Complex dymnamics

8/12/2019 Complex dymnamics

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COMPLEX DYNAMICS SIMPLIFIED

CHAPTER 1


To develop a control system for a dynamical system one must first understand

precisely how the system behaves. One can arrive at this understanding using

mathematics by performing a dynamic analysis. Dynamic analysis is performed intwo steps, often called the formulation or the modeling step and the second simply

called the solution step. In the formulation step the equations that describe the

system are developed. In the solution step, the equations are solved. One oftendistinguishes between solutions that are obtained analytically and those that are

obtained numerically. The equations, when solving them analytically, are often

approximated by constant-coefficient linear differential equations, because they

can be solved analytically with relative ease. hen solving the equationsanalytically, quantities li!e displacement, velocity, and force are expressed as

algebraic functions of time. Displacements, velocities and forces are called time

responses. The other way to solve the original equations, the numerical approach,

produces time responses in the form of computer graphs.

"ometimes the dynamical behavior of a system is relatively complicated in whichcase the motion is divided into regions of behavior called stability regions. hen

developing a control system for such a system, the control system designer usually

focuses on one stability region at-a-time. In this section, we first discuss the

development of the equations that describe the motion of systems. e#ll restrictour attention to systems that have one independent degree-of-freedom and we#ll

confine the motion to a plane. Then, we#ll find its stability regions.

1. Equations

$et#s first review how to obtain the equation that describes the motion of a planarsingle degree-of-freedom system. Toward this end, we consider a pendulum. The

pendulum is composed of a light rod pinned at one end %point O& and free at the

other end %pointA&. 't pointA, the rod is attached to a small but relatively heavysphere that has massM. The mass of the rod is neglected.

(igure )* +endulum and its (ree-ody Diagram

CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH


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The development of the equation that describes the pendulum#s motion, li!e the

equation that describes the motion of other planar bodies, begins by loo!ing at the

equations that govern the three degrees-of-freedom of a planar body, namely, the

three equations associated with a body#s rotational motion and its two translationalmotions. The equation that describes its rotational motion is usually obtained either

by summing moments about a fixed point , if one exists, or by summing momentsabout the system#s mass center C. In the case of the pendulum, for which thereexists a fixed point, we#ll sum moments about point . The two equations that

describe the translational motions of a system are obtained by summing forces in

the directions of the translational degrees-of-freedom. The three equations are*

%) )& === CyCx yMFxMFIM --

in which is the system#s angle,xCandyCare the positions of the system#s mass

center, over-dots are time-derivatives, and = dmrI /

- is the system#s mass

moment of inertia. The three degrees-of-freedom of the pendulum, , xC, andyCdepend on each other. 'ny one can be expressed in terms of the others. $et#s select

as the independent degree-of-freedom, notice that the positions xC andyC are

located at pointA, and thatxCandyCcan be expressed in terms of byxC= Lsin

andyC= Lcos. y time differentiation, the accelerations are sin/ LxC =

and cos/ LyC = . The pendulum#s mass moment of inertia is

.//

- MLdmrI == "ubstituting these results into 0q. %) )& yields

%) /& cossinsin /// MmgRMRmLmgL yx ===

0quation %) /& is 1 equations in terms of 1 un!nowns the independent degree-of-freedom and the two force reactionsRxandRy. Our attention will focus on the

first equation because it governs the motion of the pendulum. Dividing the first

equation by mL/, we get

%) 1& sinL

g=

This is a nonlinear equation, typical of the equations that govern the motion of

systems undergoing motions that are arbitrarily large. On the other hand, when the

motions are small one frequently adopts a 2small motion3 assumption to

approximate the nonlinear differential equation by a linear differential equation.

's stated earlier, the linear equation can be solved analytically with relative ease incontrast with the original nonlinear equation. The time responses obtained

analytically can be expressed in terms of the system#s parameters. (or example,the pendulum angle can be expressed as a function of time and the parameters g

and L. 'lso, by solving equations analytically it becomes possible to define

important terms li!e a system#s natural frequency and its exponential decay rate.Indeed, the simplification gained by assuming that the motion is small is not only

significant, but characteri4es how we loo! and understand how a system behaves.

It is how complex dynamics is simplified.



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2. Equii!"iu#

In any system, there are points where the system can be placed and stay there %atleast in theory&. These rest positions are called equilibrium positions. The system,

when resting at one of these positions, is said to be in static equilibrium. To find

these positions, we substitute

%) 5& -==

into the nonlinear differential equation. This results in a nonlinear algebraic

equation. In the case of the pendulum, we substitute 0q. %) 5& into 0q. %) 1& to

get

%) 6& &sin%&%in which-&% - L

gff ==

The solution to 0q. %) 6& consists of the two equilibrium positions

%) 7& .- &/%-&)%

- ==

's you might imagine, the pendulum#s first equilibrium position %at the bottom& isstable in the sense that if its position is slightly offset from equilibrium, the

pendulum will return to its equilibrium position. The moment acting on the

pendulum in the neighborhood of the first equilibrium position is a restoringmoment. On the other hand, the second equilibrium position %at the top& is

unstable, in the sense that if the pendulum is slightly offset from this equilibrium

position, the pendulum will move further away rather than return.

$. Lin%a"i&ation

0quation %) 1& can be written as

%) 8& &sin%&%in which&% L

gff ==

in which the function f was first defined in 0q. %) 6&. 0quation %) 8& is

nonlinear because the functionf is nonlinear. e can learn how a system behaves

near an equilibrium position by approximating the nonlinear differential equationof motion by a linear equation that is valid in the neighborhood of the equilibrium

position. The approximation of a nonlinear equation by a linear equation is called

lineari4ing the equation. The resulting linear equation will be accurate in theneighborhood of the equilibrium position but quite inaccurate away from the

equilibrium position.

0quation %) 8& is lineari4ed by lineari4ing the functionf%& in 0q. %) 8&. e do

this by performing a Taylor series expansion of f about , retaining in the series



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only the first two terms the linear terms. The remaining terms, that is, thenonlinear terms, are set to 4ero. The two-term Taylor series expansion of f is

%) 9& &%&%&%&% --

--

-

=

+=

ffff

In 0q. %) 9& noticed thatf%& : since the lineari4ation was about an equilibriumposition. "hortly, we#ll redefine the angle using the equilibrium position as a

reference. e#ll define

%) ;& - =

The pendulum angle is measured from the equilibrium angle .

The pendulum has two equilibrium positions and hence two stability regions to

study. (irst, consider the stability region that surrounds the first equilibrium angle

-&)%

- = . In 0q. %) 9&,

-

f

.

-&sin%

&)%-

L

g

L

g=

=

"ubstituting 0q. %)

9& and 0q. %) ;& into 0q. %) 8& yields the linear equation that describes thependulum#s motion in the neighborhood of the first equilibrium position

%) )& L

g=

e obtained 0q. %) )& by noticing that . = To obtain the linear equation in

the stability region that surrounds the second equilibrium angle =&/%- , perform

the calculation=

-

f .&sin%

&/%- L

g

L

g=

=

Then, substitute 0q. %) 9&and 0q. %) ;& into 0q. %) 8& to get

%) ))& L

g+=

The functionf%& and the two linear approximations off%& are shown in (igure /.



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0 50 100 150 200 250 300 350-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

theta (deg)

f(theta)

(igure /* The functionf and the linear approximations off

0quation %) )& is accurate in the neighborhood of the first equilibrium position

and 0q. %) ))& in the neighborhood of the second. The next step is to solve 0q. %)

)& and 0q. %) ))&. (irst, consider 0q. %) )&. Try solutions of the form

%) )/& &sin%&cos% /) tt ==

where

Lg


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%) )1& is harmonic. Therefore, the first equilibrium position is said to be stable%as expected&.

