$H(s)=\frac{3}{((s+2)(s+3))}$

Limitation:

• We know that Ω is analog frequency and its range is from $\frac{π}{T_s}$ to $\frac{-π}{T_s}$ . While the digital frequency ωvaries from -π to π. That total means from $\frac{π}{T_s}$ to $\frac{-π}{T_s}$ .ω maps from -π to π. Let K be any integer. Then we can write the general range of Ω as $(K-1)\frac{π}{T_s}$ to $(K+1)\frac{π}{T_s}$ ; But for this range also , ω maps from -π to π Thus mapping from analog from analog frequency Ω to digital frequency ωis many to one. This mapping is not one to one.
• Analog filters are not band limited so there will be aliasing due to the sampling process. Because of this aliasing, the frequency response of resulting digital filter will not be identical to the original frequency response of analog filter.
• The change in the value of sampling time $(T_s )$ has no effect on the amount of aliasing

Now,

$H(s)=\frac{3}{((s+2)(s+3))}$ ………………(i)

By using partial fraction,

$\frac{3}{((s+2)(s+3))}$ = $\frac{A}{(S+2)}$+$\frac{B}{(S+3)}$ ………………(ii)

$3=A (s+3)+B (s+2)$

Put s=-3; B=-3

Put s=-2; A=3

Put the value of A and B in equation (i)

$H(s)=\frac{3}{(s+2)}-\frac{3}{(s+3)}$

We have transformation equation

$\frac{1}{(s-P_k )} → \frac{1}{(1-e^{P_k T_S } Z^{-1} }$

$\frac{1}{(s+2} → \frac{1}{1-e^{-2(0.1)} Z^{-1}} = \frac{1}{(1-0.818 Z^{-1} }$

$\frac{1}{(s+3)} → \frac{1}{(1-e^{-3(0.1)} Z^{-1}}= \frac{1}{(1-0.740 Z^{-1}}$

$H(z)=\frac{3Z}{(Z-0.818)}-\frac{3Z}{(Z-0.740)}$