Chapter 3-3.doc

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