Taking second derivative and solving it for when it equals 0
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Here is the code that I have:
clear all; close all
m = xlsread('1 MgO.xlsx','Sheet1','B1:J18');
temp = m(:,6);
MgO = m(:,3);
plot(temp, MgO,'*')
xlabel('Temperature (K)'); ylabel('Shear Modulus (GPa)'); title('MgO')
hold on
% Line of best fit
p = polyfit(temp,MgO,3);
y = polyval(p,temp);
plot(temp,y,'-')
SSE=sum((MgO-y).^2); SST=sum((MgO-mean(MgO)).^2); Rsquared=1-SSE/SST;
text(100,100,sprintf('y = %.3x^3 + %.3fx^2 + %.3fx + %.3f, R^2 = %.3f',p(1),p(2),p(3),p(4), Rsquared))
I'm trying to figure out how to get the second derivative and find at what temperature it equals 0. temp is my xaxis and MgO is my yaxis. I tried using the diff command but it didn't really work. Could someone help please?
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Star Strider
2019-3-21
0 个投票
I am not certain what you are doing. Taking the derivative of data will significantly amplify any noise that might be present, so in general, it is best to take the derivative of a smooth signal. Also, you are fitting a 3° polynomial, so the second derivative of that (regardless of the method you use) is giong to be a linear (1°) function.
If you want to take the derivative of filtered data (eliminating as much noise as possible first), use the gradient (link) function. Use it two times to get the second derivative.
If you want to take the derivative of your fitted function ‘p’, one option is to use the polyder (link) function, again twice. You can then easily solve for the x-intercept of the linear relation this produces.
4 个评论
Christina Kersten
2019-3-21
Star Strider
2019-3-21
Yes.
If you define it as:
d2p = polyder(polyder(p));
the x-intercept is:
xint = -d2p(1)/d2p(2);
Christina Kersten
2019-3-22
Star Strider
2019-3-22
It should be:
xint = -d2p(2)/d2p(1);
That was what I intended to write, and I thought that was what I wrote.
The derivation comes from:
y = d2p(1)*x + d2p(2)
0 = dp2(1)*xint + d2p(2)
with the x-intercept defined by the point where y=0.
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Walter Roberson
2019-3-21
0 个投票
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