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ME37500FINALEXAMThursday,May5,2016
8:00–10:00amEE129
Section(circleone):Chiu(11:30am)Deng(1:30pm)Zhang(3:30m)
HWID:____________
Name:___________________________
Instructions(1) Thisisaclosedbookexamination,butyouareallowedthreesingle-sidedhand-written
8.5×11cribsheets.
(2) YoucanonlyusetheapprovedMEcalculatorfortheexam.(3) WriteyournameandHWIDonthetopofeachpage.
(4) Youmustclearlyidentifyyouranswerstoreceivecredit.InPartA,youmustshowallofyourworktoreceiveanycredit.Leavingmultiplesolutionsandderivationswithoutidentifyingtheonethatistobeconsideredwillnotreceiveanycredit.
(5) Youmustwriteneatlyandshouldusealogicalformattosolvetheproblems.Youareencouragedtoreally“think”abouttheproblemsbeforeyoustarttosolvethem.
(6) AtableofLaplacetransformpairsandpropertiesofLaplacetransformsisonthelastpageoftheexam.
PARTA PARTBProblem Score Problem Score
1.(20) 1.(10) 2.(30) 2.(10)
3.(30) 3.(15)
4.(20) 4.(10) 5.(15)
6.(10)
7.(10) 8.(10)
9.(10)
Total(A) /100 Total(B) /100
TOTAL(A+B) /200
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PART A Name: HWID:
Page 2 of 18
Problem1(20points)Considertheclosed-loopsysteminthefiguretotherightwithalooptransferfunctionL(s)oftheform
L(s)=K ⋅G(s) ,whereKisaconstantgain.TheBodediagramforG(s)(NOTincludingK)isgivenbelow
(A) (10points)UsingtheaboveinformationfindtheminimalvalueofgainKtoensurethe
magnitudeofthesteady-stateerror(e=r–y)foraunitstepdisturbance,d=1andzeroinput,r=0,islessthan0.001.
e = r − y = −y ⇒ E(s) = −Y(s) = − 11+ L(s)
⋅R(s)
eSS = lims→0
sE(s) = lims→0− s 1
1+ L(s)⋅R(s) = − lim
s→0s 1
1+ L(s)⋅ 1s= − lim
s→0
11+ L(s)
= − lims→0
11+ K ⋅ G(s)
= − 11+ K ⋅ G(0)
eSS = −1
1+ K ⋅ G(0)< 0.001 = 1
1000
From Bode plot: G(0) = 20dB = 10 ⇒ − 11+ K ⋅ G(0)
= 11+ 10K
< 0.001 = 11000
⇒ 1+ 10K > 1000 ⇒ 10> 999 ⇒ K > 99.9
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PART A Name: HWID:
Page 3 of 18
Problem1(continued)
(B) (10points)Ignoredisturbance,i.e.letd=0andletK=1.Whatisthemagnitudeofthesteady-stateerrorforinputr=sin(9t),i.e.asinusoidalinputwithfrequency9[rad/sec]andmagnitude1?(Hint:youcanapproximate1+L( jω) ≈ L( jω) .)
e = r − y ⇒ E(s) = 11+ L(s)
⋅R(s)
For r = sin(9t) ⇒ eSS(t) =1
1+ L( jω)ω=9⋅ sin(9t +φ)
eSS(t) =1
1+ L( jω)ω=9= 1
1+ L( jω)ω=9
≈ 1L( jω)
ω=9
= 1L( j 9)
From Bode plot: L( j 9) = 32dB = 39.8 or 40
⇒ eSS(t) =1
39.8≈ 0.025
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PART A Name: HWID:
Page 4 of 18
Problem2(30Points)
Giventheunityfeedbacksystemshowntotheright,theplanttransferfunctionG(s)isgivenby
G(s)= 50
s (s /5)+1( ) andthecontrollerC(s)=1.
(A) (10points)Thegaincross-overfrequencyis15.4[rad/sec].Calculatethephasemarginoftheclosed-loopsystem?
