1
$\begingroup$

Conventional methods of filtering allow one to block parts of a frequency band while passing others.

But thinking outside of conventional means, and perhaps not based on any frequency methods at all, are there filtering methods for blocking wide band noise, however passing rapid moving signal?

For example. Say I have a square wave with 'hair' on it (white-like noise). I just want to get the square wave with corners as sharply as possible - and minimum hair on the output. Outside of conventional FIR, IIR, Kalman filtering are there any nonlinear tricks?

$\endgroup$
3
  • $\begingroup$ Maybe something with curve fitting? I wonder if there's some algorithms for compression / line simplification in vector graphics software that would do something like what you want? $\endgroup$
    – Alan Wolfe
    Commented Apr 16, 2015 at 4:30
  • $\begingroup$ The answer definitely is yes. Look into techniques like PRML (Partial Response Maximum Likelyhood) which are used to decode the signal coming from the read head of a harddisk. In short, if you expect one of N signals, you compare the input to all of them and find the best match. $\endgroup$
    – MSalters
    Commented Apr 16, 2015 at 7:41
  • $\begingroup$ For the specific example of a square wave with a small amount of wideband noise, a hard limiiter -- something like sgn$(x(t))$ or sgn$(x(t)-1/2)$ (for a square wave with levels $1$ and $0$ instead of $\pm 1$) will recover the square wave almost perfectly. Some estimation of the level $A$ may be needed if the square wave takes on values $\pm A$ or $0$ and $A$. $\endgroup$ Commented Apr 16, 2015 at 13:06

1 Answer 1

1
$\begingroup$

A very common non-linear filtering technique for removing noise while preserving edges is median filtering. It basically replaces the samples inside a window by their median. Total variation denoising is another non-linear technique that works especially well for denoising piece-wise constant signals. You can find some more information about this method and Matlab software here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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