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Chapter 3
System Response
Once the system model has been developed, we can see what
predictions it makes of the system's behavior under whatever
conditions are of interest. In order to do this for a continuous-
time, dynamic model, we must solve the model's differential
equations.
3.1 Free response of a first-order model
The simplest dynamic model to analyze is a linear, first-order,
constant-coefficient, ordinary differential equation without
inputs. This model refers to the followin simple equation!
rydt
dy= "#.$-$%
where ris a constant. There are several ways of solvin this
equation, but we wish to emphasize from the start the useful
fact.
&ny constant-coefficient, linear, ordinary differential equationor coupled set of such equations of any order, without inputs,
can always be solved by assumin an eponential form for the
solution.
The eneral eponential form to be assumed for each unknown
variable is
stAety =%"
"#.(-(%
whereAandsare unknown constants. The time derivative ofy
in this form isst
sAedt
dy=
)ubstitutin the last two relations into "#.$-$% ivesstst
rAesAe =
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*or the solution to be nontrivial "nonzero for arbitrary values of
t%, the constantAmust be nonzero, andsmust be finite. Thus,
Aestdoes not equal zero, and we obtain
s=r "#.$-#%
+quation "#.$-#% is known as the characteristic equation of the
model and is a first-order polynomial equation ins. It has one
solution, the characteristic root. Thus, the solution of "#.$-$% isrtAety =%" "#.$-%
for any nonzeroA. In order to determine a value forA, an
additional condition must be stipulated. This condition usually
specifies the value of y at time to, the start of the process and is
the initial condition. +valuatin the precedin equation at t=toives
A=y(to)e-rto
&nd accordinly the solution is
y(t) = y(to)e-rtoert
or%"
%"%" ottr
o etyty
= "#.$-%
*or autonomous models "the model dy/dt =f(y,t) is autonomous
whenf(y,t)is independent of t% the oriin of the time ais may
be shifted so that to is taken to be zero without loss of
enerality. The solution is sketched in *iure #.$ for positive,
zero, and neative values of r. ecause this solution describes
the system's behavior when it is free of input's influence, this
solution is called the free response.
The time constant
If r is neative, a new constant is usually introduced by the
definition
r
$= "#.$-/%
and the solution can be written as
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0%1"%" teyty = "#.$-2%
The new parameter is the model's time constant, and it ives a
convenient measure of the eponential decay curve. To see this,
let t3 in "#.$-2% to obtain
%1"#2.1%1"%" $ yeyy = "#.$-4%
*iure #.$ *ree response of a linear first-order model.
&fter a time equal to one time constant has elapsed,yhas
decayed to #25 of its initial value. <ernatively, we can say
thatyhas decayed by /#5. If t=4,
y(4) = y(0)e-4 0.02y(0) "#.$-6%
&fter a time equal to four time constants,yhas decayed to (5 of
its initial value "or by 645%.
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Point equilibrium and stability
Often the information desired from a dynamic model simply
concerns the eistence of a point equilibrium or equilibrium
state "or simply equilibrium in common usae% and its stability.
& point equilibrium is a condition of no chane in the model's
variables. 7athematically, it is determined by solvin for the
values of the variables that make all of the time derivatives
identically zero. *or "#.$-$%, this implies that
0=ry
If ris zero, an infinite number of equilibrium values foryeists.
If ris nonzero, the only point equilibrium isy=0. Thus, ifyis
initially zero, it will remain zero unless some cause "other than
those already described by the model% displacesyfrom its
equilibrium value.
Ifydisturbed from equilibrium, a natural question to ask is,
8oesyreturn to its equilibrium9 The answer is determined by
the stability characteristics of the equilibrium. & stableequilibrium is one to which the model's variables return and
remain if slihtly disturbed. :onversely, an unstable equilibrium
is one from which the model's variables continue to recede if
slihtly disturbed. There is a borderline situation, a neutrally
stable equilibrium, defined to be one to which the model's
variables do not return and remain but from which they do not
continue to recede.
