A continuous-time filter with transfer function  $\mathrm{H}\left(\mathrm{s}\right)=\frac{2\mathrm{s}+6}{{\mathrm{s}}^2+6\mathrm{s}+8}$  is converted to a discretetime filter with transfer function $\mathrm{G}\left(\mathrm{z}\right)=\frac{2{\mathrm{z}}^2-0.5032\mathrm{z}}{{\mathrm{z}}^2-0.5032\mathrm{z}+\mathrm{k}}$  so that the impulse response of the continuous-time filter, sampled at $2\mathrm{H}\mathrm{z}$, is identical at the sampling instants to the impulse response of the discrete time filter. The value of $\mathrm{k}$ is ________

$\mathrm{H}\left(\mathrm{s}\right)=\frac{2\mathrm{s}+6}{{\mathrm{s}}^2+6\mathrm{s}+8}$ and

$\mathrm{G}\left(\mathrm{z}\right)=\frac{2{\mathrm{z}}^2-0.5032\mathrm{z}}{{\mathrm{z}}^2-0.5032\mathrm{z}+\mathrm{k}}$                                     

Now poles of $\mathrm{H}(\mathrm{s})$ are at ${\mathrm{s}}^2+6\mathrm{s}+8=0$

$\Rightarrow (\mathrm{s}+4)(\mathrm{s}+2)=0$

$\Rightarrow \mathrm{s}=-4,-2$

Now poles will get mapped to z plane according to $\mathrm{z}={\mathrm{e}}^{\mathrm{s}{\mathrm{T}}_{\mathrm{s}}}$

We have, sampling period ${\mathrm{T}}_{\mathrm{s}}=1/2\mathrm{s}\mathrm{e}\mathrm{c}$

and ${\mathrm{s}}_1=-4$ and ${\mathrm{s}}_2=-2$

$\Rightarrow {\mathrm{z}}_1={\mathrm{e}}^{-4\times 1/2}={\mathrm{e}}^{-2}$

and ${\mathrm{z}}_2={\mathrm{e}}^{-2\times 1/2}={\mathrm{e}}^{-1}$

Thus, ${\mathrm{z}}_1$ and ${\mathrm{z}}_2$ must be poles of $\mathrm{G}(\mathrm{z})$

⇒ Denominator of $\mathrm{G}(\mathrm{z})$ will be $(\mathrm{z}–{\mathrm{z}}_1)(\mathrm{z}–{\mathrm{z}}_2)$

$\Rightarrow {\mathrm{z}}^2-\left(\frac{1}{\mathrm{e}}+\frac{1}{{\mathrm{e}}^2}\right)\mathrm{z}+\frac{1}{\mathrm{e}}.\frac{1}{{\mathrm{e}}^2}$

$\Rightarrow {\mathrm{z}}^2–0.5032\mathrm{z}+0.0498$

Comparing with ${\mathrm{z}}^2–0.5032\mathrm{z}+\mathrm{k}$