题海 · pdf-page · p.381
(2) 令 \(e(t)=\sin\omega t\),则有
\[E(z)=\mathscr{Z}[\sin\omega t]=\mathscr{Z}\left[\frac{\mathrm{e}^{\mathrm{j}\omega t}-\mathrm{e}^{-\mathrm{j}\omega t}}{2\mathrm{j}}\right]=\sum_{n=0}^{\infty}\frac{\mathrm{e}^{\mathrm{j}\omega nT}-\mathrm{e}^{-\mathrm{j}\omega nT}}{2\mathrm{j}}z^{-n}\]
\[=\frac{1}{2\mathrm{j}}\left(\frac{1}{1-\mathrm{e}^{\mathrm{j}\omega T}z^{-1}}-\frac{1}{1-\mathrm{e}^{-\mathrm{j}\omega T}z^{-1}}\right)=\frac{1}{2\mathrm{j}}\left[\frac{(\mathrm{e}^{\mathrm{j}\omega T}-\mathrm{e}^{-\mathrm{j}\omega T})z^{-1}}{1-(\mathrm{e}^{\mathrm{j}\omega T}+\mathrm{e}^{-\mathrm{j}\omega T})z^{-1}+z^{-2}}\right]\]
\[=\frac{z\sin\omega T}{z^{2}-2z\cos\omega T+1}\]
根据复数位移定理可知
\[\mathscr{Z}[\mathrm{e}^{-at}\sin\omega t]=E(z\mathrm{e}^{aT})=\frac{z\mathrm{e}^{aT}\sin\omega T}{z^{2}\mathrm{e}^{2aT}-2z\mathrm{e}^{aT}\cos\omega T+1}\]
(3) 将 \(E(s)\) 展成部分分式
\[E(s)=\frac{1}{9}\cdot\frac{1}{s}-\frac{1}{9}\cdot\frac{1}{s+3}-\frac{1}{3}\cdot\frac{1}{(s+3)^{2}}\]
对上式逐项取拉普拉斯反变换,可得
\[e(t)=\frac{1}{9}-\frac{1}{9}\mathrm{e}^{-3t}-\frac{1}{3}t\mathrm{e}^{-3t}\]
则有 \(\qquad E(z)=\mathscr{Z}\left[\frac{1}{9}\cdot 1(t)\right]-\mathscr{Z}\left[\frac{1}{9}\mathrm{e}^{-3t}\right]-\mathscr{Z}\left[\frac{1}{3}t\mathrm{e}^{-3t}\right]\)
对上式各项 z 变换
\[\mathscr{Z}\left[\frac{1}{9}\cdot 1(t)\right]=\frac{1}{9}\mathscr{Z}[1(t)]=\frac{1}{9}\cdot\frac{z}{z-1}\]
\[\mathscr{Z}\left[\frac{1}{9}\mathrm{e}^{-3t}\right]=\frac{1}{9}\cdot\frac{z\mathrm{e}^{3t}}{z\mathrm{e}^{3t}-1},\qquad \mathscr{Z}\left[\frac{1}{3}t\mathrm{e}^{-3t}\right]=\frac{1}{3}\cdot\frac{Tz\mathrm{e}^{3t}}{(z\mathrm{e}^{3t}-1)^{2}}\]
可得 \(\qquad E(z)=\frac{1}{9}\cdot\frac{z}{z-1}-\frac{1}{9}\cdot\frac{z\mathrm{e}^{3t}}{z\mathrm{e}^{3t}-1}-\frac{1}{3}\cdot\frac{Tz\mathrm{e}^{3t}}{(z\mathrm{e}^{3t}-1)^{2}}\)
(4) 将 \(E(s)\) 展成部分分式
\[E(s)=\frac{1}{2s}-\frac{1}{s+1}+\frac{1}{2(s+2)}\]
对上式逐项取拉普拉斯反变换,可得
\[e(t)=\frac{1}{2}-\mathrm{e}^{-t}+\frac{1}{2}\mathrm{e}^{-2t}\]
则有 \(\quad E(z)=\mathscr{Z}\left[\frac{1}{2}\cdot 1(t)\right]-\mathscr{Z}[\mathrm{e}^{-t}]+\mathscr{Z}\left[\frac{1}{2}\mathrm{e}^{-2t}\right]\)
\[=\frac{1}{2}\left(\frac{z}{z-1}-\frac{2z}{z-\mathrm{e}^{-T}}+\frac{z}{z-\mathrm{e}^{-2T}}\right)=\frac{(1-2\mathrm{e}^{-T}+\mathrm{e}^{-2T})z^{2}}{2(z-1)(z-\mathrm{e}^{-T})(z-\mathrm{e}^{-2T})}\]
7-3 求下列各函数的 \(z\) 反变换:
(1) \(E(z)=\dfrac{2z(z^{2}-1)}{(z^{2}+1)^{2}}\);\(\qquad\)(2) \(E(z)=\dfrac{2z^{2}}{(z+1)^{2}(z+2)}\);
(3) \(E(z)=\dfrac{z}{(z-\mathrm{e}^{-aT})(z-\mathrm{e}^{-bT})}\);\(\qquad\)(4) \(E(z)=\dfrac{z}{(z-1)^{2}(z-2)}\);
(5) \(E(z)=\dfrac{(1-\mathrm{e}^{-aT})z}{(z-1)(z-\mathrm{e}^{-aT})}\)。
解 (1) 采用幂级数法。
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