When using the Laplace-transform

$$H\left(s\right)=\sum _{t=-\infty}^{\infty}h\left(t\right){s}^{-t}$$

1

This "method" is to basically become familiar with the Laplace-transform pair tables and then "reverse engineer".

Example 1

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

When dealing with linear time-invariant systems the z-transform is often of the form

$$\begin{array}{rcl}\hfill H\left(s\right)& \hfill =\hfill & \frac{B\left(s\right)}{A\left(s\right)}\hfill \\ \hfill & \hfill =\hfill & \frac{\sum _{k=0}^{M}{b}_{k}{s}^{-k}}{\sum _{k=0}^{N}{a}_{k}{s}^{-k}}\hfill \end{array}$$

2

$$H\left(s\right)=\frac{{a}_{0}}{{b}_{0}}\frac{\prod _{k=1}^{M}1-{c}_{k}{s}^{-1}}{\prod _{k=1}^{N}1-{d}_{k}{s}^{-1}}$$

3

If $M<N$ then $H\left(s\right)$ can be represented as

$$H\left(s\right)=\sum _{k=1}^{N}\frac{{A}_{k}}{1-{d}_{k}{s}^{-1}}$$

4

$$H\left(s\right)=\sum _{r=0}^{M-N}{B}_{r}{s}^{-r}+\frac{\sum _{k=0}^{N-1}{b}_{k}^{\text{'}}{s}^{-k}}{\sum _{k=0}^{N}{a}_{k}{s}^{-k}}$$

5

Example 2

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

Khan Lecture on Partial Fraction Expansion

When the z-transform is defined as a power series in the form

$$H\left(s\right)=\sum _{t=-\infty}^{\infty}h\left(t\right){s}^{-t}$$

6

Example 3

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

$$\begin{array}{rcl}\hfill H\left(s\right)& \hfill =\hfill & {s}^{2}(1+2{s}^{-1})(1-\frac{1}{2}{s}^{-1})(1+{s}^{-1})\hfill \\ \hfill & \hfill =\hfill & {s}^{2}+\frac{5}{2}s+\frac{1}{2}+-{s}^{-1}\hfill \end{array}$$

7

One of the advantages of the power series expansion method is that many functions encountered in engineering problems have their power series' tabulated. Thus functions such as log, sin, exponent, sinh, etc, can be easily inverted.

Example 4

Suppose $$H\left(s\right)={\mathrm{log}}_{t}(1+\alpha {s}^{-1})$$ Noting that $${\mathrm{log}}_{t}(1+x)=\sum _{t=1}^{\infty}\frac{{-1}^{t+1}{x}^{t}}{t}$$ Then $$H\left(s\right)=\sum _{t=1}^{\infty}\frac{{-1}^{t+1}{\alpha}^{t}{s}^{-t}}{t}$$ Therefore $$H\left(s\right)=\{\begin{array}{l}\frac{{-1}^{t+1}{\alpha}^{t}}{t}\text{if}t\ge 1\\ 0\text{if}t\le 0\end{array}$$

Without going in to much detail

$$h\left(t\right)=\frac{1}{2\pi i}\underset{r}{\oint}H\left(s\right){s}^{t-1}ds$$

8

The Inverse Laplace-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. All nevertheless assist the user in reaching the desired time-domain signal that can then be synthesized in hardware(or software) for implementation in a real-world filter.