Find value of signal A at a given value of signal B (twice for every cycle)?

In this particular task, value of signal A at a given value of signal B is required. Here, time axis is not available and interpolation is required as csv file does not include all the data points. Minima and Maxima were relatively easy to find than this.

For example, whenever B = 70, it should give 50 values of signal A for these 25 cycles in a table/matrix. I tried using interpolation function but it gives only one value and not 50 values. yes, something is missing in the code!!

I tried this code which gives only one value:

X=dlmread('file1.txt');
A=X(:,1);
B=X(:,2);  
xx = interp1(B,A, 70);

Later, seperation of those value is required based on if they were found on positive going edge or negative.

[Matt J]

Your code doesn't contain a variable called "B" nor does it show how the plot was generated.

[Himanshu Bhadja]

I have updated the variable names in the description. And also added a text file which is used to generate the plot

Answer 1

[Matt J]

Download this,

https://www.mathworks.com/matlabcentral/fileexchange/78008-tools-for-processing-consecutive-repetitions-in-vectors

and then,

[starts,stops]=groupLims( groupTrue(B>=70) , 1);

Avalues=A([starts,stops]);

[Himanshu Bhadja]

This worked but it find points from the input file only. As i mentioned in the question, not all the data points are available in file. So if the signal does not have exact value 70 then it will take any nearest number and respective value of another signal. But i would like to have interpolated values so that it could give value of signal B when signal A is exactly at 70

[Matt J]

Yes, but once you know the nearest values, you can interpolate.

for i=1:numel(starts)
       I=starts(i)-1:starts(i);
    Avalues(1,i)=interp1(B(I), A(I),70);
        I=stops(i):stops(i)+1;
    Avalues(2,i)=interp1(B(I), A(I),70);
end

Answer 2

[Bruno Luong]

You can fit a model to the interpolated data to remove the jump due to interval leap

ab=load('file1.txt');

a=ab(:,1);
b=ab(:,2);
[bb,i]=unique(b);
b2afun=@(b) interp1(bb,a(i),b);

bquery=20:120;
aquery=b2afun(bquery);

plot(bquery,aquery);
xlabel('b')
ylabel('a')

[Bruno Luong]

Actually your data seems to behave differently when a/b increases or decreases (first plot), similar to hysteresis. So here I separate interpolation in two branches

ab=load('file1.txt');
a=ab(:,1);
b=ab(:,2);
plot(a,b)

[~,locv] = findpeaks(-b);
[~,locp] = findpeaks(+b);
if  locp(1) <  locv(1)
    locv = [1; locv];
else
    locp = [1; locp];
end
if  locp(end) >  locv(end)
    locv = [locv; length(b)];
else
    locp = [locp; length(b)];
end

loc = [locp; locv];
[loc, is] = sort(loc);
ip = is <= length(locp);
[~,i] = histc((1:length(b))',loc);
n = length(loc);
icrop = max(min(i,n-1),1);
ivalid = i>=1 & i<=n-1;
asc = ivalid & ~ip(icrop) & ip(icrop+1);
dsc = ivalid & ip(icrop) & ~ip(icrop+1);

[asc_b,j]=unique(b(asc));
ia = find(asc);
ia = ia(j);
asc_b2afun=@(b) interp1(asc_b,a(ia),b);

[dsc_b,j]=unique(b(dsc));
id = find(dsc);
id = id(j);
dsc_b2afun=@(b) interp1(dsc_b,a(id),b);

bquery=20:120;
asc_aquery=asc_b2afun(bquery);
dsc_aquery=dsc_b2afun(bquery);

plot(bquery,asc_aquery,bquery,dsc_aquery);
xlabel('b')
ylabel('a')
legend('ascending interpolation', 'descending interpolation', 'location', 'best')