Let's say I have a signal, $x(t)$, defined as such,

$ x(t) = \begin{cases} 0 &\text{if} \,\,\, t < -\alpha/2 \\ \frac 1\alpha t+\frac12 &\text{if} \,\,\, -\alpha/2 \leq t \leq \alpha/2 \\ 1 &\text{if} \,\,\, t > \alpha/2 \end{cases} $

Thus, $x(t)$ is some signal that exhibits a jump from $0$ to $1$ over some timescale $\alpha$ around $t = 0$. For $\alpha = 0$, $x(t)$ has a discontinuous jump or an edge.

Now, let's add Gaussian Noise with $\sigma = 1$ to $x(t)$. Thus, $x(t)$ has a signal-to-noise of 1.

Now, say, I want to use some sort of edge-preserving smoothing filter on the noisy $x(t)$ to determine $\alpha$. It is obvious that any smoothing filter will wash away this edge a little and make it difficult to distinguish between a signal with $\alpha = 0$ and a signal with $\alpha = \alpha_{\mathrm{min}}$ for some minimum alpha. But there should be some minimum $\alpha$ that is distinguishable from a sharp edge.

For example, a Gaussian smoothing filter of some scale will make it impossible to distinguish between $\alpha$ less than that scale, but should make it possible to detect $\alpha$ larger than the scale of the filter.

What is the best possible performance a smoothing filter can have with regards to edge-preservation in this case? That is, what is the best possible smoothing filter that I can use for this low signal-to-noise regime that will be able to distinguish between $\alpha = 0$ and $\alpha = \alpha_{\mathrm{min}}$ for the smallest possible $\alpha_{\mathrm{min}}$.

Now, obviously, in practice we have discrete signals and $x(t)$ will be sampled at some sampling rate. So, hopefully, the answer to this question will address this. There's no real use to answering this question as if it was referring to a continuous signal.

  • $\begingroup$ Are you aware of Gibb's phenomenon? $\endgroup$ – Marcus Müller Jul 21 '17 at 22:42
  • $\begingroup$ I am aware of Gibb's phenomenon but I am unsure of how to use it to understand my problem. Could you please elaborate? $\endgroup$ – XYZT Jul 23 '17 at 0:10
  • $\begingroup$ The signal is not square-integrable. One cannot say the SNR is equal to 1 $\endgroup$ – Laurent Duval Dec 20 '17 at 21:04

This is Gibb's phenomenon (just not for the common edge case).

Hence, with a linear filter, the better you remove noise (which is, low-pass filtering), the more you remove the edges (which inherently have energy everywhere in the spectrum).

There's no minimum case – the noise bandwidth you allow is always the same as the signal bandwidth you allow, and thus, complementary, the amount of signal energy (and hence, edge precision) you cut off.

Among the things you can do is apply a clever signal model. Does your signal always look like your $x(t)$? Great, so you only need to estimate one parameter ($\alpha$) to fully describe it!

  • $\begingroup$ If I knew I was working with some small limited bandwidth, then I would have some sort of minimum. I am trying to figure out how to determine such a minimum. $\endgroup$ – XYZT Jul 23 '17 at 13:14
  • $\begingroup$ I don't understand. There is no minimum, as I tried to explain. $\endgroup$ – Marcus Müller Jul 23 '17 at 18:06
  • $\begingroup$ This question was not meant to be limited to linear filters. $\endgroup$ – XYZT Mar 27 '18 at 19:30

I am not sure if I am really answering your question correctly - but the point is, any linear filter by all means will do worse to preserve edges while removing noise.

However, if only you must have the linear filter. If you can go beyond linear filter you can device other techniques that can do better.

For example, you can remove noise to reasonable extent through any low pass filter, you will reduce the sharpness of the rise of pulses. However, this output you can feed to a simple threshold based function which will re-create sharp edges as soon as signal jumps at sufficiently high rate. This of course, assume that you know the exact signal shape a prior. It wont work if signal shape changes arbitrarily.

BTW: Look at variety of Analog to Digital Circuits for quantization there are similar techniques. Similarly there are many clock synchronisation circuit which uses PLLs to essentially 're-construct' the original clock pulses even as the signal is filled with lots of distortions.

I am not sure, this path is applicable to your case - but you can certainly look for non-linear filters.

  • $\begingroup$ I was actually expecting to use a non-linear filter. However, I don't know enough about them to determine what the limit is. $\endgroup$ – XYZT Mar 27 '18 at 19:30

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