MOTION IN A STABILITY REGION (PART I)
CHAPTER 4
MOTION IN A STABILITY REGION (PART I)
When motion is confined to one independent degree-of-freedom, the linearized
equation that governs the motion is
mx  cx  kx  f
(4 – 1)
In this section, we analyze Eq. (4 – 1) is more detail than in Chapter 1. The
system’s free motion (f = 0) is analyzed first and then its forced motion. The
analysis performed for the free motion is called a transient analysis. The transient
analysis is independent of the non-homogeneous term that appears on the right side
of the differential equation. The non-homogeneous term is often a force but it can
also arise as a result of prescribing the displacement at a point in the system. The
right side of the differential equation is generally called the excitation. After
analyzing the free system, this section turns to the forced system. It is shown how
the time dependence of an excitation affects a system’s time response. We start
with the constant excitation. Static loads, weight forces, and prescribed
displacements are the most frequent examples. The constant excitation change’s a
system’s equilibrium position. Next, we consider the harmonic excitation. Systems
that contain unbalanced rotating elements, like a washing machine, milling
machines, and rotating shafts, are systems that are acted on by harmonic
excitations. The harmonic excitation causes a system to undergo harmonic motion.
1. Free Undamped Motion
Fig. 4- 1 The mass-spring-damper system
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
First consider the free undamped system (See Fig. 4 – 1). Letting f = 0 and c = 0 in
Eq. (4 – 1) yields
(4 – 2)
mx  kx  0
Equation (4 – 2) is a homogeneous constant-coefficient linear differential equation.
As with any constant-coefficient linear differential equation, the solution is a
combination of complex exponential functions. We start by looking at the single
complex exponential function
(4 – 3)
x  e st
where s is a complex number that needs to be determined. Substitute Eq. (4 – 3)
and its second time derivative into Eq. (4 – 2) to get
m( s 2 e st )  ke st  0
Dividing by e st
(4 – 3)
ms 2  k  0
The values of s for which x = e st satisfies the differential equation are
k
(4 – 4a, b)
s  i n ,  n 
m
where i   1. The two solutions are
(4 – 5)
x1  e int
x 2  e int
These two solutions may seem a bit odd; after all they’re complex. The complex
solutions are actually just building blocks from which the real solution is
constructed. Recall that e i n t and e i n t in Eq. (4 – 5) are complex harmonic
functions of the form (See Fig. 4 – 2)
(4 – 6)
e int  cos  n t  i sin  n t
e int  cos  n t  i sin  n t
Fig. 4 -2
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
The real solution is a linear combination of the two complex solutions. From Eq.
(4 – 5) and Eq. (4 – 6)
(4 – 7)
x  A1 x1  A2 x 2
 A1 (cos  n t  i sin  n t )  A2 (cos  n t  i sin  n t )
where A1 and A2 are constants. By rearranging terms, the real solution is written as
(4 – 8)
x  B cos  n t  C sin  n t
in which B = A1 + A2 and C = i(A1 – A2). Equation (4 – 8) is called the transient
solution or the homogeneous solution of Eq. (4 – 1). The constants B and C depend
on whether the system was initially displaced, was initial moving, or both. The
constants B and C can also be written as
(4 – 9)
B  A cos 
C  A sin 
Substituting Eq. (4 – 9) into Eq. (4 – 8)
(4 – 10)
x  A cos( n t   )
Fig. 4 - 3
using the trigonometric identity cos  cos  n t  sin  sin  n t  cos( n t   ). We
see in Eq. (4 – 10) that the transient solution of an free undamped system is a
harmonic function. It’s amplitude is A, its natural frequency is n, and its phase
angle is . (See Fig. 4 – 3). The natural frequency is found from Eq. (4 – 4b). Its
standard units are rad/s. The natural frequency is also sometimes expressed in
terms of cycles per seconds which is the same as a Hertz, abbreviated Hz. Since 1
cycle is 2 radians it follows that 1 Hz  2 rad/s and so 1 Hz is about six times
larger than 1 rad/s. The system’s natural period is
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
Tn 
(4 – 11)
2
n
m
k
 2
2. Free Damped Motion
Now consider the free damped system (See again Fig. 4 – 1). Letting f = 0 in Eq.
(4 – 1) yields
(4 – 12)
mx  cx  kx  0
The procedure for solving Eq. (4 – 12) is the same as the procedure followed for
solving Eq. (4 – 2). Start by assuming a solution in the form of a complex
exponential function. Substitute Eq. (4 – 3) into Eq. (4 – 12) to get
m(s 2 e st )  c( se st )  ke st  0
Dividing by e st yields the quadratic equation
(4 – 13)
ms 2  cs  k  0
Its roots are
2
(4 – 14)
1 
c
k
 c 
s
 c  c 2  4mk  
 
