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+if you have a transfer function like,
+
+        b2·s² + b1·s + b0
+H(s) = ———————————————————
+        a2·s² + a1·s + a0
+
+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 equation:
+
+             (b2·x)² - (2·b2·b0 - b1²)·W·x·y + (b0·W·y)²
+|H(j·ω)|² = —————————————————————————————————————————————
+             (a2·x)² - (2·a2·a0 - a1²)·W·x·y + (a0·W·y)²
+
+(analog)        x = ω²
+                y = 1
+                W = ω0²
+
+(digital)       x = sin(ω∕2)²
+                y = cos(ω∕2)²
+                W = tan(ω0∕2)²
+
+whereas         ω is the physical 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