response.txt

@@ -1,26 +1,26 @@
 if you have a transfer function like,
 
-        b₀·s² + b₁·s + b₂
+        b2·s² + b1·s + b0
 H(s) = ———————————————————
-        a₀·s² + a₁·s + a₂
+        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 incredibly simplified equation:
+you can find its magnitude response with this equation:
 
-             (b₀·x)² - (2·b₀·b₂ - b₁²)·W·x·y + (b₂·W·y)²
+             (b2·x)² - (2·b2·b0 - b1²)·W·x·y + (b0·W·y)²
 |H(j·ω)|² = —————————————————————————————————————————————
-             (a₀·x)² - (2·a₀·a₂ - a₁²)·W·x·y + (a₂·W·y)²
+             (a2·x)² - (2·a2·a0 - a1²)·W·x·y + (a0·W·y)²
 
 (analog)        x = ω²
                 y = 1
-                W = ω₀²
+                W = ω0²
 
 (digital)       x = sin(ω∕2)²
                 y = cos(ω∕2)²
-                W = tan(ω₀∕2)²
+                W = tan(ω0∕2)²
 
-whereas         ω is the input frequency in rads/sec
-                ω₀ is the center frequency in rads/sec
+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