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Unformatted text preview: CHAPTER 10 3—101. A simplifiedwa diagram
of the Bystan is shown to the right.
The transfer flmctim C(s)/R(s} is cm 3 .61”: 12(3) 1 4 (cl 4 (mantisI  G4" 36102. lsinpnﬂad block diagrams for the system an aim balav. C
“a ‘oll IIIQﬁm+G. +6Wﬂl R ' ¢ﬁd§+M) c “ﬁm+ﬁnﬁ*ﬁ&%+9ﬁ6*¢ﬁﬁ The ma: Win: C(s}/R(s) is 0‘53 _ G163w? + H1)
R(s
J 1 + GZHZ + GzGaﬂa‘ + G3”1"3 + G16253 + G16:3”1 B—lOB. — t:tnum1,den1): = [10]; == {1 2 0]: = tf‘num23den2);
series(sysl,syszi;
feedbackisysg.{lll Transfer function: The closed—loop transfer function obtained is cs3! = 53 + 10
N3) 32 + 73 + 10 13—10—4. A MATLAB program to »
solve this prob]. is sham to >>
the right. 30
The closed—loop transfer ,,
fmctim obtained is ' >>
>> >) C(s) 105 + 50 ,9
3(3) 53 + 83'2 + 375 + 50 >> >>
>> A MA'I‘IAB program to solve this problem is given below. "Hmw___m____._m_ﬁ— tftnum1,den1); 10; [1 3 2 0]:
tf(num2,den2);
seriestsysl,sy32);
[0.5 1); 1; tftnum3.den3): sys = feedbacktsysg,syah) Transfex function: 5‘3 + 8 5‘2 + 37 s + 50 munu— ————————p —— —————.———.—..._ .——.——__._. H_ ____ _  . B—IO—S. Define the input impedance and feedback impedance as Z and z ,
respectively. as sham in the figure below. ’ 1 2 'men Because each Operational amplifier of the system involves negative feedv
back. the differential_ input voltage of each amplifier is zero. or e' = 0
and e" = 0. The voltages 151(3) and £05) are given by 51(8) = ZINE) = 1211(5) 28
Renal
E‘s)  RZCZI + 1
M80:
‘0‘“ “.51
3(3) R3
Therefore.
E (a) R R +
he 2’328 lifetu 1,
E (3) R3. 12102: 121123 R2023 The control action is proportional plus integral. Talia2 +1:
Since a speed ombroller is usually designd lllzh that
1: a1 ha s elfa2 baak )1 the transfer function Y(s)/E(s) become ——.."3‘_————‘1:2 seesine 13106. If the engine speed increases. the sleeve of the flyban gamma: moves upward. his mt acts as the input; to the hydraulic cmtrqnsr.
A positive error signal (upward motion of the sleeve) causes the pane: pistm
tomvedomward. reducesthe fuel valve Opening. anddecreasestheengine‘
speed. Referring to Figure (a) shown below, a block diagram for the system
canbedrawnasslminﬂgure (b)mnextpage. Figure (I)  Figure (b) mus. the control action of this speed controller is proportimuplusmteg—
r‘al. 3107. For the first—order systeni 913i, 1 .
Obts) 'ra,+1 theateprosponsecurveisanmoxmtialcuno. SomtimIIamstantTm_ be determined from such an expmential curve easily. From Figure III98' the time constant '1‘ 13 2 a.
If this mm is placed in a bath. the taparature of which is increasing at a rate of lo'c/min = 1/6'0/3. or
' 1
9b ' 6 t + a more a is a constant. than the steadystate error can be determined as
follows: Noting that , ' _ = _. .ﬂEL
13(3) obts) 0(a) 613(5) [1 ob“)
_ . y= —29
Gb(s_)(1 4—25 + J) Obcs) 23 + 1
we obtain
e 11111 SH ) 11m 252
= s =
as 3+0 940 26 + 1 813(5)
where
= _L_1_. a _ l + 6!!
819(8) 6 92 + s 652
'marafore,
, ______28"’ ____1+6as=_2_.'_1_=_1_.
833 $210 25 + 1 652. 1 6 3 C Thus. the steadystate error is 1/3'C.
For a mamaorder. system: £14. 1 s =’
Gibbs) (115 + 1)(Tzs + 1) A typical response curve; when this thermoeta: is placed in a bath held at a constant temperatum, i ; .J_ J . . .r. 9.214. The closedloop transfer tumtic: or the system is C(s) 3 10(5 + 1)
R(B) 32 + 105 + 10 10(3 + J.)
