chapter 3-2.doc

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3.5 Step response of a second-order model

The step function models any rapid change in the input from one

level to another. In this section, we treat the step response of

second-order systems.

Trial solution method

Consider the following model:

)(2

2

tukxdt

dxc

dt

xdm =++ (.!-")

where u(t)is the input and m, c, k, are constants. If this model is

sta#le, any constant input u will produce a steady-state response

such that $22

==dt

dx

dt

xdand

ukx = (.!-2)

%et u(t)#e a unit-step input, so that u(t)=$, t


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cossin)$( bBBadt

dx+=

or

kxB "

)$(sin = (.!-+)

)$("

,")$(-cosdt

dx

bkx

bk

aB += (.!-!)

These euations can #e solved forBand in a manner similar

to that used to find the free response. /or the case of 0ero initial

conditions, we o#tain

++

+= ")sin("

)(22

bteb

ba

ktx at (.!-1)

a

b=tan (.!-)

where is in the third uadrant (remem#er, we are assuming

a3$, #3$). In terms of the parameters and n, this solution#ecomes

++

= ")"sin(

"

"")( 2

2

tek

tx ntn (.!-4)

2"tan

= (.!-5)

where is in the third uadrant. If the initial conditions are not0ero, we can simply add the free response to the result.

/or an overdamped or critically damped system, the method is

essentially the same. 6 function of the form of the free response

is added to the steady-state. /or an overdamped case, this gives

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keAeAtx tsts "

)( 22"

" ++=

wheres1ands2are the distinct negative roots. /or 0ero initial

conditions, this gives

)"("

)( 2

2"

""

2"

2 +

= tsts ess

se

ss

s

ktx (.!-"$)

The result for the critically damped case is found in a similar

way. /or 0ero initial conditions, the unit-step response is

[ ] ")"(")( "" += ts

etsk

tx (.!-"")

The preceding solutions are for a unit-step input. If the

magnitude of the step were 7 instead, it is easy to see that the

resulting response can #e o#tained #y multiplying #y 7. This

result is due to the linearity properties of the differential

euation (.!-").

The step response is illustrated in /igure .5, with plot of the

normali0ed response varia#le kxas a function of the normali0edtime varia#le tn . The plot gives the response for several values

of the damping ratio, with n held constant. hen "> , the

response is sluggish and does not overshoot the steady-state

value. 6s is decreased, the speed of response increases. The

critically damped case "= is the case in which the steady-state

value is reached most uic8ly #ut without oscillation.

6s is decreased #elow ", the response overshoots and

oscillates a#out the final value. The smaller is, the larger theovershoot and the longer it ta8es for the oscillations to die out.

6s is decreased to 0ero (no damped), the oscillations never

die out.

&ecause the a'es of /igure .5 have #een normali0ed #y kandn , respectively, the plot shows only the variation in the

response as is varied, with kand n held constant3

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/igure .5 9ormali0ed unit step response of the model

)(2

2

tukxdt

dxc

dt

xdm =++

%et us see how each of the parameters affects the step response.

/igure ."$ shows what happens when and kare held

constant while n is changed. The effect of increasing n is to

speed up the response and ma8e the overshoot occur earlier./igure ."" shows two cases in which and n are held

constant while kis changed. The effect of increasing kis to

decrease the magnitude of the response.

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/igure ."$ nit-step response for ;$., k;", and varia#le n .

/igure ."" nit-step response for ;$., n;", and varia#le k.

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/igure ."2 nit step response for c;".+, k;", and varia#le .

/inally, /igure ."2 shows three cases for which cand kare

fi'ed while is varied #y changing m. The plot shows that as

is decreased, the response #ecomes more oscillatory, the

overshoot #ecomes larger and occurs later. This occurs eventhough the damping constant cis held fi'ed.

Transient-response specifications


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value of the response differs from some desired value, a steady-

state error e'ists.

/igure ." Transient performance specifications #ased on step

response.

