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Date Created: 09/23/15
The Damped Compound Pendulum A Second Order System Course Objective Simulation using Working Modelquot Exponentially Decaying Sinusoid de ned by con and Z Series Nd dill QSWJS Natural Frequency rads Damping ratio Period sec Angle rad 9 chon 050 0 1 1n g2 i1n X1 2A 17 ll g2 N XN1 Zita 1g2 28 T Damped Compound Pendulum Equations of Motion quot i 3 i I x quot l lex 039 Linearized 2nd order differential C ge i E equation assumes small angles dt i i i l L Bar length m I 39 L i i d Pivot to CG distance m l m 9 i L mL Mass of pendulum kg i 90 J Moment of Inertia kgm2 Nms C Viscous damping coeffioient 2127 System Identification by Matching Coef cients Compare 1 and 3 2gwn wjt90 mLigd90 J J Yields 4A 48 Now can create a model for simulation Optical Encoders and Quadrature 39P w oder Incremental optical encoders generate two data signals that are electrically 90 out of hase With each other The term quadrature refers to this 90 phase relationship CCW AB CW AB From To From To 10 11 10 00 00 10 00 01 01 00 01 11 11 01 11 00 MEM 351 Dynamic Systems Lab Control Design 1 Poleplacement Recall OpenLoop Transfer Function Motorized Propeller HS Tor us 1 ms Km q ltsgt e 2 c L Volts Nm MmVolts s J s J radsec 21 um v i 5 i u II wana 14 Step oam Scope radrturdeg 2 vmt step mput R W mm a 45 Need for control Recall Pole Locations lttgt System Stability lm Lecture 2 If the system is A controllable then poles of the closedloop system may be placed at any desired location If our desired pole locations are represented by s1 52 sn then our desired characteristic equation 06d is adS51XS52quot39SSnO 1 ad252alsa020 2 How can we force this to be our system s characteristic equation Pole Placement Characteristic Equation OpenLoop System u2V 55 AozxBozu y ColxDolu Y gt ngzl il ai Ei X Aolx Bolu y Colx Dolu as dets1 A Feedback System using Pole Placement 4sz i ihig U r2V 56 Aolx Bolu y ColxDolu What is the characteristic equation of this Pole Placement Equations In standard state space form the characteristic equation is 3 C 2 Aolx Bolu 055 detsI A01 0 Our input is now uzr Kk 3 C 2 A01x Bozo Kx r2V u 55 Aolx Boil y y Colx Dolu Simplifying we have 32 A0 BOIKX 301739 The new characteristic equation is oats detSI A01 30110 0 asszalsa0O 3 Pole Placement Equations One then matches the coefficients in 2 with those of 3 to yield values for gains ad S2061S060 0 Kk1kn as s2 a1k2sa0k1 0 Pole Placement Control An Example Pendulum Desired Poles Suppose one wants a settling time of ts 167 sec and a damping ratio 4 0707 rs 4 mn This results in poles for the damped compound pendulum S jan l z SL2 s i Sim 24 i 124 Substituting the desired poles in 2 yields adG24j2 G24 j2 adzsz485lL52 m Desired characteristic equation System characteristic equation Pendulum State Space Realization Lecture 3 Motorized Propeller Ts Tor ue J Vs K q s e Volts m Nm 2 c ngd 39 NmNoIts S J S J radsec Compound Pendulum ngd T 6 then 9mL gd K J J 6 quotV Given J J Putting this in state space form yields open loop 0 1 J39Cl ngd 6 x1 0 u x2 7 7 x2 KmJ y1 0x10x16 x2 Pendulum Characteristic Equation The characteristic equation of the closed loop system is calculated from 0 1 0 r2V r u ring l lelu y 39 T 7 7 as s detlts1 A01 30110 0 which becomes s 1 detngdKmc1 SCKmk S J J Pole Placement Gains Setting the coefficients in equations 4 and 5 equal 05d 52 48S1152 a 2 S2 CKmkzjs nga Kmk1 J 1 yields 48J c CKmk2 48 k2 K ngdKmk121152 k121152J ngd Km Pole Placement Step Reepense 20 Volts in Compound Pendulum I I I I r quot 39 1E 39 quot H 1 391 1 1882 theta 184 deg Amplitude deg D l l l I l I 1 2 3 4 5 5 Time sec Slight overshoot and reduced settling time inline with desired response MEM 351 Dynamic Systems Lab Control Design 2 PID PID Controller Background More than 50 of the industrial controllers in use today utilize PID or modified PID control schemes Can be utilized as a control technique without knowledge of the plant s mathematical model Components W used to handle the present Integral handles the past by integrating the error over time Derivative handles the future by taking the first derivative over time PID Controller A PID controller in block diagram form xA01xBolu y yC01xDolu has the following transfer function KpsKi dez K GcsKp 1de s s ernor How it Works xA01xBolu y y C0lxDolu sensor Steps 1 999 Compute the error desired actual Multiply the error by proportional constant KIo Integrate the error and multiply by Ki Take the derivative of the error and multiply by Kd Sum together