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Bilinear Transformation Method
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  • Bilinear Transformation Method (BLT method) - Relation belween L.T and z.T is $$ S = \frac{2}{T} \frac{z-1}{T} \text { os } z=\frac{1+5 T / 2}{1-S T / 2} $$

$H(S)=\frac{2 S-t}{S^{2}-3 S+2}$

design digital filter using BLT method with $T=0.5 \mathrm{s}$

$H(z)=H(S)$ $$ S=\frac{2}{T} \frac{z-1}{Z+1}=4\left(\frac{z-1}{z+1}\right) $$ $=\frac{8\left(\frac{z-1}{z+1}\right)-1} {16\left(\frac{z-1}{z+1}\right)^{2}-3 \times 4\left(\frac{z-1}{z+1}\right)+2}$

multiply both $\mathrm{N} \& \mathrm{D}$ by $(\mathrm{z}+1)^{2}$

$=\frac{8(z-1)(z+1)-(z+1)^{2}}{16(z+1)^{2}-12(z-1)(z+1)+2(z+1) 2}$

$=\frac{8 z^{2}-8-z^{2}-2 z-1}{16 z^{2}-32 z+16-12 z^{2}+12+2 z^{2}+4 z+2}$

$=\frac{7 z^{2}-2 z-9}{6 z^{2}-28 z+30}$

Divide both $\mathrm{N} \& \mathrm{D}$ by $ \mathrm{z}^{2}$

$\frac{\frac{7}{6}-\frac{1}{5} \mathrm{z}^{-1}-\frac{3}{2} \mathrm{z}^{-2}}{1-\frac{14}{3} \mathrm{z}^{-1}+5 \mathrm{z}^{-2}}$ enter image description here

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