This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EE 429 EXAM II 18 April 2007
Last Name (Print): Sal. [£3 9 Q S First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO INSTRUCTIONS 1. You have 2 hours to complete this exam.
This is a closed book exam. You may use two 8.5” x 11” note sheets. Calculators are allowed. P9330 Solve each part of the problem in the space following the question. If you need more space, continue your solution
on the reverse side labeling the page with the question number; for example7 Problem 1.2 Continued. N 0
credit will be given to solutions that do not meet this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a
grade of ZERO will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise
and clear; your complete English sentences shOuld convey what you are doing. To receive credit, you must
show your work. Problem 1: (25 Points) 1. (13 points) Figure 1 shows the block diagram of a datasampled control system. ms) m9 Figure 1: Sampleddata system with command input R(s) and controlled output Y(s). o (3 points) Which dynamic block(s) in Figure 1 implicitly contain a ZOH DAC? 0 (10 points) Without performing any calculations, explain in a single sentence whether or not is possible to ﬁnd
the transfer function Y(z)/R(z). If the answer is yes, compute Y(z)/R(z). If the answer is no, ﬁnd an expression
for Y(z) that involves R(s). . Ms) an». 6(9 contain a... 20H 06c. because. Hw— MM: 5" 1"“ “wk;
are. SumPiQﬂ signals. o Because. this} ‘6 sqmphl bgcvm, if .3 El£er=ee0J we. Can Comp/LE \ﬂ‘vmﬁ).
Fnsf *FmL erprcuiom ‘Far' the. Sampler Mir{:5 auQ dash"? “P915 Ms): RC5) ~ am «(gang :7 n" = R“ are? =% M3 =R® ‘ “‘93”
3(5) = M5) 9‘15 — «(gang z; 3*: o“ 4“—c.’8*==9 3(a) — Ma) 91;)  era/5(a)
chi = so) may 1; Tm) = am» 5‘5) :; yaw,ma 13(2) El'mowte— BL 53 hum}. 8620 = 7(2)/6(%) : % = K _ _ ‘ H ate) = "6”) we)
A c 3 (13 6: Via) Am wow 61%) DH, 6m
Ehmonuk't (
_______,_______
a7; (.9 I + GL1) WE) =3 IE)... :1 H 6(a) 3TH?)
23 Y(%) ‘''
N. 63) out) 66%) Ni) breasts) 1’ en.) ‘ ____________________._.
“a 1 + 6(1.\ + Na 6H“) 2. (12 points) Consider a continuoustime plant with the transfer function representation 012(5) : where Td = 1.4 ln(4) sec. Using the method of residues, compute the zeroorder hold discretetime equivalent transfer
function representation using the sample period T = ln(4) sec. 141.: LHT' = nT+AT “:9 Z rnT's 2 lZe'MS}
e  ’ _ ~n _ = ——a—
so“) = g—E—L E z—‘ 235(wM] Z} 50*” I
2e. : C *C\
ZﬁmT‘J l* 2 re$\&s2$% Mac)?” arty?» 0
______~, = 
i 3(S'I'I') Po‘es+
Zemfi
M—
N244)
Co, I‘aSIuoJQ— “5506wb¢&4 Wu‘b“ 1:0 l 2.
2am» = ~1s
 ( a ’———_’—' ‘ﬂrﬂ ‘—¢
(1° " [KCMO he r0
—mT‘
c VeSlo‘uQ, assocm'lzﬂr “4th 7‘" ’11!  18
‘J = ‘T
)s Problem 2: (25 points)
1. (13 points) A discretetime system has the characteristic equation
Q(z) = z“ —1.223 + 0.07;:2 + 0.32 — 0.08 = 0. Using the Jury stability test, determine whether or not the discretetime system is BIBO stable. Show the steps used
to construct the Jury stability table, as well as the conditions examined for determining stability. '3 . 2° z' is" 23 2‘ 2° 5' 2" Z 2“
Go a1 qt. 03 01 “0.03 0.3 0.07 “L?— I l "'2 0.07 0.3 o.03 be b b!— 173 o_9736 LI7L 0.075'6 'O.207
b3 51 bu be O.2.o1 —o.075'6 "‘75 vo.¢r73e
00 GI C1. o'cmgelg 4473375 0.315‘0202
ac q 49 a.
“o L :
5°: ac C.“ L) = as [32: i l’3 \GH 43
Chi “0 a“ q) q\ q?—
bo b3 b0 bl c : be bl
O  = 2 b by
b: b0 b} l». J
Therm are, n+1 = 5' (gng'trmn‘ﬁo £0 fe5t GU] .. lI.Z+ 0.074(1300? : L‘l)“ 90'!) = IH‘L +0.07 ‘03 '0.03 :: [.870 > 0
laol<an => 0.03 4.1 \/
lbol3 lb“! =5 0.9V36 v 0.2.07 \/ leol> [Co2., =7 0.74:“: > 031502.07, / aféf'ﬁa‘é W syfkm I“ A) #‘ue ConSﬁrq/h'éj arO/ S 61(50 séaiaie. 2. (12 points) The natural response of a closedloop system is observed to have the form > o (3 points) Identify the mode in the natural response that dominates the transient response characteristics
explain your reasoning in one or two short sentences. 3k ynatUc) = c1(0.1)'° + 02 (0.5)k cos (3 where the coefﬁcients c1 and 62 are realvalued. o (9 points) Specify the characteristic equation of the closed—loop system in the form Q(Z)=Zn+awn—1zn'1+~+a1z+ao :0 by specifying the value of n and the coeﬂicients 04’.  ’39, mole ,and ab It tﬂh—G) ’00»er 153 relax to taro than the. ﬂick
' 719% sari—em .5 thing amoer allU» po‘e) rou‘iéeg 45$
17
(gin/3 ,‘ “i /3
1P1: OJ} 2?; ': 0.52, ’ 2P3 :. ﬁr; .= O.r2_
. ﬂ as
In feo‘LaHinmr’ 4:me pom; 2.pz any 2?; am, (gr—[arena
I17. 5/1, t 5
2P1“ 05°9‘7g'yg)*05'/5\(1E)=  J.. ['3'
2P3 ' 39’; ‘I a"?
UNIV. thQ. Dad: 32% a"; YaSU/ﬁs L’ E 21174— 1—
am: (2—ap.3(z—z,;)ce%,3\ = (330.0(2 , ; 7)( j —, \l (2—0.0(22' “12: + _.  ..L. 7 _.
 a 2% + {12.
“ME.1 +0445? 0‘ H1)
+03% — 0.02.5” Problem 3: (25 points) Consider the closedloop datasampled system in Figure 2 where the controller and plant transfer functions are S
Gp(s) : s + 1 and the sample rate is T = ln(2) sec. Figure 2: Sampleddata system using cascade compensation. 1. (9 points) Determine the loop transfer function B(z)/E(z) and specify the system type. 2. (8 points) For a unitstep—input r(k) = 110(k), estimate the time, in seconds7 for the system to reach steadystate as four
times the dominant time constant of the closed—loop system. 3. (8 points) Without calculating e(k), determine the steady—state error for the unitstep input. I. Kczu “ E“) am YCz) {gm/5(a) caves we “w “a 2. W2) _ arrange) , 527.5— r<c a =
2: '+6¢'*75p(*) Zo.s'+$aY.r =  1'11"“ ﬂ pola 65%» 676 6%  5.0 F
739m. nwtvrJZ, response. #— the Sachem ’us the. ﬁrm. c4676]. I’QQ. Gofre$ponj~uy 5,glunq, pale, “as a. fume, cooIEcw't‘ t = " I?" = ‘ LN“) = 3.802 Sec.)
‘ £954) +
all So the vstem wll rem}, SEQ/"$430669 I” “ °‘“
VI 1:. lfL sec. In response, '50 a. unit’sbpmﬂfé.
Eu) __ 1
Mt) I+Go69
._ 2 6.5
6%  3.0 swam is Sim“ ‘“
.3 v uQ. FTO" {wart 2. we, hm.» ﬁnd: "the. doSOQ—pavf ‘6
{he po‘e, ,afas mulL 1L; umﬁ. clrck' ‘ﬁhe. '9 theorem, 2 Qf (2 E( = Lag) 91%)
e“ 297 4) 4") 31/21 (*4) W55) = ,Q.:n(4)_i'2_'°_5L i = ‘~°5 f, 1.
a" 6%J‘.o 3/? 65“ 2 e55 =Ji Problem 4: (25 points) 1. (13 points) The forward difference rule approximates the area f m) = 'e(T)dT under the curve e(t) using samples e(kT) as z(k + 1) : TZeaT), 1120
where k = 0, 1, 2, . . . is an integer and T is the sample time. o (7 points) Show that
X(z) » T
E(z) _ z — 17 and hence the integration transfer function can be approximated as
1 . T
g = . ( m o (6 points) Provide a sketch that shows where the forward difference rule maps the lefthalfplane in the s—domain
into the z—domain. K'l
><(<+:) = 1 Zeccr) + T'e(KT) = x(K3+T2(¥) L’o The. 2—trqns'RJfm e‘F‘ the, «QU'E' OJfraﬁdrm y‘alﬂo 2 X(%‘) = x112) + T 5(a) :3) Ma) From abduction CD) aalbove. 5133, 0y. E=ST+I Doha that 5=0 maps to 2:0 9.) dXL) 4', the, SvPIQnQ, maps gn‘éo 6U point 4/0? QM royEegQ. .25 (29. 223 =1. % «.49. any pom'b “JV"W the Ir. (a)
IMCS) . (5.6%)
Recs) ’9 1. 9 “a «M Le? Poles cm W
ﬁvplaﬂom Outslbe the, um'l: arc/e. I.” ﬂu. 2. (12 points) In a system identiﬁcation experiment the measured plant response y(k) is estimated as
we) = «WM, where the 1—byn regressor vector ¢T(k) contains known measurement data and the nbyl parameter estimate 9 is
chosen to minimize the performance index J = Dye) we»? ( 0) Suppose that n = 2, N = 4, and W2) = (1 0)
W3) = (0 1)
$00 = (1 1)
while
11(2) = 1
31(3) = 2
21(4) = 0 Q4 points) Using the given data set, and without solving for é, how can you determine if a least—squares estimate
0 exists, and whether or not this estimate is unique? 0 (8 points) Determine the leastsquares estimate éLs. The. flararmpter‘ eatm'l'g, é; thGE Mlmrmzﬂé T saintsLe; 4.‘
'1 A A
( 2 Maﬁa») 9 = 2 (Magus) => ,9 e z b
K=Z K:]_ A Sb;
1? the. (hurtful. A 75 nonb'ﬂaulurj A. uﬂlbve, évL‘b’m ﬁx 6 ext '
IF F} is slrvaJuyJ by": lo 75 7a ‘U'Q Calumn {pace 2! '41 than
an m'l'nmi‘e 0vm‘3QJ‘ 04:— so(u‘broo.s' elus'é3 othermw—J 0... L5 Sakiw” does not axisb. In ‘U‘n‘s Prdble": 9 = ¢’(7—D¢TCZD + ¢(3)¢Th) +—¢M¢Tm
= l (I 03 o (o I) ‘ ( l l\
(‘3) + (l) 'l (I)
.. l 0 a l l = (7.. 1
ﬂ 0 :3 4’ <0 I) + (I I) I z 10 Because 49:4: (9) : :_ (290)—1600) = 3 75C; the, muffir\x )9 ZS mmsrryulay any WQ, Can 09 ‘1!“va 5 ob  gzﬁb, We dlreu‘py' have, 4‘) w‘n'g E 7.5 a
‘2. .1 ) ( ( J
b ﬁfty“) aaacz) +¢c3 5M3) + 05 ‘1); ‘I = mm mm +< = ( 11 TRANSFORM PAIRS Laplace Transform Time Function E (s) e(t)
% 710(t)
g1? tuo(t)
S—i—a 6““u0(t)
swim (1 — 6"”) U0“)
32(sa+a) (t — Lift“) U0“) 12 z—Transform z (aT—1+e‘“T)z+(1—e_aT—aTe““T)
a(z—1)2(z——e—a ) ...
View
Full
Document
This note was uploaded on 07/23/2008 for the course EE 482 taught by Professor Schiano during the Spring '08 term at Penn State.
 Spring '08
 SCHIANO

Click to edit the document details