C++, determine the part that have the highest zero crosses - c++

Question

I’m not specialist in signal processing. I’m doing simple processing on 1D signal using c++. I want really to know how I can determine the part that have the highest zero cross rate (highest frequency!). Is there a simple way or method to tell the beginning and the end of this part.
This image illustrate the form of my signal, and this image is what I need to do (two indexes of beginning and end)
Edited:
Actually I have no prior idea about the width of the beginning and the end, it's so variable.
I could calculate the number of zero crossing, but I have no idea how to define it's range
double calculateZC(vector<double> signals){
int ZC_counter=0;
int size=signals.size();
for (int i=0; i<size-1; i++){
if((signals[i]>=0 && signals[i+1]<0) || (signals[i]<0 && signals[i+1]>=0)){
ZC_counter++;
}
}
return ZC_counter;
}

Answer 1

Here is a fairly simple strategy which might give you some point to start. The outline of the algorithm is as follows
Input: Vector of your data points {y0,y1,...}
Parameters:
Window size sigma.
A threshold 0<p<1 defining when to start looking for a region.
Output: The start- and endpoint {t0,t1} of the region with the most zero-crossings
I won't give any C++ code, but the method should be easy to implement. As example let us use the following function
What we desire is the region between about 480 and 600 where the zero density higher than in the front. First step in the algorithm is to calculate the positions of zeros. You can do this by what you already have but instead of counting, you store the values for i where you met a zero.
This will give you a list of zero positions
From this list (you can do this directly in the above for-loop!) you create a list having the same size as your input data which looks like {0,0,0,...,1,0,..,1,0,..}. Every zero-crossing position in your input data is marked with a 1.
The next step is to smooth this list with a smoothing filter of size sigma. Here, you can use what you like; in the simplest case a moving average or a Gaussian filter. The higher you choose sigma the bigger becomes your look around window which measures how many zero-crossings are around a certain point. Let me give the output of this filter together with the original zero positions. Note that I used a Gaussian filter of size 10 here
In a next step, you go through the filtered data find the maximum value. In this case it is about 0.15. Now you choose your second parameter which is some percentage of this maximum. Lets say p=0.6.
The final step is to go through the filtered data and when the value is greater than p you start to remember a new region. As soon as the value drops below p, you end this region and remember start and endpoint. Once you are finished walking through the data, you are left with a list of regions, each defined by a start and an endpoint. Now you choose the region with the biggest extend and you are done.
(Optionally, you could add the filter size to each end of the final region)
For the above example, I get 11 regions as follows
{{164,173},{196,205},{220,230},{241,252},{259,271},{278,290},
{297,309},{318,327},{341,350},{458,468},{476,590}}
where the one with the biggest extend is the last one {476,590}. The final result looks (with 1/2 filter region padding)
Conclusion
Please don't be discouraged by the length of my answer. I tried to explain everything in detail. The implementation is really just some loops:
one loop to create the zero-crossings list {0,0,..,1,0,...}
one nested loop for the moving average filter (or you use some library Gaussian filter). Here you can at the same time extract the maximum value
one loop to extract all regions
one loop to extract the largest region if you haven't already extracted it in the above step