(a + 1.1270)(s + 8.8730) For the unitstep input, we have at) g _ 10{s + 1) (a + 1.12701“ + 8.8730) 3
1 0.1455 1.1455 +
s a + 1.1270 ' a :i 8.8730 Mil—“mm 3:10—9. Since Hp is specifieé! as 0.05,; we have n
H a eJfxt = 0.05 P II 2.99.5
"‘5 t;ch = (2.99920 ;2)
Solving for the damping ratio 5 we obtain ' 01' Rani ting, 5 a 0.69
The settling‘tina 1.3 is specified as 2 seconds. So we hum
t. = 4 = 2 01'. “manprom . *Hﬂ ﬁmnnm mu—wnuu— qnm—  — ——.__ — n _—__‘m 310—10. The closed—loop transfer mum's of the system is 0(a) 100 Rts)' s3 4 232+ 10: + 100 100
F (s + 4.5815)(s  1.2907 + j4.4901)(s  1.2907 — 34.4901) This systa: is unstable became two music—conjugate closed—loop poles
are in the right half plane. '1!) visualize the unstable response. we my
enter the following mm program into the mater._ The resulting
unstable respcnse curve is shown below. . 'I'ommesystamstnbls. it isnsosmrytondmstMmnofthe
system or add an appropriate cmpsnsator. >> num [0 0 D 100];
>> den [1 2 10 100]: >> t = 0:0.01:6:
>> y = steptnum,den,tJ: >> plot{t,y) >> grid >> title(’Unit3tep Response of Unstable System'l
>> xlabe1{'t (seCJ') ' )) ylsba1('c{t}'l UnitStep Response of Unstable System 510—11. I C(B) 16 . Rte) .52 + (0.8 + 16105 4; 16 I Fran the chancteristic polyncmial, we find
(On=4. 2$wn=2xo.5x4=o.a+16k 'Hence
k=0.2
The rise tunaY tr ie'obtained from
. 7(_
Wu
Since . .
wdwn/152f4ho.25=3.46
41d 4::
= 11—=31n"10.866=—
[6 en ﬁn 3 .
wehave
..L.
tr=_72_ﬂ.c_=o.5053
3.46 '
mepeaktimetpisobtainedaa
' =£'=_3_°.1_4._=o,9075 (dd 3.46 _ Hp—e "3453
the settling time 115 is ta=__£_=__$___=25
_5wn 0.51:4 131042. The closedloop transfer function for the system 13
' cm) K 0.51! 3(3) 232 + a + Exhs + K 32 + 0.50. + lath): + 0.5K Fran this equation. we obtain can =J0.5K, 2 3'4)“ _ 0.5(1 + KKh) Since the damping ratio 5' is specified as 0.5.. we get 6d,, = 0.5(1 + xxh) 'I‘hérefore. we have 0.5(1 + ﬁfth] =1/o.sx The settling time is Specified as
4 4 16 t =——= 2
3 Scan 0.25(1+Rﬁh) 1+axh 52
Since the feede transfer {metAm 6(a) is
K
mm= 23+1 _L= K 1
1 +_ ER]: 3 23+1+I<Kh a
28+ 1
the static velocity error constant RV is
Kv=1imsG(s)=1ims K ...1__= K'
'9' 5+0 28414th 5 11“
This value‘ must be equal to' or greater than 50. Hanna,
' K
1+K'Kh a 50 Thus. the cmditions to be satisfied can be mined as follow: 0.5(1 + min) = 1051! (1) 16 $2 (2)
1 +KKh
I g} 50 (3}
1 +KKh o<xh<1 ‘4’ From Equations (1)_and (2).r we gwt B$1+KKh=JEI 32 IA
a: Prom Equation (3) we obtain K
"562141311 =‘/2K xgsooo Ifuectmaeli=500mthenweget 1+KKh=fZK=100
99 Kh = .= 0.0193 Ms,uadetem1mdaaetofva1uesofkandxhasfonm:
K = 5000. Rh = 0.0193
with these values of K and Rh. all spacificntims are satisﬁed. 3:10:91 c‘vs) . x (1 +133)
I 4“— RM le + xpa + Tda) Since Ms)  1/52, the output C(s) is obtained as K + K Ta: 1 1 1
as) _ J z 2 52 2 K Td 5
s + KpTda + KP a a + J2 s + J
Since the system is urﬂerdanped, C(s) can be written as
C(S) ' "12 "' T 2 l K T
_ 2.1 J a (=+%:d—)"+JJ 5:; )5 d
_  rd
’Ihe inverse Laplace transform of C(a) gives {I ‘
c(t)=ti 1 e’gﬁtsin IRlﬁa t‘
K 2T 2 J é—ﬁl— =_1_.