Gain cross-over frquency ω gc = 15.4 [rad/sec]
At ω gc, !G( jω gc) = 0− 90° −!( jω gc / 5+ 1) = 0− 90° − tan−115.4 5
1
⎛
⎝⎜
⎞
⎠⎟
= 0− 90° − 72° = −162°
Phase Margin PM = 180° +!G( jω gc) = 180° − 162° = 18° = 0.314 [rad]
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PART A Name: HWID:
Page 5 of 18
Problem2(Continue)
(B) (10points)Designaleadcompensatoroftheform
CL(s)=sω1
+1sω2
+1whereα =ω2
ω1>1
thatwillcreateaphaselead(φm)of55°atthegaincross-overfrequencyof15.4[rad/sec].Youwillneedtofindω1,ω2,andα.
For a lead compensator, amount of maximum phase lead φm is
φm = sin−1α − 1α + 1⎛⎝⎜
⎞⎠⎟ ⇒ α =
1+ sinφm1− sinφm
For φm = 55° ⇒ α =1+ sin55°1− sin55°
= 1+ 0.8191− 0.819
= 1.8190.181
= 10.06 ≈ 10
ωm =ω gc = 15.4=ω1 α ⇒ ω1 =ωmα= 15.4
10= 4.87
ω2 =α ⋅ω1 = 10 ⋅ω1 = 48.7
⇒α = 10,ω1 = 4.87 [rad/sec], and ω2 = 48.7 [rad/sec]
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PART A Name: HWID:
Page 6 of 18
Problem2(Continue)
(C) (10points)Withthecontrollerdesignedinpart(B),thenewclosed-loopsystem’sgaincross-overfrequencyisat40[rad/sec]withaphasemarginof50°.Iftheactualplanttransferfunctionincludedatime-delayofTD[sec],i.e.
G(s)=G(s) ⋅e−TDs = 50
s (s /5)+1( )⋅e−TDs
DeterminethelargesttimedelayTDtheclosed-loopsystemwouldremainstable.(Hint:considertheadditionalphaselagintroducedbythetimedelayandbesuretouseconsistentunitsforthephaseangles.)
Gain cross-over frequency ω gc = 40 [rad/sec]
Phase margin PM = 50°
To ensure stability, additional phase lab due to delay =TDω
need to be less than the phase margin at ω gc, i.e.
TDω gc < PM ⇒ 40 ⋅TD < 50° =50180
⋅ π = 0.873
⇒ TD <50π
40 ⋅ 180= 0.022 ⇒ TD < 0.022 [sec]
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PART A Name: HWID:
Page 7 of 18
Problem3(30Points)
Givenastate-spacerepresentationofasingleinputsingleoutputsystem:
!x = 0 1−6 −5
⎡
⎣⎢
⎤
⎦⎥x+ 01
⎡
⎣⎢
⎤
⎦⎥u
y = 1 2⎡⎣ ⎤⎦x
(A) (10points)Thedesiredclosed-loopsystemshouldhavea2%settlingtimelessthan1secondandapercentovershootlessthan4.32%.Onthecomplexplanebelow,clearlyidentify(intersectionwithreal-axisandtheappropriateangles)theregionforacceptableclosed-looppoles.
2% settling time: 4ζωn
< 1 ⇒ ζωn > 4
⇒ Real part of the roots < -4
%OS = 100e−ζ π
1−ζ 2 = 4.32%
⇒ζπ
1−ζ 2= − ln 4.32
100⎛⎝⎜
⎞⎠⎟
⇒ζ
1−ζ 2= 1 ⇒ ζ 2 = 1
2
⇒ ζ = 12= 0.707
⇒ sin−1ζ = 45°
45° -4
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PART A Name: HWID:
Page 8 of 18
Problem3(Continue)
(B) (10points)Forthesystem,
ifthedesiredclosed-looppolesistobe−5± j5 3 ,findthestatefeedbackgainvectorK,whereu=−K⋅x+r,andristhereferenceinput
