I am trying to implement low pass filter from this example.

What is the cut-off frequency for this type of filter? Is it $F_s \frac{1-\alpha}{2\pi\alpha}$, where $F_s$ is sampling frequency?


The discrete recurrence relation given on the linked page is

$y[n] = (1-\alpha)y[n-1] + \alpha x[n]$

with $x[n]$ being the input samples and $y[n]$ being the output samples. So taking the Z-transform

$Y(z) = (1-\alpha)z^{-1}Y(z) + \alpha X(z)$

$H(z) = \dfrac{Y(z)}{X(z)}= \dfrac{\alpha}{1-(1-\alpha)z^{-1}}$

To find the -3 dB corner of the filter, in terms of the normalized digital radian frequency, $\Omega$ in the range $[0, \pi]$, one solves

$\left|H(z=e^{j\Omega_{3dB}})\right|^2 = \dfrac{1}{2} = \left|\dfrac{\alpha}{1-(1-\alpha)e^{-j\Omega_{3dB}}}\right|^2$

$ = \dfrac{\alpha^2}{\left|1-(1-\alpha)\cos(-\Omega_{3dB}) -j (1-\alpha)\sin(-\Omega_{3dB})\right|^2}$

which results in this equation

$\left[1-(1-\alpha)\cos(\Omega_{3dB})\right]^2+\left[(1-\alpha)\sin(\Omega_{3dB})\right]^2 = 2\alpha^2$

After some more algebra and trigonometry, I come up with

$\Omega_{3dB} = \cos^{-1}\left[1-\dfrac{\alpha^2}{2(1-\alpha)}\right]$

and since

$f_{3dB} = \dfrac{\Omega_{3dB}}{\pi} \cdot \dfrac{F_s}{2}$

I come up with

$f_{3dB} = \dfrac{F_s}{2\pi}\cos^{-1}\left[1-\dfrac{\alpha^2}{2(1-\alpha)}\right]$

Here's some Octave/MatLab code so you can verify the result on a plot; just change your Fs and alpha as required:

Fs = 10000.0
alpha = 0.01

b = [alpha];
a = [1 -(1-alpha)];

[H, W] = freqz(b, a, 1024, Fs);

f3db = Fs/(2*pi)*acos(1-(alpha^2)/(2*(1-alpha)))

plot(W, 20*log10(abs(H)), [f3db, f3db], [-40, 0]);
grid on

Update in response to comment:

The graph is a plot of the filter's magnitude response in dB (on the y-axis) vs. input frequency in Hz (on the x-axis). The vertical line designates the -3 dB corner frequency.

Since you want to specify your -3 dB corner frequency to find the value for $\alpha$, let's start with this equation from above:

$\left[1-(1-\alpha)\cos(\Omega_{3dB})\right]^2+\left[(1-\alpha)\sin(\Omega_{3dB})\right]^2 = 2\alpha^2$

and after some algebra and trigonometry, one can get a quadratic equation in $\alpha$

$\alpha^2 +2(1- \cos(\Omega_{3dB}))\alpha - 2(1- \cos(\Omega_{3dB})) = 0$

which has solutions

$\alpha = \cos(\Omega_{3db}) - 1 \pm \sqrt{\cos^2(\Omega_{3dB}) -4\cos(\Omega_{3dB}) +3}$

of which only the $+$ answer from the $\pm$ can yield a positive answer for $\alpha$.

Using the above solution for $\alpha$ and the relation

$\Omega_{3dB} = \dfrac{\pi}{F_s/2}f_{3dB}$

one can find $\alpha$ given $f_{3dB}$ and $F_s$.

Here is an updated Octave script:

Fs = 40000
f3db = 1

format long;
omega3db = f3db * pi/(Fs/2)

alpha = cos(omega3db) - 1 + sqrt(cos(omega3db).^2 - 4*cos(omega3db) + 3)

b = [alpha];
a = [1 -(1-alpha)];

[H, W] = freqz(b, a, 32768, Fs);

plot(W, 20*log10(abs(H)), [f3db, f3db], [-40, 0]);
xlabel('Frequency (Hz)');
ylabel('Magnitude Response (dB)');
title('EWMA Filter Frequency Response');
grid on;

W2 = [0:75] * pi/(Fs/2); % 0 to 75 Hz
H2 = freqz(b, a, W2);
W2 = W2 / (pi/(Fs/2));

plot(W2, 20*log10(abs(H2)), [f3db, f3db], [-20, 0]);
xlabel('Frequency (Hz)');
ylabel('Magnitude Response (dB)');
title('EWMA Filter Frequency Response near DC');
grid on;
| improve this answer | |
  • $\begingroup$ Thank you. Could you explain how the script works - what does it plot on X any Y axis? For instance, if I want cut-off frequency at about 1Hz where sampling frequency is 40kHz, how do try getting that? Thanks. $\endgroup$ – jurij Apr 23 '17 at 9:56
  • 2
    $\begingroup$ I got the same result from my calculations. $\endgroup$ – Olli Niemitalo Apr 24 '17 at 8:03
  • $\begingroup$ you can roll this back, Andy. but i still think a little manipulation can make this result simpler. but i haven't found it yet. $\endgroup$ – robert bristow-johnson Sep 22 '17 at 5:35
  • $\begingroup$ often i have found this trig identity useful: $$ \cos(\theta) = 1 - 2 \sin^2 \left(\tfrac12 \theta \right) $$ $\endgroup$ – robert bristow-johnson Sep 22 '17 at 5:48

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