IIR filter design in digital domain using the 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 $3\textrm{ dB}$ attenuation at the corner frequency i.e. calculating $k$ and $\alpha$ in:

$$H(z) = k\frac{\left(1+z^{-1}\right)}{\left(1-\alpha z^{-1}\right)}$$

My issue is that I distinctly remember deriving the parameters using a $6\textrm{ dB}$ 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 $\omega = 0/\pi$ to calculate the gain term $k$ such that there is a $0\textrm{ dB}$ 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 appreciated.

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

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

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

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

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

$\lvert H\left(f_c\right)\rvert^2 = \frac{1}{2}$

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

To simplify the arithmatic you can solve this equation:

$$ \left(\frac{\lvert H(f_0)\rvert}{\lvert H(f_c)\rvert}\right)^2 = 2 $$

This eliminates the factor $k$.

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

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