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6

You didn't post a sample image, so I'll use the Lena test image: img = ColorConvert[ExampleData[{"TestImage", "Lena"}], "Grayscale"]; fft = Fourier[ImageData[img]]; fft = RotateLeft[fft, Floor[Dimensions[fft]/2]]; (RotateLeft is basically equivalent to *(-1)^Table[i + j, {i, IRow}, {j, ICol}] in your code: it shifts the 0/0 frequency to the center of the ...


1

Although this Answer does not use Floquet analysis, it does cast the equations into a more useful form and provide a sample numerical solution. X[t] = {{x1[t], x2[t]}}\[Transpose]; b[t] = b1 E^(-I δ t) + b2 E^(I δ t); a[t] = a1 E^(-I δ t) + a2 E^(I δ t); M[t] = {{0, FullSimplify[Conjugate[b[t]], t ∈ Reals && δ ∈ Reals && b1 ∈ ...


0

I'm not sure if this helps. I think the problem can be traced back to this simple case. Let f[n_] := Sin[n \[Pi]]/(\[Pi] n) This should Simplify[] to DiscreteDelta[n]. But at least I have found no way to accomplish this. Let us investigate the function. For Mathematica f[n] it is zero for any integer because of the Sin: Simplify[f[n], n \[Element] ...


0

Why do you need to change the interval? f-odd-extention (fox) fox[x_] := Piecewise[{ {-3 - x, x < -1}, {-x^2 - 1, -1 <= x < 0}, {x^2 + 1, 0 <= x <= 1}, {3 - x, 1 < x} }] which looks this Plot[fox[x], {x, -3, 3}] The Fourier series approximation to fox can be computed for 20 terns with fs[x_] = FourierSeries[f[x], ...


3

The answer is that Fourier does not know your time values and does not see an even function. Thus if we generate your new input data and plot we get inputData2 = h[tListSym]; ListLinePlot[inputData2, PlotRange -> All] The problem is thus how to tell Fourier that the first half of your input is for negative x values. We could change the phase of the ...


6

I'm pretty sure if you do ListLinePlot[data] you'll see your first data is actually this plot and not the square wave you expect: Generate your second plot with a Piecewise: Plot[Piecewise[{{1, 0 < x < 2}, {0, 2 < x < 3}, {1, 3 < x < 5}, {0, 5 < x < 6}, {1, 6 < x < 8}, {0, 8 < x < 9}}], {x, 0, 9}, Exclusions -> ...


7

It looks like you need some basic facts on numerical Fourier transforms. I am going to assume that you are going from the time domain to the frequency domain. The number of points in the time domain equals the number of points in the frequency domain. As the data is sampled in the time domain i.e. is a set of equally spaced points, then in the frequency ...



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