A = 0 1−6 −5
⎡
⎣⎢
⎤
⎦⎥ , B = 01
⎡
⎣⎢
⎤
⎦⎥ , K = K1 K2⎡⎣
⎤⎦
A− BK = 0 1−6 −5
⎡
⎣⎢
⎤
⎦⎥−
01
⎡
⎣⎢
⎤
⎦⎥ K1 K2⎡⎣
⎤⎦ =
0 1−6− K1 −5− K2
⎡
⎣⎢
⎤
⎦⎥
det[sI−(A− BK)]= s2 +(5+ K2)s +(6+ K1) = 0
Desired CL Characteristic Eq: DCL(s) = (s + 5)2 +(5 3)2 = s2 + 10s + 100= 0
DCL(s) = det[sI−(A− BK)]= s2 + 10s + 100= s2 +(5+ K2)s +(6+ K1)
⇒5+ K2 = 106+ K1 = 100
⇒K2 = 5K1 = 94
⇒ K = 94 5⎡⎣⎤⎦
!x = 0 1−6 −5
⎡
⎣⎢
⎤
⎦⎥x+ 01
⎡
⎣⎢
⎤
⎦⎥u
y = 1 2⎡⎣ ⎤⎦x
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PART A Name: HWID:
Page 9 of 18
Problem3(Continue)
(C) (10points)Findthesteady-stategainoftheclosed-looptransferfunctionfromreferenceinputrtotheoutputyforthestate-feedbackcontrolyoufindinpart(B).(Hint:calculatesteady-stategainoftheclosed-looptransferfunction,youmaybeabletofinditinthestandardcontrollablecanonicalform)
Mathod 1:Closed-loop system:
!x = 0 1−6− K1 −5− K2
⎡
⎣⎢
⎤
⎦⎥x+ 0
1⎡
⎣⎢
⎤
⎦⎥r
= 0 1−100 −10
⎡
⎣⎢
⎤
⎦⎥x+ 01
⎡
⎣⎢
⎤
⎦⎥r
y = 1 2⎡⎣⎤⎦x
⇒ GCL(s) =2s + 1
s2 + 10s + 100⇒ Steady-state gain = GCL(0) =
1100
= 0.01
Mathod 2:Closed-loop system:
GCL(s) = C sI− A+ BK⎡⎣ ⎤⎦−1
B = 1 2⎡⎣⎤⎦Adj[sI− A+ BK]det[sI− A+ BK]
01
⎡
⎣⎢
⎤
⎦⎥
= 1 2⎡⎣⎤⎦
x 1x s
⎡
⎣⎢
⎤
⎦⎥
det[sI− A+ BK]01
⎡
⎣⎢
⎤
⎦⎥ =
2s + 1s2 + 10s + 100
⇒ GCL(s) =2s + 1
s2 + 10s + 100⇒ Steady-state gain = GCL(0) =
1100
= 0.01
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PART A Name: HWID:
Page 10 of 18
Problem4(20points)
Considerthefollowingfirstordersystemanditsstatespacerepresentation:
G(s)= Y(s)
U(s) =4s+2 ⇔
!x =−2 ⋅ x+4 ⋅uy = x
(A) (10points)Designastate-feedback+integralcontrol,i.e.
u(t)=−k ⋅ x+kI ⋅ (r(τ )− y(τ ))dτ0t∫ ,
sothattheresultingclosed-looppolesareat−5± j5 3 .FindkandkI.
!x = −2x + 4u
y = x
Let u = −Kx + KI xI and xI = (r − y)dτ∫⇒ !xI = r − y = r − x
Augmented syste:
!x!xI
⎡
⎣⎢
⎤
⎦⎥ = −2 0
−1 0⎡
⎣⎢
⎤
⎦⎥
xxI
⎡
⎣⎢
⎤
⎦⎥+ 4
0⎡
⎣⎢
⎤
⎦⎥ −K KI⎡⎣
⎤⎦
xxI
⎡
⎣⎢
⎤
⎦⎥+ 0
1⎡
⎣⎢
⎤
⎦⎥r
= −2 0−1 0
⎡
⎣⎢
⎤
⎦⎥+
−4K 4KI0 0
⎡
⎣⎢
⎤
⎦⎥
⎡
⎣⎢
⎤
⎦⎥
xxI
⎡
⎣⎢
⎤
⎦⎥+ 0
1⎡
⎣⎢
⎤
⎦⎥r
= −2− 4K 4KI−1 0
⎡
⎣⎢
⎤
⎦⎥
xxI
⎡
⎣⎢
⎤
⎦⎥+ 0
1⎡
⎣⎢
⎤
⎦⎥r
Closed-loop Characteristic Eq:
det(sI− AE) =s + 2+ 4K −4KI
1 s= s2 +(2+ 4K)s + 4KI
Desired CL Char. Eq. DCL(s) = (s + 5)2 +(5 3)2 = s2 + 10s + 100= s2 +(2+ 4K)s + 4KI
⇒ 2+ 4K = 104KI = 100
⇒ K = 2KI = 25
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PART A Name: HWID:
Page 11 of 18
Problem4(continued)
(B) (10points)Assumethatinpart(B)theclosed-looptransferfunctionbetweenthereferenceinputrandtheoutputyassociatedwiththestate-space+integralfeedbackcontrolis
GCL(s)=
Y(s)R(s) =
64s2+8s+64 .
