\[
\text{Res}\left[\frac{z^n}{(z-\mathrm{e}^{-aT})(z-\mathrm{e}^{-bT})}\right]_{z\to \mathrm{e}^{-bT}} = \lim_{z\to \mathrm{e}^{-bT}}\left[\frac{z^n}{(z-\mathrm{e}^{-aT})}\right] = \frac{\mathrm{e}^{-bnT}}{\mathrm{e}^{-bT}-\mathrm{e}^{-aT}}
\]
\[
e(nT) = \frac{\mathrm{e}^{-anT}}{\mathrm{e}^{-aT}-\mathrm{e}^{-bT}} + \frac{\mathrm{e}^{-bnT}}{\mathrm{e}^{-bT}-\mathrm{e}^{-aT}} = \frac{\mathrm{e}^{-anT}-\mathrm{e}^{-bnT}}{\mathrm{e}^{-aT}-\mathrm{e}^{-bT}}
\]
则有
\[
e^*(t) = \sum_{n=0}^{\infty} \frac{\mathrm{e}^{-anT}-\mathrm{e}^{-bnT}}{\mathrm{e}^{-aT}-\mathrm{e}^{-bT}}\delta(t-nT)
\]
(4) 采用反演积分法
\[
E(z)z^{n-1} = \frac{z^n}{(z-1)^2(z-2)}
\]
有 \(z_1=z_2=1, z_3=2\) 三个极点,则有
\[
\text{Res}\left[\frac{z^n}{(z-1)^2(z-2)}\right]_{z\to 1} = \lim_{z\to 1}\frac{\mathrm{d}}{\mathrm{d}z}\left[(z-1)^2\frac{z^n}{(z-1)^2(z-2)}\right]
\]
\[
= \lim_{z\to 1}\left[\frac{nz(z-2)-z^n}{(z-2)^2}\right] = -n-1
\]
\[
\text{Res}\left[\frac{z^n}{(z-1)^2(z-2)}\right]_{z\to 2} = \lim_{z\to 2}\left[(z-2)\frac{z^n}{(z-1)^2(z-2)}\right] = 2^n
\]
\[
e(nT) = 2^n - n - 1
\]
相应的采样函数为
\[
e^*(t) = \sum_{n=0}^{\infty}(2^n-n-1)\delta(t-nT) = \delta(t-2T)+4\delta(t-3T)+11\delta(t-4T)+\cdots
\]
用幂级数法验证
\[
E(z) = \frac{z}{(z-1)^2(z-2)} = \frac{z^{-2}}{1-4z^{-1}+5z^{-2}-2z^{-3}}
\]
\[
= z^{-2}+4z^{-3}+11z^{-4}+26z^{-5}+\cdots
\]
故有
\[
e^*(t) = \delta(t-2T)+4\delta(t-3T)+11\delta(t-4T)+26\delta(t-5T)+\cdots
\]
(5) 因为
\[
\frac{E(z)}{z} = \frac{(1-\mathrm{e}^{-aT})}{(z-1)(z-\mathrm{e}^{-aT})} = \frac{1}{z-1}-\frac{1}{z-\mathrm{e}^{-aT}}
\]
所以
\[
E(z) = \frac{z}{z-1}-\frac{z}{z-\mathrm{e}^{-aT}}
\]
查 \(z\) 变换表,可知 \(e(nT)=1-\mathrm{e}^{-anT}\),则有
\[
e^*(t) = \sum_{n=0}^{\infty}(1-\mathrm{e}^{-anT})\delta(t-nT)
\]
7-4 求下列 \(E(z)\) 的脉冲序列 \(e^*(t)\):
(1) \(E(z)=\dfrac{z^2}{(z\mathrm{e}-1)^2}\); (2) \(E(z)=\dfrac{10z(z+1)}{(z-1)(z^2+z+1)}\)。
解 (1) 采用反演积分法,可得
\[
e(nT) = \text{Res}\left[E(z)z^{n-1}\right]_{z\to \mathrm{e}^{-1}} = \frac{1}{1!}\lim_{z\to \mathrm{e}^{-1}}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{\mathrm{e}^{-2}(z-\mathrm{e}^{-1})^2 z^{n+1}}{(z-\mathrm{e}^{-1})^2}\right] = \mathrm{e}^{-2}(n+1)\mathrm{e}^{-n}
\]
\[
e^*(t) = \sum_{n=0}^{\infty}\mathrm{e}^{-2}\left[(n+1)\mathrm{e}^{-n}\right]\delta(t-nT) = \mathrm{e}^{-2}+2\mathrm{e}^{-3}z^{-1}+3\mathrm{e}^{-4}z^{-2}+4\mathrm{e}^{-5}z^{-3}+\cdots
\]
幂级数法验证