B2 The steady—state error a” for a. unit mp input. 1: as, a: 11m [rtt] ~ c(t)] = m: [t — c(t)] 13"" tMu
‘ 2 2
 _§j x 1'
=11. 1 a JJtain éiﬁto
t'” x xz'rz '
J “2 The steadystate error can also be obtained by use of the final value
theorem Since the error 31ml 3(3) is ' I 1 '1'
H6) = Rm  C(s) = Ma) [1  cm]: .35 1 . .E( + :1"
12(3) Jo + xp(1 4. 1d.) we obtain the studyauto error a“ as e“: 11: e(t)'z 11: 355(3) = 11:: TEE£5— : 0
15". soo soo 3 (Jo + xp 4 Tarps) WH‘— ham—unh— —  ——— — nu—ﬂnmu—II— .. —w——u—n n—“m 310—14. The characteristic 'equ'atioen is ——.—.—§—.——.‘—I+ 1 I: 0
3(3 + 1H: + 5) s3+632+55+x=0
‘meRoubh array for this equation is 1 5 6 x
30—1: 0 6 K Forthesystamtobestable.ﬂareshmndbenosignchangesintrnﬂrst
collm.‘1hiarcquires 30K>0. K>O
mnee.weg'ettherangeofgainxforstabilitytobe 30>K>0 31015. Since the system is of higher order (5th order), it is easier
to find the range of gain K for stability by first plotting the root loci
and then finding critical points (for stability) on the root loci. The open—loop transfer fmotim Gts) can be written as Mez + 2; + 4)
s(s.+ 4103+ tins:2 + 1.43 + 1) '
1((82 + 25 + 4)
s5 + 11.454 + 3933 + 43.632 + 245 Gts) = The MM'IAB program given next will generate in plot of the root loci for
the system. The resulting rootlocus plot is shown below the 1mm program . >> nun: [0 0 O 1 2 4] ; >> den [1 11.4 39 43.6 24 0] ; >> rlocustnum,den) >} v = [—B 2 —5 5}; axistv}; axist'equal'}
>> title ( ' Root—Locus Plot ’)  Based on thisplot. it can be seen that the systh is umditionally stable.
A11 critical points for stability lie on the jwaxie. To determine the crossing points of the root loci with the jwaxis
substitute a = ju) into the characteristic equation which is  35+11.4s4+3933+43.632+24s+K(sZ+25I4)=0 Then 1
(MN + 11.4(ij4 + 39(‘jw)3 + (43.5 + mum)? + (24 + 2mm +m=o This equation can be rewritten as [11.4504 — (43.5 + xij + 4K] + jth — 39:03 + (24 + now] = o By equating the real part and imaginary part equal to zero. respectively.
we obtain ' .  11.4w4—(43.6+K)w2+4l(=0 (1)
505539w3+(24+2xm=o (2) Equation (2) can be written as w =0
or _
w439w2+24+2x=o (3)
From Equation' (3) we obtain
K=.—w4'+:9w224 I m By substituting Equation (4) into Equation (1); we get
11.4 :04  [43.6 + 5(w4‘+ 39402 — 24mm? — 2:04.
+ 78.102  48 = 0
which can be simplified to
4:6 — 20.2414 +92.3a52 96 =0 The roots of this last equation can be easily obtained by use of. the ram
program as given below. >>a=[1 o 420.2 o 92.5 o
>>roots(a) ans 3 3.7553
v3.7553
2.1509
1.2130
2.1509
—1.2130 The rootlows branch in the upper half plane that goes to infinity crosses.
the jw axis at u; = 1.2130, w = 2.1509, and a): 3.7553. The gain values at these crossing points are obtained as follows: ——————————————————‘1'21304 + 39 x 131302 ‘ 24 = 15.61 for a) = 1 2130 K:
2 K  = 57.51 for w = 2.1509 2 = 163.55 mm: 3.7553 K = 2 Based on the K wlues above. we obtain the range of gain K for stability as
follows: The system is stame if  ‘ 15.61 > K > 0 163.56 > K > 67.51 _______m—n__————u——— ﬁn". 310—16. A mums program to plot the root loci and asymptotes {01' the
following system K G(s)H(s) = ————'—""_‘—"'—" 3(5 + 0.5)(52 + 0.63 + 10)
K s4 + 1.153 + 10.332 + 53 is given below and the resulting rootlocus plotsare sham on the next page.