 


2m 
m
 2m
 2m 
Let’s now distinguish between three levels of damping depending on whether the
quantity under the square root is positive, zero, or negative. The roots are
    i
2
2
d d  n  


(4 – 15) s   

2
2
      n

in which
k
n 

m
n  
n  
underdamped
critically damped
n  
overdamped

c
2m
Fig. 4 – 4
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
The three levels of damping are referred to as under-damped, critically damped,
and over-damped. The function x  e st is complex in under-damped systems and
real otherwise. Figure 4 – 4 shows x  e st in the complex plane for under-damped
systems.
A. Under-damped
In under-damped motion the two solutions to Eq. (4 – 12) are
(4 – 16)
x1  e ( id )t  e t e id t  e t (cos  d t  i sin  d t )
x 2  e ( id )t  e t e id t  e t (cos  d t  i sin  d t )
The general solution is a linear combination of these complex solutions.
x  A1 x1  A2 x 2
 A1e t (cos  d t  i sin  d t )  A2 e t (cos  d t  i sin  d t )
Regrouping terms,
(4 – 17)
x  e t ( B cos  d t  C sin  d t )
in which B and C are constants. We see in Eq. (4 – 17) that the transient solution of
a free under-damped system is a damped harmonic function. Its natural damping
rate is  and its damped natural frequency is d (See Fig. 4 – 5). The damped
natural period is
(4 – 18)
Td 
2
d
2

k  c 


m  2m 
2
Fig. 4 – 5
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
B. Critically damped
In critically damped vibration, the two solutions to Eq. (4 – 2) are1
x1  e t
(4 – 19)
x 2  te t
The general transient solution is a linear combination of the two complex
solutions. From Eq. (4 – 19)
x  e t ( B  Ct )
(4 – 20)
in which B and C are constants (See Fig. 4 – 6).
C. Over-damped
In over-damped vibration the two solutions to Eq. (4 – 12) are
(13.2 – 21)
x1  e 1t
1     2   n2
x 2  e  2t
 2     2   n2
The general transient solution is (See Fig. 4 – 7)
x  Be 1t  Ce  2t
(4 – 22)
Fig. 4 – 6
Fig. 4 – 7
1
The critically damped solutions are obtained through a limiting process from the under-damped
solutions by letting d tend to zero.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
3. Free Time Response
The transient time response of an undamped system and the transient time
response of a damped system were given in Eq. (4 – 8), Eq. (4 – 17), Eq. (4 – 20),
and Eq. (4 – 22). In each case, the response was expressed in terms of two
unknown constants B and C. The two constants are determined from the system’s
two initial conditions; it’s initial displacement x0 = x(0) and its initial velocity
v 0  x (0).
First, consider the undamped response. From Eq. (4 – 8)
x 0  B cos( n  0)  C sin(  n  0)  B
v 0   n B sin(  n  0)   n C cos( n  0)   n C
so the constants are
(4 – 23)
B  x0
v
C 0
n
The transient time response of an undamped system is
(4 – 24)
v
x  x0 cos  n t  0 sin  n t
n
Following the same steps, the transient time responses of the damped systems are
v  x 0
x  e t ( x0 cos  d t  0
sin  d t )
d
(4 – 25)
x  e t [ x0  (v0  x0 )t ]
v   2 x 0   1 t v 0  1 x 0   2 t
x 0
e

e
 2  1
1   2
under - damped
critically damped
over - damped
The general forms of the responses were shown in Figs. 4 – 3, 5, 6, and 7.
4. Comparison of Limiting Cases
The under-damped, critically damped and over-damped responses given in Eq. (4
– 25) are identical to each other in limiting cases. When the natural rate of decay
tends to zero, the damped cases become the undamped case given in Eq. (4 – 8).
When the natural frequency of oscillation tends to zero, the over-damped and
under-damped cases become critically damped. When the natural rate of decay and
the natural frequency of oscillation tend to zero, the cases all converge to x(t) =
x0+v0t. These limits utilize the identities
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
(4 – 26)
lim
sin( t )
 0
lim
 0

