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Gdansk University of Technology, Faculty of Ocean Engineering and Ship Technology

Modeling and Control of Dynamic Systems, Laboratory, M. H. Ghaemi

Exercise 5Nonlinearity Effect and Nonlinear Systems

OBJECTIVEThe objective of this exercise is analyzing the effect of nonlinearities upon system

response both in qualitative and quantitative manners. It includes an introduction toclassical nonlinear elements observed in dynamic systems, their influence on the controlsystem response, linearization and comparison of linear and nonlinear systems withtaking into account stability and quality of control system.

REQUIREMENTS

Prerequisite knowledge:You are expected to know the basic nonlinearities such as : backlash, deadzone, ratelimiter, relay, delay and saturation. Additionally, following topics are Prerequisite:

1.

system nonlinearity and its sources,2. dynamic linearization,3. steady state characteristics,4. stability of nonlinear systems.

LAYOUT1. Recall the plant of Exercise 1. Examine the effect of nonlinearities caused by the

following nonlinear elements:

backlash dead zone saturationSelect the parameters of each elements so that to observe their effects.

2. Recall the closed-loop control system you have designed under Exercise 2. Add adelay to the plant transfer function. Change the time delay to find the critical value,when the closed-loop system becomes unstable.

3. Again consider the open-loop system (Exercise 1 or Exercise 2). Suppose that thefirst spring stiffness is a function of displacement and the first damper coefficient is a

function of speed:

)exp()(

)exp()(

1

2

1

xCxz

xCxk

z

k

&& =

=

where

1

1

zC

kC

z

k

=

=

Linearize the system under MATLAB/SIMULINK around the nominal point f=500 N.

Make the necessary models under SIMULINK.

MATLAB/SIMULINK

The following new MATLAB commands and functions are necessary to be understood and

applied in this exercise:

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trimlinmodlinmod2

REPORTIn addition to those points mentioned under Layout, your report should include:1. Plots which show the effect of nonlinearities and short discussion on their physical

meaning,2. Comments on stability of the nonlinear system and effect of time delay for both

configuration: delay in the plant and delay on the feedback line,3. Hand-on calculation of steady state characteristics,

4. Hand-on linearization of the open-loop system,5.

Comparison of nonlinear and linear system responses to step input.

EXAMPLE

Part One: Nonlinearity Effect

We recall the examples of Exercise 1 and 2 to examine the effect of nonlinearities uponthe open-loop system response. Then we will try to find the critical parameters of theapplied nonlinear element for closed-loop system.

The following figure gives the prepared model to examine the effect of saturation on theopen-loop system when 1 N external force is provided as the input signal (external force,f). Let us neglect in this case the effect of disturbance.

Mux

Step

1 N

Scope

Saturation

+/- 0.5

1

m.s +z.s+k2

Mass_Damper_Spring1

1

m.s +z.s+k2

Mass_Damper_Spring

The following plot compares the results for open-loop system with and withoutsaturation:

0 5 10 15 20 250

1

2

3

4

5

6

7

8x 10

-3

Time [sec]

Amplitude[m]

Open-loop system response with and without saturation

with saturation

without saturation

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For time delay equal or greater than 0.00578 [sec.], the control system will be unstable.Of course, it is possible to change the PID controller parameter so that the system backsto the stability area. But it is not the case we try to do under this exercise.

If a time delay equal 0.06 [Sec] is added to the feedback line, system will be unstable:

1

Out1

zs+k

ms +zs+k2

g2Sum1Sum

PID

PID Controller

Kp=8000, Ki=0, Kd=500

Delay

0.06 [sec]

zs+k

1

1/gd

2

In2

1

In1

0 5 10 15 20 25-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

Time [sec]

Amplitu

de[m]

Closed-loop response with feedback delay equal 0.06

Such a time delay on the feedback line can be occurred when we try to measure theregulated signal and then send it to the controller for comparison with the command

signal and necessary changes. Therefore, we have to pay attention on selection ofmeasuring device and transducers/transformers selected for. This example shows themeaning of feedback speed. Thee are all modeled in file Ex5_comp.mdl.

