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Notebook[{

Cell[CellGroupData[{
Cell["Biorthogonal Spline Filters", "Title"],

Cell["\<\
Wavelet Workshop
June 7-10, 2006
University of St. Thomas\
\>", "Subtitle"],

Cell["Objectives", "Subsubtitle"],

Cell[TextData[{
  "The purpose of this notebook is to introduce you to some basic \
wavelet-based denoising methods.\n\nThe notebook also contains a \
module-writing lab that makes up part of ",
  StyleBox["Computer Session Three",
    FontWeight->"Bold"],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell["WaveletFunctions", "Subsubtitle",
  InitializationCell->True],

Cell[TextData[{
  "This cell initializes every time you open the notebook.  It loads the ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " package ",
  StyleBox["WaveletFunctions",
    FontFamily->"Courier"],
  " for use in subsequent computations."
}], "Text",
  InitializationCell->True],

Cell[BoxData[{
    \(<< WaveletFunctions`WaveletFunctions`\), "\[IndentingNewLine]", 
    \(<< LinearAlgebra`MatrixManipulation`\), "\[IndentingNewLine]", 
    \(\(\(<< Graphics`Graphics`\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(<< Statistics`ContinuousDistributions`\), "\n", 
    \(\(\(<< Statistics`DescriptiveStatistics`\)\(\n\)
    \)\), "\[IndentingNewLine]", 
    \(\(Needs["\<Miscellaneous`Audio`\>"];\)\n\[IndentingNewLine]\
\[IndentingNewLine]\), "\[IndentingNewLine]", 
    \(\(flashdir = "\<E:/WaveletWorkshop06/Images/\>";\)\), "\
\[IndentingNewLine]", 
    \(\(sounddir = "\<E:/WaveletWorkshop06/Sounds/\>";\)\), "\
\[IndentingNewLine]", 
    \(\(Print["\<The flash drive directory is \>", 
        flashdir, "\<.\>"];\)\[IndentingNewLine]\), "\[IndentingNewLine]", 
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"\<http://cam.mathlab.stthomas.edu/pvf/Math316/TestImages/\>";\)\), "\
\[IndentingNewLine]", 
    \(\(Print["\<The URL for images is \>", imgurl, "\<.\>"];\)\)}], "Input",
  InitializationCell->True]
}, Open  ]],

Cell[CellGroupData[{

Cell["Help on WaveletFunctions", "Subsubtitle"],

Cell[TextData[{
  "If you ever need help with ",
  StyleBox["WaveletFunctions",
    FontFamily->"Courier"],
  ", go to ",
  StyleBox["Help",
    FontSlant->"Italic"],
  ", then ",
  StyleBox["Help Browser",
    FontSlant->"Italic"],
  ", and click on ",
  StyleBox["AddOns & Links",
    FontSlant->"Italic"],
  ".  If you scroll down you will find ",
  StyleBox["WaveletFunctions",
    FontFamily->"Courier"],
  ".  "
}], "Text"],

Cell[CellGroupData[{

Cell["The Spline Filter", "Section"],

Cell[TextData[{
  "The biorthogonal spline filter is defined by expanding the function ",
  Cell[BoxData[
      \(\(\@2\) \(cos\^N\) \((\[Omega]/2)\)\)]],
  " into an odd-length Fourier series if N is even or by expanding the \
function ",
  Cell[BoxData[
      \(\(\@2\) \(e\^\(i\[Omega]/2\)\) \(cos\^N\) \((\[Omega]/2)\)\)]],
  "into an even-length Fourier series if N is odd.\n\nAs we learned in the \
lecture, the result is a list of binomial coefficients multiplied by ",
  Cell[BoxData[
      \(\@2\)]],
  " and divided by ",
  Cell[BoxData[
      \(2\^N\)]],
  ".    In the cells below, we show two ways to obtain this filter.\n\nFirst, \
we can write the cosine function in terms of the complex exponential and \
expand it:"
}], "Text"],

Cell[BoxData[{
    \(\(cs[w_] := \((E^\((I*w)\) + E^\((\(-I\)*w)\))\)/
          2;\)\), "\[IndentingNewLine]", 
    \(\(\(f[w_, n_] := \ 
      Sqrt[2]*E^\((I*w*Mod[n, 2]/2)\)*cs[w/2]^n\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(Expand[f[w, 2]]\), "\[IndentingNewLine]", 
    \(Expand[f[w, 3]]\), "\[IndentingNewLine]", 
    \(\)}], "Input"],

Cell["\<\
Next, we simply need to peel off the coefficients from the Fourier series to \
obtain our filter.  We can do this using the Table command and the \
Coefficient command.  

