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# dependence of peak resolution on moving average

## dependence of peak resolution on moving average

(OP)
Hello,

I didn't know where to post, the exact topic would be signal processing, this is the closest forum I've found.

For those who aren't familiar with the definition, the resolution I'm talking about would concern peaks (my case is time domain but could be other), typically more or less symmetric, in a signal that's otherwise flat. This resolution is an adimensional number obtained by dividing the abscissa of the peak (local maximum) by the width of the peak across half the maximum height (according to the attached image, xc/w1). So when you dimensionalize it back, it tells you how apart must two peaks be so as not to be confused into one single peak.

I have a signal (a chromatogram of a DMA scan) and I want to calculate the resolution of its peaks. The first problem is that there's much noise superimposed (it's jagged), so I feel I need to smooth the signal, for the interpolation of the point at half maximum height to have any significance. In the past we have also fitted the data to a Gaussian, what it looks like at first sight, but examining it closely, specially converting the ordinate to the logarithmic scale, revealed that the distribution must actually be nothing like a Gaussian. Of course we could theorize which distribution can be expected to be found and see if we're right in practice and thus refine the true distribution... But I opted for using a moving average (which is optimal for reducing noise according to
http://www.dspguide.com/ch15/2.htm
) and running with it. Besides I'm not really knowledgeable in statistic distributions.

I get what I want, the peaks are smoothed and my algorithm works in finding the true resolution of the smoothed peak, while it would get non representative results from the jagged raw peak. But the moving average also flattens the peaks somewhat (as little as possible in exchange of the noise removed, according to the link above), and so the resolution found is smaller than what I would get from a curve fitting, that wouldn't flatten the peak in the least.

This isn't a show stopper, since if we always use the same moving average, we know what standard we're using. But I would like to get this right, and besides I'd like the results to be more in accordance with past results gotten from fitting to Gaussian.

I have found the delay in the abscissa a moving average causes (=0.5*[n-1], 'n' being the number of steps counted into the average). But I haven't found this.

What I want is the functional dependence on 'n' of the resolution. I'll start trying to find this out myself, but if someone is faster than me or knows this in advance I'll be grateful.
Replies continue below

### RE: dependence of peak resolution on moving average

Bit of philosophy first. The relationship B>=1/T is not saying that resolution is LIMITED, it is saying that there is no general meaning to a frequency resolution finer than 1/T. You can spend a lot of time devising clever refinements that appear to get around 1/T, but they introduce other errors, for general signals.

If you add some new knowledge to your bare signal then you can 'beat'  1/T. I think there is a reasonable case that the human ear/brain can tune in much faster than a DSP chip, if it knows it is listening to music, for example.

Cheers

Greg Locock

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### RE: dependence of peak resolution on moving average

(OP)
Thanks Greg. I had reached a similar conclusion, when I verified that the moving average added an equal delay to the upwards and downwards slopes, so it doesn't widen the peak (when this isn't ridiculously thin in comparison to the moving average). The resolution is not affected when the maximum ordinate isn't decreased, when the top of the peak has enough points--when the sampling ratio is high enough, in relation to the moving average size.

We often do this resolution calculation by hand at a guess, but the only better automatic way than a moving average, that I can think of, would be curve fitting. Anyway we can get something by hand only when the peak is smooth and doesn't need smoothing in the first place.

### RE: dependence of peak resolution on moving average

(OP)
Thanks Cloa, I'm sorry I didn't notice that. But I think I'll consider this solved.

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