Part One: Linearization and Nonlinear System ResponseAssume that spring stiffness is a function of displacement and damper coefficient is afunction of speed. For example:

)exp()(

)exp()(

xCxzz

xCxkk

z

k

&& ==

==

The differential equation of the plant can be presented as follows:

[ ] [ ] )()()()( txmtxftxftf kz &&& = where

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[ ] [ ]{ }[ ] [ ]{ } )()(exp)(

)()(exp)(

txtxCtxf

txtxCtxf

kk

zz

=

= &&&

We will try to linearize our mass-damper-spring system using trim, linmod and

linmod2functions. But before that, let us solve the problem hands-on.The associated linear system could be presented by linearization of [ ])(txfz & and [ ])(txfk :

[ ] [ ]

[ ] [ ] )()1()exp()()(

)()()1()exp()()(

0

0

0

0

txxxCtxx

ftxf

txCtxxxCtxx

ftxf

kk

k

zzz

z

+=

=

=+=

= &&&&&

&

&

(note that xx &&= ). This leads to the following linear differential equation:

[ ] )()()1()exp()()(0

tftxxxCtxCtxm kz =+++ &&&

where

0xxx =

and index 0 denotes a selected working point in steady state condition.The steady state characteristic of the plant is:

ssssssk fxxC =)exp(

Assume that kC and zC are given as 100 and 50, respectively. The steady state

characteristic of the plant is shown in the next figure.

Let us select the nominal point of f=100 N which gives 57.00 x . The linear differential

equation about this nominal point could be rewritten as:

)()(.6772)(50)(100 tftxtxtx =++ &&& Consider that f increases by 5% to 105 N. The response of the linearized system in

comparison with the nonlinear system is presented in the following figure:

0 50 100 150 200 250 300

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

External Force [N]

Displacement[m]

steady state characteristic of the plant

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0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Time [sec]

Displacement[m]

Step Response

Linear

Nonlinear

The response of linear system ( x ) is given again below:

0 5 10 15 20 250

0.005

0.01

0.015

0.02

0.025

0.03

Time [sec]

Deltax

[m]

Linear System Step Response

Please note that if we want to plot x, we have to add x0to x .Now, we do the same job using MATLAB/SIMULINK. At first we build our nonlinearsystem (plant) under SIMULINK:

x

xprim

xprim1

Out1

1

s

Integrator1

1

s

Integrator

1/m

Gain

u(1)*u(2)*exp(u(2))

Fcn2

u(1)*u(2)*exp(u(2))

Fcn1

Cz

Constant1

Ck

Constant

1

In1

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We call this model as Ex5_nl. Secondly, by using trim function we find steady stateparameters for our plant. Trimenables steady state parameters to be found that satisfycertain input, output and state conditions. [x,u,y,xprim]=trim('Ex5_nl',[.57;0],105)

x =0.5829-0.0000

u =104.4171

y =0.5829

xprim =

1.0e-010 *-0.0000

-0.1127Thirdly, linmodobtains linear model: [a,b,c,d]=linmod('Ex5_nl',x,u)a =

0 1.0000-2.8354 -0.5000

b =

00.0100

c =1.0000 0

d =0

Once again, the step response can be evaluated:

sys_l=ss(a,b,c,d);

[y_l,t_l]=step(sys_l);Then the results of hand-on and MATLAB calculations can be compared. The transferfunction of hand-on calculation is

776.25.0

01.0)(

2 ++=

sssGL

When the transfer function calculated by MATLAB is: [num,den]=ss2tf(a,b,c,d); g_l=tf(num,den)g_lTransfer function:

0.01

-------------------s^2 + 0.5 s + 2.835

The results are almost the same.

* * *

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