Figuring out the frequency components using FFTPlotting Fourier spectrum versus frequency of a signalWhat's the correct way to shift zero frequency to the center of a Fourier Transform?What do the X and Y axis stand for in the Fourier transform domain?Identifying the three frequencies using fourier transformPlotting the frequency spectrum of a data series using FourierFinding the discrete Fourier transform for a simple oscillator?How to deal with highly oscillatory integrand when using “NIntegrate” and have a precise resultDetecting beat frequency using FFTIdentifying the three frequencies using fourier transform

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Figuring out the frequency components using FFT


Plotting Fourier spectrum versus frequency of a signalWhat's the correct way to shift zero frequency to the center of a Fourier Transform?What do the X and Y axis stand for in the Fourier transform domain?Identifying the three frequencies using fourier transformPlotting the frequency spectrum of a data series using FourierFinding the discrete Fourier transform for a simple oscillator?How to deal with highly oscillatory integrand when using “NIntegrate” and have a precise resultDetecting beat frequency using FFTIdentifying the three frequencies using fourier transform






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


The function defined below has three frequencies 2, 4 and 10. How to use the Fourier transform (FFT) to show these frequencies?



 myfun[t_] = 
1/(2 Sqrt[
2]) ([Sqrt]Abs[
3 + Cos[4 t] -
4 Cos[2 t] (-1 +
E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2) +
4 E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2])


Edit: The suggested duplicate deals with a simple function. It really doesn't work for this complicated scenario. Would appreciate if someone gets it done.










share|improve this question











$endgroup$













  • $begingroup$
    No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
    $endgroup$
    – Henrik Schumacher
    8 hours ago











  • $begingroup$
    Thanks,@HenrikSchumacher. Could this be shown using FFT?
    $endgroup$
    – Patrick
    8 hours ago






  • 1




    $begingroup$
    Possible duplicate of Identifying the three frequencies using fourier transform
    $endgroup$
    – ciao
    8 hours ago






  • 1




    $begingroup$
    Try this one or this one or...
    $endgroup$
    – march
    8 hours ago






  • 1




    $begingroup$
    @Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
    $endgroup$
    – march
    7 hours ago


















1












$begingroup$


The function defined below has three frequencies 2, 4 and 10. How to use the Fourier transform (FFT) to show these frequencies?



 myfun[t_] = 
1/(2 Sqrt[
2]) ([Sqrt]Abs[
3 + Cos[4 t] -
4 Cos[2 t] (-1 +
E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2) +
4 E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2])


Edit: The suggested duplicate deals with a simple function. It really doesn't work for this complicated scenario. Would appreciate if someone gets it done.










share|improve this question











$endgroup$













  • $begingroup$
    No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
    $endgroup$
    – Henrik Schumacher
    8 hours ago











  • $begingroup$
    Thanks,@HenrikSchumacher. Could this be shown using FFT?
    $endgroup$
    – Patrick
    8 hours ago






  • 1




    $begingroup$
    Possible duplicate of Identifying the three frequencies using fourier transform
    $endgroup$
    – ciao
    8 hours ago






  • 1




    $begingroup$
    Try this one or this one or...
    $endgroup$
    – march
    8 hours ago






  • 1




    $begingroup$
    @Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
    $endgroup$
    – march
    7 hours ago














1












1








1





$begingroup$


The function defined below has three frequencies 2, 4 and 10. How to use the Fourier transform (FFT) to show these frequencies?



 myfun[t_] = 
1/(2 Sqrt[
2]) ([Sqrt]Abs[
3 + Cos[4 t] -
4 Cos[2 t] (-1 +
E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2) +
4 E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2])


Edit: The suggested duplicate deals with a simple function. It really doesn't work for this complicated scenario. Would appreciate if someone gets it done.










share|improve this question











$endgroup$




The function defined below has three frequencies 2, 4 and 10. How to use the Fourier transform (FFT) to show these frequencies?



 myfun[t_] = 
1/(2 Sqrt[
2]) ([Sqrt]Abs[
3 + Cos[4 t] -
4 Cos[2 t] (-1 +
E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2) +
4 E^(-0.018` t) (Cos[9.99999594999918` t] +
0.0009000003645002213` Sin[9.99999594999918` t])^2])


Edit: The suggested duplicate deals with a simple function. It really doesn't work for this complicated scenario. Would appreciate if someone gets it done.







fourier-analysis






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago







Patrick

















asked 8 hours ago









PatrickPatrick

112 bronze badges




112 bronze badges














  • $begingroup$
    No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
    $endgroup$
    – Henrik Schumacher
    8 hours ago











