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Distributed Control Systems

by: Nelle Braun DDS

Distributed Control Systems ECE 7750

Nelle Braun DDS
Utah State University
GPA 3.77


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This 46 page Class Notes was uploaded by Nelle Braun DDS on Wednesday October 28, 2015. The Class Notes belongs to ECE 7750 at Utah State University taught by Staff in Fall. Since its upload, it has received 38 views. For similar materials see /class/230397/ece-7750-utah-state-university in ELECTRICAL AND COMPUTER ENGINEERING at Utah State University.

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Date Created: 10/28/15
ECE7750 Distributed Control SystemsFlSPl Disturbance Propagation and packet loss in Vehicle Control Zhongmin Wang Center for SelfOrganizing and Intelligent Systems Dept ofElectrical and Computer Engineering Utsn Slaw University Feb 7 2005 lllnhSlale References I Pete Seiler Aniruddha Pant and Karl Hedrick Disturbance Propagation in Vehicle Strings IEEE Trans on Automatic Control Vol 49 No 10 Oct 2004 I Pete Seiler and Raja Sengupta Analysis of Communication Losses in Vehicle Control Problems Prceedings of the American Control Conference Arlington VA June 2001 14911496 llthme String Stability in Automated Highway SystemsAHS I The throughout of the tmnsportation network can be greatly increased if all the vehicles are controlled to follow its predecessor with a safe but tight dis tance at a desired velocity I The disturbance in preceding vehicles will propagate to following vehicles It can cause actuator s umtion in following vehicles and make the whole system unstable This problem is called string stability problem in platoon I String Stability problem falls into the category of networked controls sys tem problems if each vehicle can obtain the other vehicle s information by wireless communication I It is a example to show how to design a local controller to make the whole system sta e xmmmnous on AmoMAnc oomOL vol Av no In 0mm 2mm m 1 AH yum m 2 Con ediudbazklnvps v mny Having m sum smmxmu a mum swung polity madzl wnh mmda mnmuxdymlmcs 4 mm 3p my M p m n H m NM h39w polity m smuth mm a 5 nqmnd m mum a smug mahde ngn my hmgdamam sxyul u w mm m Lapin Tim can by u v m m wm 1m m mrpxupagmanmmgh w smug of mm Mam x n w Wm c m llthme Some research result If vehicles use only relative spacing information With its predecessor I It is Widely known that string stability can not be obtained if each vehicle Wants to maintain a constant distance behind their predecessors by any linear controller I If the spacing is proportional to the vehicle s velocity string stability may be achieved by linear controller I If each vehicle use different controllers the string stability may be achieved I It shows that ifthe pledece o1 39 leadel 39 can be used the string stability can be achieved Ilthme Error propagation using relative spacing Assumption I All the vehicles have same model Hs I Hs is linear SISO With two integmtors I All vehicles have the same control law I The desired spacing is constant I Initial position 9610 71396 I No extm input disturbance lllnhSlale Error propagation using relative spacing con t The spacing error dynamics are Ems mew sltsgtXoltsgt lt1 Ems HM TltsgtEHltsx lt2 1 H5Ks I 55 is the sensitivity function Ts is the complementary sensitivity inc tion There is a classjgmqmmeukm l wl and TQLJ small cm mam maskx swim Mk Mqumswwmmmuz kigummnm r 4 mm m 4 at W A my mum W a m is m lllnhSlale Error propagation using relative spacing con t There is atheorem that implies that there is a frequency w suchthat Tjw gt 1 Theorem 01 Assume that Hs is a rational tmnsferfunction with at least two poles at the origin If the associated feedback system is stable then the comple mentary sensitivity function must satisfy flmTUdew igto w I This integral relation is similar to the more common Bode sensitivity integml I Since Hs is strictly proper TQM gt O as w gt 00 and hence ln TQw lt O at high frequencies As a result theorem 01 implies that TQLJM gt 1 for some frequency w lllnhSlale Error propagation using relative spacing can t Example Assume 1 2 1 Hltsgt 77m L 52015 1 005s 1 Mm He FTTSEDETTEIM 093rads M WE Uam m mama m A Emmy sum M a smug m m m Mm lllnhSlale Following with predecessor and leader information To make the string stable with a linear controller We need to add leader informa tion in the controller The controller is designed as follows 16 U48 Kp8Ezs KzsXos X48 g This controller tries to keep the errors With respect to the preceding vehicle and With respect to the lead vehicle small The the error dynamics are l M 9605 Szp5X05 3 Home 1 H8Kp8 KM We can easily design Ks and Kps so that HTp lt 1 E45 E14 TgpsE11s 2 g i g N 4 lEIrlS Following with predecessor and leader information con t frequency content of propagating erroIs is at nuate e Spee 2 te ng Emu In the above 6mmM a ge l emmgiSalyo ieeeap605 Thus all E V f as quot quotquotquot D 1 7 em 15 7o 75 55 5 1n 5 20 25 7 Tune sec Fig 5 fu uwmgsuztw k 9 m i a h The u39rhszcingerrur 5 given by r 1 4x 1 a meme r142 vehmxe made m we see wme r142 spacing em dynamics or do meme as 10 whee we have de ned Analysis of Disturbance and Communication Losses in Vehicle Control Problems an introduction A wireless communication system is needed if the leader s information is neces sary We investigate the effect of mndom packet loss in the feedback loop due to a nonideal communication network Characteristics of communication network I The communication network is between Controller and System out I The packet loss is a mndom independent mndom process I Delay introduced I Assume no network jitter 12715 n 3 m 1v 1 i tions W 2 o a 2L 7 1 for n y i e N These restrictions just say that the probabilit r of jumping into a ukov chain must When 1 the piam use the following nocanion Am A 13m 13 and s i Fist we de ne several fozms of stability fox discrete time jump linear systems 10 De nition 1 The swam given by s with u 2 o 11 1 mean5111mm Mable M55 11 21190 IimhnmEllizleZI for every initial state 0 2 stuchlzstiwlly mm ss 1f for Every Initial mm man E 22 WHIP lt 00 3 manentially mean Iguana stable EMSS 1 for en ery inmal state 1060 there exists L onstanh 0 lt a lt 1 an gt 0 such man 3 u E Hmong lt awzou 4r ulmnst sumly stable if for mm mm mm 206 01limHm Mk n 1 It is shown in 10 that for System a the rst three de nitions of stability are actually equivalent and any one implies almost sure slabilit Furthermore the authors of 1a infer to the equivalenthqqgns of ms 55 and EMS A is co d n dynamic on that one fails to mist 1 k 1 uk I where 100 e Rquot 3915 script n is used to dim Again for 30 i 5 to denote the State nununuex We should n 39 mea de nitions given in Sect given by 7 A where the subscript cl 2 MQQ mm 1 quotmainquot quotinquot kw mm M 1quot mm Elven bV m1 n n and llthme System model The system model for each vehicle is 9576 1 AZ UC BuUc wk wk 112k 900106 1 9061206 1 chc is the communicated data 90 is aBernoqui process With P9k 0 pandP9k117p I When apacket is lost the controller Will use the latest received information I We assume the controller has knowledge of 905 I We only investigates a string With two vehicles This problem is investigated under the framework of discrete time Markovian Jump Linear SystemsMJLS A necessary condition to make sure the system is stable is given by a LMI formulation 14715 Questions 15715 Lecture Notes for ECEMAE 7360 Robust and Optimal Control Fall 2003 Jinsong Liang July 17 2003 What will we study 0 Optimal Control Theory The calculus of variations a little bit Solution of general optimization problems Optimal closed loop control LQR problem Pontryagin s minimum principle What will we study cont 0 RIOTS Introduction to numerical optimal control Introduction to RIOTS Background Usage and demo Warm up An Static Optimization Problem min LIu 1 E R u E Rm subject to fxu O f E R Define Hamiltonian function Hwu A Mm ATfxu A e Rquot Necessary conditions 8H 0 8A f 8H 2 L96 jg 0 81 811 Lufg Ao 8U What is an optimal control problem System model fzut 5105 E R ut E Rm Performance index cost function JuT gbzTT ALxtuttdt Final state constraint wTT 0 w 6 RP Some COSt fUI ICtiOI I examples 0 Minimum fuel problem J 2 AT u2tdt O o Minimum time problem T J 1dt to o Minimum energy problem J xTTgtRxT f Momma uTtRutdt O 5 A little bit of calculus of variations o Why calculus of variation We are dealing with a function of functions LutT called functional rather than a function of scalar variables LIu We need a new mathematical tool 0 What is 6106 a Relationship between dxT 61T and dT Relationship between dacT 623T and dT xt xh quot39 quot 39 dacT 633T abTdT dxT llAa 1quot L ide d xt xt dT T TdT Leibniz s rule T Jzt 2 Lo hmttdt T CL hxTTdT t h xt t6zdt 0 where hm Solution of the general optimization problem JuT IIITT ALwtuttdt 1 92315 f90ut 1305 E R HO 6 Rm 2 wTT 0 3 Using Lagrange multipliers Mt and 1 to join the constraints 2 and 3 to the performance index 1 T J wTTvaxTTt Man U tATtfx u t irdt 0 Note 1 constant Mt function Define the Hamiltonian function Hxut Lxut ATfxut then T J xTgtT uTwxTT t Hay m ATdrldt O Ugng Lennnzs nnetheincnyhentin J aszafuncUon ofincn rnentsin xpxu1h and tis df m wgv dw cm winWT wTTdv H ATirdtT 4 ft Hgax Hgdu ATM HA i2T6gtdt TO eHthate the vaHatkanin integrate by parts T T ATdirdtz gtT6zT A ATdacdt to 0 10 Remember dxT 61T iTdT so M m My MdelT cit thv HdtlT Jim T t Hx AT6x Hfau HA 22T6Aidt 0 According to the Lagrange theory the constrained minimum of J is attained at the unconstrained minimum of J This is achieved when dJ O for all independent increments in its arguments Setting to zero the coefficients of the independent increments du 612 6a and 6A yields following necessary conditions for a minimum 11 System model it fxut t Z to to fixed Cost function T mm xTgtT t Lltxutdt O Final state constraint wmm 0 State equation 8H m a f Costate adjoint equation 85 aT aL A fgt 81 5 12 Stationarity condition aH aL afT 8U 8U 8U Boundary transversality condition six wgv MTITdMT Cbt My HITdT 0 Note that in the boundary condition since dxT and dT are not independent we cannot simply set the coefficients of dxT and dT equal to zero If dxT 0 fixed final state or dT 0 fixed final time the boundary condition is simplified What if neither is equal to zero 13 An optimal control problem example temperature control in a room It is desired to heat a room using the least possible energy If 6t is the temperature in the room 6a the ambient air temperature outside a constant and ut the rate of heat supply to the room then the dynamics are 9 a0 0a bu for some constants a and b which depend on the room insulation and so on By defining the state as wet 606 6a we can write the state equation it az bu 14 In order to control the temperature on the fixed time interval 0T with the least supplied energy define the cost function as M gum g OT u2rtgtdt for some weighting 3 The Hamiltonian is H 1 gt ax bu The optimal control ut is determined by solving irzHAZ ax I bu 5 A Hx 2 LA 6 o Hu 2 u bk 7 15 From the stationarity condition 7 the optimal control is given by Mt bgtt 8 so to determine ut we need to only find the optimal costate At Substitute 8 into 5 yields the state costate equations d ax b2gt 9 A 2 ca 10 Sounds trivial Think about the boundary condition 16 From the boundary condition m My MTITdMT git thv HTdT 0 dT is zero dxT is free and there is no final state constraint So MT 2 3 371 2 s1T 10 So the boundary condition of 1 and A are specified at to and T respectively This is called two point boundary value TPBV problem Let s assume MT is known From 10 we have Mt 6 TtgtT 17 So i az b2gtTeaTt Solving the above ODE we have 5106 2 1Oeat ATeaTsmhat Now we have the second equation about xT and MT 1T xOeaT Exam e M 2a Assuming 120 2 00 MT can now be solved 20a5 ACT 2 2a b251 e 2aT 18 Now the costate equation becomes 1Oase t X t aeaT 5b2sinhaT Finally we obtain the optimal control 100LlseaIt aeaT 5b2sinhaT ut 2 Conclusion TPBV makes it hard to solve even for simple OCP problems In most cases we have to rely on numerical methods and dedicated OCP software package such as RIOTS 19 Closed loop optimal control LQR problem Problems with the optimal controller obtained so far 0 solutions are hard to compute o open loop For Linear Quadratic Regulation LQR problems a closed loop controller exists 20 System model i Atx Btu Objective function Jul 2 xTTSTxT AfaToam uTRtudt where ST and Q06 are symmetric and positive semidefinite weighting matrices Rt is symmetric and positive definite for all t e t0T We are assuming T is fixed and the final state MT is free State and costate equations d Act BR lBTA A QxATA 21 Control input ut R1BTgt Terminal condition MT STxT Considering the terminal condition let s assume that 1205 and Mt satisfy a linear relation for all t e t0T for some unknown matrix St Mt Stxt To find St differentiate the costate to get A 2 Sq Si 2 Sq 3mm BR1BTSac 22 Taking into account the costate equation we have 3x 2 ATS SA SBR lBTS on Since the above equation holds for all 106 we have the Riccati equation 3 2 ATS SA SBR lBTS Q t g T Now the optimal controller is given by w R1BTS06J06 and K06 R lBTS06x06 is called Kaman gain Note that solution of 806 does not require 1206 so K06 can be computed off line and stored 23 Pontryagin s Minimum Principle a bang bang control case study So far the solution to an optimal control problem depends on the stationarity condition 290 0 What if the control ut is constrained to lie in an admissible region which is usually defined by a requirement that its magnitude be less than a given value Pontryagin s Minimum Principle the Hamiltonian must be minimized over all admissible u for optimal values of the state and costate HIut g HIugtt for all admissible u 24 Example bang bang control of systems obeying Newton s laws System model CL1 1 1 2 CL1 2 U Objective function time optimal JuTOTldt Input constraints ut s 1 25 End point constraint womm lt 323 gt o The Hamiltonian is H 1 A1122 AQU where A A1A2T is the costate Costate equation A1 O gtgt1 constant A2 gt1 gt A205 is a linear function of t remember it Boundary condition A2TuT 1 26 Pontryagin s minimum principle requires that A3tut s A3tut How to make sure A tut is less or equal than A tut for any admissible ut Answer ut sgnAEt 8 What if A306 2 0 Then u is undetermined Since A306 is linear it changes sign at most once 50 does ut 27 Since rigorous derivation of ut is still a little bit complicated an intuitive method using phase plane will be shown below Going backward x1 O and x2 O in time from T with ut 1 or ut 1 we obtain a trajectories or switching curve because switching of control if any must occur on this curve Why 28 10 u1 Switching Curve 50 40 30 20 10 O 10 20 30 40 50 References F L Lewis and V L Syrmos Optimal Control John Wiley amp Sons Inc 1997 A E Bryson and Y C Ho Applied Optimal Control New York Hemisphere 1975 J T Betts Practical Methods for Optimal Control Using Non linear Programming SIAM 2001 I M Gelfand and S V Fomin Calculus of Variations New York Dover Publications 1991 30


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