running average vs lowpass filter

You need to digitize the signal fast enough to prevent aliasing (creating low beats of higher frequency components. But you can simulate a 1 Hz roll off RC filter by recycling a stored value through a scaled sum with the samples.

The relations between time constant and sine wave frequency is a 1 Hz corner corresponds to a 1Hz/(2*pi) time constant or about 1/6th second. So if you take 100 samples per second, and every time one comes in, you add 1/17th of it with 16/17ths of the stored result (and the sum becomes the stored result), the stored result is the filtered value with about (1/100)*17 sec = .17 sec = 1.07 Hz roll off. The scaling and addition will need to be done with at least 13 bits of precision, so assume two byte math. Not too hard to fit into 1/100 sec for lots of processors.

Hi,

I have a 40Hz -3dB RC lowpass filter and I would like to digitize it to a 1Hz -3dB lowpass filter. I see two methods to do this, by using an

cheers, Jamie

Hi Jamie,

A running average filter has a sin(x)/x response : Basically it is not a good filter for signals at frequencies like 1,5Fs, 2,5Fs, etc, where Fs is the inverse of the averaging total period. It is quite easy to understand in fact : If you input signal is a sine at a frequency which is an integral multiple of the averaging frequency, then the averaging is perfectly working, howver if the frequency is in between then the averaging is not so good...

Definitely a good low pass filter (FIR, IIR, etc) is far better but more expensive than an averaging. An intermediate solution exists : See Cascaded Integrate Comb filter (CIC).

Friendly yours,

-- Robert Lacoste ALCIOM - The mixed signal experts

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It depends- usually at ultra low frequencies you only want the 1Hz bandwidth while the signal remains within say 10% of the bandlimited average but you want a more rapid response to large deviations like 25% or more. This type of thing falls under the IIR data smoothing- which can be a single multiply and add- but you drop the smoothed output and start over.

In article , John Larkin wrote: [...]

You can get other factors in the form of N/K where K is a power of two and N has a low number of bit set in it. You do it sort of like this:

A = X A = A SHR 1 A = A + X A = A SHR 3

This gives you a 3/16 factoring.

A simple average sliding along the data is called a "box-car filter"

A Bartlett filter is a lot better than a running average and only takes a bit more computing power.

X = input Y = output Z = other variable n = Now K = Width of 1/2 filter

Assuming the X values go into a circular buffer in incrementing order. You do this:

Z = Z + X(n) - 2*X(n-K) + X(n-2K) Y = Y + Z

The trick here is to remember to zero everything out at start up and this only works for integers. Floaters will drift away.

You can do even better if you cascade two stages of filter-decimate code. The advantage is that you can make the width of the box-car or Bartlett filters place the zeros at exactly where you want the aliased signals not to be trouble.

An IIR filter takes less room in memory for a given amount of lowpassing. It has the disadvantage of a recovery tail.

I've just been working on a similair project, and I'll second the recommendation for using a substantially higher sampling rate than needed and then running a CIC filter to decimate it down to a useful data rate. In my application I'm then running a short FIR filter after that to place zeroes specifically on top of 60 Hz and the alias of 120 Hz.

Check out Donadio 2000 on the subject at

If anyone would like I've got a quick and dirty Java program I threw together to do some modeling on this that I can post.

-- Rob

Robert Lacoste wrote:

You have a 40 Hz analog filter, and you would like to create a signal from it that's filtered with a time constant of 1 Hz as if the 40 Hz filtering didn't happen? So in other words you want to unfilter the input, and then filter it differently? Do I have that right?

Would you be averaging it digitally? If so, I don't see a gain in averaging instead of filtering.

A digital low-pass filter that mimics an RC filter is

V_out = (V_new + (T/dt)*V_last) / (1 + T/dt) V_last = V_out

where V_new is the new sample, V_last holds the previous output value, dt is the sampling time, and T is the RC filter time constant. And in the interest of saving cycles, somewhere outside of the loop you should calculate the constants once,

A = 1/(1 + T/dt) B = (T/dt)/(1 + T/dt)

Then V_out = A*V_new + B*V_last.

If you want to unfilter the incoming signal, you might try an initial filter time constant of -0.025 seconds, but I've never tried it that way.