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A single-pole IIR low-pass filter can be defined in discrete time as y += a * (x - y), where y is the output sample, x is the input sample and a is the decay coefficient.

However, the definition of a varies. On Wikipedia, it's defined as 2πfc/(2πfc+1) (where fc is the cutoff frequency).

But here, a is defined as follows: 1 - e^-2πfc.

Their graphs look similar, but which one is more accurate?

The blue graph is the Wikipedia formula, the green is the second one, fc is the x-axis. GeoGebra graph

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  • $\begingroup$ Possible duplicate of Exponential weighted moving average time constant $\endgroup$ – Matt L. Dec 12 '18 at 9:04
  • $\begingroup$ I disagree. There are two conflicting formulas given, but the question you linked covers how one of them can be derived. Edit: However, there's a note that the formulas are only an approximation, which might imply that both of those are usable. $\endgroup$ – Mark Dec 12 '18 at 9:07
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    $\begingroup$ See my answer below, both are indeed approximations derived from the corresponding analog filter, and only one of them is useful, provided that the cut-off frequency is much smaller than Nyquist. $\endgroup$ – Matt L. Dec 12 '18 at 12:17
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The given single-pole IIR filter is also called exponentially weighted moving average (EWMA) filter, and it is defined by the following difference equation:

$$y[n]=\alpha x[n]+(1-\alpha)y[n-1],\qquad 0<\alpha<1\tag{1}$$

Its transfer function is

$$H(z)=\frac{\alpha}{1-(1-\alpha)z^{-1}}\tag{2}$$

The exact formula for the required value of $\alpha$ that results in a desired $3\;\textrm{dB}$ cut-off frequency $\omega_c$ was derived in this answer:

$$\alpha=-y+\sqrt{y^2+2y},\qquad y=1-\cos(\omega_c)\tag{3}$$

Even though it should be easy enough to compute $\alpha$ from $(3)$, there are several approximative formulas floating around the internet. One of them is

$$\alpha\approx 1-e^{-\omega_c}\tag{4}$$

In this answer I've explained how this formula is derived (namely, via the impulse invariant transformation of the corresponding continuous-time low pass filter). This answer compares the approximation $(4)$ with the exact formula, and it is shown that $(4)$ is only useful for relatively small cut-off frequencies (of course, small compared to the sampling frequency).

In the wikipedia link given in the question, there is yet another approximative formula:

$$\alpha\approx\frac{\omega_c}{1+\omega_c}\tag{5}$$

[Note that in all formulas of this answer $\omega_c$ is normalized by the sampling frequency.] This approximation is also derived from discretizing the corresponding analog transfer function, this time not via the impulse invariant method, but by replacing the derivative by a backward difference:

$$\frac{dy(t)}{dt}{\huge |}_{t=nT}\approx\frac{y(nT)-y((n-1)T)}{T}\tag{6}$$

This is equivalent to replacing $s$ by $(1-z^{-1})/T$ in the continuous-time transfer function

$$H(s)=\frac{1}{1+s\tau}\tag{7}$$

which results in

$$H(z)=\frac{1}{1+(1-z^{-1})\tau/T}=\frac{\frac{1}{1+\tau/T}}{1-\frac{\tau/T}{1+\tau/T}z^{-1}}\tag{8}$$

Comparing $(8)$ to $(2)$ we see that

$$\alpha=\frac{1}{1+\tau/T}\tag{9}$$

Since the (continuous-time) $3\;\textrm{dB}$ cut-off frequency is $\Omega_c=1/\tau$, we obtain from $(9)$

$$\alpha=\frac{\Omega_cT}{1+\Omega_cT}\tag{10}$$

Equating the discrete-time cut-off frequency $\omega_c$ with $\Omega_cT$ in $(10)$ gives the approximation $(5)$.

The figure below shows the actually achieved cut-off frequency for a given desired cut-off frequency for the two approximations $(4)$ and $(5)$. Clearly, both approximations become useless for larger cut-off frequencies, and I would suggest that approximation $(5)$ is generally useless.

enter image description here

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  • $\begingroup$ How about this approximation (sorry, just in coefficient format ): x = 1.0/(fs*T); % fs in Hz, T in μs a0 = 1.0; a1 = -(1 + x + 0.5 * x^2); % Taylor approx b0 = a0 + a1; % gain b1 = 0.0; $\endgroup$ – Juha P Mar 26 at 21:18

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