大三下课程自控原理lesson293state space

1、9.3State-space Establishing ofLinear SystemGeneral Methodology:2.3.4.5.From Differential Equations of System From Transfer Functions of System From State-variable Diagram of SystemLinear Transformation of State space11.From Physics Mechanism of System9.3.1 From Physics Mechanism of SystemEx.9-2 Mech

2、anism system composed by mass, spring(弹簧) and damper(阻尼器) without gravity(重力).from Newtons lawkfd 2 ydym+ f+ ky = F(t)dtmdt 2y(t)in which, F(t) is Input, y(t) is Output.F(t)Then, if the original displacement and velocity are available, the systems solution of the certaininput is available as well.2S

3、elect the displacement and velocity as the state variablesx1 = y,= v = yx2Input is:u(t) = F(t)= x2ìx1ïState-equations:íïîfm+ 1 u m2mState space representation:éùéxé0ù01éx&ùùú = êf úêú + ê 1 úu11&#

4、234;kê-múm ûêúmëx&2 û-xë 2 ûëëû0éx1ùy = 1êx úë 2 û3Ex.9-3 The State-space representation of Mechanism without gravity, the input is the pull F, the outputs are the displacement y1 and y2.From Newtons first

5、Law, we have physicsrelationship of m1 and m2:= F(t) - k2 ( y2 - y1)= k2 ( y2 - y1 ) + f2 (k1m2 &y&2m1 &y&1f1m1select 4 independent state variables:y1 (t)x1 = y1 ,ìx1 = x3x2 = y2,xk2f 2ïxm2= xïï24y (t)k22=(x - x ) +F (t)321mm11k2t) -ï(4m422éêê&

6、#234;êêêêë0 ù0000k210f1 + f 2é0êùú1 úéx1 ùéx&1ùúêxê0êx&úúúúêk + k2f 2êú2 ú + ê0úF-2=1úêxêx&úúmmmmêú1 ú&

7、#234;1113 ú3 úê1êúúëx4 ûëx&4 ûk2k2f 2f 2ê m ú-úûë2 ûmmmm2222éx1ù0ùêxúBAé yùé10100úê2 úêú = ê10ûêx3 úë y2 ûë0êx

8、úë 4 ûC5J9.3State-space Establishing ofLinear SystemGeneral Methodology:1.From Physics Mechanism of System3.4.5.From Transfer Functions of System From State-variable Diagram of SystemLinear Transformation of State space72.From Differential Equations of System9.3.2From Differential Equ

9、ations of SystemØ Methodology：Establish the differential/difference equations by the physics mechanism of system;Establish the state equation focusing on the equations and a group of state-variables;Establish the output function based on the relationshipbetween systems outputs and states.8Ø

10、; State-variable Selectionü Selection of state variable is not unique.ü Methodology:F 1. Select variable in the initial conditions or related.F 2. Select characteristic variable of independent storage components (energy or information) with certain physical meaning, such as the electric cu

11、rrent i of inductance, thevoltage uc of capacitor, and velocity v of mass, etc.9Scenario (1): No derivatives(微分) of input u contained in n-order linear differential equationsAssume the dynamic process of the SISO control system is described as follow:y(n) + a y(n-1)+L+ ay& + ay = bun-11ny(n) , y

12、(n-1) , y, yDerivatives of outputInputuIf initial conditions y(0),y(0),y(n-1)(0) of output and the input u(t) of t0 are known, the behavior of system at anytime can be obtained.10ìx1 = yïx= y&ïíM2Select state variables:ïx= y ( n-2)ïn-1ïx= y ( n-1)în= x2= x

13、ìx&1ïx&ïíM23then:ïx&= xïn-1nïîx&n = -an x1- an-1 x2-La1 xn + bu y ( n ) + a y ( n -1)1+ L + an -1 y& + an y = bu therefore:11State space:x = Ax + Bu y = Cxin which, the matrices:éêêêê00M010M001M0LL00ùú

14、3; x1 ùé0ùêúê0úê x2 úê M úúê úêM úx =Mú,A =,B =ê úúêM úêëbúûL1êúêúëxn ûê- a-a ú-a-aLë1 ûn-1n-2nC=100012Ø Then, draw

15、 the following block diagram (state variablediagram) among the state variables.ü The output of each integrator corresponds to each state variable.ü The state equations are decided by the relationship of I/O.ü The output equation is on output part.u13y (n) + a y (n-1) + ay + a y = bu1n

16、 -1nbEx.9-4assume the differential equation of the systems dynamic&y&& + 6 &y& + 11 y& + 6 y = 6uprocess isin which, u and y are input and output.Try to find the state space representation of the system.=y& , x3=&y&,Select state-variables:x1= x2y, x2ìx&1&

17、#239;x&= xí23ïx= -6+ 6u&î 33é x1 ù0 ù é x1 ùé0ùé 010-11ê xú = ê 01 ú ê x ú + ê0ú uê2 úú ê2 úêêë-6ê úêë6úûStandard Form:x = Ax +

18、 Bu y = Cxêë x3 úû-6úû êë x3 úûé x1 ùBA0 0 ê x ú2y = 1êúêë x3 úûC14Scenario(2): Derivatives of input u contained in n-order linear differential equation systemn-order linear differential equation repr

19、esentation:y(n-1) +L+ an-1 y& + an y = bo u(n) + b1 u(n-1) +L+ bn-1u& + bnuy(n) + a1Reference scenario(1):ìx = x12ïx= xï23ïí= xxïn-1nïx= -a y - ay - a y(n-1) + b u(n) + b u(n-1)+ bu + b un-1n-1nn1o1nï= -a x - ax - a x+ b u(n) + b u(n-1)+ bu + b uï

20、în-12n-1n11no1nHowever, the derivative of input u is still contained instate equation, which is INCONSEQUENCE(不合理).15ìx1 = yïx= y&ï 2íMïx= y ( n-2)ï n-1ïîx= y ( n-1)nAnalysis: if input u is a limitary Step signal: Step Function, uwill be the Impulsive

21、 Function , u(i)(i=2,3,) will be the higher-orderimpulsefunction,andstatetrajectorywillhaveinfinitejump at t0.Hence, we cannot choose the output y and its derivatives to be state variables of the system. Such group of state variables cannot decide the future state of the system based on the known sy

22、stem input and original state conditions.The principle of state variable selection:No derivative of the input/operation function could be included in any differential function in the system state equations representedby first-order differential equation sets.16Select the state variables:ìx = y

23、- b uì10= y - b0u - b1u = x1 - b1uïïx2íïx= y - b u - b u - b u = x - b uí301222ïîïïîxn- b0u- b1u- bn-2u - bn-1u = xn-1 - bn-1u= y(n-1)(n-1)(n-2)-= x2 + b1uìx1ïx= x + b uï232ïtheníx= x + buïïn-1n-1nï= ?

24、9;î xn?How to find the relationship between xn and other states: x1,x2, xn-117? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?n-1 stateequationsThe input is contained inthe state variablesSolution: xn=f(x1,x2,xn-1,u)- b- b u-L- bu& - b= y(n-1)(n-(n-2)1)state equation of x :xuun-2n-1nn01= y(n) - b

25、u(n) - b u(n-1)01-L- bn-2u&& - bn-1u&derivative of xn: differential equation: and then:substitute y(n) into xn:x&ny(n) + a y(n-1)+ + ay + a y = b u(n)+ bu(n-1)+ bu + b un-1n-11no1n= -a y(n-1)- - ay - a y + b u(n)+ bu(n-1)+ bu + b uy(n)n-1n-11no1n- b- b- b= y(n)(n-(n)1)xuuun-1n01= (-a

26、 y(n-1)y(n-2)+ b u( n-1)- a- ay - a y + b u(n)+ bu + b u)n-1n-112n01n- bu- b u- b(n-(n)1)un-101= -a y(n-1)y(n-2)- a- ay - a yn-112n+ (b - b+ (b - b- b(n-+ (b)u + b u)u(n)1)un-1n-10011n18ìx1 = y - b0uïx= y - b u - b u201ïíx3 = y - b0u - b1u - b2uïïstate variables:ïx

27、= y(n-1)- b(n- b u(n-2)- bu - b-1)uuî nn-2n-101ì y = x1 + b0uï y = x + b u + b u201ïderivatives of y:í y = x3 + b0u + b1u + b2uïï(n-ï= x + b(n-+ b u(n-2)+ bu + b+1)1)yuuîn-2n-1n01substitute output y and it derivatives: y,.y(n-1) into xn:19xn = -an x1- an-

28、1 x2- a2 xn-1- a1 xn+(b - b+ (b - b - a b+ (b - b- a b - a b)u(n-1)u(n-2)+)u(n)0+(b a1bn-1 - a2 bn-2- an-1b1- an b0 )uPrinciple:No derivative of input u(t) contained in state equations.ìb0 = b0ïb= b - a b1110ïthus:íb2ïï= b2 - a1b1 - a2 b0ïîbn- a1

29、bn-1- an-1b1- an b0= bn-20State-space of the system is:= x2 + b1uìx1ïx= x + b u232ïíïx= x + buïn-1n-1- an-1 x2na1 xn + bnu= -an x1-ïîxnx = Ax + BuRewrite the system to the matrix representation: y = Cx + DuIn which:éb1ùé0ùú10M0-an-101M

30、0-an-20LL00é xùêb ú2ê01úêêúêúê x2 úê M úêx úA = êMB= êMúx =MúêMúê0úLL1êêë- anúêúêëb n úûë n û-a1 úûC = 1D

31、 = b0= b00021Ex.9-5 Assume the dynamic equation of a control system canbe written as the differential equation:&y&& + 6 &y& + 11y& + 2 y = 11u& + 6utry to give its state space description.Solution: compare with the standard differential equationy(n) + a y(n-1)+ bu(n-1)+ a

32、y + a y = b u(n)+ bu + b un-1n-11noa3 = 2,1n= 6,= 0,a2 = 11,b1 = 0,a1b0we have:b2 = 11,b3 = 6b0= b0 = 0= b1 - a1b0 = 0b1and the coefficients:b= b - a b - a b= 11221120b3= b3 - a1b2 - a2 b1 - a3 b0= -60x1 = y x2 = x1x3 = x2 -11u22State equations are:x = x2x2 = x3x3 =+11u-2 x23State space description

33、of matrix:éx&1 ùé00 ùé10êx&ú = ê01úêê2 úêêë- 2úêêëx&3 úû- 11- 6úûêëé x1ù00ê xúy = 1ê2 úêë x3 úû23The standard for

34、m:y(n) (t) + a y(n-1) (t) + ay(t) + a y(t)n-11n= b u(n) + b u(n-1)+ bu '+ b u(t)n-101nwhile if b0=0, we can select another group of state variable as follow:xn = y= xi +1+ an-i y - bn-iui = 1, 2,3, n -1xi= xn + a1 y - b1u= xn-1- a1xn + b1uxn-1xnn-2 = xn-1+ a2 y - b2uxn-1 = xn-2- a2 xn + b2uxx1 ?

35、x3 = x2x2 = x1- an-2 xn- an-1xn+ bn-2ux2 = x3 + an-2 y - bn-2ux1 = x2 + an-1 y - bn-1u+ bn-1u24y = xnOutput equation:Furthermore:= xn + a1 y - b1u= y + a1 y - b1uxn-1xn-2 = xn-1+ a2 y - b2u= y + a1 y - b1u + a2 y - b2ux2 = x3 + an-2 y - bn-2u= y(n-2)+ a y( n-3)- b u( n-3)+ ay( n-4)- b u( n-4)+ ay -

36、bun-2n-2112225x1 = x2 + an-1 y - bn-1u= y(n-1) + a y( n-2) - b u( n-2) + a y(n-3) - b u( n-3) + ay - bu1122n-1n-1calculate derivate of x1bring y(n) into x1 according to:y(n) (t) + a y(n-1) (t) + bu(n-1)+ ay(t) + ay(t) = b u(n)+ bu + b un-1n-11n01n26x&1 = -an xn + bn ux = y(n) + a y(n-1) - bu(n-1

37、) + a y(n-2) - b u(n-2) + ay - bu11122n-1n-1x1 = x2 + an-1 y - bn-1u= y(n-1) + a y(n-2) - b u(n-2) + a y( n-3) - b u( n-3) + ay - bu1122n-1n-1x = Ax + buy = Cx + duMatrix description:-ané0ù001000é bnùê1-an-1 úêúêA = ê0êêêë0C = 0&#

38、250;êbn -1 úb =-an-2 úêúúúúûêúë b1û-a100110d = 027Ex.9-5(II) Differential equation of control system is&y&& + 6 &y& + 11y& + 2 y = 11u& + 6utry to give its state space description.Solution: standard differen

39、tial equationy(n-1) +L+ an-1 y& + an y = bo u(n) + b1 u(n-1) +L+ bn-1u& + bnuy(n) + a1a1 = 6,b0 = 0,a2 = 11,b1 = 0,a3 = 2,b2 = 11,b3 = 6select state variable28x3 = yx2 = x3 + a1 y - b1u x1 = x2 + a2 y - b2ux3 = yx2 = x3 - 6 yx1 = x2 -11y +11uState space description:-2 ù é x1 ù

40、é x1 ùé0é 6 ù0ê x ú = ê10-11ú ê x ú + ê11ú uê2 úêêë0ú ê2 úêúêë 0 úûêë x3 úû-6 úû êë x3 úû1é x1 ù1ê x

41、2 úy = 00êúêë x3 úû: for a certain system, selection of state variables is not unique.29Ex.9-6The equations of a 2input/2output 2nd-order system are:y1 + a1 y1 + a2 y2 = b1u1 +y2 + a3 y2 + a4 y1 = b4u2try to give its state space description.Solution: find derivativ

42、e of y1, y2 with highest-ordery1 = -a1 y1 + b1u1 - a2 y2 + b2u1 y2 = -a3 y2 - a4 y1 + b4u2calculate their integrationy1 = òò(-a1 y1 + b1u1 )+= ò (-a1+ b1u1 )d1=( a1+ b1u1 )1y2 = ò(-a3 y2 - a4 y130x1 = y1x2 = y2Select state variables:From the equation of y1:= -a1 y1 + b1u1x&1=

43、 -a1 x1b1u1Select another state variable:x3 = ò (-a2 x2 + b2u1+ b3u= -a2 x2 + b2u1 + b3u2x&3andFrom the equation of y2:= -a3 x2 - a4 x1 + b4u2x&2x1 = -ax1+ x3 + b1u1The equation set:=+ b4u2+ b3u2a43 x22u12x& 31= -a2312x1 = -ax1+ x3 + b1u1=+ b4u2+ b3u2a43 x22u12x& 31= -a22Rewrite

44、 the equations by the matrixes:éx&1 ùéêêêë01ú- ax&a0ê2 ú43x&3 úû- a2The output matrix equation:éùx1é yùé1ù010êx ú1=ê y úê00úê2 úë 2 ûëû

45、34;xúë 3 û3233J1q1 = Mc + MD - k(q1 -q2 ) - b(q1 -q2 )J2q2= k(q1 -q2 ) + b(q1 -q2 )349.3State-space Establishing ofLinear SystemGeneral Methodology:1.2.From Physics Mechanism of SystemFrom Differential Equations of System4.5.From State-variable Diagram of SystemLinear Transformation o

46、f State space353.From Transfer Functions of System9.3.3 From Transfer Functions of SystemTransfer FunctionsState Space1. Transfer Functions to State Space+ b sn-1b sn+Lbs + bY (s)= 01n-1n sn + a sn-1+L+ as + aU (s)n-11nY (s)U (s)E(s)Y (s)U (s)36b sn + b sn-1 +Lbs + b01n-1n1sn + a sn-1 +L+ as + a1n-1

47、nb sn + b sn-1 +Lbs + b 01n-1n sn + a sn-1 +L+ as + a1n-1nE(s)Y (s)U(s)+ a sn-1U (s) = (sn+ as + a )E(s)n-11n+ b sn-1Y (s) = (b sn+bs + b )E(s)n-1= e(t)= e(t)01nìx1 = x2ìx1ïxïx= xïï223Select state variables:Mííïïïxïx= e(n-1) (t)= xî n&

48、#238; n-1nu = xn + a1xn+ a2 xn-1+ an-1x2+ an x1y = b0 xn + b1xn+ b2 xn-1+ bn-1x2+ bn x137b sn + b sn-1 +Lbs + b01n-1n1sn + a sn-1 +L+ as + a1n-1nu = xn + a1xn+ a2 xn-1+ an-1x2+ an x1y = b0 xn+ b1xn10+ b2 xn-101+ bn-1x2+ bn x1é x1 ùé0ù é x1 ù00é0ùê x ú

49、;ê0ú ê x úê0úê2 ú = êú ê 2 ú + ê ú uêúêú êúê úê x úê-a-a ú ê xúê1 úB-a-aë n ûën1 û ë n ûë ûn-1n-2Aé x1 &#

50、249;é x1 ùCê x úê x ú)êúbLb )ê2 ú2y = b (-a- aL- a+ b u + (bê M úê M úê x ún-1n-10n10n1ê x úDën ûën ûIf b0=0,the output equation will be simplified.38For the situation that the derivative of

51、 input u is included in the differential equations.Homework: Comparing with P24Ex.9-7 Transfer function of a control system is:+ 4s + 1s2Y (s)=s3+ 9s2 + 8sU (s)transform it to the state space representation.Solution:a1 = 9,a2 = 8,a3 = 0,b0 = 0,b1 = 1,b2 = 4,= 1b3Thestateequationis:10- 8é x&

52、1 ù0ùé x1ùé0é0ùê x&ú = ê01úê x ú + ê0úuê2 úêêë0úê2 úê úêë1úûêë x& 3 úû- 9úûêë x3 úûTheoutput equa

53、tionsis:é x1 ù41 êxúy = 1ê2 úê x úë3 û39Ø Serialization of the Transfer Function:The Numerator is:The Denominator is:Ifare m zero-points of G(s)，andare n pole-points of G(s).then G(s) is：40Therefore the system is composed serially by n items bel

54、ow:The structure of the system can be described by figure (a)。(a)of which the first block is transformed as follow:Thus its structure block will be recomposed as in figure (b).41(b)(m=n-1)Assume the outputs of integral items are required state variables The state equations of the system are:42And th

55、e matrix representation:43Ø Parallel of the Transfer Function:The Denominator:is the Characteristic Equations of the system. Assume there are n characteristics roots:G(s) can be decomposed by the summation of n fractions:In which，, is called Residue (留数) of pole pi.44The parallel connection dia

56、grams:(b) Parallel connection (No repeated roots)(a)45The state-equations of figure (b):The output equation:Matrix representation:PS: the system matrix A is a diagonal matrix.46If the denominator of G(s) Den(s)=0 has repeated roots:is the only q times repeated root, we have G(s):In which：i =1,2,qj=q

57、+1,q+2,n47Parallel Connection (Repeated roots)48Select the state-variables as the output of the integral items in thediagram:49System matrix A is Jordan Standard Form(约当).50Ex.9-8 find the parallel connection of the follow system:Solution：the Denominator:The state equation of the system by parallel

58、connection:512. State Space to Transfer FunctionsState Space (Dynamic Equations) VS Transfer FunctionsØ State Space denote both the input/output relationship and the internal state variables of the system;Transfer functions present the input/output relationship only.Ø From Transfer Functio

59、ns to State Space: system realization process, which is complicated and non-unique.From State Space to Transfer Functions: simple and uniqueprocess.52v For SISO system: State-space to TFThe State-space representation of a SISO system:ì x = Ax + Buí y = Cx + Duîx, x Î Rn´1; A

60、Î Rn´n; B Î Rn´1;C Î R1´n; D is a scalar quty.Assume the initial condition zero-input, and use Laplace Transform:sX (s) = AX (s) + BU (s)Y (s) = CX (s) + DU (s)X (s) = (sI - A)-1 BU(s)The Transfer Function is:G(s) = Y (s) = C(sI - A)-1 B + DU (s)53Ex.9-8 the state-space

61、 of the system is:éx&1 ù = é 01 ùéx1 ù + é0ùuêx& úê-1- 3úêxúê1úë 2 ûëy = 1ûë 2 ûë ûéùx10êx úë 2 ûFind its transfer function:Solution:write the related mat