  • $begingroup$
    Thanks,@HenrikSchumacher. Could this be shown using FFT?
    $endgroup$
    – Patrick
    8 hours ago






  • 1




    $begingroup$
    Possible duplicate of Identifying the three frequencies using fourier transform
    $endgroup$
    – ciao
    8 hours ago






  • 1




    $begingroup$
    Try this one or this one or...
    $endgroup$
    – march
    8 hours ago






  • 1




    $begingroup$
    @Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
    $endgroup$
    – march
    7 hours ago

















  • $begingroup$
    No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
    $endgroup$
    – Henrik Schumacher
    8 hours ago











  • $begingroup$
    Thanks,@HenrikSchumacher. Could this be shown using FFT?
    $endgroup$
    – Patrick
    8 hours ago






  • 1




    $begingroup$
    Possible duplicate of Identifying the three frequencies using fourier transform
    $endgroup$
    – ciao
    8 hours ago






  • 1




    $begingroup$
    Try this one or this one or...
    $endgroup$
    – march
    8 hours ago






  • 1




    $begingroup$
    @Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
    $endgroup$
    – march
    7 hours ago
















$begingroup$
No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
$endgroup$
– Henrik Schumacher
8 hours ago





$begingroup$
No, it has more than 3 frequencies due to the present of the factor E^(-0.018 t)` and the square roots.
$endgroup$
– Henrik Schumacher
8 hours ago













$begingroup$
Thanks,@HenrikSchumacher. Could this be shown using FFT?
$endgroup$
– Patrick
8 hours ago




$begingroup$
Thanks,@HenrikSchumacher. Could this be shown using FFT?
$endgroup$
– Patrick
8 hours ago




1




1




$begingroup$
Possible duplicate of Identifying the three frequencies using fourier transform
$endgroup$
– ciao
8 hours ago




$begingroup$
Possible duplicate of Identifying the three frequencies using fourier transform
$endgroup$
– ciao
8 hours ago




1




1




$begingroup$
Try this one or this one or...
$endgroup$
– march
8 hours ago




$begingroup$
Try this one or this one or...
$endgroup$
– march
8 hours ago




1




1




$begingroup$
@Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
$endgroup$
– march
7 hours ago





$begingroup$
@Patrick. Note that your frequencies will not be 2, 4, and 10 as you think. First of all, since the Sin[10 t] is squared, the relevant angular frequency is actually 20. Second of all, those are particular angular frequencies that show up. Instead, you should see dominant peaks at $2/2pi$, $4/2pi$, and $20/2pi$, which you will see if you use one of the linked answers. The one using Periodogram' is easy to use, and the other one shows the best way to use Fourier` (the FFT) to get a power spectrum.
$endgroup$
– march
7 hours ago











2 Answers
2






active

oldest

votes


















4














$begingroup$

First lets plot your function to see what it looks like.



Plot[myfun[t], t, 0, 100]


Mathematica graphics



This is only between 0 to 100 seconds. What we see is some very high frequencies but also a constant frequency which we can estimate at about 0.3 Hz. If we plot at later times



Plot[myfun[t], t, 500, 600]


Mathematica graphics



We can see that the starting high frequencies have died away and the frequency at about 0.3 Hz remains.



Now we can do a Fourier analysis. As you have some high frequencies we need to sample at a very fast sample rate. I have guessed at the sample rate but hope it is good enough.



sr = 100; (* Sample rate *)
data = Table[myfun[t], t, 0, 1000, 1/sr];
nn = Length@data


Plotting again shows that I seem to have captured the high frequency data



ListLinePlot[data[[1 ;; Round[100 sr]]]]


Mathematica graphics



You should probably check by looking at a shorter time interval.



Now for the Fourier analysis.



ft = Fourier[data, FourierParameters -> -1, -1];
freqs = Table[f, f, 0, sr (nn - 1)/nn, sr/nn];


The second line of code is to make the frequencies that correspond to the data in the Fourier transform. See here for details.



ListLogPlot[Transpose[freqs, Abs[ft]][[1 ;; 10000]], 
PlotRange -> All, Joined -> True]


Mathematica graphics



We can see a dominant peak at about 0.3 Hz and lots of other peaks that make up your high frequencies.



