Skip to main content

Fourier in Mathematica different from octave



First I create a table of sin(x) within 0, 10



In[97]:= Table[Sin[1.0 x], {x, 0, 10}]
Out[97]= {0.,0.841471,0.909297,0.14112,-0.756802,-0.958924,-0.279415,0.656987,0.989358,0.412118,-0.544021}

Then I try to apply fft on this data (wolfram Link)


In[100]:= Fourier[{0.,0.841471,0.909297,0.14112,-0.756802,-0.958924,-0.279415,0.656987,0.989358,0.412118,-0.544021}]//MatrixForm
Out[100]//MatrixForm= (
0.425489 +0.I
0.570407 -0.270821I
-0.860518+1.09804I
-0.0342615+0.218599I

0.0450821 +0.095328I
0.0665458 +0.0283187I
0.0665458 -0.0283187I
0.0450821 -0.095328I
-0.0342615-0.218599I
-0.860518-1.09804I
0.570407 +0.270821I)

I do the same using octave


fft([0.;0.841471;0.909297;0.14112;-0.756802;-0.958924;-0.279415;0.656987;0.989358;0.412111])

ans =

1.41119 + 0.00000i
1.89183 + 0.89821i
-2.85402 - 3.64178i
-0.11363 - 0.72501i
0.14952 - 0.31617i
0.22071 - 0.09392i
0.22071 + 0.09392i
0.14952 + 0.31617i

-0.11363 + 0.72501i
-2.85402 + 3.64178i
1.89183 - 0.89821i

Which one of them is correct ?


UPDATE


using FourierParameters as suggested, However wolfram alpha online gives different result


In[101]:= Fourier[{0.,0.841471,0.909297,0.14112,-0.756802,-0.958924,
-0.279415,0.656987,0.989358,0.412118,-0.544021},
FourierParameters->{1,-1}] // MatrixForm

Out[101]//MatrixForm= (
1.41119 +0.I
1.89183 +0.898211I
-2.85402 -3.64178I
-0.113633 -0.725012I
0.149521 -0.316167I
0.220708 -0.0939226I
0.220708 +0.0939226I
0.149521 +0.316167I
-0.113633 +0.725012I

-2.85402 +3.64178I
1.89183 -0.898211I)

But why these are different ?



Answer



Read the docs for Fourier. There are different conventions, and Mathematica lets you choose which convention you want using the FourierParameters option. For example, to reproduce Octave's (and MATLAB's) convention, use


dat = Table[Sin[1.0 x], {x, 0, 10}]
Fourier[dat, FourierParameters -> {1, -1}]

This is what the docs call the "signal processing convention". The meaning of the options is given in the help file under details and options where it says:



enter image description here


So for example, with $a=1$, the term outside the sum is $1/n^0 = 1$ and there is no scaling on the forward Fourier transform. With $a=-1$, the scaling would be $1/n$. Similarly for $b$, though it appears in the exponent of the e in the sum. $b=-1$ means the forward transform has a minus sign in the exponent while $b=1$ means it has a plus sign.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...