2013-04-12 01:37:55 -07:00
|
|
|
|
if you have a transfer function like,
|
|
|
|
|
|
|
2013-04-12 04:02:02 -07:00
|
|
|
|
b2·s² + b1·s + b0
|
2013-04-12 01:37:55 -07:00
|
|
|
|
H(s) = ———————————————————
|
2013-04-12 04:02:02 -07:00
|
|
|
|
a2·s² + a1·s + a0
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
|
|
|
|
|
whereas s would be (1 - z⁻¹)∕(1 + z⁻¹)∕e^(j·ω) in the bilinear transform,
|
2013-04-12 04:02:02 -07:00
|
|
|
|
you can find its magnitude response with this equation:
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
2013-04-12 04:02:02 -07:00
|
|
|
|
(b2·x)² - (2·b2·b0 - b1²)·W·x·y + (b0·W·y)²
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|H(j·ω)|² = —————————————————————————————————————————————
|
2013-04-12 04:02:02 -07:00
|
|
|
|
(a2·x)² - (2·a2·a0 - a1²)·W·x·y + (a0·W·y)²
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
|
|
|
|
|
(analog) x = ω²
|
|
|
|
|
|
y = 1
|
2013-04-12 04:02:02 -07:00
|
|
|
|
W = ω0²
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
|
|
|
|
|
(digital) x = sin(ω∕2)²
|
|
|
|
|
|
y = cos(ω∕2)²
|
2013-04-12 04:02:02 -07:00
|
|
|
|
W = tan(ω0∕2)²
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
2013-04-12 04:02:02 -07:00
|
|
|
|
whereas ω is the physical frequency in rads/sec
|
|
|
|
|
|
ω0 is the center frequency in rads/sec
|
2013-04-12 01:37:55 -07:00
|
|
|
|
|
|
|
|
|
|
and the phase? maybe some other time
|
|
|
|
|
|
note: I'm no math genius and there's probably an error in here
|
