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Linear Control Systems

by: Isaac Hauck

Linear Control Systems EEL 3657

Isaac Hauck
University of Central Florida
GPA 3.7


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Class Notes
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This 36 page Class Notes was uploaded by Isaac Hauck on Thursday October 22, 2015. The Class Notes belongs to EEL 3657 at University of Central Florida taught by Staff in Fall. Since its upload, it has received 49 views. For similar materials see /class/227671/eel-3657-university-of-central-florida in Electrical Engineering at University of Central Florida.

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Date Created: 10/22/15
The Design of Feedback Control System Objective I The design of compensators using root locus plots to achieve the desired system performance I Powerful lead and lag controllers are introduced and used in several design examples ltPerformance of a feedback control systemgt I Damping ratio natural frequency I Setting time overshoot I Steadystate error 9 Pole positions ltDesign of a Control System compensatorgt I The selection of a suitable components and parameters I Device may be an electric mechanical hydraulic pneumatic or some other type of device or network I Root locus methods for the case of splane alter and reshape the root locus 9 the roots of the system will lie in the desired position in the splane Compensator Process Ys Rs Gas I 39l Gs I l I I Hs I Cascade Compensation k TS zpk ltC0ntrol Des igngt w 7 2 2 x 2gmw K design constant Unit step input response 4 2 PO100e 4 V1 4 Ts CW7 ltCascade Compensationgt 39 Cascade with the unalterable process in order to provide a suitable loop transfer function G069 Gs Hs 39 Compensator Gcs is chosen to alter the shape of the root M KI SZl G S 2N l C N OCUS HSpj 39 Design judicious selection of the poles and zeros of the compensators ks z Exgt consider the firstorder compensator G s C s p 1 Design problem the selection of z p and k in order to provide a suitable performance 2 PhaseLead compensator when IZl lt IPI I If the pole is neglible that is pl gtgt z and zero occurred at the origin of the splane we have a differentiator so that 650 gs thus a compensator network is a differentiator type network 3 PhaseLag Compensator when IZI gtP I If the zero is negligible that is p ltlt z and pole occurred at the origin of the splane we would have an integrator Thus a compensator network is an integratortype network Z Gcsk Root Locus Design 0 Use root locus as a tool to design control system cascade compensators including Lead compensator Lag compensator Leadlag compensator Lead and Lag compensator Will be discussed brie y cascade com pe isator Rs Cs Lead compensator 52 31 G K 39 655 KHP 0ltzltp 5s N3 X 3 2 l o The zero near the origin can be used to improve the transient response 0 Usually speeds up the system response higher bandwidth 0 Little or no effect on steadystate response Example oflead compensator GcsKp Hsl GcsKS1 Hsl s4 l G 1 S ssls2 GPO m Root locus of system with lead compensator Root locus of uncompensated system Lag compensator 52 655K 0ltpltz G Ks2 O x P S HI 37271 Pole near or at origin improves the steadystate response of the system 0 Can destabilize some systems be careful Tends to slow down system response smaller bandwidth Example oflag compensator Gclts1ltp Hlts1 chFKHOIS s01 Hs1 G S 1 1 ss1s2 GPO SHDH2 Root locus of Root locus of system with lag uncompensated compensator system Very little effect on