1、现代控制理论基础12 2 线性系统的状态空间运动分析线性系统的状态空间运动分析2.1 2.1 线性定常系统的齐次解线性定常系统的齐次解2.2 2.2 矩阵指数函数矩阵指数函数2.3 2.3 线性定常系统的非齐次解线性定常系统的非齐次解2.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵2.5 2.5 线性时变系统的运动线性时变系统的运动2.6 2.6 线性连续系统的时间离散化线性连续系统的时间离散化2.7 2.7 线性离散系统的运动分析线性离散系统的运动分析现代控制理论基础22.1 2.1 线性定常系统的齐次解线性定常系统的齐次解Axx 0()tteAxx0)0()t tte(
2、Axx线性定常系统的齐次状态方程(Homogeneous state equation)给定初始状态x(0)=x0,必有如下形式的解(solution)如果定义区间为t0,),且x(t0)=x0时,解为 其中,eAt或eA(t-t0)称为矩阵指数函数(matrix exponential function),是nn维矩阵。现代控制理论基础32.1 2.1 线性定常系统的齐次解线性定常系统的齐次解10()kkkkbb tb tkb ta bb tb tb t221231223102210332010!kkkbabbaba bbaba bbaba bkk1122111133 首先回顾一下标量微分方
3、程的求解过程上式对任意t0均成立。比较两边 tk 的系数,可得t=0时可得0()xb00()kkx tbb tb tb t212 设方程具有如下形式的解xax代入方程可得因此方程的解为2 2()()(0)2!=(0)kkatx tata ta txke x111现代控制理论基础42.1 2.1 线性定常系统的齐次解线性定常系统的齐次解)(0221232132ttttbbbAbbb010323021201!bAAbbbAAbbbAAbbAbbkkkkk1111212133证明证明(1 1)直接法证明)直接法证明上式对任意t0均成立。比较两边 tk 的系数,可得t=0时可得00)(xbx00)(k
4、kkttttbbbbx2210式中,b b0,b b1,等为待定常向量 将解的形式设为如下的向量幂级数Axx 代入方程现代控制理论基础52.1 2.1 线性定常系统的齐次解线性定常系统的齐次解022022)!()!()(xAAAIbAAAIxkkkktktttkttt1211212 202!atkkkeata ta tk 1112 20!tk kketttk112AIAAA0()tteAxx标量指数函数定义为仿此,定义称eAt为矩阵指数函数。所以现代控制理论基础62.1 2.1 线性定常系统的齐次解线性定常系统的齐次解)()(0sssAXxX0)()(xXAIss01)()(xAIX ss01
5、1)()(xAIxsLt(2 2)Laplace变换法证明变换法证明Axx 3221sasasas1)(3221)(ssssAAIAI112 2()!tLstte12AIAIAA回顾标量关系式仿此有拉氏变换11()tLseaa现代控制理论基础72.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATION (2.1)xaxIn this section,we shall obtain the general solution of the linear time-invariant state equation.We shall first consider
6、 the homogeneous case and then the nonhomogeneous case.Solution of Homogeneous State Equations.Before we solve vector-matrix differential equations,let us review the solution of the scalar differential equationIn solving this equation,we may assume a solution x(t)of the form0()(2.2)kkx tbb tb tb t21
7、2By substituting this assumed solution into Equation(2.1),we obtain10()(2.3)kkkkbb tb tkb ta bb tb tb t221231223现代控制理论基础82.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONIf the assumed solution is to be the true solution,Equation(2.3)must hold for any t.Hence,equating the coefficients of the equal p
8、owers of t,we obtainThe value of b0 is determined by substituting t=0 into Equation(2.2),or102210332010!kkkbabbaba bbaba bbaba bkk11221111330()xb0现代控制理论基础92.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONHence,the solution x(t)can be written asWe shall now solve the vector-matrix differential equati
9、on (2.4)xAx By analogy with the scalar case,we assume that the solution is in the form of a vector power series in t,or2 2()()(0)2!=(0)kkatx tata ta txke x111where x=n-vector,A=nn constant matrix()(2.5)kktttt2012xbbbb现代控制理论基础102.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONBy substituting this ass
10、umed solution into Equation(2.4),we obtainIf the assumed solution is to be the true solution,Equation(2.6)must hold for all t.Thus,by equating the coefficients of like powers of t on both sides of Equation(2.6),we obtain0()(2.6)kkkkttk tttt221231223bbbbAbbbb010323021201!bAAbbbAAbbbAAbbAbbkkkkk111121
11、2133现代控制理论基础112.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONBy substituting t=0 into Equation(2.5),we obtainThus,the solution x(t)can be written as0()0 xb2 2()()(0)!k kttttk112xIAAAxThe expression in the parentheses on the right-hand side of this last equation is an nn matrix.Because of its simil
12、arity to the infinite power series for a scalar exponential,we call it the matrix exponential and write2 2!k kttttek112AIAAAIn terms of the matrix exponential,the solution of Equation(2.4)can be written as()(0)tteAxx现代控制理论基础122.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATION (2.7)xaxLaplace Transfor
13、m Approach to the Solution of Homogeneous State Equations.Let us first consider the scalar case:Taking the Laplace transform of Equation(2.7),we obtain()(0)atx te xwhere X(s)=Lx.Solving Equation(2.8)for X(s)gives()(0)()(2.8)ssxsXaX1(0)()()(0)xssaxsaXThe inverse Laplace transform of this last equatio
14、n gives the solution现代控制理论基础132.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATION()()(2.9)tt xAxThe foregoing approach to the solution of the homogeneous scalar differential equation can be extended to the homogeneous state equation:Taking the Laplace transform of both sides of Equation(2.9),we obtain
15、where X(s)=Lx.Hence,()(0)()sssXxAX(-)()(0)ss IA XxPremultiplying both sides of this last equation by(sI-A)-1,we obtain1()()(0)ssXIAx11()()(0)(2.10)tLsxIAxThe inverse Laplace transform of X(s)gives the solution x(t)Thus,现代控制理论基础142.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONNote thatHence,the inv
16、erse Laplace transform of(sI-A)-1 gives3221)(ssssAAIAI2 23 311()(2.11)!tttLste23AAAIAIA(The inverse Laplace transform of a matrix is the matrix consisting of the inverse Laplace transforms of all elements.)From Equations(2.10)and(2.11),the solution of Equation(2.9)is obtained as()(0)tteAxxThe import
17、ance of Equation(2.11)lies in the fact that it provides a convenient means for finding the closed solution for the matrix exponential.现代控制理论基础152.2 2.2 矩阵指数函数矩阵指数函数0!Atk kketttk2112IAAA21()!ttL eLts2 22AAIAIA2.2.1 2.2.1 矩阵指数函数的定义矩阵指数函数的定义IA0ekkkttkttAAAIA0!e121222.2.2 2.2.2 矩阵指数函数的性质矩阵指数函数的性质 性质性质1
18、1 矩阵指数函数满足如下关系式证明证明 由矩阵指数函数的定义式当t=0时,即可得证。设A为nn维矩阵,则矩阵指数函数e eAtAt定义为现代控制理论基础162.2 2.2 矩阵指数函数矩阵指数函数s)s(eeeAAAtt)(e)(!)(!)()!()!()()!()!(eeststststststssttstststssttAAAAAAIAAAIAAIAAI332232233222222231213121213121212121性质性质2 2 设t和s为自变量,则必成立证明证明 根据定义式证毕现代控制理论基础172.2 2.2 矩阵指数函数矩阵指数函数ttAA e)(e1IAAA0eee)()
19、(ttst性质性质3 3 eAt必有逆,且其逆为e-At,即证明证明 由性质2,有 令s=-t和性质1,得由上式可以看出,eAt与e-At互为逆矩阵,故结论得证。ststAAAeee)(ttttttstAAAAAAAeeeeeee)()(现代控制理论基础182.2 2.2 矩阵指数函数矩阵指数函数tttBABAeee)(3322223222323322312131213121tttttttttt)(!)(!)()()(!)(!)()(!)(!)(e)(BABBABBAABABABAABBAABABAIBABABABABABAIBABABAIBAee()()!()()(33)!ttttttttt
20、tt2 23 32 23 3222322331111232311223ABIAAAIBBBIA BAABBAA BABB性质性质4 4 对于nn维方阵A和B,如果AB=BA,则证明证明 根据定义式现代控制理论基础192.2 2.2 矩阵指数函数矩阵指数函数32222223121ttttt)(!)(!eee2)(ABBABABABABABAABBABABA0 0tttBABAeee)(tttBABAeee)(将上述两式相减,得显然,只有AB=BA,才有即现代控制理论基础202.2 2.2 矩阵指数函数矩阵指数函数AAAAAtttteeedd33223121ttttAAAIA!etttttttAA
21、AAAIAAAAe!edd)21(2122232AAAttteedd性质性质5 5 对于矩阵指数函数eAt,有证明证明 根据定义将上式逐项对t求导,有同理现代控制理论基础212.2 2.2 矩阵指数函数矩阵指数函数),diag(2211nnaaaA)e,e,diag(ee2211tatatatnnA222222221122112100021000210000001000100011tatatatatatatknnnnkkkt!e0AAtatatakkknnkkkkkknntaktaktakeee!00000000010001000122112211性质性质6 6 设A为对角线阵,则eAt也必为
22、对角线矩阵,且证明证明 根据定义现代控制理论基础222.2 2.2 矩阵指数函数矩阵指数函数ttttneeee000000211PPAn000000211APP性质性质7 7 设A具有互不相同的特征值l1,l1,ln,则eAt必可 经线性非奇异变换化为对角线型,即证明证明 因A具有互不相同的特征值,故可经线性变换为其中,P为使A对角线标准化的变换阵。现代控制理论基础232.2 2.2 矩阵指数函数矩阵指数函数knkkk00000021个1111APPAPPAPPPAPkkkkttkttAAAIA0!e12122000!eee!e!nk kk tk k ttk tk knk tk tk tk12
23、121100001000000100APP=对上式的一般项,有根据定义式0e!tkkktk111APPP A P则有故现代控制理论基础242.2 2.2 矩阵指数函数矩阵指数函数kkkAttkttAAAI0!e121221010A2221ttAtAAI!e221010!2110101001tt2.2.3 2.2.3 矩阵指数函数的计算矩阵指数函数的计算(1 1)级数求和法)级数求和法例例 试计算A的矩阵指数函数eAt解解 22!2110!211ttttttee011现代控制理论基础252.2 2.2 矩阵指数函数矩阵指数函数)(e11AIAsLt3210Asssss213231)()(1AI2
24、211221221112112ssssssssttttttttt22222222eeeeeeeeeA(2 2)Laplace变换法变换法例例 试用Laplace变换法计算矩阵指数函数eAt解解 现代控制理论基础262.2 2.2 矩阵指数函数矩阵指数函数PPPAAAIPAPPAPPAPPIA APPtkkkkttktttktte)!()(!)(!)(e112211221121121111-APPAPPttee(3 3)化为)化为约当约当标准形法标准形法若矩阵P -1AP已化为约当标准形,则由下式可以直接将eAt计算出来 现代控制理论基础272.2 2.2 矩阵指数函数矩阵指数函数1PPAttt
25、tnllle000e000ee2111000!11PPAttttntttttnt111111eeee)(eeellllll1)当A的特征值互异时2)当A的特征值为重根时现代控制理论基础282.2 2.2 矩阵指数函数矩阵指数函数3210A)(22123321AI11221112,211111121PP2001ttttttttttttt222222222111200211100eeeeeeeeeeeee121PPAll例例 试用化为约当标准形法 求矩阵指数函数eAt。解解 所以A的特征值因A是友矩阵,且特征值互异,所以所以现代控制理论基础292.2 2.2 矩阵指数函数矩阵指数函数4521000
26、10A)()(212544521001223AI2,1321421211101211012312131P1211321210P2000100111APPJtttttt200000eeeeeJttttttttttttttttttttttttttttt22222222214388342222453222232e)e(e)e(e)e(e)e(e)e(e)e1(e)e1(e)e(eeeePPJA例例 试用化为约当标准形法 求矩阵指数函数eAt。解解 因为A为友矩阵,因此所以现代控制理论基础302.2 2.2 矩阵指数函数矩阵指数函数nnnnaaaf111AI)(0 0IAAAAnnnnaaaf111)
27、((4 4)待定系数法)待定系数法 待定系数法是利用Cayley-Hamilton定理,首先将eAt化为A的有限项,然后确定待定的系数。则矩阵A必满足其自身的特征方程,即IAIAIAIAIAI)()adj()()(1nnnn 12211)adj(AIIAIAInnnn)(12211证明证明 因为其中于是1)Cayley-Hamilton定理 设A为nn维矩阵,其特征多项式为现代控制理论基础312.2 2.2 矩阵指数函数矩阵指数函数IAIAIAInnnnnaaa 111121)()(0 0AAAAAAIAAAnnnnnnnnnnaaa )()(211121111令两边l幂次项的系数相等,得对上
28、述关系式,从上到下依次右乘An,An-1,A,I,然后将等式两边各项分别相加,得所以f(A)=0 0IIIIAAAnnnnnnnnnaaa11111112)()(将上式展开,得现代控制理论基础32该式表明,An可表示成An-1,An-2,A,的线性组合式中 为t的标量函数。2)矩阵指数函数的多项式表达式2.2 2.2 矩阵指数函数矩阵指数函数1110nnttttAAIA)()()(e)()()(tttn 110,IAAAnnnnaaa-111IAAAAAAIAAAAAAAAAnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaaa1112321122121121112121
29、1)()()()()()(11证明证明 根据Cayley-Hamlton定理,有上式表明An1也可由An-1,An-2,A,的线性组合表示。利用Cayley-Hamlton定理,可以将eAt的无穷幂级数化为 A的有限项表达式,其中A的最高次幂不高于(n-1)。即现代控制理论基础332.2 2.2 矩阵指数函数矩阵指数函数10niimimAA同理,An2以及更高次幂都均可以由An-1,An-2,A,的线性组合表示。写成一般形式 0!1ekkkttkAA0100100!1!1!1ekkkiniikkniikikkkttktktkAAAA0!1)(kkkiitkt(0,1,1)in,这样,对于幂级数
30、将上式括号中的级数用一个t的标量函数i(t)来表示,即现代控制理论基础342.2 2.2 矩阵指数函数矩阵指数函数tttnnnnnnnnttteee)()()(2111212222112111101110 AItnnnntnntnnnttttttttte)()()(e)()()(e)()()(1110121210111110213)矩阵指数函数的计算根据特征值互异和相重两种情况,系数i(t)的计算方法.设矩阵A的特征值li互不相同,则 证明证明 根据Cayley-Hamlton定理,li和A均满足特征方程所以现代控制理论基础352.2 2.2 矩阵指数函数矩阵指数函数ttttnn-nnnnnn
31、ntttntnnnnnttttt11111eee!1e)!(1e)!()(2!)()()()()()()(2t3-3-2111121211211111210221111210211001100010000tnnttt1e)()()(111110tnnttntt1e)()()()(21112112tnnttnntt1e)()()()(23111322162t nnttn1e)(!)(111.设矩阵A有n n重特征值l1,则证明证明 已知对l1求导一次,有再对l1求导一次,有对l1求导(n-1)次,有现代控制理论基础362.2 2.2 矩阵指数函数矩阵指数函数3210A()()=012IAtttt
32、tttttt2221212111211eeeeeeee)()(2110llAIA)()(e10ttt321112220)ee(00)ee(tttttttttttt22222222eeeeeeee例例 试用待定系数法计算 矩阵指数函数eAt解解 由特征方程解出特征根l1=-1,l2=-2现代控制理论基础372.2 2.2 矩阵指数函数矩阵指数函数452100010A20)()()()(eAAIAAttttiniit21010()12eeee()1e1e()ee()e4e()eetttttttt tttttttttttt 11311212211122223300 1221 132211 21e(e
33、e)()ee ttttttt 221 0 001020 1 03220010 0 125 4Aee()ee()ee()ee()ee()ee()ee()ee()ee()eetttttttttttttttttttttttttttttt 222222222200123221125421235422818 1122438834例例 试用待定系数法计算 矩阵指数函数eAt解解 A的特征值已计算,为l1=l2=1,l3=2现代控制理论基础382.3 2.3 线性定常系统的非齐次解线性定常系统的非齐次解 00 xxBuAxx)(tttt00)d(ee)()(BuxxAAtttttt000)d(ee)()()
34、(BuxxAA非齐次状态方程(Nonhomogeneous state equation)其解必唯一,且具有如下形式如果初始时刻(initial time)是t0(此时x(t0)=x0),则方程的解为(solution in terms of x(t0)第一项为初始状态的转移项,称为齐次解;第二项为控制作用引起的响应项,称为强迫运动项。(The first term is response to the initial condition and the second term is response to the input u(t)现代控制理论基础392.3 2.3 线性定常系统的非齐次解
35、线性定常系统的非齐次解 xAxBuBuxAxxAAAt-t-t-te)e(dd)(ee()e()dtttt00-A-AxButtt00)d(e)()(eBuxxAA-tttt00d)(e)(e)()(BuxxAA(1)(1)直接法证明直接法证明(Direct method)对上式两边左乘e-At,得将上式由0t进行积分,有再对上式两边左乘eAt,且因e-AteAt=I,从而证明了将非齐次方程写为现代控制理论基础402.3 2.3 线性定常系统的非齐次解线性定常系统的非齐次解 ()()()ssss0XxAXBU()()()sss0IA XxBU()()()()ssss110XIAxIABU001
36、1xxAIAtsLe)()()()e()dtt-Lss110AIABUBu(2 2)Laplace变换法证明变换法证明(Laplace transform approach)整理得用(sI-A)-1左乘上式两边,可得考虑到(Laplace变换中的卷积分convolution integral)BuAxx取Laplace变换,有对方程1122()(),()(),L f tF s L f tF s若112120()()()()tLF s F sf tfd则现代控制理论基础412.3 2.3 线性定常系统的非齐次解线性定常系统的非齐次解 00,03210 xxxx)(1uttttttttt22222
37、222eeeeeeeeeAtttt00)d(e)(e)()(BuxxAA例例 求系统状态方程的非齐次解,其中u(t)=1(t)解解 该系统的矩阵指数函数已求得,为tttttttttttttttttdeeeeeeee)(eeeeeeee)()()()()()()()()(02222222210222202222txxttttxxxxxxxx220102010220102010122(12(21(12(2)e)e)e)e1现代控制理论基础422.3 2.3 线性定常系统的非齐次解线性定常系统的非齐次解 0()(),(0)u tKtxx即当时,0()AtAtx te xe BK(3 3)特定输入下的
38、状态响应特定输入下的状态响应1 1、脉冲响应、脉冲响应(Response of impulse function)2 2、阶跃响应、阶跃响应(Response of step function)0()1(),(0)u tKtxx即当时,10()()AtAtx te xAeI BK3 3、斜坡响应、斜坡响应(Response of ramp function)0()1(),(0)u tKttxx即当时,210()()AtAtx te xAeIA t BK现代控制理论基础432.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATION (2.12)xaxbuS
39、olution of Nonhomogeneous State Equations.We shall begin by considering the scalar caseLet us rewrite Equation(2.12)asMultiplying both sides of this equation by e-at,we obtainxaxbu-()()()()atatatdex tax tex te bu tdtIntegrating this equation between 0 and t gives-0()(0)()tataex txebud0()(0)()tatatax
40、 te xeebudorThe first term on the right-hand side is the response to the initial condition and the second term is the response to the input u(t).现代控制理论基础442.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONLet us now consider the nonhomogeneous state equation described byBy writing Equation(2.13)as (2
41、.13)xAx+Bu()()()ttt xAx=Buand premultiplying both sides of this equation by e-At,we obtaind()()e()e()d-t-t-tetttttAAAxAxxBuIntegrating the preceding equation between 0 and t gives()()e()dttet00-AAxxBu(-)()()()dttttee00AAxxBuorwhere x=n-vector,u=r-vector,A=nn constant matrix,B=nr constant matrix.现代控制
42、理论基础452.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONLaplace Transform Approach to the Solution of Non-homogeneous State Equations.The solution of the nonhomogeneous state equationcan also be obtained by the Laplace transform approach.The Laplace transform of this last equation yieldsor xAx+Bu()()
