Modeling Analog Filters for Software Implementation

Question

I am curious about software implementation of analog filters. For example, let's say I have a moog low pass filter. What should I do to turn this device into satisfactory, analog sounding software.

I don't actually want programming advice(doesn't mean I won't be grateful if given), what I want is to know what coefficients, parameters I need to extract from this device to write this program and how am I supposed to extract them.

If I want to write a program that behaves exactly the same as the device, what method would you suggest?

Do you think changing the variables such as cutoff, resonance, eg amount etc. continuously on the device and collect values ​​for all combinations of parameter configurations while sending some sort of input to the device would get me through response of the filter? If so, how should I use these collected values ​​while recreating the analog device programmatically?

All of the assumptions I made above may be wrong, any guidance is welcomed. Thanks in advance.

PS: Looks like I wasn't clear enough about the linear/non-linear characteristics. EDIT: The desired modeling implied in this question includes the non-linear characteristic as well as linear.

Answer 1

Modeling real analog components with their non-linear behavior can be a challenge to do digitally. I suggest starting with wave digital filters. Quoting from the linked source:

A Wave Digital Filter (WDF)is a particular kind of digital filter based on physical modeling principles. Unlike most digital filter types, every delay element in a WDF can be interpreted physically as holding the current state of a mass or spring (or capacitor or inductor). WDFs can also be viewed as a particular kind of finite difference scheme having unusually good numerical properties. ... WDFs have been applied often in music signal processing.

In addition to linear applications there are ways to model non-linear devices with this structure as you would need for your application. A basic component of filters and amplifying structures are three terminal devices. A particularly recent reference is Wave Digital Modeling of Nonlinear 3-terminal Devices for Virtual Analog Applications.

As an example, this paper presents as an example modeling this amplifier circuit:

arriving at the following wave digital implementation:

You may also be interested in Enhanced Wave Digital Triode Model for Real-Time Tube Amplifier Emulation(sorry for the paywall) and Virtual Analog Modeling in the Wave-Digital Domain.

Some additional resources I found related to the original question:

• Generalized Moog Ladder Filter- Part I–Linear Analysis and Parameterization
• Generalized Moog Ladder Filter- Part II– Explicit Nonlinear Model through a Novel Delay-Free Loop Implementation Method

Answer 2

If it were a linear filter, it'd be easy. The following doesn't apply to your Moog filters, apparently.

This approach is covered extensively in the literature, so the short answer is to get a book on DSP*, read up through the chapter on implementing a DSP filter from an analog prototype, then do what it says.

There's several different methods, each of which is good in its way.

1. Use the bilinear transform, possibly with frequency warping. This is covered in every book, and if you're sampling fast enough, it works very well.
2. Take your anti-alias and reconstruction filters into account, and design digital filters that give you a similar outcome in the frequency domain.

If I want to write a program that behaves exactly the same as the device, what method would you suggest?

You can't -- you can only come close enough. But then, if you had a dozen "identical" devices, even pro-quality ones, they wouldn't really be the same either.

Coming close enough involves sampling fast enough, getting your anti-aliasing and reconstruction filters right, and not screwing up your data paths in the digital realm.

"How fast do I sample?" -- you start with a guess (I'd go with 96 or 192ksps, as a handy 2x or 4x of a typical sampling rate). Then you design prototype filters and look at their frequency response, then you run actual audio through them and run them by the shiniest golden ears you can and you get an opinion.

* I.e. "Understanding Digital Signal Processing" by Rick Lyons.

Answer 3

I try to describe the linear filter aspects of the 4-pole Moog.

The 4-pole Moog filter is a cascade of four 1-pole LPF filters that have resonant frequency adjustable and a feedback path that has the feedback gain adjustable (on the patchboard Moog synths, this feedback gain knob was labeled "Regeneration"). The resonant frequency filters is adjustable with voltage control (they're a "VCF") that had the control voltage scaled the same as the VCO. If the Regeneration knob was cranked up high enough, the filter naturally oscillated and then the only thing that keeps the amplitude from increasing without bound in a runaway manner is the nonlinearity of the system.

