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I Robust linear estimation

  1. Dec 26, 2016 #1
    Hi and thanks to everyone for his/her attention. I have to minimize a function that depends on several parameters. The aim of minimizing that function is to actually guess these parameters, which are unknown. The thing is that the author of the pdf from which I have to make the calculations, does not specify very well how to carry it out. There are 3 parameters the function depends on (one of them nonlinear), and the author says that first off we have to figure out the nonlinear parameter, by trisecting a determined interval, and afterwards guess the spare ones by means of another function that will also have to be minimized by robust linear estimation. So here is where my doubts come, how should robust linear estimation be carried out in that context? What do yk and xk stand for?.

    Here is the part of the pdf in which that function is shown:

    you can find the whole pdf here:http://www.roulettephysics.com/wp-content/uploads/2014/01/Roulette_Physik.pdf

    that function is in the 13 page.

    thank you all for your attention and I hope you can help me :)
     
  2. jcsd
  3. Dec 27, 2016 #2
    No, you don't have to figure out the nonlinear parameter first and guess the linear parameters afterwards. You need to calculate the linear parameters with a linear method for each iteration step within the non-linear method. This way you can reduce the three-dimensional minimization problem

    [itex]S\left( {\eta ,\bar \Omega _f^2 ,\varphi } \right) \Rightarrow Min.[/itex]

    to the one-dimensional problem

    [itex]S\left( \varphi \right) = S\left( {\eta \left( \varphi \right),\bar \Omega _f^2 \left( \varphi \right),\varphi } \right) \Rightarrow Min.[/itex]

    which is much easier to solve even if the funktions [itex]\eta \left( \varphi \right)[/itex] and [itex]\bar \Omega _f^2 \left( \varphi \right)[/itex] need to be calculated by linear minimization for every value ##\varphi##.

    However, I'm not sure if the parameters ##\eta## and ##\bar \Omega _f^2## are really linear because they are within absolute values.
     
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