Impulse Invariance Transformation

Let $\mathrm{h}(\mathrm{t})$ and $\mathrm{H}(\mathrm{s})$ be the Impulse response and Transfer function of the analog filter.

$\therefore h(t)=L^{-1}\{H(s)\} \cdots(1)$

Let $\mathrm{H}(\mathrm{s})$ be a rational system with $\mathrm{N}$ distinct poles.

$\therefore H(s)=\sum_{i=1}^{N} \frac{A_{i}}{s-p_{i}},$ where $A_{i}$ are the partial fraction coefficients. $\cdots(2)$

$\therefore$ From $(1)$ and $(2), h(t)=L^{-1}\left\{\sum_{i=1}^{N} \frac{A_{i}}{s-p_{i}}\right\}$

$=\sum_{i=1}^{N} L^{-1}\left\{\frac{A_{i}}{s-p_{i}}\right\}$

$=\sum_{i=1}^{N} A_{i} e^{p_{i} t} u(t),$ where $u(t)$ is the CT Unit Step function.

Impulse response of the digital filter is obtained by uniform sampling of the Impulse response of the analog filter. Let T be sampling period.

$\therefore \mathrm{t}=\mathrm{n} \mathrm{T},$ where $\mathrm{n}$ is the sampling instant.

$\therefore$ Impulse response of the digital filter

$h(n)=\left.h(t)\right|_{t=n T}$

$=\left.\sum_{i=1}^{N} A_{i} e^{p_{i} t} u(t)\right|_{t=n T}$

$=\sum_{i=1}^{N} A_{i} e^{p_{i} n T} u(n T)$

$\therefore$ Transfer function of the digital filter

$H(z)=Z\{h(n)\}$

$=Z\left\{\sum_{i=1}^{N} A_{i} e^{p_{i} n T} u(n T)\right\}$

$=\sum_{i=1}^{N} A_{i} Z\left\{\left[\left(e^{p_{i} T}\right)^{n}\right] u(n T)\right\}$

$=\sum_{i=1}^{N} \frac{A_{i}}{1-e^{p_{i} T} z^{-1}} \cdots(3)$

(2) and (3) represents Transfer Function in analog and digital domain. They must be equal.

$\therefore \sum_{i=1}^{N} \frac{A_{i}}{s-p_{i}}=\sum_{i=1}^{N} \frac{A_{i}}{1-e^{p_{i} T} z^{-1}}$ and

$\therefore \frac{1}{s-p_{i}}=\frac{1}{1-e^{p_{i} T} z^{-1}}$

$\therefore \frac{1}{s-p_{i}}=\frac{z}{z-e^{p_{i} T}} \cdots(4)$

Given:

$H(s)=\frac{1}{s+1}$

Here, $s=-1$ is the pole in s-plane.

$\left.\text { From }(4), H(z)=\frac{z}{z-e^{-1 \times 1}} \text { (Assuming } \mathrm{T}=1\right)$

$\therefore H(z)=\frac{z}{z-0.3679}$

$\therefore$ Transfer function of the digital filter

$H(z)=\frac{z}{z-0.3679 z}$