Convert analogue filter into IIR digital filter with system function. The digital filter is to have resonant frequency $ω_{r} = \frac{π}{2}$ (use by bilinear transformation)

$H (s) = \frac{s + 0.1}{{(s + 0.1)}^{2} + 16}$

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LMRC Assistant Manager Electrical (Operations) : 2018 Paper

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

Answer 1

To convert analog filter into IIR digital filter using bilinear transformation, the condition is

$s = \frac{2}{T} (\frac{z - 1}{z + 1})$

$Ω_{r} = \frac{2}{T} \tan (\frac{ω_{r}}{2})$

ω_r is the resonant frequency.

At max magnitude of the filter (occurs at poles)

Poles: (s + 0.1)² + 16 = 0

⇒ s = -0.1 ± j4 = σ ± jΩ

⇒ Ω = 4

$\Rightarrow 4 = \frac{2}{T} \tan (\frac{\frac{π}{2}}{2})$

⇒ T = 0.5

$H (s) = \frac{s + 0.1}{{(s + 0.1)}^{2} + 16}$

Now replace s with $s = \frac{2}{0.4} (\frac{z - 1}{z + 1}) = 4 (\frac{z - 1}{z + 1})$

$H (z) = \frac{4 (\frac{z - 1}{z + 1}) + 0.1}{{(4 (\frac{z - 1}{z + 1}) + 0.1)}^{2} + 16}$

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