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Remove LTPDARepositoryManager implementation * * * Remove GUI helper
author Daniele Nicolodi <nicolodi@science.unitn.it>
date Mon, 05 Dec 2011 16:20:06 +0100
parents f0afece42f48
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  <p style="font-size:1px;">&nbsp;</p>

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  "100%" cellpadding="0" cellspacing="0">
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      <td valign="baseline"><b>LTPDA Toolbox</b></td><td><a href="../helptoc.html">contents</a></td>

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      "ltpda_training_topic_5_3.html"><img src="b_prev.gif" border="0" align=
      "bottom" alt="Fitting time series with polynomials"></a>&nbsp;&nbsp;&nbsp;<a href=
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  <h1 class="title"><a name="f3-12899" id="f3-12899"></a>Non-linear least squares fitting of time series</h1>
  <hr>
  
  <p>
	
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    <tr>
      <td valign="baseline"><b>LTPDA Toolbox</b></td><td><a href="../helptoc.html">contents</a></td>

      <td valign="baseline" align="right"><a href=
      "ltpda_training_topic_5_3.html"><img src="b_prev.gif" border="0" align=
      "bottom" alt="Fitting time series with polynomials"></a>&nbsp;&nbsp;&nbsp;<a href=
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      "bottom" alt="IFO/Temperature Example - signal subtraction"></a></td>
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  <h1 class="title"><a name="f3-12899" id="f3-12899"></a>Non-linear least squares fitting of time series</h1>
  <hr>
  
  <p>
	

<p>

  Non-linear least square fitting of time-series exploits the function <tt>ao/xfit</tt>.   
</p>

<h2> Non-linear least square fitting of time series </h2>

<p>
  During this exercise we will:
  <ol>
    <li> Load time series data
    <li> Fit data with <tt>ao/xfit</tt>
    <li> Check results
    <li> Refine the fit with a Monte Carlo search
  </ol>
</p>

<p> 
  Let us open a new editor window and load test data.
</p>

<div class="fragment"><pre>
    a = ao(plist(<span class="string">'filename'</span>, <span class="string">'topic5/T5_Ex05_TestNoise.xml'</span>));
    a.setName(<span class="string">'data'</span>);
    iplot(a)

</pre></div>

<p>
  As can be seen this is a chirped sine wave with some noise.
  <div align="center">
    <IMG src="images/ltpda_training_1/topic5/ltpda_training_5_4_1.png" align="center" border="0">
  </div>
  We could now try the fit. The first parameter to pass to <tt>xfit</tt>
  is a fit model. In this case we assume that we are dealing with a linearly 
  chirped sine wave according to the equation: <br/>

  <br/>
  <div align="center">
    <IMG src="images/ltpda_training_1/topic5/ex4_chirpsine.gif" align="center" border="0">
  </div>
  <br/>
  The previous function can be stored within a <tt>smodel</tt> analysis object to pass to the fitting machinery:
<div class="fragment"><pre>
    mdl = smodel(plist(<span class="string">'Name'</span>, <span class="string">'chirp'</span>, ...
              <span class="string">'expression'</span>, <span class="string">'A.*sin(2*pi*(f + f0.*t).*t + p) + c'</span>, ...
              <span class="string">'params'</span>, <span class="string">{'A','f','f0','p','c'}</span>, ...
              <span class="string">'xvar'</span>, <span class="string">'t'</span>, ...
              <span class="string">'xunits'</span>, <span class="string">'s'</span>, <span class="string">'yunits'</span>, <span class="string">'m'</span>));

</pre></div>
  
  We need to specify a starting guess for the model parameters. 
  The output of <tt>ao/xfit</tt> is a <tt>pest</tt> analysis objects containing fit parameters.

</p>

<div class="fragment"><pre>
    plfit1 = plist(<span class="string">'Function'</span>, <span class="string">mdl</span>, ...
              <span class="string">'P0'</span>, [5,9e-5,9e-6,0,5]);

    params1 = xfit(a, plfit1);

</pre></div>

<p>
  Once the fit is done. We can evaluate our model to check fit results.
</p>

<div class="fragment"><pre>
    b = eval(params1, plist(<span class="string">'xdata'</span>, a, <span class="string">'xfield'</span>, <span class="string">'x'</span>));
    b.setName;    
    iplot(a,b)

</pre></div>

<p>
  As you can see, the fit is not accurate. One of the great problems of
  non-linear least square methods is that they easily find a local minimum of
  the chi square function and stop there without finding the global minimum.
  There are two possibile solutions to such kind of problems: the first one is
  to refine step by step the fit by looking at the data; the second one is to
  perform a Monte Carlo search in the parameter space. This way, the fitting machinery
  extracts the number of points you define in the <tt>'Npoints'</tt>
  key, evaluates the chi square at those points, reoders by ascending chi square, selects
  the first guesses and fit starting from them.
</p>

<div class="fragment"><pre>
    plfit2 = plist(<span class="string">'Function'</span>, <span class="string">mdl</span>, ...
              <span class="string">'MonteCarlo'</span>, <span class="string">'yes'</span>, ...
	      <span class="string">'Npoints'</span>, 1000, ...
	      <span class="string">'LB'</span>, [1,5e-5,5e-6,0,2], ...
              <span class="string">'UB'</span>, [10,5e-4,5e-5,2*pi,7]);
    
    params2 = xfit(a, plfit2);
    
    c = eval(params2, plist(<span class="string">'xdata'</span>, a, <span class="string">'xfield'</span>, <span class="string">'x'</span>));
    c.setName;    
    iplot(a,c)    

</pre></div>

<p>
  The fit now looks like better...
  <div align="center">
    <IMG src="images/ltpda_training_1/topic5/ltpda_training_5_4_3.png" align="center" border="0">
  </div>
  
  Let us compare fit results with nominal parameters. <br/>
  Data were generated with the following set of parameters:
</p>

<div class="fragment"><pre>
    A  = 3
    f  = 1e-4
    f0 = 1e-5
    p  = 0.3
    c  = 5
</pre></div>

<p>

  Fitted parameters are instead:
</p>

<div class="fragment"><pre>
    A  = 3.02 +/- 0.05
    f  = (7 +/- 3)e-5
    f0 = (1.003 +/- 0.003)e-5 
    p  = 0.33 +/- 0.04
    c  = 4.97 +/- 0.03
</pre></div>
<p>
  The correlation matrix of the parameters, the chi square, the degree of freedom, the covariance matrix are
  store in the output <tt>pest</tt>. Other useful information are stored in the <tt>procinfo</tt> (processing information)
  field. This field is a <tt>plist</tt> and is used to additional information that can be
  returned from algorithms. For example, to extract the chi square, we write:
</p>

<div class="fragment"><pre>
    params2.chi2
    1.0253740840052
</pre></div>
<p>
  And to know the correlation matrix:
</p>

<div class="fragment"><pre>
    params2.corr
    Columns 1 through 3

                         1         0.120986348157139       -0.0970894969803509
         0.120986348157139                         1        -0.966114904879414
       -0.0970894969803509        -0.966114904879414                         1
        -0.156801230958825        -0.848296014553159         0.717376893765734
       -0.0994358284166703         0.187645552903433        -0.169496082635319

    Columns 4 through 5

        -0.156801230958825       -0.0994358284166703
        -0.848296014553159         0.187645552903433
         0.717376893765734        -0.169496082635319
                         1        -0.199286767157984
        -0.199286767157984                         1
                       </pre></div>

<p>
  Not so bad!
</p>



















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