Every electronic circuit has an analogous mechanical circuit whose motion is governed by an equivalent differential equation. Analog synthesizers are electronic circuits, so analog synthesizers have mechanical analogues -- analog analogue synthesizers.

The heart and soul of any classic subtractive synth is its filter section. Most use a ladder circuit of RC filters. This takes a raw waveform with a lot of harsh high frequency content (like a sawtooth or triangle wave), and turns it into something more musical.

![](/img/rc_ladder.jpg#c)

Two rungs (pictured) give a two pole filter, and four, well, four. I'd like to make a mechanical analog of one such rung. If things go well, I can make multiple and chain them together, just like in the electronic circuit.

## Will it work?

In a normal analog synthesizer, voltage is used to represent air pressure (or more directly, the displacement of a speaker cone). In my analog analogue synthesizer, voltage will be replaced with force. The common electrical components then each have their own analog:

- resistance <---> damping

- capacitance <---> springiness

- inductance <---> mass

So each rung of the ladder circuit will be, in my mechanical version, a damped oscillator.

To be more precise, a normal RC circuit is described by the following ODE:

$$rc \dot{y}(t) + y(t) = x(t)$$

Here \\(r\\) is the resistance in ohms, \\(c\\) the capacitance in farads, \\(x(t)\\) is the input voltage, and \\(y(t)\\) the output voltage (i.e. what we'll need to solve for).

A driven massless damped oscillator is described by this equation:

$$r \dot{y}(t) + k y(t) = x(t)$$

Here \\(r\\) is the damping coefficient (in Newtons per meters per second), \\(k\\) the spring constant (in Newtons per meter), \\(x(t)\\) the input force, and \\(y(t)\\) the displacement of the oscillator.

For appropriate values of the constants these two systems will clearly have the same behavior. So there's hope!

In order to keep forces and velocities within reasonable ranges, I'll aim to operate 100 times slower than normal audio. So a concert A, instead of being 440 Hz, will be just 4.4.

## How would it be made?

Since we'll represent the waveform-to-filter using force, the input has to be a force control servo. I know this can be done either via current measurement, or with a load cell looped into the controller. How hard could it be?

The damping could be provided with a viscous fluid. But, I don't know of viscous fluid dampers that can adjust their viscosity on the fly, which we'll need to do if we want to sweep the cutoff frequency of our filter. Which obviously we do. I only know of one other way to get a resistance force directly proportional to velocity: eddy braking. So we'll need an electromagnet (vary the current to vary the resistance) and an aluminum rail. What could go wrong?

The capacitor, luckily, just beomes a spring. Phew.

Now a normal RC filter contains no inductor. It would be difficult to make my mechanical filter's moving parts massless<sup>[citation needed]</sup>, but luckily this isn't strictly necessary. All RC circuits contain some self-inductance, as all mechanical circuits contain some mass. I'll just try to keep my filter's mass to a minimum relative to the damping and spring factors.

Finally, instead of measuring the output voltage across the capacitor, we'll be measuring the output force across the spring. So we need a load cell.

All in all, here's a simple sketch of the main components for one stage of the filter. If all goes well I'll build multiple of these.

![](/img/mechanical_design.jpg#c)

## Frequency Response Analysis

Since our mechanical oscillator won't be massless, we might as well model it with some mass. At least then we know what to expect. This gives us a second order ODE:

$$m \ddot{y}(t) + r \dot{y}(t) + k y(t) = x(t)$$

(The symbols are defined as above, except for the addition of the mass \\(m\\).)

Let the input function be \\(x(t) = A \sin(\omega t)\\). Then our ODE is solved by:

$$y(t) = \frac{-A r \omega \cos(\omega t) + A (k - m \omega^2) \sin(\omega t)}{r^2 \omega^2 + (k - m \omega^2)^2}$$

It has an amplitude of

$$\frac{A}{\sqrt{r^2 \omega^2 + (k - m \omega^2)^2}}$$

For completeness, the first derivative is

$$\dot{y}(t) = \frac{A \omega (k - m \omega^2) \cos(\omega t) + A r \omega^2 \sin(\omega t)}{r^2 \omega^2 + (k - m \omega^2)^2}$$

and its amplitude

$$\frac{A \omega}{\sqrt{r^2 \omega^2 + (k - m \omega^2)^2}}$$

The output of our filter is the force exerted by the spring, or \\(k y(t)\\). I also care about the maximum displacement, the maximum velocity, and the maximum force exerted by the damping system.