This answer does not deal with the nonlinear description.

In the analog, continuous-time universe, a simple 1-pole filter looks like:

$$ H_1(s) = \frac{1}{1 + \frac{s}{2 \pi f_0}} $$

where $f_0$ is the -3 dB corner frequency of the 1-pole filter. Without feedback, four of these identical filters in series have a transfer function that looks like:

$$\begin{align} H_4(s) &= \big( H_1(s) \big)^4 \\ \\ &= \left( \frac{1}{1 + \frac{s}{2 \pi f_0}} \right)^4 \\ \end{align}$$

Now, if you put in a little negative feedback with "regenerative gain" $G_\mathrm{R}$, in the Laplace Transform domain, the signal path is described by:

$$ Y(s) = H_4(s) \big(X(s) - G_\mathrm{R} Y(s) \big)$$

where $X(s)$ is the transform of the input and $Y(s)$ is the transform of the output. We want to solve for $Y(s)$.

$$\begin{align} Y(s) + H_4(s) G_\mathrm{R} Y(s) &= H_4(s) X(s) \\ \\ Y(s) \big(1 + H_4(s) G_\mathrm{R} \big) &= H_4(s) X(s) \\ \\ Y(s) &= \frac{H_4(s)}{1 + H_4(s) G_\mathrm{R}} X(s) \\ \end{align}$$

The transfer function is whatever multiplies $X(s)$.

$$\begin{align} H(s) &= \frac{Y(s)}{X(s)} \\ \\ &= \frac{\left( \frac{1}{1 + \frac{s}{2 \pi f_0}} \right)^4}{1 + G_\mathrm{R} \left( \frac{1}{1 + \frac{s}{2 \pi f_0}} \right)^4 } \\ \\ &= \frac{1}{\left( 1 + \frac{s}{2 \pi f_0} \right)^4 + G_\mathrm{R} } \\ \\ \end{align}$$

The poles of $H(s)$ are whatever values of $s$ that make the denominator go to zero.

$$\begin{align} 0 &= \left( 1 + \frac{s}{2 \pi f_0} \right)^4 + G_\mathrm{R} \Big|_{s = p} \\ \\ \left( 1 + \frac{p}{2 \pi f_0} \right)^4 &= -G_\mathrm{R} \\ \\ \left( 1 + \frac{p_n}{2 \pi f_0} \right)^4 \left(e^{-j 2\pi \frac{n}{4}} \right)^4 &= -G_\mathrm{R} \qquad \qquad \qquad n \in \mathbb{Z} \\ \\ \left( 1 + \frac{p_n}{2 \pi f_0} \right)^4 &= (-1)G_\mathrm{R} \left(e^{j 2\pi \frac{n}{4}} \right)^4 \\ \\ 1 + \frac{p_n}{2 \pi f_0} &= (-1)^{\frac{1}{4}}\big(G_\mathrm{R}\big)^{\frac{1}{4}} e^{j 2\pi \frac{n}{4}} \\ \\ 1 + \frac{p_n}{2 \pi f_0} &= e^{j \frac{\pi}{4} } \big(G_\mathrm{R}\big)^{\frac{1}{4}} e^{j 2\pi \frac{n}{4}} \\ \end{align}$$

because $ \left(e^{-j 2\pi \frac{n}{4}} \right)^4 = 1 $ for any integer $n$.

So the normalized pole is:

$$ \frac{p_n}{2 \pi f_0} \ = \ -1 \ + \ \big(G_\mathrm{R}\big)^{\frac{1}{4}} \ e^{j 2\pi (\frac{1}{8}+\frac{n}{4})} \qquad \qquad \qquad n \in \mathbb{Z}$$