    SYSTEMS

    BASED ON THE INPUT AND OUTPUT

    CHARECTARISTICS OF THE SYSTEM

    LINEAR SYSTEMS

    NON LINEAR SYSTEMS

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    2/37

    LINEAR SYSTEMS The systems which satisfy

    Homogeneity Principle

    Superposition Principle Y1(t)=A1x1(t)

    Y2(t)=A2x2(t)

    A1x1(t)+ A2x2(t)=y3(t)

    Then if y3(t)= Y1(t)+ Y2(t)

    Then the system is said be a linear otherwisenon linear system.

    Most of the systems are generally non-linear,but in the range of operation of control systemcan be made to linear systems

    For example

    Spring, Motors, Damper etc.

    These non linear systems are representedin linear differential equations ------------------------------------

    Transient , Steady State response andstability

    Classical methods- Time Domain andFrequency Domain methods

    State space analysis

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    3/37

    Non Linear systems and Classification

    The systems which does not satisfy principles of linearity

    The non linearities are due to many factors, and arebasically classified as Inherently non linearities or Incidental non linearities

    Which present in the component inherently

    Intentional Non Linearities To make the systems response in desired form we intensionally

    introduce some non linearities in the system The following are some of the non linearities

    Saturation

    Friction

    Backlash

    DeadZone or DeadBand

    Hysterisis Relays

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    4/37

    Non Linear systems to linear systems

    Operating a component in restricted region Examples:

    Relays, Amplifiers,Thermionic & Semiconductor devices

    Electric Drives: 1500 rpm 1500 +/- 50

    Automatic Voltage controllers: +/- 5% of the specified Voltage

    Some components linear in working range Example: Spring

    Advantages of linearisation

    Easy to construct mathematical models Easy to design and analyse by use of transfer function basedtechniques(i.e. Time domain and frequency domain)

    The non linearties are small than the results obtained from linearanalysis do not differ from actual results

    BEHAVIOR OF NON LINEAR SYSTEMS:

    Jump resonance hormonic oscillations

    Limit cycles ete

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    5/37

    Jump Resonance In the frequency response of non-

    linear systems, the amplitude ofthe response (output) may jumpfrom one point to another fromincreasing or decreasing nature offrequency.This phenomenon iscalled jump resonance

    Example: A second order systemwith spring damper

    M

    K = K 1x + K2x3

    The frequency response curves of

    a) K= 0, b) K>0 Hard spring c) K

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    6/37

    Sub Hormonic oscillations

    If a system is excited with a sinusoidal

    function, the output is also a sinusoidal withsame frequency but with a different

    amplitude and phase angle.

    Depends upon system parameters, initial

    conditions,amplitude & frequency of forcingfunction

    Linear System

    A Sin(wt) B Sin (wt+)

    Non-Linear System

    A Sin(wt) (AnSin (wt)+Bn cos(wt)

    Where n= 0 to

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    7/37

    Limits Cycles

    The response of the non-linear systems(o/p)

    may exhibit oscillations with fixed frequency andamplitude

    These oscillations are called Limit Cycles

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    8/37

    Characteristics of non linear

    systems The output response to a particular test signal does not

    give any information to other inputs

    Highly sensitive to input amplitude

    Requires more information in addition to the usual

    mathematical model i.e. Amplitude of input Initial conditions etc.

    Non-Linear Systems posses Limit Cycles

    Jump Resonance Harmonics & Sub Harmonics etc.

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    9/37

    Singular points When the all the phase

    variables does not

    changes with

    respective to time i.e

    reaching to a constant

    point (may be origin,

    equilibrium point

    Phase trajectory of a second order system for

    different values of and for initial condition x1(0)

    =1

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    10/37

    Phase Trajectories The state equation of second order system is

    dX1/dt= f1(x1,x2) ---(1)dX2/dt= f2(x1,x2) ----(2)

    Where X1, X2state variables

    dX2/dX1= f2(x1,x2)/ f1(x1,x2)------- (3)

    Equations 3 provides slope of the phase trajectory atevery point in phase plane except at singular points

    (Which is indeterminate)

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    11/37

    Phase trajectories can be

    drawn Analytically

    equation (3) is integrated

    Numerical method

    trajectory for given set of initial conditions

    Graphically

    Popularly known method are Isoclines method

    Delta Method.

