How to use the inv-laplace transform to find solution to an inv-z-transform problem

Question

(This relates to the study of signals and systems)

You know: $H(z) = \frac{1}{1-\sqrt{2}z^{-1}+z^{-2}}$

Say for some reason you wanted to be able to use the inverse Laplace transform to solve inverse z-transform problems. What is H(s)? Do you simply replace z with s?

Assuming you can get the inverse Laplace transform of H(s), what needs to be done to the result of inv_laplace(H(s)), if anything, so that the answer matches the inverse z-transform of H(z)?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over 1 - \root{2}z^{-1} +z^{-2}} = {z^{2} \over z^{2} - \root{2}z + 1}.\quad r = {\root{2} \over 2} + {\root{2} \over 2}\,\ic = \expo{\pi\ic/4}\quad \mbox{and}\quad \bar{r}\quad}$ are the

roots of $\ds{z^{2} - \root{2}z + 1 = 0}$.

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With $\ds{\quad c > \root{2}/2\quad\mbox{and}\quad t > 0}$: \begin{align} &\int_{c - \infty\ic}^{c + \infty\ic}{z^{2} \over z^{2} - \root{2}z + 1}\, \expo{tz}\,{\dd z \over 2\pi\ic} = \int_{c - \infty\ic}^{c + \infty\ic}{z^{2} \over \pars{z - \bar{r}}\pars{z - r}}\,\expo{tz}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ {\bar{r}^{2} \over \bar{r} - r}\,\exp\pars{\bar{r}t} + {r^{2} \over r - \bar{r}}\,\exp\pars{rt} = {-\ic \over -\root{2}\ic}\,\exp\pars{\bar{r}t} + {\ic \over \root{2}\ic}\,\exp\pars{rt} \\[5mm] = &\ \root{2}\,\Re\bracks{\exp\pars{tr}} = \bbx{\ds{% \root{2}\exp\pars{{\root{2} \over 2}\,t}\cos\pars{{\root{2} \over 2}\,t}}} \end{align}