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