These three cases can be illustrated by a ball rollin on a surface"*iure #.(%. In *iure #.(a, if the ball is displaced but still
within the valley ";slihtly displaced;%, it will roll back and
forth around the bottom, transferrin enery between kinetic and
potential. If there is friction, the ball finally comes to rest after
the friction has dissipated all the enery. Thus, the equilibrium
state is the state in which the ball resets at the bottom, and it is
stable if friction is present. & frictionless surface will allow the
ball to oscillate forever about the bottom, and in this case, the
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equilibrium is neutrally stable. If the ball is displaced so much
that it now lies on the level surface, the equilibrium can no
loner be considered stable even with friction. The term local
stability is sometimes used to emphasize the restriction to small
displacements, while lobal stability implies stability with
respect to all displacements.
*iure #.( stability eamples. "a% all in a hole. 6b% ball on an
infinite hill. "c% all on a finite hill.
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*or the ball precariously balanced on top of the hill "*iure
#.(b%, this equilibrium is lobally unstable "and thus locally
unstable as well% for hypothetical hill without a bottom if the
friction is not reat enouh to stop the ball. The absence of such
a hill in reality points out that in interpretin instability, the
assumptions of the model structure must be kept in mind. Thus,
in a model of form of "#.$-$%, equilibrium aty31 is lobally
unstable mathematically if reutral stability of the first type
"oscillatory behavior% is an abstraction that does not eist in
physical systems unless a power source is present to sustain the
oscillations. Otherwise, dissipative forces, such as friction, act to
create a stable situation.
The first-order model has one characteristic root, and, as we will
see, a second-order model has two roots, etc. The local stability
properties of an equilibrium for a linear model are determinedfrom the characteristic roots and by an equivalent but
approimate process called linearization for a nonlinear model.
& linear model is lobally stable if it is locally stable, but
determinin lobal stability is frequently difficult for nonlinear
models.
The characteristic root of "#.$-$% is iven by "#.$-#%, and from
*iure #.$, we can etract the followin eneral statement!
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An equilibrium of the first-order linear model is globally stable
if and only if its characteristic root s is negatie and is neutrally
stable if and only if s is !ero. "ther#ise, the equilibrium is
unstable.
Parameter estimation
In many applications, a system can be described by "#.$-$%, but
the parameter rcannot be computed from basic principles. If
past measurements ofyat various times are available, a value
for rcan be estimated by the followin technique. If the first
data point is taken to be at to=0, the natural loarithm of both
sides of "#.$-% ives
ln y(t) = ln y(0) $ rt "#.$-$1%
If the loarithms of the measurements ofyare plotted versus t,
the transformed data should cluster about a straiht line if
"#.$-$% is an accurate model of the process "and if the
measurement error is small?%. This situation is shown in fiure
#.# for a neative value of r. & straiht line can be fitted by eye
if reat accuracy is not required, and the value of ris estimatedfrom the slope of this line. If the points have some scatter, the
method of least squares can be used to fit a line to the data.
=ith the estimated value of r, predictions can then be made
about future system behavior.
*iure #.# @oarithmic transformation of data.
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%&am'le .
*or the series circuit model shown below, lety=and 31.
Then,
1=+ ydt
dy*+
The characteristic root iss = -/*+, and thus the time constant
is 3*+. Typical values of*and +are $1,111 and 1.$ , ,
respectively. These ive a time constant of 1.11$s. If the input
voltae becomes zero but the circuit remains closed, the
capacitor voltae decays by 645 in 1.11s.
%&am'le .2
:ompute the time constant for a tank-pipe system as shown in
the followin fiure, assumin that the flow is laminar. The tank
contains fuel oil at 0owith a mass density of .2 slugs/ft
and a viscosity of 0.02 lb-sec/ft2. The outlet pipe diameter
is in., and its lenthis 2 ft.The tank is 2 ftdiameter.
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*or the iven system we have!
h*
gq
dt
dhA =
$ "#.$-$$%
where qis a volume flow rate. Aere, the tank area isA =ft2.
The laminar pipe resistance*can be calculated as!
-,,
sec,.$4-/4
%$(0$"%4(.$"
%(%"1(.1"$(4$(4
ft
lb
/
0*
===
and
sec011$2#,.1,.$4-/4
(.#( (ft*
g==
Thus,
$11$2#.1 qhdt
dh+=
and the time constant is 3 01.11$2#3 $4$( sec 3 #1.( min.
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3.2 Step response of the first-order model
7any of the models developed in :hapter ( can be put into the
followin eneral form!
brydt
dy+= "#.(-$%
where is the input. The solution obtained in )ection #.$ applies
when no input is present-that is, when =0- and this solution
represents the intrinsic behavior of the system. Aere, we bein
the analysis of "#.(-$% for the commonly encountered input
functions of time.The simplest of these perhaps is the step function depicted in
*iure #.. &s its names implies, it has the appearance of a
sinle stair step. efore the initial time t=0, the step function
has a constant value, usually zero. &t t=0, Bumps
instantaneously to a new constant value,1, the manitude,
which it maintains thereafter. The step function is an
approimate description of an input that can be switched O> in
a time interval that is very short compared to the time constantof the system.
*iure #. )tep input of manitude1.
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& common notation for the unit step function "1=% is us(t).
&ccordinly, the step with manitude 7 is denoted by1us(t).
Solution for the stable case
There are several ways of obtainin the closed-form solution of
"#.(-$% with a step input, if rand bare constants. *or now, we
will restrict ourselves to the stable case "rC1%, in which case
"#.(-$% can be written as
bydt
dy+=
$"#.(-(%
The simplest way of proceedin at this point is by employinsome physical insiht to obtain the form of solution. If the input
flow rate qiin *iure (.#( is a step input, the tank heiht and
thus the outflow rate will increase until the outflow rate equals
the input rate and the tank comes to an equilibrium state. )ince
this system is stable and its intrinsic behavior is iven by "#.$-
%, we assume that its response to a step input can be written in
the form
(
0
$%" +e+ty t
+= "#.(-#%
where +and +2are undetermined constants. >ote that as t ,
(+y as we would epect. )ubstitutin into "#.(-(% for t
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b1eb1yty t += 0D%1"E%" "#.(-%
which is plotted in *iure #. for positive values of b,1, and
y(0).
*iure #. )tep response of a first-order system.
Free and forced response
The solution iven by "#.(-% can be decomposed in several
ways. =e may think of the solution as bein composed of two
parts! one resultin from the ;intrinsic; behavior of the system
"the free response% and the other from the input "the forcedresponse%. =ritten this way, the solution is
%$"%1"%" 00 tt eb1eyty +=
where the first term represents the free response, while the
second term represents the forced response.
The principle of superposition for a linear model implies that the
total or complete response results from the sum of the free and
the forced responses. =e could have found the complete
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solution by addin the free response from "#.$-% to the forced
response iven by "#.(-% withy(0) =0. This approach is
sometimes useful for complicated "hiher order% models. In
summary, the free response represents that part of the system
behavior stimulated by the initial value ofy, and the forced
response represents the effect of the input.
Transient and steady-state responses
=e can also think of the response as bein comprised of a term
that eventually disappears "the transient response% and a term
that remains "the steady-state response%.
*or "#.(-%, the transient response is 0D%1"E teb1y
and the steady-state response isb1
*or a stable system, the free response is always part of the
transient response, and some the forced response miht appear
in the transient response.
eutrally stable and unstable cases
If ris zero in "#.(-$%, the solution iven by "#.(-% does not
apply, because is positive by definition. In this neutrally
stable case,dt
dyis a constant for t
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reulatin "feedback% nature of the term ryon the riht-hand
side of "#.(-$%.
If ris positive "unstable case%, an approach similar to that used
for the stable case produces the followin solution!
r
b1e
r
b1yty
rt += D%1"E%" "#.(-/%
The response increases eponentially.
%&am'le .
"a% =e want to find the motor torque necessary to drive a loadinertia of $. oz-in-sec(at a constant speed of $1 rad0sec. The
load's resistin torque is due entirely to dry friction and is
measured to be (/ oz-in. The tentative motor choice has has an
armature inertia of 1.$ oz-in-sec(, a rotational loss coefficient
of 2 oz-in0Frpm, and a maimum internal dry friction torque of
$ oz-in. "These data are available from manufacturer's
catalos%.
"b% Once the motor torque is determined, find how lon it willtake to reach the desired speed startin from zero.
)olution
"a% The inertias of the connectin shaft and ears are assumed to
have been lumped with the load inertia. If we nelect the
twistin of the shaft, we can also lump the armature and load
inertias to et a model similar in form to that shown in the
followin *iure, where the applied torque is the motortorque mminus the total dry friction torque . The dampin
coefficient c is iven by the rotational-loss coefficient. Thus, the
needed values are
3= .4 $ 0. = . o!-in-sec2
= m-25 - = m-2 o!-in
c= o!-in/6r'm = 0.0 o!-in/rad/sec
&t steady state, the torques must be balance, so that
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m-2 = 0.0 (0)
=here the torque on the riht is the dampin torque eperienced
at the desin speed of 0 rad/sec. The required motor torque
therefore is m= 2. o!-in.
*iure to eample #.#
"b% If this torque is suddenly applied and maintained, the step
response may be used. The system time constant is
((12.1
--.$===
c
3 sec
and it will take approimately 44 sec to reach the desired speed
if the motor torque is held constant.
%&am'le .4
:onsider eample #.( with inflow to a 2-ftdiameter tank. If the
input flow rate of fuel oil qis 20 gal/min, how lon will it take
to raise the oil level from to 0 ftif the -inoutlet line remains
open9
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)olution!
)ince .4 gal= ft, the input flow rate is 0.044 ft/sec, and
this is the manitude of the step. The tank model "#.$-$$% when
put into the standard form of "#.(-(% ives
y=h =qb=/A=/ =*A/g = 2
+quation "#.(-% can be transformed loarithmically to find the
time required to reach 0 ftif h(0)= ft.
$4$(%
--.(-$
--.(-$1ln"
--.(-%$4$(%"#.1%"$
"
t
ftb1
=
==
The time required is
t= 2 sec = . min
& by-product of this analysis is the final heiht, 2.ft, that
would be attained if the input flow rate were maintained
constant at 20 gal/min.
%&am'le .
*or the previous eample, how lon would it take for the heiht
to reach 0 ftif the outlet pipe were closed9
)olution!
In this case, the outflow termgh/*in "#.$-$$% is zero, and the
neutrally stable case applies. *rom "#.(-%, the step response is
%1"%1#.1"$.#
$%" htth +=
=ith h(t) = 0and h(0)=, the time required is
/#41#.1
%$.#%"$$1"=
=t sec = 0.5 min.
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%&am'le .5
&n electrical transducer system which consists of a resistance in
series with a capacitor. =hen subBect to a step input of size 7it
is found to ive an output of potential difference across the
capacitor , which is iven by the differential equation!
7dt
d*+ =+
=hat is the response of the system and how does vary with
time9
)olution!
%$"
0*+t
e7
=
3.3 Free response of a second-order model
The free response of second-order linear continuous-time
models is more varied than that of first-order models, because
oscillatory as well as eponential functions can occur.
Reduced and state !ariable formsThe application of physical laws can produce a second-order
model in the form of either a sinle second-order equation or
two coupled first-order equations. The mass-sprin-damper
model f8&dt
d&c
dt
&dm =++
(
(
is an eample of the first form. The 8:
motor model is an eample of the second form, which is called
the state variable form. The two forms are equivalent, but each
has its own advantaes for analyzin system behavior. The state
variable form is more convenient for use with numerical-simulation techniques. The reduced form is more convenient for
findin the response analytically, for low-order models with
relatively simple inputs.
It is easy to convert from one form to another. The model
f8&dt
d&c
dt
&dm =++
(
(
can be converted to state variable form by
definin as a new variable the time derivative of the variable .
@et the new variable be denoted . since dtd&= , v is obviously
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the speed of the mass m. The two variables whose derivatives
appear in the state variable form are called the state variables of
the system. Generally, there is no unique choice for the state
variables.
ecausedt
d&= , the model f8&
dt
d&c
dt
&dm =++
(
(
can be written in
state variable form as!
dt
d&= "#.#-$%
8&cfdt
dm = "#.#-(%
In order to solve these equations, it should be obvious that weneed to know the startin value of the two state variablesH that
is,&(0)and (0)3 %1"dt
d&. In order to solve the reduced form
f8&dt
d&c
dt
&dm =++
(
(
, we need the same information.
:onversion from state variable form to reduced form requires
one the equations to be differentiated and substituted into the
other to eliminate the unwanted variable. *or the 8: motor
model, suppose we desire a reduced model in terms of the speed. To obtain this, differentiate "(.-% to yield
dt
dc
dt
di6
dt
d3 a
2
=
(
(
)olve "(.-#% for dia/dtand substitute into the precedin
equation to obtain
dt
dcc
dt
d3
6
*
0
6
dt
d3
2
2
+= D"E
(
(
*inally, solve "(.-% for ia, and substitute into the precedin
equation to eliminate ia. This ives
dt
dc6c
dt
d3
6
*
0
6
dt
d3 e
2
2
+= D%"E(
(
or
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666c*dt
dc0*3
dt
d03 22e =++++
%"%"
(
(
"#.#-#%
Table #.$ ives the eneral relationship between the two forms
for the second-order model.
Table #.$ :onversion from state variable form to reduced form! a second-
order case.
)tate variable form
fb&a&adt
d&
fb&a&adt
d&
(((($($(
$($($$$$
++=
++=
+quivalent reduced form$. In terms of the variable&
fbabadt
dfb
&aaaadt
d&aa
dt
&d
%"
%"%"
$((($($
$($$((($$$
(($$(
$
(
+
=++
(. In terms of the variable&2
fbabadt
dfb
&aaaadt
d&aa
dt
&d
%"
%"%"
($$$($(
(($$((($$(
(($$(
(
(
+
=++
Solution from the reduced form
The mass-sprin-damper system is often used to illustrate the
free response of a second-order system. Its model can be written
as
f8&dt
d&c
dt
&dm =++
(
(
"#.#-%
The free response can be obtained by usin the same trial
solution as in section #.$, namely,
stAet& =%" "#.#-%
whereA andsare constants to be determined. This approach is
used to emphasize that the free response of any linear constant-
coefficient equation can always be obtained by eponential
substitution.
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8ifferentiatin "#.#-% twice ives
st
st
Aesdt
&d
sAedt
d&
(
(
(
=
=
)ubstitute these into "#.#-% withf=0, and collect terms to et
1%" ( =++ stAe8csms
&s with the first-order model in )ection #.$, a eneral solution
is possible only if 1stAe . Thus,
1(
=++ 8csms "#.#-/%
This is the model characteristic equation. It ives the followin
solution for the unknown constant s!
m
m8ccs
(
,( = "#.#-2%
=e see immediately that two characteristic roots occur. +ach
root enerates a solution of the form of "#.#-%. If the roots are
distinct, the free response is a linear combination of these two
forms. @etsands2denote the two roots. The free response is
tstseAeAt& ($ ($%" += "#.#-4%
It is easy to show that this solves "#.#-% withf=0, by computin
dt
d&and (
(
dt
&d, substitutin these epressions into "#.#-%, and
notin that bothsands2satisfy "#.#-/%.
