27 lines
1.0 KiB
Plaintext
27 lines
1.0 KiB
Plaintext
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if you have a transfer function like,
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b₀·s² + b₁·s + b₂
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H(s) = ———————————————————
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a₀·s² + a₁·s + a₂
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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 incredibly simplified equation:
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(b₀·x)² - (2·b₀·b₂ - b₁²)·W·x·y + (b₂·W·y)²
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|H(j·ω)|² = —————————————————————————————————————————————
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(a₀·x)² - (2·a₀·a₂ - a₁²)·W·x·y + (a₂·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 input 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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