alexgrover

Multi Poles Zero-Lag Exponential Moving Average

Introduction

Based on the exponential averaging method with lag reduction, this filter allow for smoother results thanks to a multi-poles approach. Translated and modified from the Non-Linear Kalman Filter from Mladen Rakic 01/07/19 https://www.mql5.com/en/code/24031

The Indicator

length control the amount of smoothing, the poles can be from 1 to 3, higher values create smoother results.

Difference With Classic Exponential Smoothing

A classic 1 depth recursion (Single smoothing) exponential moving average is defined as y = αx + (1 - α)y which can be derived into y = y + α(x - y)

2 depth recursion (Double smoothing) exponential moving average sum y with b in order to reduce the error with x, this method is calculated as follow :

  • y = αx + (1 - α)(y + b)
  • b = β(y - y) + (1-β)b

The initial value for y is x while its 0 for b with α generally equal to 2/(length + 1)

The filter use a different approach, from the estimation of α/β/γ to the filter construction.The formula is similar to the one used in the double exponential smoothing method with a difference in y and b

  • y = αx + (1 - α)y
  • d = x - y
  • b = (1-β)b + d
  • output = y + b

instead of updating y with b the two components are directly added in a separated variable. Poles help the transition band of the frequency response to get closer to the cutoff point, the cutoff of an exponential moving average is defined as :

Cf = F/2π acos(1 - α*α/(2(1 - α)))


Also in order to minimize the overshoot of the filter a correction has been added to the output now being output = y + 1/poles * b

While this information is far being helpful to you it simply say that poles help you filter a great amount of noise thus removing irregularities of the filter.

Conclusion

The filter is interesting and while being similar to multi-depth recursion smoothing allow for more varied results thanks to its 3 poles.

Feel free to send suggestions :)

Thanks for reading
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Komentar

Very nice creations, thanks for your efforts and generosity =D
+1 Jawab
@ICEKI, Thank you for your support as always (how many times did i said that ?) :)
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ICEKI alexgrover
@alexgrover, LOL you're welcome no problem =D
+1 Jawab
Thank you for this line.
+1 Jawab
alexgrover aaahopper
@aaahopper, You are welcome :)
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