From 097b5e519bb6e08fbe1fd27cf69ef8fee579a350 Mon Sep 17 00:00:00 2001
From: notwa <cloningdonor@gmail.com>
Date: Fri, 12 Apr 2013 08:37:55 +0000
Subject: [PATCH]

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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(ω∕2)²
+                y = cos(ω∕2)²
+                W = tan(ω₀∕2)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