Hope that helps.






share|improve this answer











$endgroup$






















    1














    $begingroup$

    Periodogram and Spectrogram are also useful:



    data = Table[myfun[t], t, 0, 100000, .1];
    Periodogram[data, SampleRate -> 10, PlotRange -> All, Frame -> True,
    FrameLabel -> "Hz", "dB"]


    enter image description here



    Spectrogram[data, SampleRate -> 10, PlotRange -> 0, .5, 
    FrameLabel -> "Time", "Hz"]


    enter image description here






    share|improve this answer









    $endgroup$

















      Your Answer








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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      $begingroup$

      First lets plot your function to see what it looks like.



      Plot[myfun[t], t, 0, 100]


      Mathematica graphics



      This is only between 0 to 100 seconds. What we see is some very high frequencies but also a constant frequency which we can estimate at about 0.3 Hz. If we plot at later times



      Plot[myfun[t], t, 500, 600]


      Mathematica graphics



      We can see that the starting high frequencies have died away and the frequency at about 0.3 Hz remains.



      Now we can do a Fourier analysis. As you have some high frequencies we need to sample at a very fast sample rate. I have guessed at the sample rate but hope it is good enough.



      sr = 100; (* Sample rate *)
      data = Table[myfun[t], t, 0, 1000, 1/sr];
      nn = Length@data


      Plotting again shows that I seem to have captured the high frequency data



      ListLinePlot[data[[1 ;; Round[100 sr]]]]


      Mathematica graphics



      You should probably check by looking at a shorter time interval.



      Now for the Fourier analysis.



      ft = Fourier[data, FourierParameters -> -1, -1];
      freqs = Table[f, f, 0, sr (nn - 1)/nn, sr/nn];


      The second line of code is to make the frequencies that correspond to the data in the Fourier transform. See here for details.



      ListLogPlot[Transpose[freqs, Abs[ft]][[1 ;; 10000]], 
      PlotRange -> All, Joined -> True]


      Mathematica graphics



      We can see a dominant peak at about 0.3 Hz and lots of other peaks that make up your high frequencies.



      Hope that helps.






      share|improve this answer











      $endgroup$



















        4














        $begingroup$

        First lets plot your function to see what it looks like.



        Plot[myfun[t], t, 0, 100]


        Mathematica graphics



        This is only between 0 to 100 seconds. What we see is some very high frequencies but also a constant frequency which we can estimate at about 0.3 Hz. If we plot at later times



        Plot[myfun[t], t, 500, 600]


        Mathematica graphics



        We can see that the starting high frequencies have died away and the frequency at about 0.3 Hz remains.



        Now we can do a Fourier analysis. As you have some high frequencies we need to sample at a very fast sample rate. I have guessed at the sample rate but hope it is good enough.



        sr = 100; (* Sample rate *)
        data = Table[myfun[t], t, 0, 1000, 1/sr];
        nn = Length@data


        Plotting again shows that I seem to have captured the high frequency data



        ListLinePlot[data[[1 ;; Round[100 sr]]]]


        Mathematica graphics



        You should probably check by looking at a shorter time interval.



        Now for the Fourier analysis.



        ft = Fourier[data, FourierParameters -> -1, -1];
        freqs = Table[f, f, 0, sr (nn - 1)/nn, sr/nn];


        The second line of code is to make the frequencies that correspond to the data in the Fourier transform. See here for details.



        ListLogPlot[Transpose[freqs, Abs[ft]][[1 ;; 10000]], 
        PlotRange -> All, Joined -> True]


        Mathematica graphics



        We can see a dominant peak at about 0.3 Hz and lots of other peaks that make up your high frequencies.



        Hope that helps.






        share|improve this answer











        $endgroup$

















          4














          4










          4







          $begingroup$

          First lets plot your function to see what it looks like.



          Plot[myfun[t], t, 0, 100]


          Mathematica graphics



          This is only between 0 to 100 seconds. What we see is some very high frequencies but also a constant frequency which we can estimate at about 0.3 Hz. If we plot at later times



          Plot[myfun[t], t, 500, 600]


          Mathematica graphics



          We can see that the starting high frequencies have died away and the frequency at about 0.3 Hz remains.



          Now we can do a Fourier analysis. As you have some high frequencies we need to sample at a very fast sample rate. I have guessed at the sample rate but hope it is good enough.



          sr = 100; (* Sample rate *)
          data = Table[myfun[t], t, 0, 1000, 1/sr];
          nn = Length@data


          Plotting again shows that I seem to have captured the high frequency data



          ListLinePlot[data[[1 ;; Round[100 sr]]]]


          Mathematica graphics



          You should probably check by looking at a shorter time interval.



          Now for the Fourier analysis.



          ft = Fourier[data, FourierParameters -> -1, -1];
          freqs = Table[f, f, 0, sr (nn - 1)/nn, sr/nn];


          The second line of code is to make the frequencies that correspond to the data in the Fourier transform. See here for details.