transient response but better steadystate response 51 GsKp Hs1 GcsK Hs1 1 s G S ss 1s 2 Gp S m Root locus of gad d3an thhat uncompensated eSta quot26M 6 2 1 2 1 Design of Feedback Control System using Root Locus 39 Lead and lag compensators controllers Rs G45 gt Gps liq Ys I IIs r Compensator Gcs is chosen to alter the shape of the root locus Kfi a GAS NN Hspj 9 Design judicious selection of the poles and zeros of the compensators ks Z Exgt the firstorder compensator Ge 2 S p 9 The selection of z p and k in order to provide a suitable performance kQz Go S S p 4 PhaseLead compensator when IZI ltP I If the pole is neglible that is I pl gtgt Izl and zero occurred K at the or1g1n of the splane GE S s 9 a differentiatortype 5 PhaseLag Compensator when IZI gtP I If the zero is negligible that is P ltlt IZI and pole occurred at the origin of the splane 9 an integratortype Z M g i N U 4 LI 9 Design Example of PhaseLead Compensator List the system speci cations and translate them into desired root location for the dominant roots Sketch the uncompensated root locus and determine whether the desired root locations can be realized with the uncompensated system If the compensator is necessary place the zero of the ohase lead network directly below the desired root location or to the left of the first two real poles Determine the pole location so that the total angel at the desired root location is 180 and therefore is on the compensated root locus Evaluate the total system gain at the desired root location and then calculate the error constant Repeat the steps if the error constants is not satisfactory ltLead compensatorgt k1 GS 2 S s z S p IPI gt Z GcS Required TS s 4 sec PO s 35 P0 100e 4 W1 42g 035 42 032 TS 4 s4 wnzl 6 cos1 o 5032 9713 1 Select a desired dominant root location 6tan39 2 64 gt g aw 64 045 2 Place the zero of Gcs at s Z 1 3 Angle criterion at desired root 2 zeros Z poles r180 r i1 i3 zero at 1 gt 90 a poles at 0 gt 116 9011161116 9p 480 6p38 4 Find the pole of Gcs tan 38 3 078 a a 256 the pole is located 256 left from 1 S p 356 GcS W M sp s356 GCG k1s1 77s2s356 2 k1 7 s s356 SH r 39 7 j Deaimd hae J Med root can gt v gt V 039 II I Dashudmolloca on b Uncompensawdmotlocns in El Desired 6 Addition 6f d 39Ima un of new pole i Design Example of PhaseLag Compensator Obatin the root locus for the uncompensated system Determine the transient performance speci cations for the system and locate suitable dominant root locations on the uncompensated root locus that will satisfy the speci cations Calculate the loop gain at the desired root location and thus the system error constant Compare the uncompensated error with the desired error constant and calculate the necessary increase that must result from the polezero ratio of the compensator 0t With the known ratio of polezero combinations of the compensators determine a suitable location of the pole and zero of the compensators so that the compensated root locus will still pass through the desired root locations Locate the poles and zeros near the origin of the splane in comparison to wn ltExamplegt Y A for input 2 s A ex limi KV lims HUSGQ Kv Hm K G S ss2 52 aw 22p W Required 4 045 kV 20 1 from uncompensated RL for 4045 roots should be located S12 ilij2 cos 1 045 64 2 calculate Kof S 1 i j2 K 7ss 2 5 57121 KV lims 2 5 since H0 5S2 gt do not satisfy KV 20 g UHE 1016 HOOK loans 9 uncnm sated 3 calculate x of Example 133539 5mm 7 Kvc0mp 7amp7 Kvunc0mp 25 2 P 8 4 select the pole and zero of Gcs near the on39gin set Z 01 a 7 01 p 8 700125 39 a 8 