Re: [Discuss-gnuradio] curve fitting data points

[mike revnell]

I had a go at fitting a quadratic using the Scilab leastsq function.
I ended up with the following:

  y = -3.6330888 + 0.3255432 * x + 0.0781971 * x^2

for whatever units the xdata and ydata arrays were in another message
with the python code that seems to be doing much the same.

The following scilab code computes the coefficients and plots the 
original and
fitted function on the same plot. The fit misses a bit because the last 
pair of
points seems to falling off quite a bit.

--------------------------------------------------------------------------
xdata = [5.357, 5.457, 5.797, 5.936, 6.161, 6.697, 6.731, 6.775, 8.442, 
9.769, 9.861];
ydata = [0.376, 0.489, 0.874, 1.049, 1.327, 2.054, 2.077, 2.138, 4.744, 
7.068, 7.104];
xm=xdata';
ym=ydata';
m=max(size(xdata))
wm=ones(m,1);

x0=[1;1;1];

function y = func( p, x )
 y = p(1) + p(2) * x + p(3) * x^2
endfunction

function e = errfunc( p, xm, ym, wm )
 e = wm .* ( func( p, xm ) - ym )
endfunction

[f,xopt,gropt]=leastsq(list(errfunc,xm,ym,wm),x0)

plot( xdata, ydata )
x=(5:.5:10)
plot(x,func(xopt,x))

-----------------------------------------------------------------------------------------------------------

cswiger wrote:

> This is for the mathematicians out there - what is a simple
> working algorithm for creating a function model to fit an
> arbitrary number of data points. What I have for a first
> approximation, simple linear (y=mx+b) actually works better
> than nothing, but there's room for improvement.
>
> I set one frequency input X1=3.9e6 and an arbitrary antenna control
> output Y1=1800 and adjust the physical antenna for peak signal.
> Then I change the frequency to X2=4e6 and adjust the antenna
> output for peak signal, which happens to be Y2=1580. From there
> you can get the slope m=(Y2-Y1)/(X2-X1) and the y intercept
> b  = Y1 - m * X1.   After that calibration the antenna tune
> output is m * freq + b, works fine.   Change the frequency and
> the antenna peak tracks ok, especially near the two calibration
> points, but going very far out or if the points are too far
> apart noticible error creeps in.
>
> What is a method similar to that for 3 points? or n p  x
>  p
> oints?
> I want to collect (x1, y1), (x2, y2), (x3, y3)  then use
> a simple python formula to get a workable y for any given x.
>
> Thanks
>
> --Chuck
>
>
>
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