상태변수 방정식

Example 상태 방정식과 시스템 응답 예제

Example • Views 386 • Comments 0 • Last Updated at 1 year ago  
  • 시스템 응답
  • 상태방정식

상태변수 방정식과 시스템 응답 예제

  • 다음 시스템의 응답을 구하여라.

A=[01−2−3]B=[01]C=[10]D=[0] A = \begin{bmatrix} 0&1\\ -2&-3\end{bmatrix} \quad B = \begin{bmatrix} 0\\ 1\end{bmatrix} \quad C = \begin{bmatrix} 1&0\end{bmatrix} \quad D = \begin{bmatrix} 0 \end{bmatrix}

x(t0)=0 x(t_0) = 0

eAt = L−1  [(sI − A)−1] = Φ(t) e^{At} = \mathcal{L}^{-1}  [(sI - A)^{-1}] = \Phi(t)

x(t)=Φ(t−t0)x(t0)+∫t0t Φ(t−τ)Bu(τ)dτ=∫t0t Φ(t−τ)Bu(τ)dτ \begin{aligned} x(t) &= \Phi(t-t_0) x(t_0)+ \int_{t_0}^{t}  \Phi (t-\tau ) B u( \tau ) d \tau \\ &= \int_{t_0}^{t}  \Phi (t-\tau ) B u( \tau ) d \tau \end{aligned}

 Φ(t−τ)= L−1  [(sI − A)−1] = L−1(([s00s]−[01−2−3])−1)= L−1([s−12s+3]−1)= L−1(1s(s+3)+2[s+31−2s])= L−1(1(s+1)(s+2)[s+31−2s])= L−1([s+3(s+1)(s+2)1(s+1)(s+2)−2(s+1)(s+2)s(s+1)(s+2)])= L−1([2s+1−1s+21s+1−1s+2−2s+1+2s+2−1s+1+2s+2])= [2e−(t−τ)−e−2(t−τ)e−(t−τ)−e−2(t−τ)−2e−(t−τ)+2e−2(t−τ)−e−(t−τ)+2e−2(t−τ)] \begin{aligned}  \Phi (t-\tau ) &=  \mathcal{L}^{-1}  [(sI - A)^{-1}]  \\ &=  \mathcal{L}^{-1} \left( \left( \begin{bmatrix}s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix}0 & 1 \\ -2 & -3 \end{bmatrix} \right)^{-1} \\ \right) \\ &=  \mathcal{L}^{-1} \left( \begin{bmatrix}s & -1 \\ 2 & s+3 \end{bmatrix} ^{-1} \right) \\ &=  \mathcal{L}^{-1} \left( \dfrac{1}{s(s+3)+2} \begin{bmatrix}s+3 & 1 \\ -2 & s \end{bmatrix} \right) \\ &=  \mathcal{L}^{-1} \left( \dfrac{1}{(s+1)(s+2)} \begin{bmatrix}s+3 & 1 \\ -2 & s \end{bmatrix} \right) \\ &=  \mathcal{L}^{-1} \left( \begin{bmatrix}\dfrac{s+3}{(s+1)(s+2)} & \dfrac{1}{(s+1)(s+2)} \\ \dfrac{-2}{(s+1)(s+2)} & \dfrac{s}{(s+1)(s+2)} \end{bmatrix} \right) \\ &=  \mathcal{L}^{-1} \left( \begin{bmatrix}\dfrac{2}{s+1}-\dfrac{1}{s+2} & \dfrac{1}{s+1}-\dfrac{1}{s+2} \\ \dfrac{-2}{s+1}+\dfrac{2}{s+2} & \dfrac{-1}{s+1} +\dfrac{2}{s+2} \end{bmatrix} \right) \\ &=  \begin{bmatrix} 2e^{-(t-\tau)}-e^{-2(t-\tau)} & e^{-(t-\tau)}-e^{-2(t-\tau)} \\ -2e^{-(t-\tau)}+2e^{-2(t-\tau)} & -e^{-(t-\tau)}+2e^{-2(t-\tau)} \end{bmatrix} \\ \end{aligned}

x(t)=∫0t Φ(t−τ)Bu(τ)dτ=∫0t [2e−(t−τ)−e−2(t−τ)e−(t−τ)−e−2(t−τ)−2e−(t−τ)+2e−2(t−τ)−e−(t−τ)+2e−2(t−τ)][01]u(τ)dτ=∫0t [e−(t−τ)−e−2(t−τ)−e−(t−τ)+2e−2(t−τ)]u(τ)dτ= [∫0t(e−(t−τ)−e−2(t−τ))u(τ)dτ∫0t(−e−(t−τ)+2e−2(t−τ))u(τ)dτ] \begin{aligned} x(t) &= \int_{0}^{t}  \Phi (t-\tau ) B u( \tau ) d \tau \\ &= \int_{0}^{t}  \begin{bmatrix} 2e^{-(t-\tau)}-e^{-2(t-\tau)} & e^{-(t-\tau)}-e^{-2(t-\tau)} \\ -2e^{-(t-\tau)}+2e^{-2(t-\tau)} & -e^{-(t-\tau)}+2e^{-2(t-\tau)} \end{bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix}u(\tau) d \tau \\ &= \int_{0}^{t}  \begin{bmatrix} e^{-(t-\tau)}-e^{-2(t-\tau)} \\ -e^{-(t-\tau)}+2e^{-2(t-\tau)} \end{bmatrix} u(\tau) d \tau \\ &=  \begin{bmatrix}\int_{0}^{t}( e^{-(t-\tau)}-e^{-2(t-\tau)}) u(\tau) d \tau \\ \int_{0}^{t}( -e^{-(t-\tau)}+2e^{-2(t-\tau)}) u(\tau) d \tau \end{bmatrix} \\ \end{aligned}

y(t) =[10] [∫0t(e−(t−τ)−e−2(t−τ))u(τ)dτ∫0t(−e−(t−τ)+2e−2(t−τ))u(τ)dτ]+0u(t) =∫0t(e−(t−τ)−e−2(t−τ))u(τ)dτ\begin{aligned} y(t) & = \begin{bmatrix} 1&0 \end{bmatrix}  \begin{bmatrix}\int_{0}^{t}( e^{-(t-\tau)}-e^{-2(t-\tau)}) u(\tau) d \tau \\ \int_{0}^{t}( -e^{-(t-\tau)}+2e^{-2(t-\tau)}) u(\tau) d \tau \end{bmatrix} + 0u(t) \\ & = \int_{0}^{t}( e^{-(t-\tau)}-e^{-2(t-\tau)}) u(\tau) d \tau \end{aligned}

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