Nonlinear Servo Simulation

download Nonlinear Servo Simulation

of 13

Transcript of Nonlinear Servo Simulation

  • 7/29/2019 Nonlinear Servo Simulation

    1/13

    NONLINEAR SERVO SIMULATION - BACKLASH

    Heikki Koivo

    EXAMPLE: Consider a linear servo model given in Fig. 1 in SIMULINK form withan open-loop transfer function

    2

    ( ) 10( )

    ( )

    X sG s

    E s s s= =

    +

    whereX(s) is the Laplace transform of the output andE(s) is the Laplace transform ofthe input. Assume (as you would with a transfer function) that the initial conditionsare zero. Simulate the closed-loop unit step response.

    Fig. 1. Linear servo system to be simulated.

    Move the cursor on top ofStepand click twice.

    1

  • 7/29/2019 Nonlinear Servo Simulation

    2/13

    Note that theStep-function is initially zero and at time t=1 it will change to the finalvalue, which by default is one. It is worthwhile to repeat that the stepping time is t=1.

    There are two alternatives to simulate the system in SIMULINK. Either useSimulation from menu or pressStart/Pause Simulationbutton. Both of these areshown in the Figure below.

    Click Scope twice to see the result from simulation. This is shown in the Figurebelow.

    2

  • 7/29/2019 Nonlinear Servo Simulation

    3/13

    To autoscale the result click binoculars button.

    If initial conditions are to be included, the SIMULINK model must be represented ina different way, which is called the state-space model.

    The open-loop transfer function

    2

    ( ) 10( )

    ( )

    X sG s

    E s s s= =

    +

    can be written as

    3

  • 7/29/2019 Nonlinear Servo Simulation

    4/13

    2( ) ( ) 10 (s s X s E s+ = )

    and further

    2

    2 10d x dx edt dt+ = .

    Solve the highest order derivative (acceleration) or

    2

    210

    d x dxe

    dt dt= + .

    Further, let .(0) 2, (0) 1x x= =

    In configuring the system, first set up the two integrators in series. Then collect to theinput of the first integrator what is on the right hand side of the equation. The result isshown in the Figure below. Because it is not straightforward to give initial conditionsat a different time than t=0 s, change the step time inStepto t=0.

    Note that the scope will also display ( )x t .

    Click the integrator on the right, corresponding tox, twice. Now you can adjust theinitial condition.

    4

  • 7/29/2019 Nonlinear Servo Simulation

    5/13

    Repeat in a similar fashion for (0) 1x = .

    The corresponding unit step response is as follows.

    REMARK: If only closed-loop step response is desired, then MATLAB commandsof the CONTROL SYSTEMS TOOLBOX are sufficient.

    g=tf([10],[1 1 0])

    Transfer function:10

    -------

    5

  • 7/29/2019 Nonlinear Servo Simulation

    6/13

    s 2 + s

    T=feedback(g,1)

    Transfer function:

    10------------s 2 + s + 10

    step(T)

    The result agrees with SIMULINK simulation.

    Time (sec.)

    Amplitude

    Step Response

    0 2 4 6 8 10 120

    0.2

    0.4

    0.6

    0.8

    1

    1.2

    1.4

    1.6

    1.8

    From: U(1)

    To:Y(1)

    It is also interesting to see the phase plane plot, i.e. x as a function ofx. Time iseliminated in a phase plane, but appears in the curve as a parameter. To see the phaseplane, useXY Graph.

    6

  • 7/29/2019 Nonlinear Servo Simulation

    7/13

    XY Graphdoes not have autoscale-feature. Therefore to see the whole picture adjustthe scales. The result is een below.

    The graph is not very smooth. This is due to the stored points inXY Graph. This canbe fixed by adjusting theSample time.

    7

  • 7/29/2019 Nonlinear Servo Simulation

    8/13

    The parameters chosen above produce a smooth figure.

    8

  • 7/29/2019 Nonlinear Servo Simulation

    9/13

    SERVO WITH BACKLASH

    In practical servo systems many typical nonlinearities occur. These are backlash,saturation, friction, etc. Here we will consider servo withBacklash.

    ClickBacklashblock open and click alsoHelpto better understandBacklashbehavior.

    9

  • 7/29/2019 Nonlinear Servo Simulation

    10/13

    Initial conditions in a nonlinear system are essential, because the qualitative behaviorcan change completely depending on the initial conditions. In the linear case thebehavior remains the same. Let us study this with different initial values.

    Start with .(0) 2, (0) 1x x= =

    10

  • 7/29/2019 Nonlinear Servo Simulation

    11/13

    Next try .(0) 0.5, (0) 0.5x x= =

    Next study the corresponding phase plane plots. In order to obtain a complete enoughphase plane plot, change the simulation time. First, chooseSimulationmenu and thenParameters.

    11

  • 7/29/2019 Nonlinear Servo Simulation

    12/13

    TheStop timeis changed, after some experimenting to 20 s. Other parameters can beleft as they are.

    The phase plane plot is shown below. Observe that the system with these initialconditions will reach a stable limit cycle.

    12

  • 7/29/2019 Nonlinear Servo Simulation

    13/13

    If initial conditions are changed to (0) 0, (0) 0.5x x= = the phase plane plot looks

    different.

    13