Help with interpol1. The grid vectors are not strictly monotonic increasing

Hello,

I have my StressAmp and FatigueLife -vectors that I am plotting and least Square fitting using the polyval and polyfit commands.

I only have data for Three Points as seen in my example. I am having trouble using interpol1 to get data within my least Square fit. I need to know the value on the x-axis when the value of y-axis is 6. What am I doing wrong and how can I fix it?

My error message is:

Error using griddedInterpolant

The grid vectors are not strictly monotonic increasing.

Error in interp1 (line 191) F = griddedInterpolant(X,V,method);

Error in MatlabHelpSNFatigue (line 30) K1=interp1(StressAmp,10.^fity,Amp1)

My matlab code is:

 StressAmp=[10 6 6];
 FatigueLife=[60000 264000 330000];

 %Log
 X=log10(StressAmp);
 Y=log10(FatigueLife);

 semilogx(FatigueLife,StressAmp,'*');
 ylim([0 12])
 set(gca,'YTick',[0:1:12])
 xlabel('N')
 ylabel('F')
 grid on
 hold on

coef=polyfit(X,Y,1)
fity=polyval(coef,X)
semilogx(10.^fity,StressAmp,'r')

 %interpol
 Amp1=6
 K1=interp1(StressAmp,10.^fity,Amp1)
 semilogx(K1,Amp1,'*r')

Answer 1

[Guillaume]

As the error message says, for interpolating, the known x must be strictly monotonically increasing. That means no duplicate (you have two 6) and in increasing order (so 6, 10).

I don't really understand what you're trying to do. What do you expect linear interpolation to do if you have twice the same x and two different y?

[Daniel]

Hi,

Sorry, I should have Amp1=8 and not 6.

I have plotted the least Square fit and need to know the value of a certain Point at this straight line.

I do not have the data when the value on the y-axis is 8. This is what I want. I hope it is clear now.

[Guillaume]

The problem is not with Amp1, but with StressAmp and more importantly, your understanding of linear interpolation.

linear interpolation just finds between which two known x your query point falls and then just takes the appropriate ratio of the two corresponding known y. That is, it's just:

interpolationresult  = (yafter-ybefore)*(unknownx-xbefore)/(xafter-xbefore)

If you have two ybefore for a single xbefore, what do you put in the formula above?

Anyway, since you've already performed a linear regression with polyval why don't you use polyfit to find your unknown?

logK1 = polyval(coeff, log(Amp1))

[Daniel]

Hi, Okey. Thank you. I understand better now, but using your last line and using log10 instead of log like below gives me almost what I want. This is not spot on the least Square fit (see Picture). Why is that so?

    logK1 = polyval(coef, log10(Amp1))
    semilogx(10^logK1,Amp1,'*r')