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0

If $z^n$ means $\frac{\partial ^n\text{}}{\partial t^n}$, then you can do like this: eq = 3 (z + 4)/(z^2 + 2 z + 5)//Simplify; lhs = Total@MapIndexed[#1 D[y[t], {t, #2[[1]] - 1}] &, CoefficientList[Denominator[eq], z]] // Simplify; rhs = Total@MapIndexed[#1 D[u[t], {t, #2[[1]] - 1}] &, CoefficientList[Numerator[eq], z]] // Simplify; ...

10

Looking at your plotted data you can see about 40 cycles of the dominant frequency, this tells you that the peak will appear somewhere around the 40th element of the DFT. That's in the region where your plot of the DFT is clipped, so it's no wonder you can't see the peak. Looking at the relevant part of the DFT you can see the peak quite clearly: ...

8

update Just to clean things up a bit, we can use the discussion here to make a couple functions that help extract the frequency data from this dataset. I define two functions findPeriod and reconstruct: Clear[findPeriod]; findPeriod[data_, threshold_] := Module[{fs, s1, s = {}, i, a0f, af, pf, pos, fr, frpos, fdata, fdatac, n, per}, n = ...

4

bn[a_, T_, f_] := 2/T Integrate[f Sin[(2 π n)/T t], {t, a, T + a}, Assumptions -> T ∈ Reals] bn[0, L, Piecewise[{{0, 0 < t < L/3}, {w, L/3 < t < 2 L/3}, {0, 2 L/3 < t < L}}]]

0

Your problem is in how you define kx1 and ky1. When you call f1[], x2 and y2 are substituted in the expression they are immediately visible. And since kx1 and ky1 don't explicitly depend on z and x2 and y2, these values aren't substituted. To fix this, you should define kx1 and ky1 as functions: kx1[z_,x2_] = ((2*Pi)/(λ*z))*x2; ky1[z_,y2_] = ...

2

You can try using Mathematica FourierSeries function directly. Change the plot range to see more periods. Manipulate[ Module[{x, f, g}, f = Piecewise[{{10 x, 0 <= x < 01/10}, {-(10/9) x + (10/9),1/10 <= x < 1}}]; g = FourierSeries[f, x, n]; Plot[{f, g}, {x, -lim, lim}, PlotRange -> All, ImageSize -> 300,ImagePadding -> 20] ...

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