COMPLEX DYNAMICS SIMPLIFIED
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 in two 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 often distinguishes 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. When solving the equations analytically, quantities like 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.
Sometimes the dynamical behavior of a system is relatively complicated in which case the motion is divided into regions of behavior called stability regions. When 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. We’ll restrict our 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.
Let’s first review how to obtain the equation that describes the motion of a planar single 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 (point A ). At point A , the rod is attached to a small but relatively heavy sphere that has mass M . The mass of the rod is neglected.
Figure 1: Pendulum and its Free-Body Diagram
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
The development of the equation that describes the pendulum’s motion, like the equation that describes the motion of other planar bodies, begins by looking 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 translational motions. The equation that describes its rotational motion is usually obtained either by summing moments about a fixed point 0, if one exists, or by summing moments about the system’s mass center C . In the case of the pendulum, for which there exists a fixed point, we’ll sum moments about point 0. 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:
(1 – 1)

M
0

I
0
  
F x

M  x 
C

F y

M  y 
C in which
 is the system’s angle, x
C
and y
C
are the positions of the system’s mass center, over-dots are time-derivatives, and I
0
  r
2 dm is the system’s mass moment of inertia. The three degrees-of-freedom of the pendulum,

, x
C
, and y
C depend on each other. Any one can be expressed in terms of the others. Let’s select
 as the independent degree-of-freedom, notice that the positions located at point A , and that x
C
and y
C
can be expressed in terms of x
C and y
C are
 by x
C
= L sin
 and y
C
= –L cos

. By time differentiation, the accelerations are  x 
C
 
L
  2 sin
 and  y 
C is I
0
 
L
  2 cos
 . The pendulum’s mass moment of inertia
  r
2 dm

ML
2
.
Substituting these results into Eq. (1 – 1) yields
(1 – 2)
 mgL sin
  mL
2  
R x
 
M
  2 sin

R y
 mg

M
  2 cos

Equation (1 – 2) is 3 equations in terms of 3 unknowns – the independent degreeof-freedom

and the two force reactions R x
and R y
. Our attention will focus on the first equation because it governs the motion of the pendulum. Dividing the first equation by mL
2
, we get
(1 – 3)
     g
L sin

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 “small motion” assumption to approximate the nonlinear differential equation by a linear differential equation.
As stated earlier, the linear equation can be solved analytically with relative ease in contrast with the original nonlinear equation.
The time responses obtained analytically can be expressed in terms of the system’s parameters. For example, the pendulum angle can be expressed as a function of time and the parameters g and L. Also, by solving equations analytically it becomes possible to define important terms like a system’s natural frequency and its exponential decay rate.
Indeed, the simplification gained by assuming that the motion is small is not only
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED significant, but characterizes how we look and understand how a system behaves.
It is how complex dynamics is simplified.
In any system, there are points where the system can be placed and stay there (at least 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
(1 – 4)
      
0 into the nonlinear differential equation. This results in a nonlinear algebraic equation. In the case of the pendulum, we substitute Eq. (1 – 4) into Eq. (1 – 3) to get
(1 – 5) f (

0
)

0 in which f (

)
  g
L sin(

)
The solution to Eq. (1 – 5) consists of the two equilibrium positions
(1 – 6)

0
( 1 ) 
0

0
( 2 )  
.
As you might imagine, the pendulum’s first equilibrium position (at the bottom) is stable 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 restoring moment. 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.
Equation (1 – 3) can be written as
(1 – 7)
    f (

) in which f (

)
  g
L sin(

) in which the function f was first defined in Eq. (1 – 5). Equation (1 – 7) is nonlinear because the function f is nonlinear. We can learn how a system behaves near an equilibrium position by approximating the nonlinear differential equation of 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 linearizing the equation. The resulting linear equation will be accurate in the neighborhood of the equilibrium position but quite inaccurate away from the equilibrium position.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
Equation (1 – 7) is linearized by linearizing the function f (

) in Eq. (1 – 7). We do this by performing a Taylor series expansion of f about

0
, retaining in the series only the first two terms – the linear terms. The remaining terms, that is, the nonlinear terms, are set to zero. The two-term Taylor series expansion of f is
(1 – 8) f (

)
 f (

0
)

 f
 
(
  
0
)

 f
 
(
  
0
)
0 0
In Eq. (1 – 8) noticed that f (

0
) = 0 since the linearization was about an equilibrium position. Shortly, we’ll redefine the angle
 using the equilibrium position as a reference. We’ll define
(1 – 9)
    
0
The pendulum angle
 is measured from the equilibrium angle

0
.
The pendulum has two equilibrium positions and hence two stability regions to study. First, consider the stability region that surrounds the first equilibrium angle
 ( 1 )
0

0 . In Eq. (1 – 8),
 f
 
0




(
 g
L sin

)
 ( 1 )
0

0
  g
L
.
Substituting
Eq. (1 – 8) and Eq. (1 – 9) into Eq. (1 – 7) yields the linear equation that describes the pendulum’s motion in the neighborhood of the first equilibrium position
(1 – 10)
    g
L

We obtained Eq. (1 – 10) by noticing that
     
.
To obtain the linear equation in the stability region that surrounds the second equilibrium angle
 (
0
2 )  the calculation
 f
 
0




(
 g
L sin

)
 (
0
2 )  
 g
L
.

, perform
Then, substitute Eq. (1 – 8) and Eq. (1 – 9) into Eq. (1 – 7) to get
(1 – 11)
    g
L

The function f (

) and the two linear approximations of f (

) are shown in Figure 2.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0 50 100 150 200 theta (deg)
250 300 350
Figure 2: The function f and the linear approximations of f
Equation (1 – 10) is accurate in the neighborhood of the first equilibrium position and Eq. (1 – 11) in the neighborhood of the second. The next step is to solve Eq. (1
– 10) and Eq. (1 – 11). First, consider Eq. (1 – 10). Try solutions of the form
(1 – 12)

1
 cos(
 t )

2
 sin(
 t ) where


 g / L
It’s easy to see why these are solutions to Eq. (1 – 10) by substituting them into
Eq. (1 – 10) and verifying that the left side is equal to the right side. Since Eq. (1 –
10) is linear, its solutions satisfy the linear superposition principle, that is, any linear combination of two solutions is itself a solution of the differential equation.
(If you’re skeptical, prove it for yourself.) In other words, the general form of the solution of Eq. (1 – 10) is
(1 – 13)
 
A cos(
 t )

B sin(
 t ) where A and B are constants. The constants A and B depend on the initial conditions imposed on the pendulum, that is, its initial angle and initial angular rate. Regardless of the initial conditions, notice that the time response

( t ) in Eq.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
(1 – 13) is harmonic. Therefore, the first equilibrium position is said to be stable
(as expected).
Next, turning to the second equilibrium position, let’s try to solve Eq. (1 – 11). Try solutions to Eq. (1 – 11) of the form

1
 e
 t 
2
 e
  t
Again, it’s easy to verify that these solutions satisfy Eq. (1 – 11) and, by the linear superposition principle, that the general solution to Eq. (1 – 11) is of the form
(1 – 14)
 
Ae
 t 
Be
  t where A and B depend on initial conditions. This time, notice that the solution is an unbounded function of time, regardless of the constants A and B (unless A = 0
*
).
Therefore, the second equilibrium position is described as unstable (as expected).
*
Can you describe this unusual situation in which the response in the unstable region approaches zero?
As a second example, consider a more complicated system than the pendulum. As shown in Figure 3, 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 k and the damping constant is c . The bar rotates in the horizontal plane. Let’s find the nonlinear equation that governs the bar’s motion, its equilibrium positions, and its corresponding linearized equations.
Figure 3: Bar System and its Free-Body Diagram
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
The equation that governs the motion is
( a )

M
0

I
0
  where the mass moment of inertia is
( b ) I
0

1
3
(
1
2
M ) a
2 
1
3
(
1
2
M ) a
2 
Ma
2
3
To calculate the resultant moment

M
0 let’s perform the following preliminary calculations (See Figure 3): r
A
 a (cos
 i
 sin
 j ) r
B
 a (
 sin
 i
 cos
 j ) r
D
 a ( i
 j ) d
A
 a
2
( 1
 cos

)
2  a
2
( 1
 sin

)
2  a 3

2 cos
 
2 sin
 d
B e
A e
B


 a
2
( 1
 sin

)
2  a
2
( 1
 cos

)
2 r
D / A
AD r
D / B
AD
 r
D


AD r
A r
D
AD
 r
A


1
3

2 cos

3

2 sin
1

 a 3

2 sin

2 sin


2 cos

[( 1

[( 1

  cos

) sin

)
2 cos
 i

( 1
 sin i


) j ]
( 1
 cos

) j ]
F
A
 k ( d
A
 a ) e
A
F
B
 d c dt
( d
B
 a ) e
B
The resultant moment is
M
0
 r
A

F
A
 r
B
 a (cos
 i
 sin
 j )

F
B

3
 k
2
( d
A cos


 a )
2 sin

[( 1
 cos

) i

( 1
 sin

) j ]
( c )
 a (
 sin

 a [ k ( d
A
 i a

) cos
 j )

3
 d c dt
2
( sin d

B
 a )

2 cos

[( 1
 sin

) i

( 1
 cos

) j ]
3

1
2 cos
 
2 sin

(cos
  sin

)
 c d ( d
B dt
) 1
3

2 sin
 
2 cos

(sin
  cos

)] k in which
( d ) d ( d
B
) dt
 a
Substituting Eq. ( c ) into Eq. ( a ) yields
3
 cos
  sin
2 sin
 

2 cos

Ma
2
   
3 a
2 k ( 1

1
3

2 cos
 
2 sin

)(cos

 
 sin

)
 ca
2
3
(sin


2 sin

 cos


2
)
2 cos

 
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
3
Multiplying both sides by
Ma
2
, we get the nonlinear equation of motion of the system
( e )
   f (

,
 
) in which f (

,
 
)
 k
3
M
( 1

1
3

2 cos
 
2 sin

)(cos
  sin

)

3 c
M 3
(sin


2 sin

 cos


2
)
2 cos

 
Clearly, these equations are more complicated than the nonlinear equations that describe the motion of the pendulum, Eq. (1 – 7). Not only are there more terms but it’s also a function  
.
Despite the added complications, we’ll still be able to find the system’s equilibrium positions and the linear equations that describe the motion in the neighborhood of each of the equilibrium positions. Let’s start by finding the system’s equilibrium positions. Substituting
      
0 into Eq. ( e ) yields the nonlinear algebraic equation that governs the equilibrium position:
( f ) 0
 f (

0
, 0 ) in which f (

, 0 )

3 k
M
( 1

1
3

2 cos
 
2 sin

)(cos
  sin

)
In general, the nonlinear algebraic equation can’t be solved by hand. However, it can always be solve numerically very easily by graphing the function f (

, 0 ) versus

. The function crosses the
 axis at the equilibrium positions (See Figure
4).
0.2
0
-0.2
-0.4
-0.6
1
0.8
0.6
0.4
-0.8
-1
0 50 100 150 200 theta (deg)
250 300
Figure 4: Graph of f (

) (Letting 3 k / M = 1)
350
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
Interestingly, it happens that this equation can also be solved analytically. Notice in Eq. ( f ) that 0
 f (

0
, 0 ) when either
( g ) 0

1

1
3

2 cos

0

2 sin

0
or 0
 cos

0
 sin

0
By solving these equations, or from Figure 4, it is found that the bar has the following 4 equilibrium positions:
( h )

0
( 1 ) 
0

0
( 2 ) 


0
( 3 ) 


0
( 4 ) 
3

4 2 4
Next, let’s find the linear differential equation that governs the motion in the neighborhood of each equilibrium position. Unlike the pendulum, the right side of
Eq. ( e ) is a function of
 and
 
. Therefore, the right side of Eq. ( e ) needs to be approximated by a linear function of
 and
 
. To do that, a two-variable Taylor series approximation is performed. The right side is approximated by
( i )
 f (

0
,

 
)
 f
 

0 f (

0
,
 
0
)

(
  
0
)

 f
 
(
  
0
 f
  
0
(
  
0 )
0

)

 f
 
 f
  
0
(

0
(
    
0
)
 
0
)

 f
  
0
 
Replacing Eq. ( i ) with the right side of Eq. ( e ) yields the linear equation that governs the motion in the neighborhood of each equilibrium position:
( j )
   
 f
 
0
(
  
0
)

 f
  
0
 
Equation ( j ) can be written as
( k )
   
C
  
K (
  
0
)

0 in which
( l ) K
 
 f
 
C
 
 f
  
0 0
It remains to calculate the constants K and C in Eq. ( k ). They can be obtained analytically or numerically. By differentiation of Eq. (e) the analytically obtained derivatives are:
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
( k )
K
 k
3
M


 3
1
3

2 cos

0

2 sin

0

( 1

(cos

0
 sin

0
)
2
1
3

2 cos

0

2 sin

0
)(sin

0
 cos

0
)



C

3 c
M 3

2 sin

0
1

2 cos

0
(sin

0
 cos

0
)
2
Note that Eq. ( g ) was used to simplify the expressions for the derivatives in Eq.
( k ). The constants for each of the stability regions are tabulated below.

0
K C
0 k
3
M c
3
M

4

3
2

4

6 k
M k
3
M

2 .
48 k
M c
2
M c
0 .
6
M c
2
M
The constants C and K , rather than find them analytically, could always have been found numerically. K is the negative of the slopes in Figure 4 evaluated at the equilibrium positions and C is simply the “coefficient” in front of the  
in Eq. ( e ) evaluated at the equilibrium positions.
Later in the book, when solving equations of the form of Eq. ( k ), you’ll see that the constant K is an artificial spring constant and that C is a damping constant.
Systems of the form of Eq. ( k ) will be shown to be stable when K ≥ 0 and C ≥ 0 and therefore unstable when either K < 0 or C < 0. If follows, by referring to Eq.
( l ), that the first and third equilibrium positions are stable and that the second and forth are unstable. Notice also that the two instabilities were produced by K ; C was positive for each of the equilibrium positions.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
The problems statements given in this book are posed in a general way, without referring 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 2, 4, 5, and 7.
A bar-spring-damper system is shown. Let m = 20 kg, k = 500 N/m, a = 1.5 m, and c
0
= 100 Ns/m. The system lies in the horizontal plane and it is shown in its equilibrium position. The applied moment is M.
System 1: Bar–Spring–Damper System
A shock absorbing system is composed of a rotating disk, a spring, and an immersed fluid, as shown. The unstretched length of the spring is H. Let H = 4 in,
R = 3 in, mg = 15 lb, k = 15 lb/in, and c = 0.01 lb·s/in. The system lies in the horizontal plane and it is shown in its equilibrium position. The applied moment is
M.
System 2: Shock Absorbing System
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
A stabilizer is composed of a rotating disk, a spring, and a damper, as shown. Let m = M = 0.1
kg, R = 0.05 m, k = 0.4 N/m, and c = 0.03 N·s/m. The system lies in the horizontal plane and it is shown in its equilibrium position. The applied moment is M.
System 3: Stabilizer System
A 2-disk system is composed of a pair of geared disks, a spring, and a springdamper, as shown. Let m = 1 kg k = 1 N/m, R = 1 m, and c = 0.1 N·s/m. The system lies in the horizontal plane and it is shown in an equilibrium position. In the equilibrium position shown, the springs are unstretched,
(

=

B
). The applied moment is M.

A
= 90
°
and

B
= 45
°
.
Select the angle of the right disk as the system’s independent degree of freedom
System 4: Two-Disk System
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
A roller system is composed of a disk that rolls on a circular track that is acted on by a spring and a damper. Let m = 0.1 slug, k = 15 lb/ft, c = 0.3 lb·s/ft, r = 0.25 ft, and R = 1 ft. The bar that the disk is attached to is light. The system lies in the horizontal plane and it is shown at equilibrium. The angle of the bar is
β starting at 0 in the position shown, taken positive counter-clockwise. The angle of the disk is
θ starting at 0 in the position shown, taken positive clockwise (See figure to the right). The applied moment is M
A
.
System 5: Roller System
A mechanical switch is composed of a uniform bar that rotates about point O , a spring and a damper. Let m = 0.09 slug, k
0
= 15 lb/in, c
0
= 12 lb·s/in, a = 6 in, b =
4 in, d = 5 in, and cos

0
= 0.8. The system lies in the horizontal plane and it is shown at equilibrium. The applied force is F.
System 6: Switch System
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
A disk system is composed of a uniform disk, a spring, and a damper. Let m = 10 kg, k = 10 N/m, c = 25 N·s/m, R = 30 cm, and a = 10 cm. The unstretched length of the spring is 40 cm. The applied moment is M.
System 6: Switch System
(a) Carefully draw a free body diagram of the system. In the diagram, show the system in a displaced position. Remember that the displacement is arbitrarily large. The diagram should show a coordinate system, all of the forces acting on the system, and other dimensions that are needed. The diagram should also list all of the given parameters.
(b) Develop a vector expression for the spring force and the damping force.
(c) Using vector algebra, derive the nonlinear differential equation that governs the motion of the system.
This problem begins where Problem 1 – 1 ends.
(a)
Starting with the system’s nonlinear differential equation, find the nonlinear algebraic equation that governs the static equilibrium positions of the system. Remember to set the applied forces and applied moments to zero.
(b) Find each of the equilibrium positions by solving the nonlinear algebraic equation numerically. Use MATLAB when generating the needed graph from which the equilibrium positions are found.
(c) Extra Credit: If possible, find the equilibrium positions analytically.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH

COMPLEX DYNAMICS SIMPLIFIED
This problem begins where Problem 1 – 2 ends.
(a) Write the nonlinear differential equation of motion in the general form
    f (

,
 
).
Find the stability derivatives
 f
  and
 f
   about each of its equilibrium positions.
(b) For the neighborhood around each equilibrium position ( k = 1, 2, … ), let

( t )
 
( t )
 
0
( k )
where

0
( k ) is the kth equilibrium angle write down the associated linear approximation of f
(

0
 in terms of

0
( 1 )
( t )
) and
 and
  ( t ).
Then, write down the linear differential equation governing the motion of the system in the neighborhood of each equilibrium position.
(c) Find a general form of the solution of each linear differential equation.
State which equilibrium positions are stable and which are unstable.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH