matlab教學實現(xiàn)連續(xù)時間系統(tǒng)的狀態(tài)空間分析

上下求索教/ :上下求 輸入信號(激勵信號)的作用下，對輸入信號的響應(系統(tǒng)的輸出)本課程只討論單輸入單輸出系統(tǒng)(SISO :上下求索教連續(xù)時間系統(tǒng)的狀態(tài)空間模型(StateSpace輸入信號(激勵信號輸出信號(系統(tǒng)響應系統(tǒng)的狀態(tài)矢量:(t) (t) uy(t)Cx(t) :上下求索教案例分析LCRLCR電路的基本參數(shù)：L4H,C0.01F,RuC(0)2V,i(0)0.05

i(0)0.05激勵電壓：uS(t4

uS(t)uS(t)4 ,t

y(0)2,y(0)0.04y(t)y(0)2,y(0)0.04y(t)0.15y(t)y(t)

uC(0) :上下求索教案例分析LCRi(tuC(ti(tuC(t為系統(tǒng)的一組狀態(tài)變量Ri(t)Ldi(t)u(t) duCSdi(t)Ri(t)1u(t)1 C1C duL1CLCLS輸入的激勵信號：uS(t4

, 系統(tǒng)的初始條件：uC(0)2V,i(0)0.05 :上下求索教案例分析LCRdi(t)du(t)=L1Li(t)+1 0u0LuSyy(t) 1i(t)u y(t)Cx(t)yy(t)u(t)u(t)u(t)i(t)R 1i(t)1uLSCuSy(t)Cx(t) :上下求索教案例分析LCR系統(tǒng)的狀態(tài)方程并不唯一uR(tuC(t)LduR uC(t)uSduRRRR uR(t) uC(t) uSLLL duC(t)1uRduC(t) 1uR du(t) =1LRLu(t)uRR+LuS 0 0uy(t) 1uR(t)u

y(t)Cx(t)

:上下求索教案例分析LCR0.04y(t)0.15y(t)y(t)y(0)2,y(0)x(t)Ax(t)y(t)Cx(t)xx(t)i(t) AuL1 0BL C 1D

i(0)0.05uS(t)4cos(7t)et,tLuS(t)4cos(7t)et,txx(0)i(0)u2

uC(0)x(t)ux(t)uuRLRLR A 0B LC1D0x(0)uR(0)i(0)R0.75u u(0) 2上下求 sys_ss=ss(A,B,C,根據(jù)系統(tǒng)微分方程/y=lsim(sys_ss,u,t,x0)&[y,t,x]=lsim(sys_ss,u,t, :上下求索教 :上下求 / Rs

EsmEsm1 s

01nCsnCsn1 s01ndenden 1,Cn]num[E0,E1Em1,Em

完整地表述系統(tǒng)的微分方程dnddn1r(t)dddCnr(t)dmddm1e(t)ddd :上下求索教//[A,B,C,D]=tf2ss(num,sys_ss=ss(A,B,C,sys=tf(num,sys_ss=ss[A,B,C,D]=ssdata狀態(tài)空間模型轉(zhuǎn)換為系統(tǒng)微分方程[num,den]=ss2tf(A,B,C, :上下求索教/0.04y(t)0.15y(t)y(t)y(0)2,y(0)

den[0.04,0.15,1]num[0,0,將狀態(tài)空間模型轉(zhuǎn)換為微分方程傳遞函數(shù)(系數(shù)向量 →狀態(tài)空間模型之狀態(tài)變量的初值 :上下求索教/y(0),0.04y(t)0.15y(t)y(t)y(t)Cx(t)x(t)Ax(t)y(t)Cx(t)y(0)Cx(0)y(0)Cx(0)Du(0)0 :上下求索教/y(0)Cx(0)Du(0)0u(0)Cx(0)y(0)Du(0)0 xx(0) C1D0 ) :上下求索教/u(t4y(0)2,y(0)0.04y(t)0.15y(t)y(t)u(ty(0)2,y(0)0.04y(t)0.15y(t)y(t)y(0)0,y(0)0.04y(t)0.15y(t)y(t) :上下求索教/ :上下求索教xx(0)C1D C1 x(t)Ax(t)y(t)Cx(t)x(t)y(t)連續(xù)時間系統(tǒng)的全響應零狀態(tài)響應零輸入響應：lsim基于狀態(tài)空間模型求解 :上下求索教x(t)Ax(t)y(t)Cx(t)y(t)y(t)y(0)CACC1x(0)CAx(0)3CA3 :上下求索教C1x(0)CAC1x(0)CACA3H(s)s23s (s1)(ss36s211s (s1)(s2)(s :上下求索教H(s)s2