Upload
eng-mohammed-kayed
View
223
Download
1
Embed Size (px)
Citation preview
8/13/2019 Chapter 3-3.doc
1/8
3.7 Response of higher order systems
The primary difference between first-order and second-order
linear models is that the characteristic roots of second-order
models can be complex numbers. This leads to types of behavior
not seen in first-order models and sometimes requires modified
algorithms in order to handle quantities with real and imaginary
parts. The second-order case contains all of the possible types of
behavior that can occur in linear systems. This is because a
characteristic equation with real coefficients must have roots
that fall into one of the three categories.
1. Real and distinct.2. Distinct but complex conugate pairs.
!. Repeated.
"nce we understand the behavior generated by each of the three
root cases# the superposition principle will allow us to apply the
results to a linear system of any order. $or example# the
response of a sixth-order system with a complex conugate pair
of roots# two real repeated roots# and two real distinct roots canbe analy%ed by combination of each response pattern will give
the total system response.
$or example# consider the third-order model
&'12
2
2!
!
! tfybdt
dyb
dt
ydb
dt
ydb o =+++ '!.(-1&
)ts characteristic equation is
*12
2
!
! =+++ obsbsbsb '!.(-2&
)f the three rootss1, s2# ands3are real and distinct# the free
response is an extension of the form given by '!.!-+
tststs eAeAeAty !21 !21&' ++= '!.(-!&
224
8/13/2019 Chapter 3-3.doc
2/8
)fs1ands2are complex conugates buts3is real# the free
response is an extension of '!.!-2+ namely#tsat eAbtBety !!&sin'&' ++=
'!.(-,&
wheres1= -a + ib# ands2= -a -ib. $inally# if all three roots are
repeated# then the extension of '!.!-!,& gives
tststs eAeAeAty !21 !21&' ++= '!.(-&
wheres1= s2= s3. These cover all the cases for the third-order
model. The forced response can be obtained using the trial
solution method using equation '!.-1 where the free-response
form is given by one of the preceding solutions# and the
particular solution form is obtained as outlined in /ection !..
The only situation requiring further algebraic manipulation is
the case# where any of the repeated roots are complex. This can
occur in only fourth-order models or higher. The free response
is an extension of the form given by '!.!-!, where there are
now two pairs of repeated roots#s1=s2= -a +ibands3=s4= -a
-ib. The free response is
tstststs etAeAetAeAty !!11 ,!21&' +++= '!.(-&
The algebra required to converty(t)into a useful form
containing sine functions is similar to the procedures used
following '!.!-1&.
The trial solution method can be used to solve a linear
differential equation of any order. 0owever# the algebra required
to solve for the solutions constants becomes tedious as the
systems order increases. $or example# the step response of a
third-order model requires four constants to be found# one from
the differential equation and three from the initial conditions
y'* dy/dt'* and d2y/dt2'*&. Two alternatives to trial solution
method are '1& the 3aplace transform method and '2&
simulation.
225
8/13/2019 Chapter 3-3.doc
3/8
3.8 The Routh-Hurwitz stability criterion
4erhaps the most important question to be answered by a
dynamic model concerns the stability of the system under study.
)f the proposed system is predicted to be unstable# the designer
immediately see5s to modify the system to ma5e it stable. This
will be especially true in the design of feedbac5 control systems.
Recall that the stability properties of a linear system are
determined by the roots of its characteristic equation. )f any root
lies in the right half of the complex plane# the system is
unstable. The Routh-0urwit% criterion tells us how many roots
lie in the right-half plane. )t sometimes gives enough
information to enable us to determine the location of all roots.
The second-order case
/ince the quadratic formula is relatively simple# it is not difficult
to use it to show that characteristic equation
*12
2 =++ obsbsb '!.+-1&
represents a stable system if and only if b2, b1, boall have thesame sign. )f bo6* and b2has the same sign as b1# the system is
neutrally stable# since one root iss6* and the other is negative.
)f b16* and bo/b27*# the system is again neutrally stable with
two purely imaginary roots. '"f course# if b26*# the system is no
longer second order.&
The third-order case
Third-order models occur quite frequently# and the Routh-0urwit% results are simply stated. 8onsider the equation
*12
2
!