Note that the agitation for tha W 1 K
(s + 0.275)4 Ga(s)Hh(s) = K
‘34 + 1.133 + 0.453852 + 0.08319: 4 0.005719 8' >> am [0 D O 0 1]; I >> den [1 1.1 10.3 5 0];
>> numa = [0 0 0_ 0 1]; >> dena = [1 1.1 0.4538 0.08319 0.005719];
>> subplottZZl); zlocusInum,den); '
>> v = [5 5 —5 5]; axislv); axist'equal')
3) subplot1222); :locusinum,den) >> hold Current plot held >> rlocu5{numa,dena]; >> v = {5 5 5 5]; axistv}; axisl'equal'}
>3» titlet'Plot _of Root 1.021 and Asynptotes') Root Locus 3—1017. The following mm program will generate ai rootlocus plot. Gfs) = R resulting.plot is shown below. >> num= [0
>> den  [1 0
7 >> zlocus {nun1, den] 2») v = [—4 2 3 PM of Root Loci and Asyrnplalns The openloop transfer.function 6(a) is m 2 '
s + 5 52(5 + 2) Kzts + 1)
s4 + 733 + 1032 The 3] ; axis [v] J axis {'equal') >> title ( 'RootLocus Plot: ' } Fran the plot we find that the critic'ai value of gain _K for stability
oorrespmdstotheerossingpointoftherootlocusbnnditMtgoesto
infinity and the imaginary axis. Hence, we first find the crossing
frequency and then find the corresponding gain value. The characteristic equation for this system is
s4+7s3+1052+2KB+ZK=0
By substituting a = 160 into the characteristic equation. we obtain
(J'wﬂ + 7(Jwi3 + 10(jw)2 + mij + 23 = 0
which can be rewritten as
(w4iow2+2x)+Jw(—7w2+2x) =0 By equating the real part and imaginary part of this last equation to
zero, reSpectively, we get w41.0w2+2ﬁ=0 (1)
cut7422+2K) =0 (2)
Equation (2) can be rewritten as
w = 0
or
—7w2+2x=o ‘ (3) By substituting Equation (3) into Equation (1); we rim!
6.)“  ma}2 4 “dz = 0 or
w4—3wzéo which yields #:0: (d‘o: 60:]; W"’J? SinoewJJS—is the crossing frequency with the jwazda. by substituting a) I: E into Equation (3) we obtain the critical value of. gain K for
stability. k=3.5w2= 3.5x 3=1o.5
Hence the stability range for K is m5>K>o ———_— —_———_———_m 510—13. The angle deficiency is 130' — 120' . 120" = — 60’ A lead compensator can contribute 60". Let us choose the zero of the
lead compensator at e = 1. Then, to obtain phase lead angle of 60':
the pole of the ompeneator must be located at a = 4. mos, 8+1 Ge“) = “Tr; 'I'he gain K can be detehnined from the magnitude oouditim. JR 5+1 1 ' _1
8+4 2 n
a s=—1+jf§ :8
+1
3 5=—1+j.l3
Heme the lead oomensator beams as follows: 8+1 Gc(5)=8 8+4 The feedfbrward transfer function is Bs+8
334452 The following me program will gamma a root10cm plot. The
resulting plot is Brawn below the new program. came) = >> num = [O 0
>> deaf [1 4 0 0];
a») rlocusinum,den) 2 4 4]; axis {v}; axisanual 'J Note that the closedloop transfer funotim‘is IC(s)  83 + 8 12(8) 53 + 452 + 83 + 8 The closedloop poles are located at B = 1 1 jﬁ and s = ~2. ———————.———.———_—____'_—________— _—...——.——.—..————...—__ 13—1019. The mm program giVen below generates a root—locus plot
for the given system. me resulting plot is sham below the program. >> num [0 0 1]:
>> den [1 5 4 0]: >> zlocustnum,den}
>> 1; = [6 4 5 5]; axis Iv}; axist'equal') Root LOO". Ram Note that mutantg points (0< 5(1) lie on a straight line having
angle 8 frcn the 1a) axis as sham in the figure below.
Fran the figure we obtain Noteuoothat§=0.61ineoanbedefimdby s a: —O.75a +ja where a is a miable (o (a (co). "no find the wine of K out that the
damping ratio 3 of the dcminant closed—loco poles is 0.6 can he foumd
by finding the intersection of the line a  43.753. + 3: ,
The intersection point can be determined by solving the following aim].
taneoue equations .for a. ' s = O.7Sa + ja (1) s{s + 1)(s + 41+ K = 0 (2)
By substituting Equation (1) into Equation (2).
(—0.75a + ja)(—O.75a + ja + 1)(70.75a + ja + 4) + K = 0
which can be rewritten as I (1.020103  2.1375212 — 3a + K} + 5(0.6075a3 — 7.52:2 4 4a) '= 0 By equating the red. part and imaginary part of this last equation to zero.
respectively. we obtain _ 1.8281a3 —2.1075a2 —3a 4 K = 0 .  (3)
0.63733 — 7.5.12 + 431 = 0 _ (4)
Equation (4) can be rewritten as
a = 0
0.687532 — 7.5a + 4 = 0
much can be writbm as a2 _ 10.9091a + 5.8182 = 0
or (a  0.5623)(a  10.3468) = D a = 0.5623 ' or. a = 10.3403
Fran Equation (3) we find at = 4.323133 + 2.107332 + 30. a 2.0535 for a = 0.5623 x = 1.8281a3 + 2.1875122 + 3: 3 4759.74 for a = 10.3468  since the R «an. is positive for a = 0.5523 Iandﬂnagultgivefor a = 10.3453,
we choose a a 0.5623. The required 90,111 K is 2.0535. since the characteristic equetiom with K = 2.0535 18 s(s + 1M5 + 4) ' + 2.0535 = 0
01' s3 + 552 +43 + 2.0535 s 0 the closedloop poles can be obtained by use of the following mm
program. >> p = {1 .5 4 2.0535];
>> rootsip} aﬁs =
"4.1565 —0.4217 + 0.5623i
—0.4217 — 0.56231 Thus. the cloudloop poles are located at
B = 0.4217 t j0.5623, a = 4.1565 The unit—step response of the systemwithx a 2.0535 can be obtained
by entering the following mama pct09m into the camber. The resulting
waitstep response curve is shown below the MATIAB program. ' >> num n [0 0 0 2.0535}:
>> den = [1 5 4 2.0535];
>> = 0:0.01I20; >> y  steplnum,den,t}; >> plotttryl >> grid >> titlet'UnitStep Response‘}
>> xlabell't {5ec)') >> ylabel('0utput'l UnitStep Response —————+n—.‘ "nw——m_wm—— 310—20. The closed—loop transfer function of the system 13 C(s) Ms + ms + b)
R(s) 5(52 + 1) + K(s + 8003 + b) = K(az+as+'bs+ab2 s3+s+K(52+as+bs+ab) Since the daninant closedloop poles need be located at s = 1 i jl, the
Claracteristic equation must be divisible by (s+1+j1)(s+lj1)=sz+2s+2
Hence
83+Ksz+(l+aK+ti{)s+abl{=(32+2s+2)(s+o() where s = D( is the unknown third pole. By dividing the left side
of this last equation by 52 + 23 + 2, we obtain  s3+Ksz+(1+aK+bK_)s+abK='(52+23+2)(s+K—2)
+(axaibK2K+3)s+xab—2(K—2)
Thereminder of divisionmustbezero. Hence we sat
aKihK2K‘+3=O
KabZtK2)=O since a is specified as 0.5, by substituting a = 0.5 into these two
equations; we obtain  bx = 1.5K — 3 (1) 0.5m: — 2(x — 2) = o (2)
By substituting Equation {1) into nation (2); we have o.5(1.5x ‘3) — 2(K  2) = o
01'
K = 2
men. by substituting K = 2 into Equation (1). we get
I 21: = 1.5 x 2 — 3 = 0 Hence
b=0 The PID controller: with K = 2 and b I 0 beacon; (+1 +1: I .
Gc(3)=K 5 a: J=Km—:—§E—=K(s+0.5) Thus. the controller beeanes a PI) controller. The openloop transfer
function bitcanes K(s + 0.5)
32 + l Gamma) = The closed—loop transfer fmctian (with b = 0 and K I 2) bananas as
follows: C(s) 2(5 + 05) 3‘5) '32 + 23 + 2 23+1
(s+1+j1)(s+1j1) mam—Implotforthedesigxwdsystmcanbeobtaimdbymberring
the following mmm into the when >> num = [0 1 0.5};
 >> den = [l 0 1]; >> rlocuatnum,denl
>> v = [2 1 1.5 1.5]: axistv); axist'equal'l The resulting root1m plot is 31m below. menal 15
1
3 Q5 E O =
E .05
4
45 a 4 O I ...
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This note was uploaded on 04/15/2008 for the course MECE 3338 taught by Professor Song during the Spring '08 term at University of Houston.
 Spring '08
 Song

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