What are these and how do they work?! (source, source, source)
In an early circuits class, we learn about low-pass filter topologies such as RC and LR. We also learn that we can build high-pass, band-pass, and notch filters using passive RLC components. These filters are good enough, up to the mid MHz range, but into the high MHz and GHz range our RLC components start acting very bizarrely — capacitors start developing inductance, inductors start developing capacitance, and resistors start acting more like inductors. Noticing our SMD/axial components becoming less reliable, we try using filters packaged onto chips etched in silicon, but we notice something very strange — the magnitude of our high-frequency output signal is much smaller than expected! This is because silicon is such a lossy substrate that most of our signal is lost before it even leaves the chip. Remember that in high frequency applications, the signals are not in the wires themselves but rather travel through the air around them, including through the PCB substrate.
If discrete components don’t work, and chips don’t work, then we need a new technique to design passive filters, and new techniques to analyze such circuits. These passive filters will be made of traces printed on a low-loss substrate.
tl;dr we first come up with the desired cutoff frequencies and passband, then write the transfer function, then translate that into a circuit.
We will look at the design process of 2 different classes of filters here, Butterworth and Chebyshev. Note that these names describe their transfer functions, so there are multiple circuit topologies that correspond to these transfer functions. Let $H(\omega)$ denote a circuit’s transfer function and $G(\omega)$ denote its gain.
Since these are low-pass filters with a maximally flat passband, we want $G(\omega)$ to have the property that its derivatives at $\omega=0$ are all 0, up to the (2N-1)th derivative. A lot of polynomials satisfy this property, but we end up picking
$$ G(\omega) = \frac{G_0}{\sqrt{1+\left(\frac{\omega}{\omega_c}\right)^{2N}}} \implies \frac{1}{\sqrt{1+\omega^{2N}}} $$
where $\omega$ is in rad/s, N is the number of poles (aka number of reactive components) the filter has, $G_0$ is the DC gain, and $\omega_c$ is the cutoff frequency. (If we normalize the coefficients so $G_0=1, \omega_c=1$, we get the equation on the right.) We cannot directly equate G to H, but we know they are equal in magnitude, so we can write:
$$ \displaystyle G^2(\omega) = |H(j\omega)|^2 = H(j\omega)H^*(j\omega)=\frac{G_0^2}{1+\left(\frac{\omega}{\omega_c}\right)^{2N}} \implies \frac{1}{1+\omega^{2N}}
$$
However, we want to find the poles of $H(s)$, not of $G(\omega)$. Since by Laplace transform properties, $H(-j\omega)={H^*(j\omega)}$, we substitute $s=j\omega$, so $\omega=-js$:
$$ \displaystyle G^2(\omega) = H(j\omega)H^*(j\omega)= H(s)H(-s)= \frac{G_0^2}{1+\left(\frac{-js}{\omega_c}\right)^{2N}} = \frac{G_0^2}{1+\left(\frac{-s^2}{\omega_c^2}\right)^{N}} $$
Denoting the poles of $H(s)$ as $s_k^2$, we can relate them to the Nth roots of unity: