Chapter 3-1.doc

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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. &lternatively, 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


12/34

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


23/34

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.

186


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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


30/34

=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 " $


31/34

=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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