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notwa 2013-04-12 08:37:55 +00:00
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if you have a transfer function like,
b₀·s² + b₁·s + b₂
H(s) = ———————————————————
a₀·s² + a₁·s + a₂
whereas s would be (1 - z⁻¹)(1 + z⁻¹)e^(j·ω) in the bilinear transform,
you can find its magnitude response with this incredibly simplified equation:
(b₀·x)² - (2·b₀·b₂ - b₁²)·W·x·y + (b₂·W·y)²
|H(j·ω)|² = —————————————————————————————————————————————
(a₀·x)² - (2·a₀·a₂ - a₁²)·W·x·y + (a₂·W·y)²
(analog) x = ω²
y = 1
W = ω₀²
(digital) x = sin(ω2)²
y = cos(ω2)²
W = tan(ω₀2)²
whereas ω is the input frequency in rads/sec
ω₀ is the center frequency in rads/sec
and the phase? maybe some other time
note: I'm no math genius and there's probably an error in here