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    12/37

    Construction of phase plane trajectory by

    isoclines method

    Introduction

    From equation (3)

    Slope of the phase plant at any point is S = dX2/dX1=

    f2(X2,X1)/f1(X2,X1) (4)Let any point slope of the phase trajectory is S1, then

    F2(X1,X2) = S1f1(X1,X2)

    i.e. locus of all such points in phase-plane at which the scope of

    the phase trajectory is S1.

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    13/37

    Note: for each set of initial conditions one phase

    trajectory can be drawn

    A locus passing through the points of same slopein phase plane is called Isocline

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    14/37

    Procedure to construct phase trajectories by

    isocline method.

    Step1: Phase trajectory starts at a point

    corresponding to initial conditions

    Note: For each set of initial conditions, one

    phase trajectories can be constructed. Step2

    Let S1,S2,S3,etc be the slopes of associated

    with isoclines 1,2,3 etc Let 1 = Tan

    -1(S1)

    2 = Tan-1(S2)

    3 = Tan

    -1(S3

    )

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    15/37

    Note: If a straight line is drawn at an angle from a point,

    then the slope of the line at that point is tan

    step2. Draw two straight lines from point A, one at a slope ofS1

    i.e 1=Tan-1(S1)] and the other at a slope of S2 [i.e at angleof 2=tan-1(S2)]

    step3.Let these two lines meet the isoclines 2 atp and q.

    Now we can say that the trajectory would cross the isoclines2 at midway between p & q. Mark the point B. on theisocline-2 approximately midway between p and q

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    16/37

    step4. The constructional procedure is now

    repeated at B to find the crossing point C onthe isocline-3.

    By similar procedure the crossing points onthe isoclines are determined

    A smooth curve drawn through the crossingpoints on the isoclines are determined

    A smoothe curve drawn through the crossingpoints gives the phase trajectory starting at

    point A

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    17/37

    Note 2 The accuracy of the trajectory is closely related to the

    spacing of the isoclines the phase trajectory will be more

    accurate if large number of isoclines are used which are very

    close to each other. It should be noted that using a set of

    isoclines, any number of isoclines can be constructed.

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    18/37

    (1,-1.5)

    (2,-2.5)(.75,-2.5)

    (.25,-.8)

    (1,-.63)(2,-.86)

    (2,1.2)(1,.6)

    (1,1.4)

    (1.5,2.1)

    (.5,2.5)

    (.25,1.25)

    (1,-.77)

    X1

    X2

    S1=2

    S2=1.5

    S3=.5S4=0

    S5=-2

    S6=-1S7=0.5

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    19/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    20/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    21/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    22/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    23/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    24/37

    Phase Portrait of example

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    25/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    26/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    27/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    28/37

    Constr ction of phase trajectories b

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    29/37

    Construction of phase trajectories by

    delta method Consider a second order non-liner or linear system represented

    by the equation d2X/dt2+f(X, dX/dt,t) = 0 (1)

    The equation (1) can be converted to the form shown below

    d2X/dt2+wn2(X+(X, dX /dt,t) = 0

    In the above equation is a function of x,

    dX/dt ,t but for short intervals, the change in phase variables arenegligible. Hence for a short interval, is considered as aconstant.

    d2X/dt2+wn2(X+)= 0 ----(3)

    Let us choose as the state variables as

    X1=x,x2= dx/dt /wn---(4)

    from above equation Dx/dt=wnx2= dx1/dt ----(5)

    d2X/dt2 = wn dx2/dt ---(6)

    Substituting equation 4 & 5 in 3

    dx2/dt = -wn(X1+ )

    So the state equations are dx1/dt=wnx2,dx2/dt= -wn(X1+ )

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    30/37

    From the above equations, the slopeequation over short interval can be written as

    dx2/dx1= -(x1+)/x2---(8) using above slope equation, a short segment

    of the trajectory can the drawn from theknowledge of at any point on the trajectory,

    as explained below Procedire:

    Let point A be a point on phase trajectorywith co-ordinates(X1,X2) as shown in fig

    below (usually the point A will be thestarting point of the trajectory obtained fromthe initial conditions)

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    31/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    32/37

    X1

    X2

    A(1,0)

    X2 =4|x2|x2 Center

    0(A) 0 C1

    -.1875(B) -.140625 C2

    -.325(C) -.4225 C3

    -.375(D) -.5625 C4

    -.4375(E) .765625 C5

    ..... ...... ......

    ..... ...... .

    .. . .

    B(..,-.1875)

    C2 C3C1

    C(..,-.325)

    C4

    C5D(..,-.5625)

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    33/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    34/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    35/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    36/37

  • 7/26/2019 Chp3 Non Linear Stability_latest PDF

    37/37

    Thank You