Simple low pass digital filter implementation for non-uniform sampling?

Question

I have a system sampling an analog input and I want to filter that value using a simple, first order filter. EG something of the form:

Yn = A*Xn + B*Yn-1

However I have no control over the underlying process and the sampling period is non-uniform, and varies (randomlyish) somewhere between 25ms and 29ms. In addition I want to use a 30 minute time constant, and can't keep a running buffer of 72,000 sample values.

Given that a typical filter implementation requires a uniform sampling period, is there any simple algorithm that takes into account the time difference between Yn-1 and Xn? (which I guess really means a time varying version of my "constants" A and B)

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Note that I'm inclined to simply take the average sample time over an extended period and using that to calculate my constants, however there is also the possibility of the final system not running at the same average rate as my test system, so I want to come up with a solution that is somewhat adaptable.

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answering questions

1. The filtered signal is being used as a part of a vibration monitoring system in some industrial equipment. It will be used for more of as a qualitative than quantitative purpose.
2. The customer requirements are 30 minute rolling average for the data.
3. I have zero control over the sample rate.
4. And the sample rate varies every second
5. And the sample rate defines the speed at which my code runs
6. The final system should run at about the same speed as my test system, however it could be somewhat slower, maybe down to a 35ms rate

Answer 1

You say you have no control over the input sampling rate. That is understood but you are not considering the fact that you can resample once inside your digital system AND, I'm presuming you do have control over that else why would you be asking this question. So, resample your digital data like this: -

Now "real sample" 1 took 25 ms and "real sample" 2 took 29 ms but who cares if you internally resample that data? Yes, you have to process at a much higher rate but it's not like you are having to do major recalculations each time because \$A*X_n\$ is static for 25 to 29 samples.

Answer 2

Consider the first order continuous time filter \$\frac{dy(t)}{dt} = -k y(t) + k x(t)\$

One of the many choices to discretise the above filter is

$$\frac{y_{(t)}-y_{(t-T_s)}}{Ts} = -k y_{(t-T_s)} + k x_{(t)}$$ $$ y_{(t)} = k\cdot T_s x_{(t)} + (1 - k\cdot T_s)y_{(t- T_s)}$$

A and B to be used for each sample can be found from above expression.

The above is only an ad-hoc method. There may be proper algorithms for variable sampling rate filtering available in lierature with well defined mathematical properties.

A suggestion from From DSP.SE

Remember that filters are dynamical systems and solve the corresponding ordinary differential equation (ODE) In the second method, you basically use the retrieved time step as the time step in the ODE solver. Assuming you have a second order system:

Answer 3

Here comes the math. Let me assume, your low pass filter has a time constant \$\tau\$. It takes amplitude inputs values \$a_n\$ at points in time \$t_n\$ and it shall produce ouptut values \$b_n\$ for the same points in time. The starting condition is \$a_0 = 0\$ at \$t_0 = 0\$. So I initialize

\$b_0 = 0\$

All further output values can be computed as a weighted average of the previous output and the current input.

\$b_n = w_n a_n + (1-w_n) b_{n-1}\$

The weights \$w_n\$ can be computed from the well known low pass repsonse in time to the unit step function.

\$w_n = 1 - exp(\frac{-\Delta t_n}{\tau} )\$

where

\$\Delta t_n = t_n - t_{n-1}\$

So that is basically it. Here comes an example with \$\tau = 32\$. I have used a far higher non-uniformity for the sampling period as in the given question to make the effects more visible.

When working on a restricted processor such as a tiny microcontroller, calculating the weights nonlinearily as \$w_n = 1 - exp(-\Delta t_n / \tau)\$ may be costly. In those cases a linear approximation may serve well.

\$w_n \approx min(1, \frac{\Delta t_n}{\tau})\$

I tried it with the depicted example and the results cannot be distinguished visually.

Answer 4

There are ways to do what you ask for exactly... but you don't really need to do that. A 30 minute running average will erase all of your hard work.

Start by converting to a regular 1-second sampling rate just by taking the average of all the sample values within each second.

Then apply regular-sampling methods on the result.