          ListLogPlot[Transpose[freqs, Abs[ft]][[1 ;; 10000]], 
          PlotRange -> All, Joined -> True]


          Mathematica graphics



          We can see a dominant peak at about 0.3 Hz and lots of other peaks that make up your high frequencies.



          Hope that helps.






          share|improve this answer











          $endgroup$



          First lets plot your function to see what it looks like.



          Plot[myfun[t], t, 0, 100]


          Mathematica graphics



          This is only between 0 to 100 seconds. What we see is some very high frequencies but also a constant frequency which we can estimate at about 0.3 Hz. If we plot at later times



          Plot[myfun[t], t, 500, 600]


          Mathematica graphics



          We can see that the starting high frequencies have died away and the frequency at about 0.3 Hz remains.



          Now we can do a Fourier analysis. As you have some high frequencies we need to sample at a very fast sample rate. I have guessed at the sample rate but hope it is good enough.



          sr = 100; (* Sample rate *)
          data = Table[myfun[t], t, 0, 1000, 1/sr];
          nn = Length@data


          Plotting again shows that I seem to have captured the high frequency data



          ListLinePlot[data[[1 ;; Round[100 sr]]]]


          Mathematica graphics



          You should probably check by looking at a shorter time interval.



          Now for the Fourier analysis.



          ft = Fourier[data, FourierParameters -> -1, -1];
          freqs = Table[f, f, 0, sr (nn - 1)/nn, sr/nn];


          The second line of code is to make the frequencies that correspond to the data in the Fourier transform. See here for details.



          ListLogPlot[Transpose[freqs, Abs[ft]][[1 ;; 10000]], 
          PlotRange -> All, Joined -> True]


          Mathematica graphics



          We can see a dominant peak at about 0.3 Hz and lots of other peaks that make up your high frequencies.



          Hope that helps.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          HughHugh

          7,4262 gold badges19 silver badges47 bronze badges




          7,4262 gold badges19 silver badges47 bronze badges


























              1














              $begingroup$

              Periodogram and Spectrogram are also useful:



              data = Table[myfun[t], t, 0, 100000, .1];
              Periodogram[data, SampleRate -> 10, PlotRange -> All, Frame -> True,
              FrameLabel -> "Hz", "dB"]


              enter image description here



              Spectrogram[data, SampleRate -> 10, PlotRange -> 0, .5, 
              FrameLabel -> "Time", "Hz"]


              enter image description here






              share|improve this answer









              $endgroup$



















                1














                $begingroup$

                Periodogram and Spectrogram are also useful:



                data = Table[myfun[t], t, 0, 100000, .1];
                Periodogram[data, SampleRate -> 10, PlotRange -> All, Frame -> True,
                FrameLabel -> "Hz", "dB"]


                enter image description here



                Spectrogram[data, SampleRate -> 10, PlotRange -> 0, .5, 
                FrameLabel -> "Time", "Hz"]


                enter image description here






                share|improve this answer









                $endgroup$

















                  1














                  1










                  1







                  $begingroup$

                  Periodogram and Spectrogram are also useful:



                  data = Table[myfun[t], t, 0, 100000, .1];
                  Periodogram[data, SampleRate -> 10, PlotRange -> All, Frame -> True,
                  FrameLabel -> "Hz", "dB"]


                  enter image description here



                  Spectrogram[data, SampleRate -> 10, PlotRange -> 0, .5, 
                  FrameLabel -> "Time", "Hz"]


                  enter image description here






                  share|improve this answer









                  $endgroup$



                  Periodogram and Spectrogram are also useful:



                  data = Table[myfun[t], t, 0, 100000, .1];
                  Periodogram[data, SampleRate -> 10, PlotRange -> All, Frame -> True,
                  FrameLabel -> "Hz", "dB"]


                  enter image description here



                  Spectrogram[data, SampleRate -> 10, PlotRange -> 0, .5, 
                  FrameLabel -> "Time", "Hz"]


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 3 hours ago









                  David KeithDavid Keith

                  2,2851 gold badge5 silver badges17 bronze badges




                  2,2851 gold badge5 silver badges17 bronze badges































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                      Кастелфранко ди Сопра Становништво Референце Спољашње везе Мени за навигацију43°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.5588543°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.558853179688„The GeoNames geographical database”„Istituto Nazionale di Statistica”проширитиууWorldCat156923403n850174324558639-1cb14643287r(подаци)