# Exponential moving average cut-off frequency

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?

• dsprelated.com/showarticle/182.php Apr 22, 2017 at 18:54
• I think you'll find the answer here: en.wikipedia.org/wiki/… Apr 22, 2017 at 18:59
• @MarcusMüller So is the answer here correct? dsp.stackexchange.com/questions/39063/… Apr 22, 2017 at 20:18
• how to calculate EMA high pass filter in matlab ? Dec 9, 2017 at 17:04
• If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
– MBaz
Dec 9, 2017 at 17:33

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);

figure(1);
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));

figure(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;


Additional information in response to Sep 11 '20 comment

For a stable, low pass filter, the pole, $$(1-\alpha)$$, must be inside the unit circle of the z-plane and be greater than $$0$$ on the real axis. So we must satisfy

$$0 < 1 - \alpha < 1$$

thus

$$0 < \alpha < 1$$

These are bounds on $$\alpha$$ (and not a "definition") that come from requirements for stability and a filter passband centered at DC.

However, not all valid values of $$\alpha$$ will yield a -3 dB frequency that exists in the interval from DC to the Nyquist frequency, $$\Omega \in [0, \pi]$$ or equivalently $$f \in \left[0, \dfrac{F_s}{2}\right]$$.

When the pole, $$(1-\alpha)$$, is very close to the origin of the z-plane, the filter response will be so gradual or flat, that it will never drop to -3 dB or below. Specifically, when

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

is undefined, due to

$$\left|1-\dfrac{\alpha^2}{2(1-\alpha)}\right| > 1$$

the filter response is so gradual, it will not have a drop down to -3 dB or below before the Nyquist frequency. The $$\alpha$$ at which this happens is given by the positive solution to

$$\alpha^2+4\alpha-4 =0$$

which is

$$\alpha = -2 + 2\sqrt{2} \approx 0.82843$$

EWMA filters with an $$\alpha$$ that is this high, have a pole so close to the origin of the z-plane, that the filter response is so gradual, there is no frequency attenuated by 3 dB or more.

The Wikipedia derivation of the 3 dB cutoff point is flawed and comes to an incorrect/inexact result. Specifically it directly applies a continuous time, Laplace s-domain relationship:

$$f_c = \dfrac{1}{2\pi RC} \quad \text{(Don't use this!)}$$

into a derivation of a result for a discrete time, z-domain, filter. That step invalidates the derivation of an exact result in the Wikipedia article.

As a consequence, the result of the derivation in the Wikipedia article is only an approximation and (subjectively) only good for cutoff frequencies up to about $$0.1 F_s$$.

• 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. Apr 23, 2017 at 9:56
• I got the same result from my calculations. Apr 24, 2017 at 8:03
• 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. Sep 22, 2017 at 5:35
• often i have found this trig identity useful: $$\cos(\theta) = 1 - 2 \sin^2 \left(\tfrac12 \theta \right)$$ Sep 22, 2017 at 5:48
• @GabrielStaples My answer is correct. I have updated it with additional information in response to your comment. As a general note, if an equation has a region in which it is undefined, it does not mean that the equation is wrong, it means that there is no answer there. In this case specifically, the equation is telling you that the filter has no -3 dB point for $\alpha$ values in that region, which is exactly the case. Sep 11, 2020 at 13:00

The accepted answer can't be right, as it results in an undefined cutoff frequency for any $$\alpha$$ value > ~0.8284, yet alpha values between 0 and 1 are and must be, by definition, valid. The accepted answer has this as the cutoff frequency:

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

However, take a look at just this part:

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

The input to the acos() func must be between -1 and 1, yet with an $$\alpha$$ value of 0.8285 you get −1.001201895, which means its arc cosine is undefined, since that's outside the valid range of -1 to +1, inclusive.

So, that equation is broken.

I don't understand the mathematical derivation, but Wikipedia shows here on the "Low-pass filter" page under the "Simple infinite impulse response filter" section that for the standard, basic, first-order IIR (Infinite Impulse Response) Exponentially-weighted Moving Average (EMA) filter of:

$$y_i = \alpha x_i + (1 - \alpha)y_{i-1}$$

where $$\alpha$$ is the smoothing factor, $$x_i$$ is the current sample, $$y_i$$ is the filtered value, and $$y_{i-1}$$ is the previous filtered value, the cutoff frequency, $$f_c$$, is:

$$\boxed{f_c = \dfrac{\alpha}{(1 - \alpha)2\pi\Delta T}}$$

where $$\Delta T$$ is the sample period, or $$\Delta T = 1/sample\_freq = 1/F_s$$.

So, that's the answer! That's the equation for cutoff frequency.

Here's a table of comparisons for cutoff frequency calculations using the equation in the accepted answer vs the equation from Wikipedia. Notice that the Wikipedia cutoff frequency calculation works for all $$\alpha$$ values where $$0 \le \alpha \lt 1$$, but the other equation fails for $$\alpha$$ values $$\ge \approx 0.8285$$.

View this spreadsheet on Google Sheets here. Make a copy of it to manipulate yourself if you'd like to edit it or play with the equations.

## Other References:

1. How to draw a box around an equation: https://stackoverflow.com/questions/5184977/how-to-box-a-formula-in-mathjax/5319398#5319398
• This answer did not contain satisfactory analysis to support it's assertion that the original answer was wrong, specifically why the called out arccos() equation may have been wrong. Where did the original derivation in the original answer fail? This answer also relied on a Wikipedia derivation for which you state you "[...] don't understand the mathematical derivation," in your answer, indicating that you will be unable to defend this answer. Why should a reader believe your answer, when you haven't put in the work? Sep 11, 2020 at 13:09
• The values in the table in the spreadsheet, show a discrete time filter with an $F_s = 1000$ Hz, but lists filter cutoff frequencies greater than the Nyquist frequency of $F_s/2 = 500$ Hz. Those values are clearly incorrect for a digital filter. Sep 11, 2020 at 13:19