The rise time trcan #e defined as the time reuired for the

output to rise from "$= to 5$= of its final value. >owever, no

agreement e'ists on this definition. *ometimes, the rise time is

ta8en to #e the time reuired for the response to reach the final

value for the first time. ?ther definitions are also in use. /inally,the delay time tdis the time reuired for the response to reach

!$= of its final value.

These parameters are relatively easy to o#tain from an

e'perimentally determined step-response plot. >owever, if they

are to #e determined in analytical form from a differential

euation model, the tas8 is difficult for models of order greater

than two. >ere, we o#tain e'pressions for these uantities from

the second-order step response given #y (.!-4)

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*etting the derivative of (.!-4) eual to 0ero gives e'pressions

for #oth the ma'imum overshoot and the pea8 time tp. 6fter

some trigonometric manipulation, the result is

$"sin"

" 22

=

= tekdt

dxn

tn n

/or


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percent ma'imum overshoot ;2

"A

ma'

"$$

"$$

=

=

e

ss

ss

x

xx

(.!-"1)

This is shown graphically in /igure ."+a. The normali0ed pea8

time pnt is plotted versus in /igure ."+#.

/igure ."+ Transient response specifications as functions of

damping ratio . (a) 7a'imum percent overshoot. (#)

9ormali0ed pea8 time pnt . (c) 9ormali0ed "$$= rise time n

tr.

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6nalytical e'pressions for the delay time, the rise time, and the

settling time are difficult to o#tain. /or the delay time, set

x=0.5xss=0.5/kin (.!-4) to o#tain

22 "!.$)"sin( =+ dt

tne dn (.!-")

where is given #y (.!-5). /or a given and n , tdcan #e

o#tained #y a numerical procedure, such as 9ewtonBs method.

This is easily done on a calculator especially if the following

straight-line appro'imation is used as a starting guess:

"$,.$"

+

n

dt

(.!-"4)

6 similar procedure can #e applied to find the rise time tr. In this

case, two euations must #e solved, one for the "$= time and

one for the 5$= time. The difference #etween these times is the

rise time. These calculations are made easier #y using the

following straight-line appro'imation

"$,!.24.$=5$,"$ +

n

rt (.!-"5)

If we choose the alternative definition of rise time-that is, the

first time at which the final value is crossed-then the solution for

tris easier to o#tain. *etx=xss=1/kin (.!-4) to o#tain

$)"sin( 2 =+ te n

tn

This implies that for 0, n=2, sin is in the third uadrant. Thus,

2="$$

"

2

=

n

rt (.!-2")

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where is given #y (.!-5). The rise time is inversely

proportional to the natural freuency n for a given value of .

6 plot of the normali0ed rise time n tr versus is given in

/igure ."+c.

In order to e'press the settling time in terms of the parameters

and n , we can use the fact that the e'ponential term in the

solution (.!-4) provides the envelopes of the oscillations. These

envelopes are found #y setting the sine term to " in (.!-4).

The magnitude of the difference #etween each envelope and the

final value 1/kis

2"

"

tne

k

&oth envelopes are within 2= of the final value when

$2.$" 2

te n

The 2= settling time can #e found from the preceding

e'pression. /or less than a#out $., tscan also #e

appro'imated #y noting that $2.$+ e and using the formula

n

st

+=2 (.!-22)

Thus, tsis appro'imately four time constants for less than

appro'imately $..

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Exampe !."

Typical parameter values for an armature-controlled C are:

#e=0.1, #$=2%.", &=0.0021', =0.!, c=0.0*, +=0.1

(a) /ind the differential euation relating the load speed to

the applied voltage .

(#) /ind n ,, and d for this system.

(c) 6 step input voltage of ;"$D is applied when the system is

initially at rest. Evaluate the step response.

*olution:

(a) The governing state euations for the state varia#les andiawere o#tained in Chapter 2 and converted to reduced form in

*ection .. ifferential euation (.-) is repeated here.

