An example of my data is as follows: xdata=[1150 1200 1263.3 1333.3 1429 1497.3 1567.8 1608.3 1643.5 1668.8]; ydata=[6.2 7.2 8.6 10.4 11.2 12.9 13.3 13.3 13.5 13.55];
A scatter plot shows there are arguably 2 linear sections to the graph, the first steeper than the second. I would like to find out the slope of the first linear section (always positive) and the break point (can vary). the slope the second section may be zero, positive or negative (depending on the weather).
I tried the on line suggested solution but get an error message as shown. xdata=[1150 1200 1263.3 1333.3 1429 1497.3 1567.8 1608.3 1643.5 1668.8]; ydata=[6.2 7.2 8.6 10.4 11.2 12.9 13.3 13.3 13.5 13.55]; >> coefs = @(xbreak) [ones(numel(xdata),1),min(xdata,xbreak)-xbreak]\ydata; >> errfun = @(xbreak) norm([ones(numel(xdata),1),min(xdata,xbreak)-xbreak]*coefs(xbreak) – ydata); >> xbreak = fminbnd(errfun,0,1) Error using horzcat Dimensions of matrices being concatenated are not consistent.
Error in @(xbreak)norm([ones(numel(xdata),1),min(xdata,xbreak)-xbreak]*coefs(xbreak)-ydata)
Error in fminbnd (line 228) x= xf; fx = funfcn(x,varargin{:});
Hoping someone can help me?
Kind regards, Wendy
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