강의노트 상태 변수 방정식과 시스템 응답

강의노트 • 조회수 461 • 댓글 0 • 수정 1년 전  
  • 시스템 응답
  • 상태방정식

상태 변수 방정식과 시스템 응답

u(t)=0u(t)=0인 경우

(1)

x˙(t)=Ax(t) \dot x(t) = Ax(t)

x(t)=eAtx(0) \begin{aligned} x(t) = e^{At} x(0) \end{aligned}

이면

eAt=I+At+A2t22++Antnn!+ e^{At} = I + At + \dfrac{A^2 t^2}{2} + \ldots + \dfrac{A^n t^n}{n!} + \ldots  

deAtdt=A+A2t++Antn1(n1)!+=A(I+At+A2t22+)=AeAt \begin{aligned} \dfrac{de^{At}}{dt} &= A + A^2t + \ldots +\dfrac{ A^n t^{n-1}}{(n - 1)!} +  \ldots \\&= A (I + At + \dfrac{A^2t^2}{2} +\ldots ) = Ae^{At} \end{aligned}

x˙(t)=AeAtx(0)=Ax(t) \dot x(t) = Ae^{At} x(0) = Ax(t)

sX(s)x(0)=AX(s) sX(s) - x(0) = AX(s)

X(s)=(sIA)1x(0)X(s) = (sI - A)^{-1} x(0)

x(t)=L1[X(s)]=  L1  [(sIA)1]x(0) x(t) = \mathcal{L}^{-1} [X(s)] =  \mathcal{L}^{-1}  [(sI - A)^{-1}]x(0) (1)과 비교하면

eAt=L1  [(sIA)1]=Φ(t) e^{At} = \mathcal{L}^{-1}  [(sI - A)^{-1}] = \Phi(t) → 상태 천이 행렬

L[eAt]=(sIA)1 \mathcal{L}[e^{At}] = (sI - A)^{-1}

x(t)=eAtx(0)=Φ(t)x(0) x(t) = e^{At} x(0)=\Phi(t)x(0)

상태 천이 행렬 성질

  1. Φ(0)=eA0=e0=I\Phi(0) = e^{A0}=e^0=I
  2. Φ1(t)=(eAt)1=eAt=eA(t)=Φ(t) \Phi^{-1}(t)=(e^{At})^{-1}=e^{-At}= e^{A(-t)}=\Phi(-t)
  3. Φ(t2t0)=eA(t2t0)=eAt2eAt0(eAt1eAt1)=(eAt2eAt1)(eAt1eAt0)=eA(t2t1)eA(t1t0)=Φ(t2t1)Φ(t1t0)\begin{aligned}\Phi(t_2-t_0)&=e^{A(t_2-t_0)}=e^{At_2}e^{-At_0}(e^{At_1}e^{-At_1})\\ &=(e^{At_2}e^{-At_1})(e^{At_1}e^{-At_0})=e^{A(t_2-t_1)}e^{A(t_1-t_0)} \\&=\Phi(t_2-t_1)\Phi(t_1-t_0) \end{aligned}
  4. (Φ(t))k=Φ(kt)(\Phi(t))^k = \Phi(kt)

u(t)0u(t) \neq 0인 경우

x˙(t)Ax(t)=Bu(t) \dot x(t) - Ax(t) = Bu(t)

eAt[x˙(t)Ax(t)]=eAtBu(t)  e^{-At} [\dot x(t) - Ax(t)]=e^{-At}Bu(t)

ddt[eAtx(t)]=eAtx˙(t)AeAtx(t)=eAt[x˙(t)Ax(t)] \dfrac{d}{dt}[e^{-At}x(t)] = e^{-At}\dot x(t) - A e^{-At} x(t) = e^{-At}[\dot x(t) - Ax(t)]  

ddt[eAtx(t)]=eAtBu(t) \dfrac{d}{dt}[e^{-At}x(t)]= e^{-At}Bu(t)

eAtx(t)x(0)=0teAtBu(τ)dτ e^{-At}x(t)-x(0)=\int_0 ^t  e^{-At}Bu(\tau) d\tau

x(t)=eAtx(0)+0teA(tτ)Bu(τ)dτ=Φ(t)x(0)+0tΦ(tτ)Bu(τ)dτ \begin{aligned} x(t) &= e^{At}x(0) + \int_0 ^t  e^{A(t-\tau)}Bu(\tau) d\tau \\ &= \Phi(t)x(0)+\int_0 ^t  \Phi(t-\tau)Bu(\tau) d\tau \end{aligned}

  • t=t0 t= t_0 인 경우

x(t0)=eAt0x(0)+0t0eA(t0τ)Bu(τ)dτ=Φ(t0)x(0)+0t0Φ(t0τ)Bu(τ)dτ \begin{aligned} x(t_0 ) &= e^{At_0} x(0) + \int_0^{t_0}  e^{A (t_0 - \tau )} B u( \tau ) d \tau \\&= \Phi (t_0) x(0) + \int_0^{t_0}  \Phi (t_0-\tau ) B u( \tau ) d \tau \end{aligned}

x(t)=eAteAt0eAt0x(0)+0teA(tτ)eA(t0t0)Bu(τ)dτ=eA(tt0)eAt0x(0)+0t0eA(tt0)eA(t0τ)Bu(τ)dτ+t0teA(tτ)Bu(τ)dτ=eA(tt0)(eAt0x(0)+0t0eA(t0τ)Bu(τ)dτ)+t0teA(tτ)Bu(τ)dτ=eA(tt0)x(t0)+t0teA(tτ)Bu(τ)dτ=Φ(tt0)x(t0)+t0tΦ(tτ)Bu(τ)dτ=x(tt0) \begin{aligned} x(t ) &= e^{At}e^{-At_0}e^{At_0} x(0) + \int_0^{t}  e^{A(t-\tau)}e^{A (t_0 - t_0)} B u( \tau ) d \tau \\ &= e^{A(t-t_0)}e^{At_0} x(0) + \int_0^{t_0}  e^{A(t-t_0)}e^{A (t_0 - \tau)} B u( \tau ) d \tau + \int_{t_0}^{t}  e^{A(t-\tau)} B u( \tau ) d \tau\\ &= e^{A(t-t_0)} \left( e^{At_0} x(0) + \int_0^{t_0} e^{A (t_0 - \tau)} B u( \tau ) d \tau \right)+ \int_{t_0}^{t}  e^{A(t-\tau)} B u( \tau ) d \tau\\ &= e^{A(t-t_0)} x(t_0) + \int_{t_0}^{t}  e^{A(t-\tau)} B u( \tau ) d \tau\\ &= \Phi (t-t_0) x(t_0) + \int_{t_0}^{t}  \Phi (t-\tau ) B u( \tau ) d \tau \\ &= x(t-t_0)\end{aligned}

y(t)=Cx(t)+Du(t)=CeA(tt0)x(t0)+Ct0teA(tτ)Bu(τ)dτ+Du(t)=CΦ(tt0)x(t0)+Ct0tΦ(tτ)Bu(τ)dτ+Du(t)\begin{aligned} y(t) & = Cx(t) + Du(t) \\ &= C e^{A(t-t_0)} x(t_0) + C\int_{t_0}^{t}  e^{A(t-\tau)} B u( \tau ) d \tau + Du(t) \\ &= C \Phi(t-t_0) x(t_0) + C\int_{t_0}^{t} \Phi(t-\tau) B u( \tau ) d \tau + Du(t) \end{aligned}

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