Reconciling results from Fourier with those form FourierTransform

Question

I need to use the discrete Fourier transform for function that represented as list of values.

I started with an easy task to check my understanding. I tried to get the amplitude values for 2*Sin[x]. I think they should be somwhere about 2. But using the discrete Fourier operation produces something quite different. Can someone tell me where my mistakes are?

Here is what I tried.

data = Table[2*Sin[x], {x, 0, 100}];
ListPlot[data, Joined -> True]

ListPlot[Abs[Fourier[data]], Joined -> True, PlotRange -> All]

ListPlot[
  Table[Abs[FourierTransform[2*Sin[x], x, w]] /. 
    DiracDelta -> DiscreteDelta, {w, 0.1, 10, 0.1}], 
  Joined -> True, 
  PlotRange -> All]

How can I get the same results from 'Fourier' as I get from 'FourierTransform'?

Answer 1

Paying close attention to the documentation for Fourier and FourierTransform one notes that the coefficients of the Sum/Integral terms are different; therefore, to obtain a discrete transform with amplitudes equal to those from the continuous transform, one must multiply the former by Sqrt[2 Pi / n] where n is the length of the dataset:

The continuous waveform:

DiscretePlot[
 Evaluate[Abs@FourierTransform[2 Sin[x], x, w] /. 
   DiracDelta -> DiscreteDelta], {w, 0, 2}]

and the discrete waveform:

With[{datalength = 100}, 
 ListPlot[(Sqrt[2 Pi]/Sqrt[datalength]) Abs[
    Fourier[Table[2*Sin[x], {x, 0, datalength}]]], Joined -> True, 
  PlotRange -> {{0, 2}, All}, DataRange -> {0, 2Pi}]]

Appendix: Why isn't the amplitude "right"

In the comments, the OP asks why the amplitude of the function 2 Sin[x] appears to be ~2.5 in the transformed data. Let's take a pedagogical approach here, in part to show off how one might use Mathematica to answer these types of questions on their own.

Is there a relationship between the amplitude of the wave and the height of the peak in the transform?

Hopefully, the answer is yes, but let's generate some data, plot it and see.

testdata = 
 Table[{i, 
   Evaluate[
    Abs@FourierTransform[i Sin[x], x, w] /. {DiracDelta -> 
       DiscreteDelta, w -> 1}]}, {i, 1, 10, 1}]

There is a linear relationship - yeah! What is that relationship?

Let's perform a linear least squares analysis on testdata to see what the slope of that line is. I will assume that the plot should go through zero.

LinearModelFit[testdata, x, x]["BestFitParameters"] // Chop
(* {0, 1.25331} *)

So that means that the height of the peak in the transformed data will be 1.25331 times larger than the amplitude of the time-domain function. Where does this 1.25331 come from? Taking a closer look at testdata:

We see a Sqrt[2 Pi]/2 nestled in there. Evaluating N@Sqrt[2 Pi]/2 yields:

(* 1.25331 *)

Nice.

Answer 2

This code will show a very tight correspondence between Fourier and FourierTransform. For the continuous case, note the imaginary negative infinite DiracDelta spike at w=-1 (2nd term of answer), and the positive one at w=1.

FourierTransform[2*Sin[x], x, w]
(*I*Sqrt[2*Pi]*DiracDelta[-1+w]-I*Sqrt[2*Pi]*DiracDelta[1+w]*)

Here is the original table range in your question modified slightly to fit an exact number of wavelengths. 32Pi is slightly over 100, and Pi/3 is a little more than 1. The pure function and Thread are used to break the real and imaginary components into two separate lines for plotting purposes. This link shows the code snippet for dftMost, which uses Fourier.

ListPlot[Thread[{{#,Re@#2},{#,Im@#2}}&@@@
First[dftMost@Table[{x,2Sin[x]},{x,0,32Pi,Pi/3}]]],
Joined->True,PlotRange->All]

Note the negative and positive spikes in the imaginary component of the discrete Fourier transform at w=-1 and w=1. If you use more points in the Table command, the spikes will increase toward infinity.

DeleteCases[Chop@Cases[%,Line[pts_?MatrixQ]:>pts,
Infinity],{_,0},Infinity]
(*{{},{{-1.,-40.1061},{1.,40.1061}}}*)

Here we pull out the nonzero real and imaginary components from the blue (real) and orange imaginary components in the ListPlot graph above. There are no nonzero real compoents, so that first list is empty. The only two non-zero imaginary components are at w=-1 and w=1 in the 2nd list, as previously mentioned.

Bonus: You may also like to see a case where the continuous transform isn't integrated from negative to positive infinity, but is instead windowed. By changing the number of periods of integration, n, you can watch the aforementioned frequency peaks gradually rise toward infinity.

fourierTransform[f_,{t_,tini_,tfin_},w_,a_,b_]:=
Sqrt[Abs@b/(2Pi)^(1-a)] Integrate[f Exp[I b w t],{t,tini,tfin}]
Assuming[{n>=1,(n/2)\[Element]Integers},With[{T=n 2Pi/w0},
fourierTransform[2 Sin[w0 t],{t,-T/2,T/2},w,0,1]/.w0->1//Simplify]]
(*2*I*Sqrt[2/Pi]*Sin[n*Pi*w]/(-1+w^2)*)
Plot@@{Through@{Re,Im}@%/.n->16,{w,-2,2},PlotRange->All}