We also need to know the limits on the indices of the filter.  This is pretty \
easy - if N is even, the limits are -(N-1) to (N-1) and if N is odd, the \
limits are -N+2 to N-1:\
\>", "Text"],

Cell[BoxData[{
    \(n = 2\), "\[IndentingNewLine]", 
    \(\(\(filter = 
      Table[Coefficient[Expand[f[w, n]], E^\((I*w)\), 
          k], {k, \(-\((n - 1)\)\), n - 1}]\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(n = 3\), "\[IndentingNewLine]", 
    \(filter = 
      Table[Coefficient[Expand[f[w, 3]], E^\((I*w)\), k], {k, \(-n\) + 2, 
          n - 1}]\)}], "Input"],

Cell[TextData[{
  "Alternatively, (and much easier!!), we could simply write down a table of \
binomial coefficients, multiply by ",
  Cell[BoxData[
      \(\@2\)]],
  " and divide by ",
  Cell[BoxData[
      \(2\^N\)]],
  "."
}], "Text"],

Cell[BoxData[{
    \(n = 2\), "\[IndentingNewLine]", 
    \(\(\(filter\  = \ 
      Sqrt[2]*Table[Binomial[n, k], {k, 0, n}]/2^n\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(n = 3\), "\[IndentingNewLine]", 
    \(filter\  = \ Sqrt[2]*Table[Binomial[n, k], {k, 0, n}]/2^n\)}], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Using ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " to Find the Dual Filter"
}], "Section"],

Cell[TextData[{
  "Let's consider the spline filter ",
  Cell[BoxData[
      \(h\&~\)]],
  " for N=3 and try to find the dual filter h of length 8.  Let's write down \
the Fourier series for both:"
}], "Text"],

Cell[BoxData[{\(n = 3;\), "\[IndentingNewLine]", 
    RowBox[{\(hw = Sqrt[2]*Table[Binomial[n, k], {k, 0, n}]/2^n\), 
      "\[IndentingNewLine]", "\[IndentingNewLine]", 
      StyleBox[\( (*\ Don' t\ forget\ to\ make\ h\ symmetric\ *) \),
        FontColor->RGBColor[1, 0, 0]]}], "\[IndentingNewLine]", \(y = 
        Array[h, 4, 1];\), "\[IndentingNewLine]", 
    RowBox[{\(z = Join[Reverse[y], y]\), 
      "\[IndentingNewLine]"}], "\[IndentingNewLine]", \(Hw[w_] := 
      hw . Table[E^\((I*k*w)\), {k, \(-2\), 3}]\), "\[IndentingNewLine]", \(H[
        w_] := z . Table[E^\((I*k*w)\), {k, \(-3\), 4}]\)}], "Input"],

Cell[TextData[{
  "We know that ",
  Cell[BoxData[
      \(\(H\&~\) \((\[Omega])\) \(H \((\[Omega])\)\)\&_\  + \ \(H\&~\) \((\
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\((\[Omega])\) H \((\(-\[Omega]\))\) + \ \(H\&~\) \((\(-\[Omega]\) + \[Pi])\) 
              H \((\(-\[Omega]\) + \[Pi])\) = 2\)\)]],
  ", so we can expand this equation, peel off the Fourier coefficients, set \
them equal to 2 or 0, and solve a linear system.\n\nQuick aside - to solve a \
linear system in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", the commands look like:"
}], "Text"],

Cell[BoxData[{
    \(c = {a, b}\), "\[IndentingNewLine]", 
    \(eqs = {a + b \[Equal] 2, a - b \[Equal] 4}\), "\[IndentingNewLine]", 
    \(Solve[eqs, c]\)}], "Input"],

Cell["\<\
Use Table, Expand, and Coefficient to set up a system of linear equations \
that you could solve to find the filter h.  Don't forget the constraint H(\
\[Pi]) = 0!\
\>", "Text"],

Cell[BoxData[
    \( (*\ Put\ your\ Mathematica\ code\ here\ *) \)], "Input",
  FontColor->RGBColor[0, 0, 1]],

Cell["\<\
Did you get a unique solution?  If not, add enough derivative conditions \
until you do!\
\>", "Text"],

Cell[BoxData[
    \( (*\ Put\ your\ Mathematica\ code\ here\ *) \)], "Input",
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[{
  "The correct solution for h is given in the book on page 484.\n\nYou can \
also check your solution by expanding out formula (9.108) on page 486 in the \
case where N = ",
  Cell[BoxData[
      \(TraditionalForm\`\(N\&~\)\)]],
  "= 3."
}], "Text"],

Cell["\<\
Next, go back to the top and pick the length of hw to be 3 and the length of \
h to be 9.\
\>", "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Extra Credit - Tough One!", "Section"],

Cell[TextData[{
  "An explicit form of h is given by Theorem 9.5 on pages 484-485.  Can you \
use this theorem to write a ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " module to produce h? "
}], "Text"]
}, Open  ]]
}, Open  ]]
}, Open  ]]
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