5
$\begingroup$

Is there a term to describe this IIR averaging/smoothing filter?

$ y[n] = \alpha x[n] + (1 - \alpha) y[n - 1] $

$\endgroup$
  • $\begingroup$ it should be noted that $$ 0 < \alpha \le 1 $$ for the leaky integrator to appear most like an integrator, then $$ 0 < \alpha \ll 1 $$ this is the integrator to undo the differentiator of DC blocking filter. i like to represent it in terms of the single pole value which is $$p = 1-\alpha$$ so $$ y[n] \ = \ p \, y[n-1] + (1-p) x[n] $$ and the transfer function is $$ H(z) \triangleq \frac{Y(z)}{X(z)} = \frac{(1-p)z}{z-p} $$ $\endgroup$ – robert bristow-johnson Sep 21 '17 at 23:54
  • $\begingroup$ also maybe related: dsp.stackexchange.com/questions/4699/… $\endgroup$ – user13267 Sep 22 '17 at 1:37
5
$\begingroup$

This is often called a leaky integrator, a special case of a first-order IIR lowpass filter. They are discussed in more detail in several previous questions:

$\endgroup$
3
$\begingroup$

It is called an exponentially weighted moving average (EWMA) filter. Here is a previous answer where I provided a Matlab script for computing $\alpha$ for a desired cutoff frequency: Exponential moving average cut-off frequency

$\endgroup$
  • $\begingroup$ i think Andy is correct. $\endgroup$ – robert bristow-johnson Sep 21 '17 at 23:41
  • $\begingroup$ I'm not gone on using Moving Average for describing an Autoregressive filter, though I've certainly heard the term before. $\endgroup$ – Peter K. Sep 22 '17 at 0:42
2
$\begingroup$

Wikipedia calls it "basic exponential smoothing".

I'd call it a first order smoother or first order lowpass filter.

Some combination of those words will probably work.

$\endgroup$
  • $\begingroup$ sounds to me to be a Wikipedia neologism. but i dunno. i guess it implies a DC gain of 0 dB, which is important. $\endgroup$ – robert bristow-johnson Sep 21 '17 at 23:42
  • $\begingroup$ @robertbristow-johnson not so sure about the "basic" but I'd seen it called exponential smoothing or an exponential filter before Wikipedia existed :) $\endgroup$ – hobbs Sep 22 '17 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.