43、()ssss(0)XxAXBU(-)()()sss(0)IA XxBUPremultiplying both sides of this last equation by(sI-A)-1,we obtain()()()()ssss11(0)XIAxIABU现代控制理论基础462.1 2.1 SOLVING THE TIME-INVARIANT STATE EQUATIONUsing the relationship given by Equation(2.11)givesThe inverse Laplace transform of this last equation can be obt
44、ained by use of the convolution integral as follows:()()ttsL eL es(0)AAXxBUtttt00d)(e)(e)()(BuxxAASolution in Terms of x(t0).Thus far we have assumed the initial time to be zero.If,however,the initial time is given by t0 instead of 0,then the solution to Equation(2.13)must be modified to0(-)()0()()(
45、)dtt ttttete0AAxxBu现代控制理论基础472.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 在状态空间分析中,采用状态转移矩阵可以:),)()()(0000tttttt IA2.4.1 2.4.1 状态转移矩阵状态转移矩阵(state transition matrix)对于线性定常系统 ,称之为系统的状态转移矩阵。当所观察的时 间区间为0,)时,状态转移矩阵可表示为 。)(0tt)(t的解Axx,x(t0)=x0 满足如下矩阵方程和初始条件(2)能使时变系统状态方程的解写成解析形式,从而可能建立一种对定常系统和时变系统都适用的统一的求解公式。(1)对线性系
46、统的运动给出一个清晰的描述;现代控制理论基础482)利用F F(t-t0 0),可以将系统的自由运动规律表示为 2.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 00)()(xxttt00000000 xxxAxxAxx)()()()()()(0ttttttttttt下面对线性定常系统状态转移矩阵作进一步说明:证明证明 假设x(t)=F F(t-t0)x0为解,显然满足状态方程和初始条件:3)对于线性定常系统显然有 应该指出,矩阵指数函数eAt和状态转移矩阵F F(t)是从两个不同的角度提出来的概念。矩阵指数函数eAt是一个数学函数的名称,而状态转移矩阵表征了对初始状态x0
47、的转移关系。1)状态转移矩阵 是以t)(0tt n n为自变量的维矩阵。)(e)(00ttttA现代控制理论基础492.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 5)状态转移矩阵是齐次状态方程 在初始状态为基向量时的一个基本阵。)()(ttAxx1000,0010,000121neee)()()()()()()()()()()()()()()()()(00000002221212121112122211211xtxtxtxtxxttttttxx4)(t-t0)的含义是,x(t)在 t t0 任何时刻的状态,只是初始状态x0通过变换阵(t-t0)的一种转移。而且,对于给定
48、系统,(t-t0)是唯一的,因此这种状态转移也是唯一的。以二阶系统为例。设的自由解为)()(ttAxx若取初始状态x(0)=1 0T,则x(t)=j11(t)j21(t)T 若取初始状态x(0)=0 1T,则x(t)=j12(t)j22(t)T 现代控制理论基础50例例 已知某二阶系统2.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 tttt00)d()()()(BuxxAxx 120)(xtttee)(2x11(0)xttttttteeee)(2x6)基于状态转移矩阵的系统非齐次解可写为tttttt000)d()()()(Buxx或,其不同初始状态的响应为试求该系统的状态
49、转移矩阵。现代控制理论基础512.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 (0)()(xxtt 11122222211211)()()()(eeeeeetttttttttttt122211211111222tttttttttttteeeeee)()()()()(teeeeeetttttttttt242解解 可以写出下列方程所以120)(x11(0)xttttttteeee)(2xtttee)(2x现代控制理论基础522.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 I)()(00ttI)(tttsLtAeAI)()(11IeeAA000tt)(2.
50、4.2 2.4.2 状态转移矩阵的性质状态转移矩阵的性质(properties of state-transition matrices)亦即证明证明 因为所以表示状态从t时刻又转移到t时刻,状态并没有发生变化。(1 1)自身性自身性)()()(020112tttttt()()()()()()()()()ttttttttttttt1100221121100 xxxxx)()()(0022ttttxx)()()(011202tttttt(2 2)传递性)传递性 证明证明 由解的唯一性,有又有现代控制理论基础532.4 2.4 线性定常系统的状态转移矩阵线性定常系统的状态转移矩阵 )()()()(