Noise and Filter Theory

Noise Spectra

References

1. Wikipedia contributors, "White noise," Wikipedia, The Free Encyclopedia, wikipedia.org
2. Wikipedia contributors, "Colors of noise", Wikipedia, The Free Encyclopedia, wikipedia.org

White Noise

White noise is a random signal with a constant power spectral density and the same intensity at all frequencies. Consequently, there is an equal amount of power in a frequency interval, but an increasing amount of power in an octave: the noise power doubles when moving to the next octave.

Coloured Noise

Pink noise has a $1/f$ power spectral density (decreasing at 10dB/decade or 3.01dB/octave) such that equal power is contained in frequency intervals that are proportionally wide. Therefore, a pink noise signal has equal power in each octave. Blue noise is the opposite of pink noise: its power spectral density increases at 10dB/decade (3.01dB/octave). Completing the colored noise spectra are Brownian noise (or Brown noise), decreasing at 20dB/decade (i.e. a first order LPF applied to white noise) and violet/purple noise increasing at 20dB/decade (i.e. a first order HPF applied to white noise).

Fractional Order Filters

References

1. Juraj Valsa and Jiri Vlach, "RC models of a constant phase element," Intl. J. Circuit Th. and App. 41 p59 (2013) doi
2. Lobna Said, et al., "Fractional-Order Filter Design," p358 doi

Low Pass Filter

A first order low pass filter (LPF) has a transfer function of the form

$$H(s) = \frac{\omega_0}{s + \omega_0} = \frac{1}{s/\omega_0 + 1}$$

As the frequency $s$ increases, for $s \gg \omega_0$, $H(10s) / H(s) \to 1/10$, which is equivalent to a slope of $-20dB/dec$ (in intensity units $20dB \equiv 10 \log |10|^2 = 20\log |10|$). A first order LPF can be realized as an RC circuit whose transfer function is

$$H(j\omega)=\frac{1}{j\omega RC + 1}$$

substituting $s\to j\omega$ with $\omega_0 = 1/RC$. The magnitude and phase of this transfer function are

$$\begin{align*} |H(j\omega)| &= \frac{\omega_0}{\sqrt{\omega^2 + \omega_0^2}}\\ \angle H(j\omega)&= \tan^{-1}\left(-\frac{\omega}{\omega_0}\right) \end{align*}$$

Generally, the LPF can be written (Said2018)

$$H(s)=\frac{\omega_0}{s^\alpha + \omega_0} = \frac{1}{(s / \omega^{1/\alpha}_0)^\alpha + 1}$$

with magnitude

$$|H(j\omega)|=\frac{\omega_0}{\sqrt{\omega^{2\alpha}+2\omega_0\cos(\alpha\pi/2)\omega^\alpha+\omega_0^2}}$$

as $\omega \to \infty$ , $|H(j\omega)|\to \omega_0/\omega^\alpha$ such that the slope is

$$20\log \left(\frac{\omega_0}{10^\alpha \omega^\alpha}\right)-20\log \left(\frac{\omega_0}{ \omega^\alpha}\right)=20 \log 10^{-\alpha} = -20\alpha$$

To obtain a slope of $-10dB/dec$ ($-3dB/oct$), $\alpha = 1/2$ and

$$H(s) = \frac{1}{\sqrt{s/\omega^2_0} + 1}$$

The general corner frequency is found when

$$\begin{align*} \frac{1}{\sqrt{2}} = |H(j\omega)| &= \frac{1}{\sqrt{\frac{\omega^{2\alpha}}{\omega_0^2}+2\cos(\alpha\pi/2)\frac{\omega^\alpha}{\omega_0}+1}} \\ \to 2 &= \frac{\omega^{2\alpha}}{\omega_0^2} + 2\cos(\alpha\pi/2)\frac{\omega^\alpha}{\omega_0} + 1 \\ \to 0 &= \omega^{2\alpha} + 2\cos(\alpha\pi/2)\omega_0 \omega^\alpha - \omega_0^2 \\ \to \omega^{\alpha} &= -\cos(\alpha\pi/2)\omega_0 \pm \frac{1}{2}\sqrt{\cos^2(\alpha\pi/2)\omega^2_0 + 4\omega^2_0} \\ &= \left[-\cos(\alpha\pi/2) + \frac{1}{2}\sqrt{\cos^2(\alpha\pi/2) + 4}\right]\omega_0\\ \to \omega_c &= \left[-\cos(\alpha\pi/2) + \frac{1}{2}\sqrt{\cos^2(\alpha\pi/2) + 4}\right]^{-\alpha}\omega_0^{-\alpha} \end{align*}$$