)()()(2

2

t,###c(dt

dc&(+

dt

d&+ $$e =++++

(.!-2)

/rom the given data, we o#tain

)(5.212$".2$+"!.$$$$2".$2

2

t,dtd

dtd =++ (.!-2+)

(#) Comparing the preceding with the left-hand side of the

general form (.!-") and using the appropriate formulas, we

o#tain5$2.$= 4.""=n radAs

2.+5=d radAs $$5.$= s

(c) *ince "


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55.$ radAs. Thus, the percent overshoot is appro'imately

$."+=.

The delay time can #e found appro'imately from (.!-"4). This

gives

$"+.$4.""

)5$2.$(.$"=

+=dt s

The delay time here is appro'imately !$= greater than the time

constant. The "$$= rise time is found from (.!-2") to #e

$!+.$".+5

!45.2="$$ =

=

rt s

The 2= settling time is

.+$+1.=2 == s-ts

The solution for the armature current iacan #e found #y using

the solution for )(t in (2.!-+), rewritten here as

$$21.$$$".$)(" +=+=dtdc

dtd+

#i

$

a

ifferentiate (.!-2!) to o#taindt

dand su#stitute in the

preceding euation to o#tain

2!.$)!45.2.+5sin(12.41

)!45.2.+5cos(42.+")(

."$2

."$2

++

+=

te

teti

t

t

a

(.!-21)

The solutions are plotted in /igure ."!. The pea8 current that

must #e supplied #y the power source can #e seen to #e "1.+56.

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/igure ."! *tep response of the C motor system of E'ample

.4.

Exampe !.10

/ind the current i(t)in the following&series circuit when a

step voltage is applied to the circuit.

6ssume that F;"$$, %;2.$> and C;2$ 0

*olution:

The mathematical model of the system is:

&.

/i

&.dt

di

&

(

dt

id=++

"

2

2

If we compare the preceding euation with the general second-order differential euation of:

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)(2

2

tukxdt

dxc

dt

xdm =++

then the natural freuency is given #y

"!4"$2$2

""1 ===

xx&.n >0

The damping ration&

.(

&.x

&(

+)A"(+

)A( 22

== ; $."1

*ince is less than ", the system is underdamped. The damped

oscillation freuency 2" = nd ; "!1>0

*ince i(0); $ and di/dt (0);$, the solution of the differential

euation is:

i(t) = e25.!t(c-s 15%t 0.1%sin 15%t)

Exampe !.11

Consider the following system.

The input, a torue $, is applied to a disc with a moment of

inertia+a#out the a'is of the shaft. The shaft is free to rotate at

the disc end #ut is fi'ed at its far end. The shaft rotation is

opposed #y the torsional stiffness of the shaft, an opposing

torue of -k occurring for an input rotation of - . kis a

constant. /rictional forces damp the rotation of the shaft and

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provide an opposing torue of cd -/dt, where cis a constant.

hat is the condition for this system to #e critically dampedG

*olution:

e first need to o#tain the differential euation of the system.

The net torue is

9et torue ; -- k

dt

dc$

The net torue is 2

2

dt

d+ -

, hence

2

2

dt

d+ -

; -

- kdt

dc$

or$k

dt

dc

dt

d+

-

--=++

2

2

The condition for critical damping is given when the damping

ratio "= . Comparing the a#ove differential euation with the

general form of the second-order differential euation, then

"+

22

==+k

c

Thus, for critical damping we must have +kc = .

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3.6 Ramp and sinusoidal response patterns

The step input is perhaps the most commonly used input, and we

have therefore devoted much attention to the step response. Two

other input types commonly encountered are the ramp and

sinusoidal inputs, which we now treat. ?ur purpose here: first,

to show how an understanding of the free response can #e used

along with the trail solution method to o#tain a uic8 estimate of

the general #ehavior pattern of the forced response and second,

to develop an understanding of the ramp and sinusoidal

responses for first-and second-order systems.

Trial solution method

The trial solution method used in *ections .2 and .! for step

response can #e generali0ed as follows. Ifx(t)is the forced

response to #e found, try a solution of the form

x(t); (free response form) H (particular solution form) (.1-")

where the free-response form has the same functional form as

the free response #ut with ar#itrary constants replacing theinitial conditions. The particular solution form is determined as

follows. If the order of the differential euation is n, compute all

the time derivatives of the given input function up to and

including the nth-order derivative or until the derivatives #ecome

0ero or #egin to repeat their functional form. The particular

solution form is the linear com#ination of the input function and

these derivatives. The total trial solutionx(t)given #y (.1-") is

then su#stituted into the differential euation, and thecoefficients of identical functional forms are compared. /orm

(.1-") is also used with the initial conditions to find nof the

constants. If only the forced response is desired, set the initial

conditions to 0ero. The preceding steps will generate as many

linear alge#raic euations to solve, as there are ar#itrary

constants.

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Ramp response of a first-order system

The ramp input is used to descri#e a start-up process for a

system initially at rest. The ramp response can #e used to

determine how will the systemBs output follows the start-up

command. Consider the first-order model (.2-") and assume

that b;" for simplicity. Thus,

,r3dt

d3+= (.1-2)

If the input (t)is a ramp function with a slope m, then

= mt (.1-)

The free response form isAert. &ecause (.1-2) is first order, we

need to compute only d/dt, which is a constant, d/dt;m. Thus,

the particular solution form isBHt. The total trial solution is

3(t) = Aert B t (.1-+)

*u#stituting this and (.1-) into (.1-2) givesrAert = rAert rB rt mt

Collect terms to o#tain

(rA rA)ert (r m)t rB =0

/or this euation to #e true, the coefficient of each functional

form must #e 0ero. Thus,

r m=0rB =0

Therefore, = m/randB =/r= m/r2. The constantAis found

#y setting t=0in (.1-+).

3(0) = A B

Thus,A = 3(0) B = 3(0) m/r2. To o#tain the forced response,

set3(0) =0.The solution is

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)"()(2

rter

mt3 rt = (.1-!)

/or sta#le case, A"=r and

)()( A tetmt3 += (.1-1)

9ote that we could have used a ramp input with a unit slope and

multiplied the resulting response #y the slope m to o#tain the

general solution (.1-1). This is shown in normali0ed form in

/igure ."1 along with the ramp input.

/igure ."1 Famp response of a first-order linear system. (a)

Famp input. (#) 9ormali0ed ramp input and ramp response.

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The response approaches a straight line, which passes through

the point ($, ) and is parallel to the normali0ed ramp input /m,

a unit ramp. In the steady-state condition, the normali0ed output

3/bm, at any given time, is less than the normali0ed input #y

an amount . 6lso, in this condition, the normali0ed output at

time teuals what the normali0ed input was at time (t- ). This

phenomenon is the steady-state lag time and another

interpretation of the time constant.

The time constant also indicates how long it ta8es for the

steady-state lag to #ecome esta#lished. The difference #etween

the normali0ed input and output is

)"()()( A

te

m

t3

m

t, =

/or this, it can #e seen that the steady-state difference is

(as t ). 6t t; , the difference is $.1 at t;+ , the

difference is $.54 .

Ramp response of a second-order system

The steady-state part of the solution (.1-1) could have #eeno#tained without specifying the initial conditions. %et us use this

o#servation to develop a uic8 way of determining the steady-

state ramp response of the second-order model (.-+) with a

ramp function, that is, let

ptt4 =)( (.1-)

The second-order and higher derivatives of this function are0ero, so the particular solution form isB1 B2t. 6ssume the

system is overdamped, with characteristic rootss1ands2. Then,

the total trial solution is

tBBeAeAtx ts

ts 2"2"2

")( +++= (.1-4)

*u#stituting this into (.-+) gives

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=

=+++++2

"

2"2

2)(

i

ts

iii pttkBkBcBeAkcsms i

&ecause the rootss1ands2satisfy the characteristic euation

(.-1), the preceding euation reduces to

cB2 kB1 (kB2p)t =0 (.1-5)

Thus,

kB2p=0

cB2 kB1=0

?r

k

pB =2

2" k

cpB =

The free response form is a transient, so the steady-state solution

is

)()(

k

ct

k

ptx = (.1-"$)

which passes through the pointx;$ at t=c/k. /rom this, we see

that if kdoes not eual ", the solution diverges from the ramp

input as t increases.

