除了噪声数据的局部最大值和最小值之外，还找到局部“平坦”

Question

There is a time-stress spectrum with 10,000s counts of records.

The timestamp-stress was recorded at random frequency from 10e-5 to 1HZ.

At the beginning I am trying to find:

• Local peak and valley, with fluctuation higher than certain threshold (say 0.6 unit).
• Just an extreme simple way to filt out some noise and record noticeable changes in pressure.

I searched the web and common approach is de-noise and find local maxima and minima.

On the other hand, the legacy code I have implemented in the following way (in Javascript).

  function filtSpectrum(res, minor){  
    var filted = []; //array of filted minor cycles
    var lookformax = 1; // 1 for look for max, 0 for look for min
    var currentstress;
    var mx_index = 0;
    var mn_index = 0;
  
    for (var i=0; i < arrayLength; i++){
      currentstress = res[i].psi;
      if(currentstress > res[mx_index].psi){
        mx_index = i;
      }
      if (currentstress < res[mn_index].psi){
        mn_index = i;
      }
      if(lookformax === 1){
        if(currentstress < res[mx_index].psi - minor){
          filted.push(res[mx_index]);
          mn_index = i;
          lookformax =0;
        }
      }else if (currentstress > res[mn_index].psi + minor){
        filted.push(res[mn_index]);
        mx_index = i;
        lookformax = 1;
      }
    }
    return filted;
  }

It has no issue to output the results like the following picture

Now the spectrum is represented by peak - valley type of points.

But there are some area in the middle if the picture that are flat . The oversimplification ignored the flat area and misrepresent the slope / (or frequency) of the peak and valleys.

Is there some simple algorithm to convert the original signal into peak-flat-valley kind of representation?

I am rather new to DSP. Thanks in advance for any suggestions and feedbacks.

Just like the red dashed flat line in the following image.

Answer 1

One idea would be to:

• Always select the very first and last points of your series
• Look for the point that deviates the most from the line through those two end points, and select that point as well.
• Split the series into two at that newly selected point, such that this point is the last one in the first subseries, and the first in the second subseries. Then use recursion to apply the above algorithm for these two subseries.
• Stop the recursion when the largest deviation found is smaller than the threshold. In that case only keep the end points.

Here is an implementation, with a demo on a data set that is similar to what you pictured in your question, but it is structured as a simple array of [x, y] pairs, as I couldn't completely see from your question how your data is structured (but y seems to correspond to psi ):

 function simplify(polyline, maxdy) { function recur(first, last) { let [x0, y0] = polyline[first]; let [x1, y1] = polyline[last]; let m = (y1 - y0) / (x1 - x0); let localMaxdy = 0; let choice = 0; for (let i = first + 1; i < last; i++) { let [x, y] = polyline[i]; let dy = Math.abs(y0 + m*(x - x0) - y); if (dy > localMaxdy) { choice = i; localMaxdy = dy; } } return localMaxdy < maxdy? [[x0, y0]] // Only keep first point: recur(first, choice).concat(recur(choice, last)); } return recur(0, polyline.length - 1).concat(polyline.slice(-1)); // always add last point } const ctx = document.querySelector('canvas').getContext('2d'); function drawPolyline(polyline, color) { ctx.lineWidth = 1.5; ctx.strokeStyle = color; ctx.beginPath(); for (const [x, y] of polyline) ctx.lineTo(x, y); ctx.stroke(); } // Demo let data = [[0,0],[81.5,1.5],[97.5,7.5],[98,8.5],[113,10.5],[113.5,12],[123,10.5],[127.5,12],[131.5,13.5],[132.5,15.5],[133,10],[133.5,30],[136,14.5],[138,15.5],[139.5,20],[140.5,21.5],[141,33],[149,35.5],[149.5,37.5],[152,40],[152.5,41],[159.5,45],[167.5,46],[171.5,47.5],[208,48.5],[290.5,52.5],[322,56.5],[330.5,64.5],[335,66],[345,63.5],[347.5,61.5],[352,65],[362.5,66.5],[367.5,64.5],[369.5,66],[386,65],[390,66],[391,67.5],[392,76],[394,80],[396,151.5],[396.5,154],[398,155],[400,159],[401.5,152.5],[402.5,140],[403.5,137.5],[405.5,180],[407.5,168],[409,139.5],[410,129.5],[412,126.5],[413.5,127.5],[415,123],[415.5,117],[416.5,114],[417.5,105.5],[419,102.5],[419.5,96],[420.5,91.5],[422.5,89.5],[423.5,78],[424.5,71.5],[426,69],[429,77],[432,43.5],[433,50.5],[434,52.5],[435.5,51.5],[436,54.5],[439,57],[440.5,60.5],[442.5,58.5],[444.5,58],[446,56],[447,51.5]]; drawPolyline(data, "grey"); let simple = simplify(data, 10); drawPolyline(simple, "red");

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