Asked by Harshit Jain
on 22 Apr 2013

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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Answer by 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);

Harshit Jain
on 22 Apr 2013

Slope is not zero at all the points.

Eg - only at i = 1 and n.

Answer by Image Analyst
on 22 Apr 2013

Edited by 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.

Harshit Jain
on 22 Apr 2013

Thanks. Got it

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

http://www.mathworks.com/matlabcentral/fileexchange/24443-slm-shape-language-modeling

Image Analyst
on 22 Apr 2013

Thanks for pointing that out. I didn't know that and it's nice to know.

Answer by 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');

Harshit Jain
on 23 Apr 2013

thanks for the answer. I have done it through spline. It works fine and is very easy.

Answer by Jan Simon
on 22 Apr 2013

You find in the FileExchange polynomial fits with constraints:

Harshit Jain
on 23 Apr 2013

thanks for the files.

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