>ext, turning to the second equilibrium position, let#s try to solve 0q. %) ))&. Try

solutions to 0q. %) ))& of the form

ttee

== /)

'gain, it#s easy to verify that these solutions satisfy 0q. %) ))& and, by the linearsuperposition principle, that the general solution to 0q. %) ))& is of the form

%) )5& tt BeAe +=

whereAandBdepend on initial conditions. This time, notice that the solution is an

unbounded function of time, regardless of the constantsAandB %unlessA: ?&.Therefore, the second equilibrium position is described as unstable %as expected&.

*

Can you describe tis unusua! situation in "ic te res#onse in te unstab!e regiona##roaces $ero%

A S%'on( E)a#*%'s a second example, consider a more complicated system than the pendulum. 's

shown in (igure 1, the system is a thin bar of length L and of mass M that is

connected to a linear spring and a linear damper. The spring constant is & and the

damping constant is c. The bar rotates in the hori4ontal plane. $et#s find thenonlinear equation that governs the bar#s motion, its equilibrium positions, and its

corresponding lineari4ed equations.

(igure 1* ar "ystem and its (ree-ody Diagram



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The equation that governs the motion is

%a& = -- IM

where the mass moment of inertia is

%b& 1&/

)

%1

)

&/

)

%1

) ///

-

Ma

aMaMI =+=

To calculate the resultant moment -M let#s perform the following preliminarycalculations %"ee (igure 1&*

cos/sin/1&cos)%&sin)%

sin/cos/1&sin)%&cos)%

&%&cossin%&sin%cos

////

////

+=++=

=+=

+=+=+=

aaad

aaad

aaa

B

A

'BA +i"+i"+i"

BBBAAA

A'B'B

A'A'A

addt

dcad&

A'A'

A'A'

%F%F

+i"""

%

+i"""

%

&%&%

@&cos)%&sin)A%cos/sin/1

)

@&sin)%&cos)A%

sin/cos/1

)

otice also that the two instabilities were produced by )G Cwaspositive for each of the equilibrium positions.


- ) C

M

&1

M

c1

5

M

&7

M

c/

/

M

&1

M

c7.-

5

1

M

&59./

M

c/


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PRO-LEM SYSTEMS

The problems statements given in this boo! are posed in a general way, withoutreferring to a specific system. The systems are described separately. The systems

described below pertain to problems that are given in this chapter as well to

problems that are found in Chapters /, 5, 6, and 8.

Sst%# 1 / 1' bar-spring-damper system is shown. $et m = / !g, & = 6 >s


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Sst%# 1 / $' stabili4er is composed of a rotating dis!, a spring, and a damper, as shown. $et

m = M = .)!g,R = .6 m, &: .5 >Hs


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Sst%# 1 / ' roller system is composed of a dis! that rolls on a circular trac! that is acted on

by a spring and a damper. $et m = .) slug, & = )6 lb


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Sst%# 1 / 3' dis! system is composed of a uniform dis!, a spring, and a damper. $et m = )

!g, & = ) >Hs


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P"o!%# 1 / $: D%t%"#inin5 Sta!iitThis problem begins where +roblem ) / ends.

%a& rite the nonlinear differential equation of motion in the general form

&.,% f= (ind the sta!iit (%"i4ati4%s

fand

fabout each of

its equilibrium positions.

%b& (or the neighborhood around each equilibrium position %& = ), /, J &, let&%

-&%&% &tt = where &%-

& is the &-th equilibrium angle &% &)%

-- = and

write down the associated linear approximation of f in terms of &%t and&.%t Then, write down the linear differential equation governing the

motion of the system in the neighborhood of each equilibrium position.%c& (ind a general form of the solution of each linear differential equation.

"tate which equilibrium positions are stable and which are unstable.


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Complex dymnamics