Solution:

Bilinear or Tustin Transformation:

To overcome the difficulty of the impulse invariant method, bilinear transformation may be used. Tn this method the entire $j \Omega$ axis for $-\infty\lt\Omega\lt\infty$ maps uniquely onto the unit circle for $-\pi\lt\omega\lt\pi$.

The bilinear transform provides a nonlinear one-to-one mapping of the frequency points on the jr axis in the 's' plane to those on the unit circle in the ' z ' plane, thus this procedure allows us to implement any kind of digital filter.

The digital filter derived from this method has approximately the same time domain response as the original analogy filter for any type of input.

Let us consider a simple analog filter (analog integrator) as shown in the figure,

Take L.T on both sides of $y(t)$ we get,

$ Y(s)=\frac{1}{s}[x(s)-x(0)] \text {. Assume } x(0)=0 \text {. }\\ $

Thus, $Y(s)=\frac{X(s)}{s}$.

$ \text { or } H(s)=\frac{Y(s)}{x(s)}=1 / s \cdot\\ $

Take $L^{-1}$ on both sides of equ (i), we get, $h(t)=u(t) ; \quad$ for $|t| \geq 0$. $=0$, elsewhere.

The $\%$ response to an any arbitrary input is determined by using convolution integral,

$ \text { ie; } y(t)-\int_0^t x(\tau) h(t-\tau) d \tau \cdot .-\text { (3) }\\ $

$ \begin{aligned}\\ &\left\langle S=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)=\frac{2}{T}\left(\frac{z-1}{z+1}\right)\right. \\\\ &\text { Thus } H(z)=H(s) /(8)\\ \end{aligned}\\ $

, wee know, $S=\frac{2}{T}\left(\frac{z-1}{z+1}\right)$

$ \begin{aligned}\\ &\frac{S T}{2}=\frac{Z-1}{Z+1} \\\\ &S T Z+S T=2 Z-2 \\\\ &S T Z-2 Z=-2-S T\\ \end{aligned}\\ $

$ \begin{aligned}\\ z(S T-2) &=-(2+S T) \\\\ z &=\frac{-(2+S T)}{S T-2} \\\\ &=\frac{2+S T}{2-S T}=\frac{1+T / 2 S}{1-T / 2 S} .\\ \end{aligned}\\ $

$ \begin{aligned}\\ S &=\sigma+j \Omega \text {. then, } \\\\ z &=\frac{1+\frac{T}{2}(\sigma+j \Omega)}{1-\frac{T}{2}(\sigma+j \Omega)}=\frac{\left(1+\frac{T}{2}\\ \sigma\right)+j \frac{T \Omega}{2}}{\left(1+\frac{T}{2} \sigma\right)-j \frac{T \Omega}{2}}\\ \end{aligned}\\ $

$ |z|=\left|\frac{\left(1+\frac{\pi}{2} \sigma\right)^2+\left(\frac{\pi}{2} \Omega\right)^2}{\left(1-\frac{\pi}{2} \sigma\right)^2+\left(\frac{T}{2} \Omega\right)^2}\right|^{1 / 2}\\ $

When $\operatorname{Re}(s)=\sigma \rightarrow 0$ then $|z|\gt1$ $\sigma\lt0$, then $|z|\lt1$. $\sigma=0$, then $|z|=1$

Therefore the ' $j \Omega$ ' axis maps onto the unit circle $|z|=1$, the LHS of $s$ plane maps onto the interior part of the unit circle $|z|=1$, and the RHS of 's' plane maps onto the outside the unit are $|z|=1$. As a result, of this "the stable analog filter is transformed into a stable digital, and $z=e^{j w}$ in eqn. $ \begin{aligned}\ S &=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right) \\ j \Omega &=\frac{2}{T}\left(\frac{1-e^{-j \omega}}{1+e^{-j \omega}}\right)\ \end{aligned}\ $ $ =\frac{2 e^{-j \frac{\omega}{2}}}{T_e^{-j \frac{\omega}{2}}}\left(\frac{e^{j \omega / 2}-e^{-j \frac{\omega}{2}}}{e^{j \omega / 2}+e^{-j \omega / 2}}\right)\ $ $ \begin{gathered}\ j \Omega=j \frac{2}{T} \tan \left(\frac{\omega}{2}\right) \\ \Omega=\frac{2}{T} \tan \left(\frac{\omega}{2}\right) \rightarrow \\ \frac{\Omega T}{2}=\tan \left(\frac{\omega}{2}\right) \\ \tan ^{-1}\left(\frac{\Omega T}{2}\right)=\frac{\omega}{2} \\ \omega=2 \tan ^{-1}\left(\frac{\Omega T}{2}\right) \rightarrow \end{gathered}\ $ Eqn, shows that there is a nonlinear relationship between analog frequency $(\Omega)$ and digital frequency $(\omega)$.