Signal processing by means of WaveletTransform

Question

I updated my question to explain what I want. I have the voltage-time curve from the real industial object. This curve was gotten from the digital oscilloscope:

As you can see it has hight-frequency (kHz) interference. We extracted data from the oscilloscope into the Microsoft Excel file and here it is: http://www.fileconvoy.com/dfl.php?id=g3c9169c7c6eb991d999470396248137bf77d4cabe

Using this data we reconstructed the curve (in MathCAD). As an equivalent to time-axis we have n-axis ("n" - number of the point; every point corresponds to certain moment of time; in all we have 4000 points for 8 ms):

Using Fourier Transofrm we got frequency spectrum (amp-freq curve):

Inasmuch as this signal is nonstationary, the usage of Fourier Transform is not appropriate. So we used Fast Wavelet Transform and the result was quite different (and probably correct). This is an amp-point curve which is similar to the recostructed source curve, but there are displayed amplitudes for 4,7 - 5,5 kHz frequencies (for different wavelet coefficients), which appeared exactly in these moments:

And now by some reason there is a need to process the source data in Wolfram Mathematica, also because it is far and away more powerful than MathCAD. But Mathematica is something new for me in the field of signal processing, and now I have no time to study it. There is almost one week left for us to have this task done. The main purpose of the work is to use different wavelet families and obtain the correct frequency spectrum (amp-freq curve).

We are working with power quality and electromagnetic compatibility. This task is an engeneering task.

So here is the diference between datas. Original data is on the top and the interpolated is below.

Answer 1

Once again I repeat that trying to reproduce something like the curve you are trying to get is going to be inaccurate - There are more suitable representations you can use to get the desired frequency spectrum, but you are the one asking the questions :)

First, we fetch your data

data = Import[
       "http://www.fileconvoy.com/gf.php?id=geed872d9b8a38dc6999443310.
         4661369818fbd5fa1bf3bc&sts=138977900593152145247060b494909aee48bb2a26b595048647"];

Additional function we will use:

f[list_, pos_] := Module[{x = list}, x[[All, pos]] = Sequence[]; x]

fdata = Flatten@f[data, 1];

cwd = ContinuousWaveletTransform[fdata, MorletWavelet[], {6, 20}, SampleRate -> 500000]

WaveletScalogram[cwd, ColorFunction -> "AvocadoColors", ImageSize -> 700]

ListLinePlot[Total[Abs[#]^2] & /@ Reverse@cwd[All, "Values"], PlotRange -> All,
            ImageSize -> 700, BaseStyle -> Thick, PlotStyle -> ColorData[19, "ColorList"]]

WaveletScalogram with the scale axis re-adjusted.

freq = (1000/(# MorletWavelet[]["FourierFactor"])) & /@
                (Thread[{Range[6], 1}] /. cwd["Scales"]);

ticks = Transpose[{Range[Length[freq]], freq}];

WaveletScalogram[cwd, Frame -> True, FrameTicks -> {{ticks, Automatic}, Automatic}, 
             FrameLabel -> {"Time", "Frequency(Hz)"},
             ColorFunction -> "SunsetColors", ImageSize -> 700]

Note that you have to change 500 000 to the value of your SampleRate

NB: Bear in mind that the axes are not scaled ! First observe that the number of octaves multiplied by the voices is equal to the max number on the x-axis and then you can use the relationship to scale the y-axis: $$scaledMagnitude = \frac{2 \times Magnitude}{N}$$ where $N$ is the sample size.

To illustrate the method just described, consider the signal

$$Sin[16 π x] + Sin[4 π x]$$

ListLinePlot[Abs@Fourier@Table[N@Sin[16 π x] + N@Sin[4 π x], {x, 0, 32 π, .001}], 
       PlotRange -> {{0, 900}, {0, 170}}, ImageSize -> 700]   

The frequencies are clearly distinguishable.

Now do the following:

cwt = ContinuousWaveletTransform[Table[N@Sin[16 π x] + N@Sin[4 π x], {x, 0, 32 π, .001}], 
          MorletWavelet[]]

WaveletScalogram[cwt, ImageSize -> 700, ColorFunction -> "SunsetColors"]

And again

ListLinePlot[Total[(Abs[#]^2)] & /@ Reverse@cwt[All, "Values"], ImageSize -> 700,  
    PlotRange -> All, BaseStyle -> Thick, PlotStyle -> ColorData[19, "ColorList"]]

There it is ... Do not forget to scale the axes accordingly !