Two state variables are required to describe this system's
dynamics. Therefore, the initial values of these variables must
be specified in order for the solution to be completely
determined. This means that the two constantsAandA2are
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determined by the two initial conditions. If the values of %"t&
anddt
d&are iven at t=0, then from "#.#-4%,
($
1
(
1
$%1" AAeAeA& +=+= "#.#-6%
8ifferentiatin "#.#-4% and evaluatin at t=0, we obtain
(($$%1" AsAsdt
d&+= "#.#-$1%
The solution ofAandA2in terms of %1"& and %1"dtd&
is
($
(
$
%1"%1"
ss
&sdtd&
A
= "#.#-$$%
$
($
$
( %1"
%1"%1"
A&ss
dt
d&&s
A =
= "#.#-$(%
%&am'le .
*ind the free response of "#.#-% for m=, c=, and 8=4. Theinitial conditions are $%1" =& and -%1" =
dt
d&.
)olution
The characteristic roots from "#.#-2% ares= -ands2= -4. *rom
"#.#$$% and "#.#-$(%,A=,A2= -2, and "#.#-4% ives
tt eet& (#%" =
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"scillatory solutions
The solution iven by "#.#-4% is convenient to use only when the
characteristic roots are real and distinct. If the roots are
comple, the behavior of "t% is difficult to visualize from "#.#-
4%. =e now rearrane the solution into a more useful form for
the case of comple roots.
:omple roots of "#.#-/% occur if and only if c( - mkC1. If so,
the roots are comple conBuates. *or now, assume that the real
part of the roots is neative and write the roots as
ibas +=$ "#.#-$#%
ibas =( "#.#-$%
where
m
ca
(= "#.#-$%
m
cm8b
(
, (= "#.#-$/%
*or this case, b
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A$ A2= 2A*i(A-A2) = -2A3
and%sin(cos("%" btAbtAet& 3*
at=
"#.#-($%
where
(
%1"&A* = "#.#-((%
b
a&dt
d&
A3(
%1"%1" +
="#.#-(#%
It is now apparent from "#.#-($% that the free response consists
of eponentially decayin oscillations with a frequency of bradians per unit time. The precise nature of the oscillation can be
displayed by writin "#.#-($% as a sine wave with a phase shift.
se the identity
bt9bt9bt9 cossinsincos%sin" +=+ "#.#-(%
and compare with the term in parentheses in "#.#-($%. This
shows that
*A9 (sin = "#.#-(%
3A9 (cos = "#.#-(/%
=e can define9to be positive and absorb any neative sins
with the phase anle .
Thus,%",(cos(sin (((( 3* AA99 +=+
or
3* AA9 ((( += "#.#-(2%
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since $cossin (( =+ . =ith found from this epression, is
computed from "#.#-(% with "#.#-(/% used to determine the
quadrant of .
The free response for comple roots is thus iven by
%sin"%" += btt9et& a "#.#-(4%
This is illustrated in *iure #./. The sinusoidal oscillation has a
frequency b"radians0unit time% and therefore a period of b0( .
The amplitude of oscillation decays eponentiallyH that is, the
oscillation is bracketed on top and bottom by envelopes that are
proportional to e-at. These envelopes have a time constant ofa0$= . Thus, the amplitude of the net oscillation occurrin
after t=/awill be less than #25 of the peak amplitude. *or
t:4/a, the amplitudes are less than (5 of the peak.
*iure #./ *ree response of the second-order model with
comple roots %sin"%" += btt9et& a
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%&am'le .
)olve f8&dtd&
cdt
&d
m =++
(
(
for m=, c=, 8=., andf=0. The
initial conditions are&(0) =and -%1" =dt
d&.
)olution
The characteristic roots are
is -.(-.$(
#6#=
=
and therefore a=.and b=2.. The constantsA*, A3, and canbe found from "#.#-((%, "#.#-(#%, and throuh "#.#-(2%.
Aowever, we now show that it is sufficient simply to remember
the solution form "#.#-(4%. *rom this, we see that
%cos"%sin"%1" +++= btb9ebta9edt
d& atat"#.#-(6%
and sin%1" 9& = "#.#-#1%
cossin%1" b9a9dt
d&+= "#.#-#$%
)ubstitute sin9 into the last equation to obtain
b
a&dt
d&
9
%1"%1"
cos
+
=
The present numerical values ive
/.(-.(
-.$-cos
$sin
=+
=
=
9
9
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)ince9is defined to be positive, these relations imply that1sin > and 1cos > .
