Skip to main content

plotting - Get Y-Axis of FFT in decibels



I am trying to make a plot of a sound file similar to Audacity's FFT plot which looks like this when the x-axis is on a log scale:


Audacity FFT


My plot of the same sound file (below) has the problem that the units are in absolute amplitude, not decibels. How can I fix this?


fft2



Answer



decibel is a relative unit. I'm pretty sure there is no implied standard reference in audio processing, (it looks like audiologists have a few go-to's e.g. dB HL, but I don't know what Audacity does). That said, you need a reference value. Since you're looking at FFT's the total power might be a good choice. Then decibels will tell you how strong a particular frequency is relative to the signal.


signal = RandomReal[{0,1},1000];(*A random signal*)
fft    = signal//Fourier//Abs;
fft    = fft[[1;;Ceiling[Length@signal/2]]]; (*Since we're looking at power only take the first N/2 points.*)
fftPowerRef = Total[fft^2]; (* Power is amplitude squared. *)

inDecibels = 10*Log[10, fft^2/fftPowerRef ];

In this case you will always have <0dB since a pure tone would concentrate all the power at one fft sample, and you'd be taking Log[10,1]


If you do the same thing with a signal like signal = Sin[2*pi*#/200]&/@Range[1000] you should see one main peak with most of the power.



Alternate Reference Power


Using the total power as a reference means that from song to song a value of -10dB won't have much meaning except to give you the relative strength of a particular tone in that data set. You could calibrate your plots by always referencing the same power, for example the power in pure sine wave played at some particular volume, to make the axis meaningful from sample to sample.


You could also get crazy and take the FFT power of a wave that fully utilizes the dynamic range of a particular digitizer. You can reference relative to RMS power, or peak power, etc. etc. etc. It really depends on what your end goal is.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

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}]
Post a Comment
Read more