Fractional Capacitor

To obtain an impedance proportional to $1/\sqrt{j\omega}$, define $Z_{CF}(j\omega)$ as

$$\begin{align*} Z_{CF}(j\omega)&\equiv\frac{\psi}{\sqrt{j\omega}}=\frac{\psi}{\sqrt{\omega}(\cos \pi/4 + j \sin\pi/4)}=\frac{\psi\sqrt{2}}{(1 + j)\sqrt{\omega}}=\psi\sqrt{\frac{1}{2\omega}}(1-j) \end{align*}$$

Fractional Low Pass Filter

In series with a resistor $R$ (and output across $Z_{CF}$), the transfer function becomes

$$\begin{align*} \to H(j\omega) &= \frac{1}{R/Z_{CF}(j\omega) + 1} = \frac{1}{\frac{R\sqrt{\omega}}{\psi}e^{j\pi/4} + 1} = \frac{1}{\sqrt{\omega / \omega'_0 }(1+j) + 1}\\ \sqrt{\frac{\omega}{\omega'_0} }(1+j) &= \frac{R}{\psi\sqrt{2}}(1 + j)\sqrt{\omega} \to \omega'_0 = \frac{2\psi^2}{R^2} \end{align*}$$

This yields the location of the corner frequency ($|H(j\omega_0)| = 1/\sqrt{2}$):

$$\begin{align*} |H(j\omega)| &= \left|\frac{1}{(1 + \sqrt{\omega/\omega'_0}) + j\sqrt{\omega/\omega'_0}} \right| \to \left| (1 + \underbrace{\sqrt{\omega/\omega'_0}}_a) + j\sqrt{\omega/\omega'_0}\right| = \sqrt{2} \\ \to |(1 + a) + ja|^2 &= [(1+a) + ja][(1+a) - ja] = 1 + 2a + 2a^2 = 2 \\ \to a^2 + a - 1/2 &=0 \to a = \sqrt{\frac{\omega}{\omega'_0}} = \frac{-1+\sqrt{3}}{2} \\ \to \omega_0 &= \left(\frac{-1+\sqrt{3}}{2}\right)^2\omega'_0 = \left(\frac{-1+\sqrt{3}}{2}\right)^2\frac{2\psi^2}{R^2} \end{align*}$$

Given a target $\omega_0$ and one of $R$ or $\psi$, the other impedance can be chosen.

Fractional High Pass Filter

In series with a resistor $R$ (and output across $R$), the transfer function becomes

$$\begin{align*} \to H(j\omega) &= \frac{R/Z_{CF}(j\omega)}{R / Z_{CF}(j\omega) + 1} = \frac{\sqrt{\omega / \omega'_0 }(1+j)}{\sqrt{\omega / \omega'_0 }(1+j) + 1},\quad \omega'_0 = \frac{2\psi^2}{R^2} \end{align*}$$

as above. The corner frequency ($|H(j\omega_0)| = 1/\sqrt{2}$) is then

$$\begin{align*} \frac{1}{\sqrt{2}} = \left|H(\omega_c)\right| &= \left| \frac{\sqrt{\omega_c / \omega'_0 }(1+j)}{\sqrt{\omega_c / \omega'_0 }(1+j) + 1}\right| \\ \to \sqrt{2} &= \left|1 + \frac{1}{\sqrt{\omega_c / \omega'_0 }(1+j)}\right| =\left|1 + \frac{1-j}{2\sqrt{\omega_c / \omega'_0 }}\right| \\ &= \left|\frac{(1+2\sqrt{\omega_c / \omega'_0 }) - j}{2\sqrt{\omega_c / \omega'_0 }}\right| = \left|\frac{(1+2a) - j}{2a}\right| \\ &= \sqrt{\left(1 + \frac{1}{2a}\right)^2 + \left(\frac{1}{2a}\right)^2} \\ \to b \equiv \frac{1}{2a},\quad 2&= (1+b)^2 + b^2 \to 0 = 2b^2 + 2b - 1 \to b = \frac{-1+\sqrt{3}}{2} \\ \to a = \sqrt{\omega_c / \omega'_0 } &= \frac{1}{-1 + \sqrt{3}} \to \omega_c = \left(\frac{1}{-1 + \sqrt{3}}\right)^2\frac{2\psi^2}{R^2} \end{align*}$$

Filter Approximation

Details on the filter implementation are in the notebook, including an implementation of the method described in Valsa2013. Effectively, Valsa's method generates a similar structure to the shelf-filter approximation with an extra low-pass pole at a higher frequency (at the expense of an extra RC pair). After some simulations and testing, I chose to keep the shelf-filter approximation.