IftheyouaretoimplementacontrollerC(s)intheblockdiagramtotherighttoachievethesameclosed-looptransferfunctionGCL(s),findthecontrollertransferfunctionC(s).
64s2 + 8s + 64
=C(s) 4
s + 21+ C(s) 4
s + 2
=4C(s)
(s + 2)+ 4C(s)
Solve for C(s)
⇒ 4(s2 + 8s + 64)C(s) = 64(s + 2)+ 256C(s)
⇒ 4s(s + 8)C(s)+ 256C(s) = 64(s + 2)+ 256C(s)
⇒ 4s(s + 8)C(s) = 64(s + 2)
⇒ C(s) =64(s + 2)4s(s + 8)
=16(s + 2)s(s + 8)
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PART B Name: HWID:
Page 12 of 18
Problem1(10points)–ZeroandstepresponseGiventhefollowingtwoclosed-looptransferfunctions
G1(s)=
(s+1)(s2+ s+1) and
G2(s)=(0.1s+1)(s2+ s+1) ,
whichonewillhavealargerovershootforastepinput?Brieflyexplainwhy?
G1(s) has larger overshoot
Both transfer functions have the same poles at - 12±- 3
2G1(s) have zero at -1 and G2(s) has zero at -10zero at -1 is closer to the poles, hence G1(s) will have larger overshoot.
Problem2(10points)–SensitivityGivenatransferfunction
T(s)= A
τ s2+ s+kA,
determinethesensitivityofthetransferfunctionT(s)tochangesintheparameterA,i.e.SAT(s)
SAT (s ) = AT (s )⋅ ∂T (s )∂A
= AT (s )
⋅ (τ s2 + s + AK) − AK
(τ s 2 + s + AK)2= A
A(τ s 2 + s + AK)
⋅ τ s2 + s
(τ s 2 + s + AK)2
= τ s2 + s
(τ s 2 + s + AK)=
s (τ s + 1)(τ s 2 + s + AK)
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PART B Name: HWID:
Page 13 of 18
Problem3(15points)–RouthstabilitytestConsiderthefollowingclosed-loopcharacteristicequation
1+K 1
s(s2+ s+1) =0
(A) (10points)UsetheRouthtesttodeterminetherangeofKforwhichtheclosed-loopsystemisstable.
Closed-loop Characteristic Eq.
DCL(s) = 1+ K1
s(s2 + s + 1)= 0 ⇒ s(s2 + s + 1)+ K = 0
⇒ s(s2 + s + 1)+ K = s3 + s2 + s + K = 0
Routh test:
s3 : 1 1s2 : 1 Ks : 1− K1: K
⇒ Stability ⇒ 1− K > 0K > 0
⇒ 0< K < 1
(B) (5points)Findthejω-axiscrossingpoleswhenKisatthestabilitylimit.
At statbility limit, substitute s = jω into CLCE
⇒ ( jω)3 +( jω)2 +( jω)+ K = 0
⇒ − jω 3 −ω 2 +( jω)+ K = 0
⇒Real part: K−ω 2 = 0
Imaginary part: ω −ω 3 = 0⇒ K =ω
2
ω(1−ω 2) = 0
⇒ K = 1ω = 1
⇒ jω-axis cross poles are ± j
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PART B Name: HWID:
Page 14 of 18
Problem4(10points)–RootlocusandPIDcontrolA positioning system with the following transfer function
1( )( 1)
G ss s
=+
is to be put under PID control. There can be 4 different PID controllers, namely, P control, PI control, PD control, and PID control. The following set of four Root Loci are the open-loop transfer function of G(s) with the 4 (P, PI, PD, PID) types of controllers.
