Solution:

Given this equation,

$ H_a(s)=\frac{3 s}{s^2+0.5 s+2}\\ $

To get $H(z)$ using the bilinear transformation, put

$ s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right) \text { in } H_a(s) .\\ $

$ H(z)=\left.H_a(s)\right|_{s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}=\left.\frac{3 s}{s^2+0.5 s+2}\right|_{s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}\\ $

$ =\frac{3 \times 2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}{\left[2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\right]^2+0.5\left[2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\right]+2}\\ $

$ =\frac{6\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}{\frac{4\left(1-z^{-1}\right)^2+\left(1-z^{-1}\right)\left(1+z^{-1}\right)+2\left(1+z^{-1}\right)^2}{\left(1+z^{-1}\right)^2}}\\ $

$ \begin{aligned} & =\frac{6\left(1+z^{-1}\right)}{4\left(1-2 z^{-1}+z^{-2}\right)+\left(1-z^{-2}\right)+2\left(1+2 z^{-1}+z^{-2}\right)} \\\\ & =\frac{6+6 z^{-1}}{7-4 z^{-1}+5 z^{-2}}\\ \end{aligned} $