11.3 Inverse laplace transform

When given $$H(s)=\frac{s}{s-\alpha }$$ with an ROC of $$\left|s\right|> \alpha$$ we could determine "by inspection" that $$h(t)=\alpha ^{t}u(t)$$

Find the inverse z-transform of $$H(s)=\frac{1+2s^{(-1)}+s^{-2}}{1-3s^{(-1)}+2s^{-2}}$$ where the ROC is $\left|s\right|> 2$ . In this case $M=N=2$ , so we have to use long division to get $$H(s)=\frac{1}{2}+\frac{\frac{1}{2}+\frac{7}{2}s^{(-1)}}{1-3s^{(-1)}+2s^{-2}}$$ Next factor the denominator. $$H(s)=2+\frac{-1+5s^{(-1)}}{(1-2s^{(-1)})(1-s^{(-1)})}$$ Now do partial-fraction expansion. $$H(s)=\frac{1}{2}+\frac{A_1}{1-2s^{(-1)}}+\frac{A_2}{1-s^{(-1)}}=\frac{1}{2}+\frac{\frac{9}{2}}{1-2s^{(-1)}}+\frac{-4}{1-s^{(-1)}}$$ Now each term can be inverted using the inspection method and the Laplace-transform table. Thus, since the ROC is $\left|s\right|> 2$ , $$h(t)=\frac{1}{2}\delta (t)+\frac{9}{2}2^{t}u(t)-4u(t)$$

Now look at the Laplace-transform of a finite-length sequence .

Suppose $$H(s)=\log_t(1+\alpha s^{(-1)})$$ Noting that $$\log_t(1+x)=\sum_{t=1}$$∞ -1 t 1 x t t Then $$H(s)=\sum_{t=1}$$∞ -1 t 1 α t s t t Therefore $$H(s)=\begin{cases}\frac{-1^{(t+1)}\alpha ^t}{t} & \text{if $t\ge 1$}\\ 0 & \text{if $t\le 0$}\end{cases}()$$