Thus, is in the first quadrant and
#4/.1/.(
$
cos
sintan ===
= 2.04o= 0.5 rad
*inally,
24/.(sin
$==
9
and the solution is
%#/2.1-.(sin"24.(%" -.$ += tet& t "#.#-##%
The time constant is
//2.1-.$
$== time units
and the oscillation will have essentially disappeared after
"1//2% 3 (./2 time units.
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#eha!ior $ith repeated roots
=hen the two roots of the characteristic equation are repeated
"equal%, the combination iven by "#.#-4% does not consist of
two linearly independent functions. In this case, it can be shown
that the proper combination is
tstsetAeAt& ($ ($%" += "#.#-#%
wheres=s2. This situation occurs for the vibration model
"#.#-% when c2-4m8=0. In this case,s= -c/2m.
If the repeated roots are neative, the free response "#.#-#%
decays with time despite the presence of tas a multiplier,
because es2tapproaches zero faster than t becomes infinite.
& peak can occur before the solution beins to decay, dependin
on the relative values ofAandA2, which are found from the
initial conditions as before.
%1"$ &A = "#.#-#%
%1"%1"$( &s
dt
d&A = "#.#-#/%
3.% The characteristic equation
=e will now present some convenient criteria to use in
determinin the system's stability, its time constant, oscillatory
behavior, and frequency of oscillation, if any. &ll of these
properties are determined by only the characteristic equation,
and not by the initial conditions. The latter determine amplitudesand phase shifts. ecause we have used the vibration model
"#.#-% as our eample, we will use its form of the characteristic
equation as the reference.
1(
=++ 8csms "#.-$%
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Stability
=e have now treated all three situations that can arise for the
free response of the linear second-order model with constant
coefficients. =ithf=0in the vibration model "#.#-%, the only
possible equilibrium is&=0if 18 . *rom the definition of
stability iven in )ection #.$, it can be seen that the solution&(t)
must approach zero as tbecomes infinite, in order for the
equilibrium to be stable. This occurs in real, distinct-roots and
the repeated-roots cases only if both roots are neative. In the
case of comple roots, "#.#-(4% shows that the real part of the
roots must be neative. =e can summarize all three cases by
statin that the system is stable if and only if both roots haveneative real parts. "*or the real-root cases, the imainary parts
are considered to be zero.%
>eutral, or limited, stability occurs for this second-order system
in the comple roots case if the real part is zero. The solution is
then a constant-amplitude sinusoidal oscillation about the
equilibrium. >eutral stability also occurs if one root is neative
and the other zero. )ince two repeated zero roots are impossible
here, we can eneralize this result to state that the equilibrium isneutrally stable when at least one root has a zero real part and
the other root does not have a positive real part.
:onversely, it can be seen from "#.#-4% that only one root need
be positive for&(t)to become infinite. *or the comple-roots
case, this happens if the real part is positive. Thus, for all root
cases, if at least one root has a positive real part, the equilibrium
is unstable. These observations hold true for systems of hiher
order.8iscussion of a second-order system automatically implies that
the coefficient min "#.#-% is nonzero. =ith this restriction, an
analysis of the root solution "#.#-2% will show that both roots
have neative real parts if and only if m, c, and 8are positive. If
either cor 8is zero, the equilibrium is neutrally stable.