et  e t

t cos(t )
t
 0
1
 lim
tet  te t
 lim
 2t
1
 0
5. Constant Excitation
Fig. 4 – 8
Let a damped system be acted on by the constant excitation f = F0 (See Fig. 4 – 8).
From Eq. (4 – 1)
(4 – 27)
mx  cx  kx  F0
A solution to Eq. (4 – 27) is
(4 – 28)
F
x 0
k
Substitute x and its derivatives into the left side of Eq. (4 – 27) to verify that Eq. (4
– 28) satisfies Eq. (4 – 27). A solution that satisfies Eq. (4 – 27) is called a steadystate solution to the differential equation. It is just one of the many solutions that
satisfy the non-homogeneous differential equation. Therefore, it is also called a
particular solution. The general form of the solutions that satisfy the homogeneous
differential equation (letting f = 0) was found in the previous section. For example,
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
recall when a system is under-damped, from Eq. (4 – 17), that the general form of
the transient solution is
(4 – 29)
xt  e t ( B cos  d t  C sin  d t )
Let’s now construct a general form of a solution that satisfies the nonhomogeneous differential equation. The general form is constructed by adding
together the general form of the transient solution and one steady-state solution,
written
(4 – 30)
x  xt  x s
In an under-damped system it follows from Eq. (4 – 28) and Eq. (4 – 29) that the
general form of the solution is
F
(4 – 31)
x  xt  x s  e t ( B cos  d t  C sin  d t )  0
k
Notice as time goes on that the transient part of the solution damps out leaving the
steady-state solution, and hence the name steady-state. Also, notice that the steadystate solution is eventually independent of the constants B and C, which in turn
depend on the system’s initial conditions. Hence the steady-state solution, after the
transient motion has damped out, is unique. In the case of a constant excitation, the
steady-state solutions tends to the constant
F0
.
k
The effect of the constant
excitation is to change the equilibrium position of a system.
The Linear Superposition Principle
The procedure followed above to find the general form of the solution of a
differential equation employs the linear superposition principle, and we used it
earlier in the book. Let’s now define the linear superposition principle precisely.
Toward this end, we consider two problems. First, we consider a system that is
acted on by an excitation f1 and then we consider the same system acted on by the
excitation f2. The corresponding steady-state solutions are x1 and x2. The
differential equations for each problem are
(4 – 32)
mx1  cx1  kx1  f1
mx2  cx 2  kx2  f 2
The linear superposition principle states that a steady-state solution for the same
system acted on by the excitation f = a1f1 + a2f2 is x = a1x1 + a2x2, in which a1 and
a2 are constants. The linear superposition principle is represented below
symbolically.
(4 – 33)
 f  x1
If  1
then a1 f1  a 2 f 2  a1 x1  a 2 x 2
 f 2  x2
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
The linear superposition principle is proven as follows:
mx  cx  kx  m(a1 x1  a 2 x2 )  c(a1 x1  a 2 x 2 )  k (a1 x1  a 2 x 2 )
 a1 (mx1  cx1  kx1 )  a 2 (mx2  cx 2  kx2 )  a1 f1  a 2 f 2
The general form of the solution obtained in Eq. (4 – 30) uses the linear
superposition principle; it’s a special case of it. By letting a1 = a2 = 1, f1 = 0 and f2
= F0 in Eq. (4 – 33), we get Eq. (4 – 30). The linear superposition principle is
behind the procedures that are followed when solving linear differential equations
and it’s called upon again in the following.
6. Harmonic Excitation
Fig. 4 – 9
A damped system is acted on by the harmonic excitation
(4 – 34)
f  F0 cos t
in which F0 is a constant and  is excitation frequency (See Fig. 4 – 11). To find a
steady-state solution, we start out by considering a different problem. We assume
that the system is acted on by the complex excitation
(4 – 35)
f C  F0 e it
Since e it  cos t  i sin t , the real and imaginary parts of fC are
f R  F0 cos t
f I  F0 sin t.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
Notice that the real part of fC is the same as the original excitation in Eq. (4 – 34).
In Eq. (4 – 1) we have
mx  cx  kx  f C
(4 – 36)
in which x denotes the associated complex response. By the principle of linear
superposition, if the response to fR is denoted by xR and the response to fI is
denoted by xI then the complex response x to the complex excitation fC = fR + ifI is
x = xR + ixI. This follows from Eq. (4 – 33) by letting a1 = 1, a2 = i, f1 = fR and f2 =
fI.. Thus, to find the steady-state solution to the system acted on by the original
excitation given in Eq. (4 – 34) we can first find the complex response to the
excitation given in Eq. (4 – 35) and then extract from x its real part.
Begin by trying a steady-state solution in the form
x  X 0 e it
(4 – 37)
where X0 is a complex constant that needs to be determined. Substituting Eq. (4 –
37) into Eq. (4 – 36)
(4 – 38)
m[(i ) 2 X 0 e it ]  c[(i ) X 0 e it ]  k[ X 0 e it ]
 [m 2  ci  k ] X 0 e it  F0 e it
Dividing by e it and solving for X0 yields
(4 – 39)
X0 
F0
F
 0
k
k  m 2  ic
1
 