You are asked to match the 4 different Root Loci below with their corresponding controllers by circle the appropriate controller: Root Locus Controller
(A) ⇔ P PI PD PID
(B) ⇔ P PI PD PID
(C) ⇔ P PI PD PID
(D) ⇔ P PI PD PID
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PART B Name: HWID:
Page 15 of 18
Problem5(10points)–DirectpoleplacementGiventhefollowingfeedbackcontrolsystem
UsedirectpoleplacementdeterminethepolynomialsD(s)andN(s)sothattheclosed-loopcharacteristicequationisDCL(s)=s3+4s2+5s+8=0
CLCE: s D(s)(s + 1)+ 2N(s) = s3 + 4s2 + 5s + 8
⇒D(s) = (s + a) 1st orderN(s) = (bs + c) 1st order
⇒ s(s + a)(s + 1)+ 2(bs + c) = s3 + 4s2 + 5s + 8
⇒ s3 +(1+ a)s2 +(a + 2b)s + 2c = s3 + 4s2 + 5s + 8
⇒1+ a = 4
a + 2b = 52c = 8
⇒a = 3b = 1c = 4
⇒ C(s) = s + 4s(s + 3)
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PART B Name: HWID:
Page 16 of 18
Problem6(15points)–State-spacerepresentationThefollowingthirdordersystemdifferentialequation:
!!!y(t)+4 !!y(t)+5 !y(t)+6 y(t)=3!!u(t)+2!u(t)+u(t) canberepresentedasthefollowingblockdiagramandstate-spacerepresentation.Fillintheemptyblocksandmatrixelements.Thestatevariablesarenotedintheblockdiagram.
and
!x1
!x2
!x3
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
=
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
x1
x2
x3
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
+
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
u
y = ⎡⎣⎢
⎤
⎦⎥
x1
x2
x3
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
Y(s)U(s)
= 3s2 + 2s + 1
s3 + 4s2 + 5s + 6
Let: X(s) = 1s3 + 4s2 + 5s + 6
U(s) ⇒ Y(s) = (3s2 + 2s + 1)X(s)
4
3
2
1
5
6
0 0 0 0
0 0 1
11
-4 -5 -6
1 2 3
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PART B Name: HWID:
Page 17 of 18
Problem7(10points)–ModulationanddemodulationAconstantvaluedsignalXDCundergoesamplitudemodulationusinga4kHzcarriersignalwithamplitude1.TheresultingAMsignaliseM(t)=XDC·sin(8000πt).Todemodulate,theAMsignalisfirstmultipliedbyitscarriersignal,resultingine’M(t)=XDC·sin2(8000πt).Whatarethefrequencycomponentsandtheassociatedmagnitudesofthesignale’M(t)?
e'M(t) = XDC ⋅ sin2(8000πt) = XDC ⋅12− 1
2cos(2⋅ 8000πt)⎡
⎣⎢⎤⎦⎥=
XDC2−
XDC2
cos(2⋅ 8000πt)
⇒DC component: 0 [rad/sec] magnitude: XDC
2
16000π [rad/sec] or 8000 Hz: magnitude: XDC2
Problem8(10points)–KarnaughmapThetruthtableofathreeinputoneoutputcombinationallogicisgivebythetruthtabletotherightwithA,B,andCastheinputsandYistheoutput.CompletetheKarnaughmapbelowtogenerateaminimalrepresenationofYasabinaryfunctionoftheinputsA,B,andC.
Y =B + AC + AC
00 01 11 10
0
1
ABC
A B C Y
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 10
0 1 1
1 1 1
1
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PART B Name: HWID:
Page 18 of 18
Problem9(10points)–GainmarginandphasemarginTheBodediagramofthelooptransferfunctionL(s)ofthefeedbacksystemtotherightisshownbelow.Identifythegainandphasecross-overfrequenciesandthecorrespondinggainmargin(GM)andphasemargin(PM).(YouneedtomarkontheBodediagramthetwocross-overfrequenciesandthecorrespondingmagnitudeandphaseinformationyouusedtocalculatetheGMandPM.)
Gain cross-over frequency ω gc = 0.4 [rad/sec]: PM = 180° +(−80°) = 100°
Phase cross-over frequency ωpc = 4 [rad/sec]: GM = 0−(−10dB) = 10dB= 3.16
~-80°
ωgc = 0.4 rad/s
~-10 dB
ωpc = 4 rad/s