Answer 2

Since you haven't provided any data I define something like:

data = 
  Table[ Sin[2 Pi t] 
        + 0.86 Sin[97 Pi t] Cos[46 Pi t] Sin[39 Pi t] Cos[19 Pi t] Exp[-102 (1/3 - t)^2],
        {t, 0.091, 0.519, 1/4095}];
ListLinePlot[data, PlotStyle -> Thick]

Now let's demonstrate how WaveletScalogram depends on the choice of ContinuousWaveletTransform with various wavelets, we choose a few examples with different ColorFunction, appropriate choice strongly depends on a specific purpose of signal processing and the real data one deals with.

GraphicsGrid[
  Table[{ Plot[{Re @ #, Im @ #}& @ WaveletPsi[k[[1]], x], {x, -4, 4}, 
                PlotRange -> All, Evaluated -> True, PlotStyle -> Thick], 
          WaveletScalogram[ ContinuousWaveletTransform[ data, k[[1]], {7, 12}, 
                                                        SampleRate -> 4095], 
                            ColorFunction -> k[[2]]]},
        {k, {{ MorletWavelet[],      "BlueGreenYellow"}, 
             { GaborWavelet[3],      "AvocadoColors"}, 
             { MexicanHatWavelet[2], "DeepSeaColors"}, 
             { PaulWavelet[2],       "SolarColors"}}}], 
  ImageSize -> 930]

These scalograms should be sufficient to start experimenting with your own data.

Answer 3

An alternative; we will discard the time values.

f[list_, pos_] := Module[{x = list}, x[[All, pos]] = Sequence[]; x]

data = Import[
"http://www.fileconvoy.com/gf.php?id=geed872d9b8a38dc6999443310.
     4661369818fbd5fa1bf3bc&sts=138977900593152145247060b494909aee48bb2a26b595048647"];

fdata = Flatten@f[data, 1];

cwd = ContinuousWaveletTransform[fdata, GaborWavelet[4], SampleRate -> 1000000]

freq = (1000000/(#*GaborWavelet[4]["FourierFactor"])) & /@ 
            (Thread[{Range[11], 1}] /. cwd["Scales"]);
ticks = Transpose[{Range[Length[freq]], freq}];

WaveletScalogram[cwd, ColorFunction -> "AvocadoColors", ImageSize -> 500,
                 Frame -> True, FrameTicks -> {{ticks, Automatic}, Automatic}, 
                 FrameLabel -> {"Time", "Frequency(Hz)"}]

Now that's a clear WaveletScalogram

---

data = Import[
"http://www.fileconvoy.com/gf.php?id=geed872d9b8a38dc6999443310.
     4661369818fbd5fa1bf3bc&sts=138977900593152145247060b494909aee48bb2a26b595048647"];

dwd = DiscreteWaveletTransform[data]

efrac = dwd["EnergyFraction"]

eth[x_, ind_] := If[(ind /. efrac) < 0.01, x*0., x] /; MemberQ[efrac[[All, 1]], ind]
eth[x_, ___] := x

fwd = WaveletMapIndexed[eth, dwd]

ListContourPlot[Abs@Reverse@Partition[Flatten[fwd[All, "Values"]], 4000], 
                   MaxPlotPoints -> 300, ColorFunction -> "AlpineColors"]

ListContourPlot[Abs@Reverse@Partition[Flatten[dwd[{{_}}, "Values"]], 4000], 
                   MaxPlotPoints -> 300, ColorFunction -> "AlpineColors"]

You can see the difference.

---

Okay, let's have some fun.

First, the data

data = BinaryReadList["http://www.physionet.org/physiobank/database/ptbdb/patient056/
                        s0196lre.dat"];

pdat = Take[data, 2000];

cwd = ContinuousWaveletTransform[pdat, GaborWavelet[], SampleRate -> 1000]

WaveletScalogram[cwd, ColorFunction -> "FallColors", ImageSize -> 500]

Oh, that heart ... Does not look good ...

---

Expanding on Artes' work you can do the following:

data = Table[ Cos[4 Pi t] + 0.3 Sin[55 Pi t] Exp[- 86 (1/3 - t)^2], {t, 0, 1, 1/4095}];

cwd = ContinuousWaveletTransform[data, SampleRate -> 4096]

WaveletScalogram[cwd, ColorFunction -> "RoseColors"]

Which gives us more information about the signal. And then

f = cwd["LinearScalogramFunction"]

Plot3D[f[x, y], {x, 0., 0.999756}, {y, 0.299259, 515.371}, 
                ColorFunction -> "DeepSeaColors", ImageSize -> 500]

Or

ContourPlot[f[x, y], {x, 0, 0.999755859375}, {y, 0.2992592856356853, 515.3711319499473},
      ColorFunction -> "SunsetColors", PlotPoints -> 200]