to form input u Tuning Parameters Parameters Rise Time Overshoot 8163229 88 Error P Decrease Increase small Decrease Change I Decrease Increase Increase Eliminate Small Small D Change Decrease Decrease Change Attendance Survey Results MEM 351 Dynamic Systems Lab Lecture 2 Transfer Functions Poles and Zeros Linearized 2nd order differential equation assumes small angles L Bar length m d Pivot to CG distance m mL Mass of pendulum kg J Moment of Inertia kgm2 Nms C Viscous damping coefficient w 2 What do we expect Q to be 9 2 wnt9 a 9 0 Frequency Domain Laplace Transforms 2ncl order clamped system 24 wn939 wig Laplace Transform 3293 ZgwnSQCS wig Yields 52 25am a 0 Complex roots 4 0 i 141 2 Nothing to control Torque Something We Can Control Voltage Vs applied to motor Propeller spins creating lift force Fs Lift on lever arm r creates torque Ts Pendulum angle defined by s 2ncl order damped system 171 8016 T J 9S 1 Laplace Transform TS S2 S mgd J Not actually controlling torque Voltage Something We Can Control Voltage Vs applied to motor Propeller spins creating lift force Fs Lift on lever arm r creates torque Ts Pendulum angle defined by s Motorized Propeller Ts Tor ue m Vs K q s e Volts m Nm 2 c ngd s NmVolts S J J radsec Compound Pendulum Calculating Constants Motorized Propeller Ts Tor ue J W8 K q s e Volts m Nm 2 C ngd s NmVolts s J J radsec Compound Pendulum K JL Theoretically can calculate lift force if have propeller pitch and m radius dimensions air density and motor angular velocity Experimentally apply known voltage V and pendulum will eventually reach steadystate Recall J c ngdsin6T At steadystate angular acceleration and velocity are zero The torque at this known voltage is calculated by T T 2 mL gd sm HSS And hence K 2 SS SS 111 V Propeller T T J Vs s orque m gt S 6 Volts himVolts Nm 5 J s LJ radsec Compound Pendulum OLTF 9 Km H d G010 I V S s2 C s ng Given J Km 2 0017 NmV d 0023 m J 00090 kgm2 mL 2 043 kg OpenLoop Transfer Function Motorized Laplace domain OL Transfer function C 000035 Nmsrad 3 lgl l Flle Edk View Smulatlon Funnah Tools Help D uglx gm yum g gmm Simulations ms olsmaesmsezj DATE39 cumsus Auw RE 9552 DL Step stpansa on damped enmpound pEndMum r5 dW demees 2 0 Vans Sup Input S u I n k Ready mayo ude 4 IEI 5 g quoti W V WH H H w W M W x Mva WW Ml 3 Huv uw HH M l J System Settling Time Simulation reveals long settling time This is consistent with the small damping ratio Poles of the characteristic equation reveal the large oscillations Recall from 1 egg 2 189 26015 Vs s 0039s1077 Roots of the denominator ie the poles are s1 00019 1328 Small real root will yield S2 o001932g long settling times Can be shown T V W L 1 1 1willV39i3 il i j 5x i ii l in lm Effect of System Poles 1m 39 Faster response X Unstable poles X Re L DH X More oscillatory Control Designer s Goal Side Note Create compensators that yield S12 2 o19Jr 1328 desired damping and rise time Find can and phase angle 6 In other words place poles where one wants them Homework 1 DUE TUESDAY 012207 Tedious Math Time domain differential equation 2ncl order damped system z w 26 0 n n Yields complex roots an i wny l 2 Time domain solution 60 t e ga quott A1 cosant 1 2 A2 sinant11 2 1 Small real rootwill yield long settling times Can be shown l l ni zli Evill l ill 2 lljigl li llij le MEM 351 Dynamic Systems Lab State Space Realizations Recall Last Week OpenLoop Transfer Function Motorized Propeller T T J Vs s orque m gt S e Volts himVolts le s Jis LJ radsec Compound Pendulum OLTF 9 Km H d G010 m Given J J Km 0017 NmV d 0023 m J 00090 kgm2 mL 2 043 kg Laplace domain OL Transfer function C 000035 Nmsrad Easier Math ll State Space Realization Motorized Propeller Ts Tor ue 1J Vs gt K q s e Volts m Nm 2 c ngd NmNoIts S J S J radsec Compound Pendulum ngd6 V 2 J J Suppose define two state variables One can rewrite 1 as X1 x2 d K xzz ng mijg x17mV lt3 State Space Realization continued State space form given by matrices 56 F Gu c 4 y H Ju Hence reexpressmg x1 x2 0 m d x2 x2 LJg x1 quot V Gives 33961 mogd 16 x1 0 u x2 LT 7 x2 KmJ 5 State space realization 4 transfer function 1 and differential equation are just the same representation of the realworld system As such they should all have the same characteristic equation ie same poles The characteristic equation from state space 4 is defined as 055 2 dets F 0 5 s 0 O 1 asdet 0 j ngd S J J r s 1 W 2 dct S 0039s 1077 7 1077 s0039 Which is the same as the E 189 G0 3 denominator given in 1 VS S2 0039S 1077 1 Duh That s why it s call the characteristic equation