plotting - Graphical representation of a moving sound source

Some explanation since I have got a specific and another more open-ended question: I used two microphones in a 90° arrangement (see picture below) to capture sound from a source moving around the room in a circular /elliptical fashion. The intention is to get an idea about directional hearing. (A second measurement with a KEMAR dummy-head was done as well).

arrangement microphone arrangement

I saved the measured data as .wav (48kHz) and imported the .wav file into Mathematica. (The .wav- file can be found here). That made a pretty long list - approximately 900000 entries per channel. Hence I used

 dropfunction[data_, n_] := 
 Table[Nest[Drop[#, {1, -1, 2}] &, data[[i]], n], {i, 1, 2}]

to reduce the resolution. I found by hearing/looking at ListLinePlot that n = 7 is a good catch. That leaves around 7000 entries per channel.

Here comes my first and very specific question: I plotted the absolute values of my list in a fashion like:

ListLinePlot[Abs@data[[1]], -Abs[data[2]]]

and got:

left vs. right channel

with left and right channel clearly divided along the x-axis.

Question 1: Is there a possibility to flip the plot in such a way that x-axis becomes the y-axis? (So that for instance the "loudness" of the right channel is on the right hand side). I tried PairedBarChart actually gives the kind of representation I wish, but It is just not suitable for my data.

Question 2: Does anybody else have a great idea how to graphically represent my data? I tried plotting the difference of the absolute values of both channels to get a graph showing which channel is currently louder at the positions in time/on my elliptical path. With some lowpass-filtering I got:

enter image description here

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog