Convert analogue filter into IIR digital filter with system function. The digital filter is to have resonant frequency \({\omega _r} = \frac{\pi }{2}\) (use by bilinear transformation)\(H\left( s \right) = \frac{{s + 0.1}}{{{{\left( {s + 0.1} \right)}^2} + 16}}\)

\(H\left( z \right) = \frac{{0.128 + 0.006{z^{ - 1}} + 0.122{z^{ - 2}}}}{{1 + 0.0006{z^{ - 1}} + 0.975{z^{ - 2}}}}\)

\(H\left( z \right) = \frac{{0.128 + 0.006{z^{ - 1}} - 0.122{z^{ - 2}}}}{{1 + 0.0006{z^{ - 1}} + 0.975{z^{ - 2}}}}\)

\(H\left( Z \right) = \frac{{0.128}}{{1 + 0.0006{z^{ - 1}}}}\)

\(H\left( Z \right) = \frac{{0.006{z^{ - 1}}}}{{1 + 0.0006{z^{ - 1}}}}\)

Convert analogue filter into IIR digital filter with system function.

1.	Convert analogue filter into IIR digital filter with system function. The digital filter is to have resonant frequency \({\omega _r} = \frac{\pi }{2}\) (use by bilinear transformation)\(H\left( s \right) = \frac{{s + 0.1}}{{{{\left( {s + 0.1} \right)}^2} + 16}}\)
A.	\(H\left( z \right) = \frac{{0.128 + 0.006{z^{ - 1}} - 0.122{z^{ - 2}}}}{{1 + 0.0006{z^{ - 1}} + 0.975{z^{ - 2}}}}\)
B.	\(H\left( z \right) = \frac{{0.128 + 0.006{z^{ - 1}} + 0.122{z^{ - 2}}}}{{1 + 0.0006{z^{ - 1}} + 0.975{z^{ - 2}}}}\)
C.	\(H\left( Z \right) = \frac{{0.128}}{{1 + 0.0006{z^{ - 1}}}}\)
D.	\(H\left( Z \right) = \frac{{0.006{z^{ - 1}}}}{{1 + 0.0006{z^{ - 1}}}}\)
Answer» B. \(H\left( z \right) = \frac{{0.128 + 0.006{z^{ - 1}} + 0.122{z^{ - 2}}}}{{1 + 0.0006{z^{ - 1}} + 0.975{z^{ - 2}}}}\)

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Convert analogue filter into IIR digital filter with system function. The digital filter is to have resonant frequency \({\omega _r} = \frac{\pi }{2}\) (use by bilinear transformation)\(H\left( s \right) = \frac{{s + 0.1}}{{{{\left( {s + 0.1} \right)}^2} + 16}}\)
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Convert analogue filter into IIR digital filter with system function. The digital filter is to have resonant frequency \({\omega _r} = \frac{\pi }{2}\) (use by bilinear transformation)\(H\left( s \right) = \frac{{s + 0.1}}{{{{\left( {s + 0.1} \right)}^2} + 16}}\)

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