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1.离散化模型的推导
设系统的状态空间表达式为:
x
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y
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\dot{\mathbf{x}}(t)=A\mathbf{x}+B\mathbf{u}(t)\\ \mathbf{y}(t)=C\mathbf{x}(t)+D\mathbf{u}(t)
x˙(t)=Ax+Bu(t)y(t)=Cx(t)+Du(t) 对上面两个式子分别取拉普拉斯变换得到:
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X(s)=(sI-A)^{-1}X(0)+(sI-A)^{-1}BU(s)
X(s)=(sI−A)−1X(0)+(sI−A)−1BU(s) 再对上面的式子求拉普拉斯反变换得到:
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\mathbf{X}(t)=e^{At}\mathbf{X}(0)+\int_0^te^{A(t-\tau)}B\mathbf{u}(\tau)d\tau\\ =\mathbf{F}(t)\mathbf{X}(0)+\int_0^t\mathbf{F}(t-\tau)B\mathbf{u}(\tau)d\tau\\ \mathbf{F}(t)=e^{At}
X(t)=eAtX(0)+∫0teA(t−τ)Bu(τ)dτ=F(t)X(0)+∫0tF(t−τ)Bu(τ)dτF(t)=eAt 对上面的结果进行离散化处理,设采样周期是
T
T
T，考察
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t=(k+1)T,t=kT
t=(k+1)T,t=kT时
X
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\mathbf{X}(t)
X(t)的值。
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\mathbf{X}((k+1)T)=e^{A(k+1)T}\mathbf{X}(0)+\int_0^{(k+1)T}e^{A((k+1)T-\tau)}B\mathbf{u}(\tau)d\tau\\=e^{AT}(e^{AkT}\mathbf{X}(0)+\int_0^{kT}e^{A(kT-\tau)}B\mathbf{u}(\tau)d\tau+\int_{kT}^{k(T+1)}e^{A(kT-\tau)}B\mathbf{u}(\tau)d\tau)\\=e^{AT}(\mathbf{X}(kT)+\int_{kT}^{k(T+1)}e^{A(kT-\tau)}B\mathbf{u}(\tau)d\tau)
X((k+1)T)=eA(k+1)TX(0)+∫0(k+1)TeA((k+1)T−τ)Bu(τ)dτ=eAT(eAkTX(0)+∫0kTeA(kT−τ)Bu(τ)dτ+∫kTk(T+1)eA(kT−τ)Bu(τ)dτ)=eAT(X(kT)+∫kTk(T+1)eA(kT−τ)Bu(τ)dτ) 若离散化时对输入
u
u
u采用零阶保持器,则有:
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=
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,
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T
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u(t)=u(kT),(kT\leq t |