1  
 n
2


  i 2
n


is called the damping factor and n is the system’s natural
n
frequency. The complex constant X0 can be written in polar form as
in which  
(4 – 40)
X 0  X 0 e i
in which
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
F
(4 – 41) X 0  0
k
1
  
1  
   n




2 2
  [2  ] 2
n




 2  

n 
  tan 1 
2
    
 
1  
  n  
Substituting Eq. (4 – 40) into Eq. (4 – 37) we get
(4 – 42)
x  X 0 e i e it  X 0 e i(t  )  X 0 [cos(t   )  i sin( t   )]
Equation (4 – 42) is the steady-state solution to the system acted on by the
complex force fC. The steady-state solution to the system acted on by the original
force given in Eq. (4 – 34) is its real part,
(4 – 43)
x R  X 0 cos(t   )
The steady-state solution to the system acted on by fI is
(4 – 44)
x I  X 0 sin(t   )
We see in Eq. (4 – 43) and Eq. (4 – 44) that the time response of a damped system
acted on by a harmonic excitation of frequency  is harmonic, having the same
frequency as the excitation (See Fig. 4 – 10). The amplitude of the time response is
X 0 and the phase angle is  Next we look closer at the amplitude and phase
angle of the vibration.
Fig. 4 – 10
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
Amplitude and Phase
The question arises how the amplitude and phase of the time response change with
excitation frequency for a given system. To answer this question, we examine Eq.
(4 – 42). The amplitude | X0 | / F0 and the phase  are graphed as functions of the

relative frequency
in Fig. 4 – 11 for different damping factors  . The
n
amplitude | X0 | is also called the frequency response of the system. Notice that the
frequency response, regardless of the damping factor, is always equal to F0/k when
the excitation frequency is zero. An excitation frequency of zero corresponds to a
constant excitation and hence its solution is the same as Eq. (4 – 28). At very high
frequencies, the harmonic excitation doesn’t excite the system. Notice when the
excitation frequency is equal to the natural frequency that the amplitude becomes
very large (or infinite when there is no damping). This is called resonance.
Fig. 4 – 11
Next, consider the phase angle. The phase angle represents a lag in the
displacement behind the force. Notice when there is no damping that the phase
angle is zero; the displacement and the force are in phase. The non-zero phase
angle is a result of damping; the damping causes the displacement to lag behind
the force when the excitation frequency is less than the natural frequency and to
lead in front of the force when the excitation frequency is greater than the natural
frequency.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
MOTION IN A STABILITY REGION (PART I)
PROBLEM STATEMENTS
The problems in this chapter consider systems 1 – 1 through 1- 7.
Problem 4 – 1: Simplified Free-Body Diagrams
(a) Redraw the free body diagram of the system. This time let the angle of the
bar in the diagram be small. You will now make a small-angle assumption.
In the diagram you will treat sin() as  and cos( ) as 1. This means that
small arcs become right angles. Therefore, when rotating an arc of length a
by a small angle , the diagonal length is still a and the side length is a .
(b) Using this free body diagram, find the linear differential equation that
describes the motion of the system about the equilibrium. You should find
that this linear differential equation is the same as the one found in Problem
1 – 3.
Problem 4 – 2: Impulse Response
The system is now acted on by an applied load M or F. We’ll denote it my M. Two
loads M are compared: 1) a finite pulse M1 and an instantaneous impulse M2.
When the time of the pulse is “short” enough, the response of the system acted on
by the finite pulse should look like the response of the system acted on by the
impulse. The question arises what do we mean by “short.”
The short-time pulse M1 and the impulse M2 are given by:
0  t  T0
T0  t  
A ,
M1   0
0,
M 2  A0T0 (t )
where T0 is the period of the pulse (not to be confused with step size or period).
Let T0 = Tf/10 and A0 = 1 lb (or 1 N depending on the units used). Notice for both
functions that the integral over time of the load is the same, that is
T0
0
T0
M 1 (t )dt   M 2 (t )dt  A0T0
0
(a) Find the response (t) of the system subject to each of the two loads
assuming that the system is initially at rest. Using MATLAB, plot each
response for about 6 oscillations.
(b) Look at the two responses and comment on the nature of the
approximation.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH