How to force slope to be zero at a particular point using function PolyFit?
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Q1 = 1xn;
T = 1xn;
Finding a curve fitting (T,Q1) such that slope is zero at Q1(i), where i = 1,2,...,n
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Answers (4)
Teja Muppirala
on 22 Apr 2013
If you have LSQLIN in the Optimization Toolbox, this can be done with a little bit of effort, as described here http://www.mathworks.com/support/solutions/en/data/1-12BBUC/
% Making some random data
Q1 = 0:0.01:1;
T = cos(2.1*pi*Q1)+0.2*randn(size(Q1));
plot(Q1,T,'k.');
% Polyfit without constraints
order = 4;
C1 = polyfit(Q1,T,order);
hold on;
plot(Q1,polyval(C1,Q1))
V = bsxfun(@power,Q1(:),order:-1:0); % Make Vandermonde Matrix
% Make Constraints on the derivatives
Aleft = [(order:-1:1).*Q1(1).^(order-1:-1:0) 0];
Aright = [(order:-1:1).*Q1(end).^(order-1:-1:0) 0];
Aeq = [Aleft; Aright];
beq = [0;0]; %Value of the derivative is set to zero
% Call LSQLIN with options to prevent warnings
opts = optimset('lsqlin');
opts.LargeScale = 'off';
C2 = lsqlin(V,T,[],[],Aeq,beq,[],[],[],opts);
plot(Q1,polyval(C2,Q1),'r')
hold off;
legend({'Data' 'Polyfit' 'Constrained Polyfit'},'location','best');
Matt J
on 22 Apr 2013
If the slope is zero at all i=1...n, it means you are fitting Q1 with a constant.
p=mean(Q1);
Image Analyst
on 22 Apr 2013
Edited: Image Analyst
on 22 Apr 2013
Your assignment says nothing about requiring the polyfit() function. There's no way to have the slope be 0 at "1" and "end" in general, unless you use Matt's solution. For specific functions, e.g. 4th order, you may luck out if your data happens to go in between the right elements, e.g. the two humps of a 4th order, but in general that won't happen. So, I'd probably use a spline. You might have to replicate the first and last value of the array to make sure that the slope is zero there.
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Matt J
on 22 Apr 2013
You might have to replicate the first and last value of the array to make sure that the slope is zero there.
MATLAB's SPLINE command let's you specify the endslopes directly. For more general constrained splines, there is this FEX tool
Image Analyst
on 22 Apr 2013
Thanks for pointing that out. I didn't know that and it's nice to know.
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