From 097b5e519bb6e08fbe1fd27cf69ef8fee579a350 Mon Sep 17 00:00:00 2001 From: notwa Date: Fri, 12 Apr 2013 08:37:55 +0000 Subject: [PATCH 1/3] --- gistfile1.txt | 26 ++++++++++++++++++++++++++ 1 file changed, 26 insertions(+) create mode 100644 gistfile1.txt diff --git a/gistfile1.txt b/gistfile1.txt new file mode 100644 index 0000000..042d40e --- /dev/null +++ b/gistfile1.txt @@ -0,0 +1,26 @@ +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 From b739c4b6c0eb2c5789e8762af7590db58167f4c2 Mon Sep 17 00:00:00 2001 From: notwa Date: Fri, 12 Apr 2013 08:39:43 +0000 Subject: [PATCH 2/3] --- gistfile1.txt => response.txt | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename gistfile1.txt => response.txt (100%) diff --git a/gistfile1.txt b/response.txt similarity index 100% rename from gistfile1.txt rename to response.txt From b26f4074ec6430afda3ec844eef096fd4030ddd5 Mon Sep 17 00:00:00 2001 From: notwa Date: Fri, 12 Apr 2013 11:02:02 +0000 Subject: [PATCH 3/3] --- response.txt | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/response.txt b/response.txt index 042d40e..cc4cf77 100644 --- a/response.txt +++ b/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