i P 01 G50 S s 0125 353311 lnmfgfs ofme I 7 5S 01 manna amuumemlfdma ma a G S 39 GU 333533 w 1 i39 3gm M sint Ices I d cos I smt dt 1 d SlIlt gt 05t phase lead h 1 cost gts1nt P age ag ltcos8 sin8 900 sin8 0058 7 900 3 2 G SKlts p 1 Phase lead lpl lel Numerator sz 1j gt 450 Denominator sp 10j gt 570 angGs 450 570 3930 phase lag z 2 lpl N sz10j gt570 D sp1j gt450 angGCs 3930 Stability Analysis BIBO stable if the output remains bounded with increasing time for every bounded input Stability in the splane Stable Neutral Unstable A A 1 l 1 8 lt O 05 gt 0 S S Pole Position amp SVstem Response 1w g l 1L 1 ulttgt e m gt smIn S2b2 gt b at e slnbt 5a2b2 Freguencx m amp attenuation Bel Characteristic Eguation Poles roots of Qs 9 stability LHS in the Laplace domain 9 decaying time function Y PS QS Ts 39 Denominator ofTF l GH O 9 Qs detsI A O i HS G 2 52 2Ks4 1GH 1 K 2 s6s82K s2s4 2 1GH21 K 2 Z s 6s82K s2s4 s2s4 Since Qs 1GH is a fraction numerator s26s82K should be zero in order to make Qs O Characteristics equations 1GH 0 Qs detsI A O n n l ans an1s 011s010 O G includes both process amp control R E Y m CE Qs 1GH s26s82K 0 Stable if all the poles roots of CE of the TF are in the LHS of splane 9 negative real part 6 s 6 jwd ltNecessary condition for LHS polesgt 1 N0 coef cient ai 0 2 A11 ai gt O Q Qs s2 0252 als2 a1sao s pslts p2gts p3gt 52 P1P2SP1P2S P3 53 Pl P2 P3 52 P1P2 P1P32 P2P3S P1P2P3 Extending this expansion to the nthorderpolynomial Q46 2 s an1s q1 a1s s0 an 1 2 negative of the sum of all roots an 2 sum of the products of all possible combinatio ns of roots takes two at a time an 3 2 negative of the sum of the products of all possible combinatio ns of roots taken thr ee at a time a0 1 multiplied by the prodcut of all the roots Routh Stabili Criterion necessary and suf cient o no changes in sign in the rst column 0 only evaluate the system stability Formulation for Routh Stability Criterion 1er Iran 57 I l 615 ast 9353 ast i a5 an L4 2nd nu at m a m P 15 53 as m 2 5 h E 3 cl 3 s a 95 ariynan 15 Lwalsao Sn an VIZ 1 4 Lt 6 n S art iauZan5 0177 5quot bl 72 b3 4 s4 5 3 5 51 CZ 3933 C4 is 5 l 7 S k 4 J I R39a S1 J J Z1 50 0 ml b1 v 1 an luv2 b7 1 an 0M4 a quot1 In 1 an a art l an l art 5 cl 3 an 1 In 3 C2l ail 1 an 39 bi iii 72 b1 b1 b3 xx xx xx ltcase 1gt N0 element in the rst column is zero Qss2s2s4s33s26s8 a3s3a2s2a1sa0 1i 1 16 1 1227 s23 i 2 2 2 8 1a3 6a1 Stable s 3a2 8a0 s1 b1103 s0 cl8 b i 1 1a2a3a1 a3a0a1a2 a2 a0 a 2 1 10 8 18 3 3 1 012 a2 a0 b1 0 1 cl b O kaz aob 1 gtXlt i 10 8 8 10 3 ltCase 2gt the rst element in a row is zero with at least one nonzero element in the same row Qss52s42s34s2 11s 10 S 1 2 11 b112 1 2 0 s 2 4 10 2 4 83 ME 1326 s2 c1128 c210 9 set bl 8 s d16 S 6100 b2 12 1 11 2 10 1 l 12 4 1 1 2 10 c 12 4 2 1 El E M i E C E 0 10 d 3 61 1ngs13 5 1 1251210 1z Ef 12 1 a 3 10 1 5 11 s4 X X Note that the values of the elements s3 X X circled the nal elements in the s2 X rows designated by the even powers s1 X ofs are always egual s0 9 ltCase 3gt All elements in a row are zero Q1SSZ1 Q2SS1SZ2S3SZZS2 3 2 s 12 51 11 s2 12 s 0 1 so s 0 s0 0 Auxiliary polynomial evenpolynomial factor Qa1ss2l Qa2s22 9 to continue the investigation replace the zero in an2S the row W1th the coef c1ent of ds 2 s3 12 s2 12 s1 2 s0 2 But in this case the sys is marginally stable ltcase 4gt repeated roots on the jw axis Qss5s4s32s2s1s1sj2s j2 1 1 1 1 2 1 0 0 gt Unstable mmmmmm OHNMBUI gt Same as case 3 check Qa before proceed the next step Qas s4 2s2 1s2 12 sjjjj 2 1 Find the steadysate error for a