考研851 自动控制原理
题海 · table · p.19

续表

序号 拉普拉斯变换 \(E(s)\) 时间函数 \(e(t)\) \(z\) 变换 \(E(z)\)
14 \(\dfrac{s+d}{(s+a)(s+b)(s+c)}\) \(\dfrac{d-a}{(b-a)(c-a)}\mathrm{e}^{-at}+\dfrac{d-b}{(a-b)(c-b)}\mathrm{e}^{-bt}+\dfrac{d-c}{(a-c)(b-c)}\mathrm{e}^{-ct}\) \(\dfrac{(d-a)z}{(b-a)(c-a)(z-\mathrm{e}^{-aT})}+\dfrac{(d-b)z}{(a-b)(c-b)(z-\mathrm{e}^{-bT})}+\dfrac{(d-c)z}{(a-c)(b-c)(z-\mathrm{e}^{-cT})}\)
15 \(\dfrac{abc}{s(s+a)(s+b)(s+c)}\) \(1-\dfrac{bc}{(b-a)(c-a)}\mathrm{e}^{-at}-\dfrac{ca}{(c-b)(a-b)}\mathrm{e}^{-bt}-\dfrac{ab}{(a-c)(b-c)}\mathrm{e}^{-ct}\) \(\dfrac{z}{z-1}-\dfrac{bcz}{(b-a)(c-a)(z-\mathrm{e}^{-aT})}-\dfrac{caz}{(c-b)(a-b)(z-\mathrm{e}^{-bT})}-\dfrac{abz}{(a-c)(b-c)(z-\mathrm{e}^{-cT})}\)
16 \(\dfrac{\omega}{s^2+\omega^2}\) \(\sin\omega t\) \(\dfrac{z\sin\omega T}{z^2-2z\cos\omega T+1}\)
17 \(\dfrac{s}{s^2+\omega^2}\) \(\cos\omega t\) \(\dfrac{z(z-\cos\omega T)}{z^2-2z\cos\omega T+1}\)
18 \(\dfrac{\omega}{s^2-\omega^2}\) \(\sinh\omega t\) \(\dfrac{z\sinh\omega T}{z^2-2z\cosh\omega T+1}\)
19 \(\dfrac{s}{s^2-\omega^2}\) \(\cosh\omega t\) \(\dfrac{z(z-\cosh\omega T)}{z^2-2z\cosh\omega T+1}\)
20 \(\dfrac{\omega^2}{s(s^2+\omega^2)}\) \(1-\cos\omega t\) \(\dfrac{z}{z-1}-\dfrac{z(z-\cos\omega T)}{z^2-2z\cos\omega T+1}\)
21 \(\dfrac{\omega}{(s+a)^2+\omega^2}\) \(\mathrm{e}^{-at}\sin\omega t\) \(\dfrac{z\mathrm{e}^{-aT}\sin\omega T}{z^2-2z\mathrm{e}^{-aT}\cos\omega T+\mathrm{e}^{-2aT}}\)
22 \(\dfrac{s+a}{(s+a)^2+\omega^2}\) \(\mathrm{e}^{-at}\cos\omega t\) \(\dfrac{z^2-z\mathrm{e}^{-aT}\cos\omega T}{z^2-2z\mathrm{e}^{-aT}\cos\omega T+\mathrm{e}^{-2aT}}\)
23 \(\dfrac{b-a}{(s+a)(s+b)}\) \(\mathrm{e}^{-at}-\mathrm{e}^{-bt}\) \(\dfrac{z}{z-\mathrm{e}^{-aT}}-\dfrac{z}{z-\mathrm{e}^{-bT}}\)
24 \(\dfrac{a^2b^2}{s^2(s+a)(s+b)}\) \(abt-(a+b)-\dfrac{b^2}{a-b}\mathrm{e}^{-at}+\dfrac{a^2}{a-b}\mathrm{e}^{-bt}\) \(\dfrac{abTz}{(z-1)^2}-\dfrac{(a+b)z}{z-1}-\dfrac{b^2z}{(a-b)(z-\mathrm{e}^{-aT})}+\dfrac{a^2z}{(a-b)(z-\mathrm{e}^{-aT})}\)

3. \(z\) 变换性质

\(z\) 变换有一些基本定理,可以使 \(z\) 变换的应用变得简单和方便。以下定理的阐述均略去其证明。

1) 线性定理

\(E_1(z)=\mathscr{Z}[e_1(t)],E_2(z)=\mathscr{Z}[e_2(t)],a\) 为常数,则

\[\mathscr{Z}[e_1(t)\pm e_2(t)]=E_1(z)\pm E_2(z)\]
\[\mathscr{Z}[ae(t)]=aE(z)\]

式中,\(E(z)=\mathscr{Z}[e(t)]\)

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