If the system is underdamped, the trial solution (.1-4) will #e

replaced with

tBBbtBetx at

2"

)sin()( +++= (.1-"")

and (.1-5) would still result. Thus, (.1-"$) is the steady-state

ramp response for all values of as long as the system is sta#le

( 3$).

The unit-ramp response is illustrated in /igure .". The steady

state is reached when the free-response form has disappeared.

This occurs after appro'imately four time constants, unless the

system has two real roots spaced closely together. 9o generalstatement can #e made in that case. 9ote that when k;", the

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term c/k(;c) plays the same role here as played in /igure

."1 that is, cis the steady-state output lag as well as the lag

time.

/igure ." nit ramp response of the second-order system:)(

2

2

t4kxdt

dxc

dt

xdm =++ . /or all cases, m;". The dominant time

constant ;"$, ", 2.", and $.! respectively, so all cases have

reached steady-state e'cept case ". The steady-state response is

parallel to the input ramp only when k;".

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Ta#le .2 summari0es the ramp responses for the second-order

modelBs three root types.

Ta#le .2 nit ramp response of the second-order model:

)(2

2

tdukxdt

dxc

dt

xdm =++

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Sinusoidal response of a first-order system


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6t +=t , the transient term has essentially disappeared, and the

steady-state response is

)sin()cos(sin)( +== tBtt.t3 (.1-"!)

where

"22 +

=

bA. (.1-"1)

$"22>

+

=

AbB (.1-")

= $,2A

$,$2A

tan b

b

(.1-"4)

e defineBto #e positive for convenience, and a#sor# the

reuired sin into . Thus, the steady-state response is sinusoidal

with the same freuency as the input, and its amplitude

decreases as increases. It is retarded in phase relative to the

input, and this retardation increases with the freuency. 6n

e'ample of this effect is shown in /igure ."4, where ;2, b;",

andA;". 9ote that the transient response also has a greatermagnitude at the lower freuency.

These results can #e used to find the response for related input

types. /or e'ample, if )sin( += tA, , then the phase angle is

simply increased #y . 6n important case in this category is the

cosine input, #ecause )2Asin(cos += tt .

The ramp and sine responses of the first-order model are

summari0ed in Ta#le ..

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/igure ."4 *inusoidal response of the first-order system: d3/dt

; - $.!3H (t)for (t) = sint. The steady-state is reached at

appro'imately t;4. (a) 2A= . The phase shift is 21."= and

the time shift is 4.$A == t . (#) 2= . >ere +5."= rad and

the time shift is -$.2+.

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Ta#le . Famp and sine response of the first-order model:

)(ta,3dt

d3=+

Exampe !.12

6 model of a vi#ration isolator is shown in /igure ."5. *uch a

device is used to isolate the motion of point " from the input

displacement32, which might #e produced #y a mechanism

driving a lightweight tool at point 2.


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/igure ."5 Di#ration isolator.

*olution

The sum of forces at point " must eual 0ero, since the mass at

this point is assumed to #e 0ero. Thus,

$)( ""2 =dt

d3c33#

or

2"" #3#3

dt

d3c +=

In terms of the notation for (.2-2),3=31, =32, b=#/c, and

;c/#. The amplitude reuirement can #e stated with the notation

in (.1-"!) as

B$."A;$.2

The reuirement is met e'actly ifB;$.2 at ;"$$ radAs. /rom

(.1-"), since here b;" andA;2,

2.$""$$

2

22=

+

=

B

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The result is ;$."s. The isolation reuirement is met for any#

and csuch that c/#;$."s.

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