In many of the papers it is said that the derivative filter transfer function is given by: $$H(z) = \dfrac{1}{8T}\left(-z^{-2} - 2z^{-1} + 2z + z^{2}\right)$$ But no one gave the detailed information about it. Anyone has an idea about this?

  • $\begingroup$ Please cite the specific paper so we can look at the contest. It's one of the many possible ways to build an approximate differentiator $\endgroup$
    – Hilmar
    Commented Jul 1, 2015 at 19:15
  • $\begingroup$ academia.edu/5263479/…. In this paper they used 5 point derivative filter. But I am still banging my head with transfer function $\endgroup$ Commented Jul 1, 2015 at 19:44

2 Answers 2


This is not a standard 5-point derivative formula, the corresponding transfer function of which is


The figure below shows the magnitude responses of an ideal differentiator (red), of the standard 5 point approximation given by (1) (green), and of the approximation in your question (blue):

enter image description here

As you can see, the non-standard approximation approximates the ideal differentiator only in a frequency region up to about $1/10$ of the Nyquist frequency (half the sampling frequency), whereas the standard approximation approximates the ideal differentiator quite well up to about a normalized frequency of $0.3$.

The differentiator in your question is a maximally flat lowpass differentiator. It avoids amplification of high-frequency noise. It achieves a maximally flat approximation of the ideal differentiator at $\omega=0$ and it has a given number of zero derivatives at $\omega=\pi$ (Nyquist), i.e. it is has a flat response at Nyquist, which causes high frequencies to be attenuated.

The design formula is derived in this paper. There is also a Matlab program available to design such maximally flat differentiators: link. Your specific filter is obtained by the command h=lowdiff(3,0);

  • $\begingroup$ Nice. $ a \Delta _{\pm 1} \ + (1-a) \Delta _{\pm 2} ), \ a \ 1/2 .. 4/3, $ traces out curves in between (with noise amplification factors 0.4 .. 0.95). $\endgroup$
    – denis
    Commented Jan 18, 2019 at 11:44

It seems to be some kind of weighted central-difference formula over different time intervals, maybe to minimize the effect of noise on the signal for shorter time intervals.

In your case the weighted central-difference formula can be written as,

$$ \dot{y}(t) \approx \frac{\frac{y(t+2T) - y(t-2T)}{4T} + \frac{y(t+T) - y(t-T)}{2T}}{2} = \frac{-y(t-2T) - 2y(t-T) + 2 y(t+T) + y(t+2T)}{8T}. $$


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