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The dampin& ratio
The behavior of the precedin free response for the stable case
can be conveniently characterized by the dampin ratio
"sometimes called the dampin factor%. *or the characteristic
equation "#.-$%, this is defined as
m8
c
(= "#.-(%
Jepeated roots occur if c2-4m8=0H that is if m8c (= . This value
of the dampin constant is the critical dampin constant, and
when c has this value, the system is said to be critically damped.If the actual dampin constant is reater than m8( , two real
distinct roots eist, and the system is overdamped. If m8c (< ,
comple roots occur, and the system is underdamped. The
dampin ratio is thus seen to be the ratio of the actual dampin
constant cto the critical value. *or a critically damped system,$= . +ponential behavior occurs if $> , and oscillations eist
for $= , and therefore the roots are real and
distinct. There will be no oscillatory free response. In +ample
with m=, c=, and 8=.H since the dampin ratio is $-$.1
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=e can write the characteristic equation in terms of the
parameters and n . *irst, divide "#.-$% by m and use the fact
that mcn 0( = . The equation becomes
1( ((
=++nn
ss "#.-%
and the roots are
($ = nn is "#.-%
:omparison with "#.#-$#% shows that na = and ($ = nb .
)ince the time constant
is /a,n
$= "#.-/%
The frequency of oscillation is sometimes called the damped
natural frequency, or simply damped frequency d , to
distinuish it from n .
($ = nd "#.-2%
*or the underdamped case " $
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=cos "#.-4%
*iure #.2 @ocation of the upper comple root in terms of the
parameters dn ,, , .
Therefore, all roots lyin on the circumference of a iven circle
centered on the oriin are associated with the same undampednatural frequency n . *iure #.4a illustrates this for two
different frequencies, $n and (n . *rom "#.-4%, we see that all
roots lyin on the same line passin throuh the oriin are
associated with the same dampin ratio "*iure #.4b%. The
limitin values of correspond to the imainary ais " 31%
and the neative real ais " 3$%. Joots lyin on a iven line
parallel to the real ais all ive the same damped natural
frequency "*iure #.4c%.
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*iure #.4 Graphical representation of the parameters dn ,, , and inthe comple plane. "a% Joots with the same natural frequency n lie on
the same circle. "b% Joots with the same dampin ratio lie on the same
line throuh the oriin. "c% &ll roots on a iven line parallel to the real
ais have the same damped natural frequency d . "d% &ll roots on a
iven line parallel to the imainary ais have the same time constant .
The root lyin the farthest to the riht is the dominant root.
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The dominant-root concept
)incen
$= , the distance from the root to the imainary ais
equals the reciprocal of the time constant for that root. &ll rootslyin on a iven vertical line have the same time constant, and
the reater the distance of the line from the imainary ais, the
smaller the time constant "*iure #.4d%H this leads to the
dominant root. *or a iven characteristic equation, this is the
root that lies the farthest to the riht in the s plane. Therefore, if
the system is stable, the dominant root is the root with the
larest time constant.
The dominant-root concept allows us to simplify the analysis ofa iven system by focusin on the root that plays the most
important role in the system's dynamics. *or eample, if the two
roots ares3 -20ands23 -2, the free response is of the form
tteAeAt&
(
(
(1
$%" +=
*or time measured in seconds, the first eponential has a time
constant (10$$ = sec, while the second eponential's timeconstant is (0$( = sec. The first eponential will have essentially
disappeared after 4/20sec, while the second eponential takes
about 4/2=2secto disappear. nless the constantAis very
much larer thanA2, after about 0.2sec, the solution is iven
approimately by
teAt&
(
(%"
The roots23 -2is the dominant root. The term in the free
response correspondin to the dominant root remains nonzero
loner than the other terms.
The usefulness of the dominant root in the precedin eample is
that it allows us to estimate the response time for the system.
This is the time constant of the dominant root. Obviously, the
reater the separation between the dominant root and the other
roots, the better the dominant-root approimation.
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*or a second-order system with two real roots, the dampin ratio
is a useful indicator of how much separation eists between
roots. Therefore is an indicator of the accuracy of the
dominant-root approimation for such systems. )uppose we
wish the dominant root to bes3 -b, where b is a specified
number. =rite the secondary root ass3 -nb, where n is the root
separation factor. To see how n depends on , write the
polynomial correspondin to these two roots as
1%$"%%"" ((
=+++=++ nbbsnsnbsbs "#.-6%
The dampin ratio is
n
n
nb
nb
(
$
(
%$"
(
+=
+= "#.-$1%
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