! =+++ obsbsbsb '!.+-2&
$or simplicity# assume that we have normali%ed the equation so
that b37*. The system is stable if and only if all of the following
conditions are satisfied
226
8/13/2019 Chapter 3-3.doc
4/8
!21
12 **#*
bbbb
bbb
o
o
>
>>>
'!.+-!&
The neutrally stable cases occur if '!.+-!& is satisfied except for
the following1. bo6* ' a root exists ats6*&
2. b2=bo6* 'roots are 1#* bis = &
!. b1b2-bob36* 'roots are !21 9# bbbs =
Example 3.13
: certain system has the characteristic equation
*2.; 2!
=+++ Ksss '!.+-,&
$ind the range ofK values for which the system will be stable.
/olution
$rom '!.+-!& it is necessary and sufficient that
K7*# and ;'2& -K7*
Thus# the system is stable if and only if *
8/13/2019 Chapter 3-3.doc
5/8
$rom '!.+-! this occurs when
K-2,7*
!'2& - 'K-2,&7*
"r
2,228
8/13/2019 Chapter 3-3.doc
6/8
Problems
!.1 : tan5 has the model 1!"dt
d"A += # withA6 ! ft2and6
(** sec9ft2. /uppose that the height h is initially ft when the
inflow rate !1is suddenly turned "? at a constant value of *.*ft!9sec. $ind "(t).
!.2 8onsider the model
&'1*2 t#ydt
dy=+
'1& )fy($)=* and #(t)=1 'a unit step what is the steady-
state value of the outputyss@ 0ow long does it ta5e
before ;+A of the difference betweeny($)andyssis
eliminated@
'2& Repeat 'a& withy($)6 and #(t)=1.
'!& Repeat 'a& withy($)and #(t)62*.
!.! $ind the free response and the step response of the
following models. )f the solution is oscillatory# it should be
expressed in the form of a sine function with a phase shift.
'1& 1&*'#*&*'#2+,22
===++dt
dxxx
dt
dx
dt
xd
'2& 1&*'#*&*'#212+22
===++dt
dxxx
dt
dx
dt
xd
'!& 1&*'#*&*'#2,,22
===++dt
dxxx
dt
dx
dt
xd
!., 4ut the models in 4roblem !.! into state-variable form.
!. 8onvert the following models into reduced form in terms
of the variablex1. The input isf.
'1&fxx
dt
dx
xdt
dx
+=
=
212
2
21
-!
'2&
fxxdt
dx
xxdt
dx
+=
+=
212
2
211
-!
229
8/13/2019 Chapter 3-3.doc
7/8
!. Determine ## % and d for the dominant root in each of
the following sets of characteristic roots.
'1& s6 -2# -!i
'2& s6-!# -22i
!.( $ind ## % and d for the models given in 4roblem !.!.
!.+ Biven the model
*-2&2'2
2
=+++ &dt
dx&
dt
xd
'a& $ind the values of the parameter r for which the system is
'1& /table'2& ?eutrally stable
'!& Cnstable
'b& $or the stable case# for what values of r is the system
'1& Cnderdamped@
'2& "verdamped@
!.; : certain system is described by the model
fxdtdx'
dtxd =++ ,2
2
/et the value of the damping constant c so that both of the
following specifications are satisfied. Bive priority to the
overshoot specification. )f both cannot be satisfied# state the
reason.
1. aximum percent overshoot as small as possible and no
greater than 2*A.2. 1**A rise time as small as possible and no greater than !.
!.1* $ind the forced response of the following models
'1& teydt
dy
=+2
'2& 22 tydt
dy=+
!.11 $ind the steady-state part of the ramp response of thefollowing model# wheref(t)6t. :t steady-state# what is the
230
8/13/2019 Chapter 3-3.doc
8/8
difference betweenf(t)andx(t)@ :pproximately how long will
it ta5e to reach steady-state@
fxdt
dx
dt
xd=++
+2
2
!.11 : certain system has the characteristic equation
s(s+1) + K =$
>e desire that all the roots lie to the left of the lines= -bto
guarantee a time constant no longer than 61/b. Cse the Routh-
0urwit% criterion to determine the value ofKand so that bothroots meet this requirement.
!.12 Determine the range ofKvalues that will give a time
constant of less than 192 for the following characteristic
equation
s3+ 1$s2+ 31s + K =$
!.1! : mercury-in-glass thermometer has a time constant of
1*s. )f it is suddenly ta5en from being at 2* o8 and plunged into
hot water at +* o8# what will be the temperature indicated by the
thermometer after 'a& 1*s# 'b& 2*s@
231