constant input of unity Since Hl l l l l R ess 11m S 1G s90 1Gss 1GO 2k G0 1m GcsGps 1m f k s gtO s gtOS 4S 5S2 2 Find range of K so that ess must be less than 2 percent of a constant input 1 l ess lt 1 G0 1 k 50 9 K must be greater than 49 The calculation of ess is based on the assumption of stability 9 we must ensure that the system is also stable for the range of K required The system characteristic equation is given by 2k 1G sG s 1 C p s34s25s2 or Qss34s25s22k0 The Routh array for this polynomial is then s3 1 5 2 4 22k 18 2k gt s1 T klt 9 s0 2 2k gt kgt 1 Therefore 1 lt K lt 9 K should be bigger than 49 in order to satisfy ess requirement but this choice Will make the system unstable In order to satisfy with both stability and ess requirements need a nonproportional simple gain controller 9 try a PI controller replace K with 19 for K 3 and nd K 1 range K 19 Ch apter 3 State Variable Models 1 System modeling using timedomain methods 2 Utilizing a nonunique set of variables 39 X1t X2t X110 function of time describe the future response if initial condition is kn own 3 In matrix form 10lnl al i 0W quot03 Dynmry V39 Sys Dynamics Ml b izky 0 de ne state variables x1y and X2yx1 x2y Then 1 M b 39y ky 211 k Ieryebimw X2 EHXZ 7 kx utl If output y X1 position change 2 State vector xAxBu D yCxDu For matrix equivalent Jocx gtnx1 u gt rxl y gtpx1 A gt7rgtlt7r B gt7rxr C gtpgtlt7r D gtpgtltr o in the above example x 2gtlt22gtlt1 2X11gtlt1 2X1 gt2gtlt1 gt2gtlt1 y 1X22gtlt1 WUm Solution of State Vector DE Consider lSt order DE C ax bu taking LT sXs x0 aXs bUs s aXs x0 bUs Xs amp LUs s a s a the inverse LT t xt eatx0 egg 01m Tdl39 0 Convolution of two time functions matrix form xl eA x0 jeAHburdr lt 22 kk Where 6 IAtA2f Ak39t Which converges for all nite t and any A X S1 A1x0 s A1bUs s s A 1 L 1ltIgtltsgt M eAI l Fundamental or state transition matrix Unforced response of the system From 2 equation 1 can be written as t xt tx0 j t IBurdr 0 Then the unforced sys u 0 W IX0 X1 11 in X10 X2 21 2n X20 xn nl nn xn O ZJ t i the response of the ith state variable due to an initial condition of the jth state variable Dynamic System 9 nth order DE 9 n of 1st order DE 9 matrix 9 LT Transfer Function of a SISO system x Ax 3 u input y output y Cx Taking LT sXs x0 AXs Bu s Ys CXs Then S AXs X0 Bus Xs s A 1x0sI A lBus CDsx0 CDsBus s Fundamental or state transition matrix With zero ICs Xs 2 s1 A 1Bus sBus System output Y CXs C13S BUS Therefore Y s TF Us CqgtsB CSI A IB System output for nonzero ICs Xs s X0 CDs Bus Ys CXs CCDsx0 CCDsBus in time domain yt L 1 Y ltExample Find TF amp System Output Responsegt Input ut Output v0 RiL From the circuit u 1395 Q v iC u z39 dv 1 1 0 dt L i u lt dt CL C d LLRiL vco diL 1 dt De ne x1 V6 x2 12 Then from 1 L y0 Rx Y 1 FindTF Gs SCsI AIB um I 1 1 0 0 E S E 1 0 I L 5 L H5 L L L L R 1 MALL 6 Ms L S L As detsI 7A s s jiiLgtltLj s2 sL L C Then Gs CsI 7A39IB 1e 1 1 1 1 7 i 0 1e A0 L S g L 1 R L 1 1e JR C AU 1 0 My LC 1e 1e 2 7 S LC LC x2 x1 1 squot 1 L LC L LC ltRecallgt 39 1 x17 x2 u C 39 1 R x2 x17 x2 L L y sz Mason s Gain Formula 1313 A11 L17 s 1 L27 s LC L LC PA 39G 1 1 S 17L1L2 2 Find the unforced sys response 11 0 Xs s1 A1x0 CDsx0 Given conditions x10 x20 1 C 05 L 1 R 3 R 1 0 q3113 q3123 1 Sf E q321 q3223 A0 L S L 33 1 1 s3 2 s23s2 sz232 1 s L S 31 q q 1 q q XSqsx0 11 12 11 12 S Sl2 21 22 1 Zl 22 s1s2 1 x1S S 2 1 9623 Xs 32 x0 I 1Xs 0 Complete System Response Xs 2 CDs x0 CDsBUs Ys CXs CCDsx0 CCDsBUs Tana Rare55 7mm ma 110 o


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