Going from digital to analog filters

Question

I've been learning about digital signal processing for some time, and I understand the principles in the digital world, as well as the conversion of continuous filters in the s-domain to digital filters in the z-domain.

I'd love to learn more about the process of designing analog filters and converting this into circuits that I can try and test. Since I cannot brag about my knowledge in electronics, with the exception of a little experience dabbling with Arduino projects, I don't see how an equation in the s-domain can be translated to a circuit. From what little I have found, most sources go the other way, taking a circuit and analysing it in the s-domain, and the equations that are given usually aren't explained how they came about. This makes it somewhat hard to understand the relationship between the math and circuits. As far as I understand, the sources generally have equations readily available for simple circuits, and rely mostly on theory of parallel and serial filters to derive larger circuits — is this correct? If so, is this the be all end all method of creating larger circuits, or what is the "correct" approach?

My goal with this is to 1) gain a better understanding of designing analog filters, and 2) take my own filters and make circuits for say guitar pedals and similar applications.

To be honest, I'd love to take any shortcut methods (minimal reading, maximum hands-on and results for motivation), such that I can get some circuits going right away, and then later take a deeper learning approach later on. Similarly, any "catches" or "gotya's" that I need to know about circuits, such that I don't end up blowing myself up, would be nice to know about ahead of time.

What is your recommended learning route to get started as fast as possible? Do you have any recommended material (articles, books, videos, etc.) for a speedy start?

Answer 1

The main trick is knowing how typical lumped circuit element impedances are represented in the $s$-domain. Recall the current-voltage relationship for a capacitor:

$$ i(t) = C \frac{dv(t)}{dt} $$

Transforming this to the Laplace domain yields:

$$ I(s) = CsV(s) $$

Note that voltage and current still have a proportional relationship in this domain, just like the time domain. This allows us to write a (frequency-dependent) $s$-domain impedance for the capacitor. This impedance can be used with Ohm's law and techniques like nodal analysis to calculate what a circuit's Laplace-domain output would be for a given Laplace-domain input signal (e.g. an impulse, whose Laplace transform is unity).

For a capacitor, this impedance is:

$$ Z(s) = \frac{V(s)}{I(s)} = \frac{1}{Cs} $$

This gives capacitors their known highpass characteristic; for $s=j\omega$, as $\omega$ gets larger, the impedance gets smaller. For the limiting case of $\omega=0$, $Z(0) = \infty$; a capacitor passes no DC current.

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We can follow the same procedure for an inductor, starting with its current-voltage relationship:

$$ v(t) = L\frac{di(t)}{dt} $$

Apply the Laplace transform:

$$ V(s) = LsI(s) $$

and again, we can calculate the inductor's $s$-domain impedance:

$$ Z(s) = \frac{V(s)}{I(s)} = Ls $$

As the dual to capacitors, inductors have a lowpass characteristic. For small frequencies ($s \to 0$), $Z(s) \to 0$ as well.

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Given these impedances, it becomes very straightforward to apply basic circuit analysis techniques to determine the $s$-domain transfer function of a network containing resistors, capacitors, and inductors.

Answer 2

As far as I understand, the sources generally have equations readily available for simple circuits, and rely mostly on theory of parallel and serial filters to derive larger circuits — is this correct? If so, is this the be all end all method of creating larger circuits, or what is the "correct" approach?

Pretty much. In the analog domain you still typically want to use cascaded second-order sections for accuracy, just like you do in the digital domain. The various standard biquad blocks (Sallen-Key) have been around for a long time, so you just copy and paste them, put as many blocks in your circuit as you need, and choose their component values to create the transfer function you want.

You need to group the poles and zeros so that each stage has optimum dynamic range. You don't want a huge peak in the frequency response of one stage that will clip the signal, followed by a big notch that will attenuate it back down to flat overall.

Within a stage, usually the ratios between the components are what's important, so there are an infinite number of circuits with the same transfer function. You choose the absolute values of the components for other reasons, like minimizing noise, distortion of the op-amp, power/current limits, etc.

Answer 3

So you can implement an analog transfer function that is a rational function of $s$ using a canonical form just like the Direct Form II:

but instead of a unit sample delay $z^{-1}$, you replace that with an analog integrator $s^{-1}$. the integrators are little circuits with an op-amp, capacitor, and resistor:

the transfer function of the integrator is $\frac{-1}{RC} s^{-1}$. that $\frac{-1}{RC}$ becomes a factor of the feedback and feedforward coefficients.

and you replace $x(n)$ with $x(t)$ and similarly to all other signals. for your adder (or subtractor) circuit, something like this:

the large case Vn are added and the lower case vm are subtracted and the coefficient is inversely proportional to the resistance (Rn, rm). keep in mind the minus sign you pick up from the integrators.

the transfer function turns out just like in the digital filter except it's $H(s)$ instead of $H(z)$

$$\begin{align} H(s) &= \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}}{1 + a_1 s^{-1} + a_2 s^{-2}} \\ \\ &= \frac{b_0 s^2 + b_1 s + b_2}{s^2 + a_1 s + a_2} \\ \end{align}$$

and if you have a digital filter that works pretty good for you, you can map it to an $s$-plane filter using the bilinear transform in reverse.