IIR filter design in digital domain using Magnitude Squared

Question

Does anyone have any good references for deriving parameters of an IIR Low pass/High Pass filter directly in the digital domain using the magnitude squared at the corner frequency?

I have been able to derive the parameters of a first order Low/High pass filter with 3 dB attenuation at the corner frequency i.e. calculating K and alpha in:-

H(Z) = k(1+z^-1)/(1-alpha*z^-1)

My issue is that I distinctly remember deriving the parameters using a 6 dB attenuation at the corner frequency in a DSP course I have done previously but I have forgotten the trigonometric identiftes used to finish the derivation.

The general procedure is as follows:- 1) Let w = 0/pi to calculate the gain term k such that there is a 0dB gain at 0/pi 2) Calculate the magnitude squared at the corner frequency to obtain a value for alpha in terms of the corner frequency.

The problem may be that it should be a second order filter or I am recalling the method for a band pass/stop filter but I'm not sure and it appears this method is not used very often except in the case of band pass/stop filters for parametric EQ.

I hope the question is clear and I will try to improve the structure with the responses so it will be useful for others. Any help will be apreciated

Answer 1

To solve the case that you mentioned...

You have 2 variables to determine, so you need two relationships to resolve the two variables. I'm going to use $k$ and $a$ as the variables to make this easy to type up.

$H(z) = k\frac{1 + z^-1}{1-az^-1}$

Start by considering the passband gain. Use $f = 0$ for this. Assume you want unity gain at $f_0=0$.

Assume: $H(f_0) = 1, f_0 = 0$

Substitute $e^{i2\pi f/fs}$ for z, fs is your sampling rate, set $f = 0$ and solve for $k$ to satisfy $H(f_0) = 1$

From this you get $k = \frac{(1-a)}{2}$

Now work on the gain squared at your desired corner frequency ($f_c$) to determine $a$.

$H(f_c) = -3$dB (magnitude squared will be -6dB as you've stated)

We'll work with the magnitude squared at $f_c$ and set the gain to 1/2 (-6dB).

$|H(f_c)|^2 = \frac{1}{2}$

This time substitute $e^{i2\pi fc/fs}$ for z.

To simplify the arithmatic you can solve this equation:

$$ \left(\frac{|H(f0)|}{|H(fc)|}\right)^2 = 2 $$

This eliminates the factor $k$.

You will end up with a quadradic relationship in $a$. Solving for $a$ yields:

$ a = \frac {1 - \sqrt{1-cos^2(2\pi\frac{f_c}{f_s})}} {cos(2\pi\frac{f_c}{f_s}) }$

Simplify this to:

$ a = \frac {1 - sin(2\pi\frac{f_c}{f_s})} {cos(2\pi\frac{f_c}{f_s}) }$