# How to filter a signal using a bandpass filter consisted of two moving average filters?

I want to filter a PPG signal on a microcontroller. I have limited memory and a was searching for low computational methods. I found the work of Kazuhiro Taniguchi, Earable POCER: Development of a Point-of-Care Ear Sensor for Respiratory Rate Measurement where they use moving average filters (m3, m30 and m80) in order to filter their data and obtain freequencies between certain values, specifically they create a a passband between 189 mHz and 504 mHz. Their idea is that after the initial m3 moving average(every 3 values) they create the m30 and m80 and they apply an iteration between them ( r= m30 - m80). This is their way to obtain the values corresponding to the passband of interest, eventhough, as they specify,

The moving average does not have an ideal lowpass filter function, and so some frequency elements other than those in the passband may pass, even if they are attenuated.

to obtain the window size for their moving average filters they applied the following equation: I couldn't quite understand where this equation (5) was coming from but as soon as you replace the value $$n$$ with 30 or 80 you get the values of the passband cut-off freequencies 189 mHz and 504 mHz.

When I asked for specifications on how they did it, they send me to a japanese forum that unfortunately I couldn't translate but there were two links to stackexchange, on cuttoff freequency and filter design.

I tried to adapt all these new staff on my model but couldn't get the passband wanted (0.1 Hz - 0.8 Hz) using the equation (5) from the image above with my parameters (sampling freequency of 50Hz, cut-off freqs etc...).

I don't understand what is the problem and my question is the following:

What are the moving average filter windows that I have to use and in which order to be able to filter a signal with a sampling frequency of 50 Hz in order to isolate the frequencies between the passband 0.1 Hz - 0.8 Hz?

• Your link to the original paper doesn't work. Also, your band pass filter specification lacks two information: 1. You've told us what you want to let through, but not by how much you want to suppress the things you don't want to let through ("stopband attenuation"), 2. you forgot to define where the things that you don't want to let through start (below 0.05 Hz and above 0.85 Hz? below 0.002 and above 0.8002? That makes a gigantic difference for your filter!) Oct 12 '20 at 9:25
• For the article i am trying a different link ; ncbi.nlm.nih.gov/pmc/articles/PMC6165419/pdf/… Oct 12 '20 at 9:28
• That works now! Can you edit your question to use that link (also, citing the title might be good style) Oct 12 '20 at 9:28
• also, before you invest too much time into this: your approach to build an efficient band pass out of moving averages is probably not warranted – they simply don't make good filters, and thus, your computational effort to still work with them will be larger than if you started out with a sensible design to begin with. Also, 50 Hz?! Even if the filter spec is as harsh as I could imagine it to be, and we implement it in the dumbest way possible, that should still be pretty doable with any modern microcontroller. Oct 12 '20 at 9:30
• The attenuation i want is an attenuation ratio of 0.7. and the things i don't want to let thourgh, i am kind of a rookie in filter processing so i would say the less possible? I want to do a FFT afterwards, i don't know if that helps you. I am sorry Marcus. Oct 12 '20 at 9:33

The Japanese link actually implies how to derive Eq.5, and ironically they also refer to an existing dsp.se answer at the bottom.

Derivation of Eq.5 is as follows:

Consider a moving average filter of length $$N$$, with the impulse response $$h[n]$$:

$$h[n] = \begin{cases} ~~~1/N~~~,~~~n=0,1,...,N-1 \\ ~~~0~~~,~~~ \text{otherwise} \end{cases} \tag{1}$$

Magnitude of its frequency response $$~H(\omega)~$$ (DTFT of $$h[n]$$) can be shown to be:

$$|H(\omega)| = \frac{1}{N} \left|\frac{ \sin(\frac{\omega}{2}N)}{\sin(\frac{\omega}{2})}\right| \tag{2}$$

Now, after replacing the $$\sin()$$ functions with their Taylor expansions in powers of $$\omega$$, it can be shown by polynomial long division that Eq.2 is also given by :

$$|H(\omega)| = 1 + \frac{1}{24}(1-N^2) \omega^2 + H.O.T. \tag{3}$$ where H.O.T. refers to higher order terms in powers of $$\omega$$. An approximation for small values of $$\omega$$ is obtained by neglecting H.O.T.:

$$|H(\omega)| \approx 1 + \frac{1}{24}(1-N^2) \omega^2 \tag{4}$$

Using this approximate frequency response magnitude, we can obtain an approximate cutoff frequency $$\omega_c$$ at which the magnitude $$|H(\omega_c)|$$ falls to $$1/\sqrt{2}$$ of its value at $$\omega = 0$$, which is $$H(0) = 1$$.

$$|H(\omega_c)| = \frac{1}{\sqrt{2}} = 1 + \frac{1}{24}(1-N^2) \omega_c^2 \tag{5}$$

Replace the discrete-time frequency $$w_c$$ by $$w_c = 2\pi f_c /f_s$$, where $$f_c$$ is the analog cutoff frequency in Hz, and $$f_s$$ is the sampling frequency in Hz. Finally solving the resulting algebraic expression for $$f_c$$ yields the formula that you refer to as Eq.5 in the document :

$$|H(\omega_c)| = \frac{1}{\sqrt{2}} = 1 + \frac{1}{24}(1-N^2) \left( 2\pi \frac{f_c}{f_s} \right)^2 \tag{6}\\\\$$

$$f_c = \frac{1}{\pi} \frac{\sqrt{6 - 3\sqrt{2}}}{\sqrt{N^2-1}} ~f_s ~~ =~~ \frac{0.422}{\sqrt{N^2-1}} ~f_s \tag{7}$$

Eq.7 above is the formula that provides an approximate cutoff frequency calculation for the moving average filter of length $$N$$ (order $$N-1$$). In your posted link, Eq.5 there's a slight variation at the scale of $$0.442$$ instead of $$0.422$$, probably they tried to apply some correction to the actual cutoff vs approximated one.

Note that in the derivation, we've used an approximation of the DTFT magnitude which was valid as long as $$\omega$$ was small compared to $$\pi$$. This means that the approximation will be satisfactory if $$\omega_c$$ is close to $$0$$, or in other words, $$f_c$$ is a small compared to $$f_s$$. And indeed this will be the case for high order moving average filters. And the approximation gets better as $$N$$ increases.

Using such two moving average filters with approximate cutoff frequencies $$f_{c1}$$ and $$f_{c2}$$ to create a bandpass filter will not be very satisfactory unless your out of band signal energy is insignificand after 20 to 30 dB of attenuation.

• Thank you for the answer @Fat32. I am not very good at math, but if i understand correctly, this method CAN be used for very small ω. If i am just trying to filter my signal in the same way that they did, (cut off frequencies [0.1 Hz - 0.8 Hz]), should I just change the sampling freequency to the one i use or there are further adjustements to make? And how can i check if "out of band signal energy is insignificand after 20 to 30 dB of attenuation"? Oct 12 '20 at 14:40
• Yes you can use this procedure to design your bandpas filter out of two moving average filters. Just set your sampling frequency (say 50 Hz) and use Eq.7 to solve for the required length N of the filters for fc1 and fc2. As long as they are small (such as 0.1 Hz to 0.8 Hz in your case) the approximation will be valid. Your out of band energy is totally your thing to know. The energy will be insignificand as long as your filter works satisfactorily ;i.e., signal of concern appears at its outpıut and nothing else :-) Oct 12 '20 at 14:45
• Thank you very much for your help @Fat32!! :D Oct 12 '20 at 14:52